SOA exam FM CAS

Chapter 1. Basic Interest Theory.

Manual for SOA Exam FM/CAS Exam 2. Chapter 1. Basic Interest Theory. Section 1.1. Amount and accumulation functions. c

2009. Miguel A. Arcones. All rights reserved.

Extract from: ”Arcones’ Manual for the SOA Exam FM/CAS Exam 2, Financial Mathematics. Fall 2009 Edition”, available at http://www.actexmadriver.com/

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Manual for SOA Exam FM/CAS Exam 2.


Section 1.1. Amount and accumulation functions.

Interest I

When money is invested or loaned the amount of money returned is different from the initial one.

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Interest I


I

The amount of money invested (or loaned) is called the principal or principle.

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Interest I


I


I

The amount of interest earned during a period of time is I = final balance − invested amount.

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Interest I


I


I

The amount of interest earned during a period of time is I = final balance − invested amount.

I

The effective rate of interest earned in the period [s, t] is final balance − invested amount . invested amount

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Example 1 Simon invests $1000 in a bank account. Six months later, the amount in his bank account is $1049.23. (i) Find the amount of interest earned by Simon in those 6 months. (ii) Find the (semiannual) effective rate of interest earned in those 6 months.

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Example 1 Simon invests $1000 in a bank account. Six months later, the amount in his bank account is $1049.23. (i) Find the amount of interest earned by Simon in those 6 months. (ii) Find the (semiannual) effective rate of interest earned in those 6 months. Solution: (i) The amount of interest earned by Simon in those 6 months is I = 1049.23 − 1000 = 49.23.

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Example 1 Simon invests $1000 in a bank account. Six months later, the amount in his bank account is $1049.23. (i) Find the amount of interest earned by Simon in those 6 months. (ii) Find the (semiannual) effective rate of interest earned in those 6 months. Solution: (i) The amount of interest earned by Simon in those 6 months is I = 1049.23 − 1000 = 49.23. (ii) The (semiannual) effective rate of interest earned is 1049.23 − 1000 = 0.004923 = 0.4923%. 1000

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Amount function Suppose that an amount A(0) of money is invested at time 0. A(0) is the principal. Let A(t) denote the value at time t of the initial investment A(0). The function A(t), t ≥ 0, is called the amount function. Usually, we assume that the amount function satisfies the following properties: (i) For each t ≥ 0, A(t) > 0. (ii) A is nondecreasing.

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Amount function Suppose that an amount A(0) of money is invested at time 0. A(0) is the principal. Let A(t) denote the value at time t of the initial investment A(0). The function A(t), t ≥ 0, is called the amount function. Usually, we assume that the amount function satisfies the following properties: (i) For each t ≥ 0, A(t) > 0. (ii) A is nondecreasing. In this situation, I The amount of interest earned over the period [s, t] is A(t) − A(s).

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Amount function Suppose that an amount A(0) of money is invested at time 0. A(0) is the principal. Let A(t) denote the value at time t of the initial investment A(0). The function A(t), t ≥ 0, is called the amount function. Usually, we assume that the amount function satisfies the following properties: (i) For each t ≥ 0, A(t) > 0. (ii) A is nondecreasing. In this situation, I The amount of interest earned over the period [s, t] is A(t) − A(s). I

The effective rate of interest earned in the period [s, t] is A(t) − A(s) . A(s) 11/52

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Example 2 Jessica invests $5000 on March 1, 2008, in a fund which follows t the accumulation function A(t) = (5000) 1 + 40 , where t is the number of years after March 1, 2008. (i) Find the balance in Jessica’s account on October 1, 2008. (ii) Find the amount of interest earned in those 7 months. (iii) Find the effective rate of interest earned in that period.

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Example 2 Jessica invests $5000 on March 1, 2008, in a fund which follows t the accumulation function A(t) = (5000) 1 + 40 , where t is the number of years after March 1, 2008. (i) Find the balance in Jessica’s account on October 1, 2008. (ii) Find the amount of interest earned in those 7 months. (iii) Find the effective rate of interest earned in that period. Solution: (i) The balance of Jessica’s account on 10–1–2008 is 7/12 A(7/12) = (5000) 1 + = 5072.917. 40

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Example 2 Jessica invests $5000 on March 1, 2008, in a fund which follows t the accumulation function A(t) = (5000) 1 + 40 , where t is the number of years after March 1, 2008. (i) Find the balance in Jessica’s account on October 1, 2008. (ii) Find the amount of interest earned in those 7 months. (iii) Find the effective rate of interest earned in that period. Solution: (i) The balance of Jessica’s account on 10–1–2008 is 7/12 A(7/12) = (5000) 1 + = 5072.917. 40 (ii) The amount of interest earned in those 7 months is A(7/12) − A(0) = 5072.917 − 5000 = 72.917.

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Example 2 Jessica invests $5000 on March 1, 2008, in a fund which follows t the accumulation function A(t) = (5000) 1 + 40 , where t is the number of years after March 1, 2008. (i) Find the balance in Jessica’s account on October 1, 2008. (ii) Find the amount of interest earned in those 7 months. (iii) Find the effective rate of interest earned in that period. Solution: (i) The balance of Jessica’s account on 10–1–2008 is 7/12 A(7/12) = (5000) 1 + = 5072.917. 40 (ii) The amount of interest earned in those 7 months is A(7/12) − A(0) = 5072.917 − 5000 = 72.917. (iii) The effective rate of interest earned in that period is 72.917 A(7/12) − A(0) = = 0.0145834 = 1.45834%. A(0) 5000 15/52

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Cashflows Often, we consider the case when several deposits/withdrawals are made into an account following certain amount function. A series of (deposits/withdrawals) payments made at different times is called a cashflow. The payments can be either made by the individual or to the individual. An inflow is payment to the individual. An outflow is a payment by the individual. We represent inflows by positive numbers and outflows by negative numbers. In a cashflow, we have a contribution of Cj at time tj , for each j = 1, . . . , n. Cj can be either positive or negative. We can represent a cashflow in a table: Investments Time (in years)

C1 t1

C2 t2

··· ···

Cn tn

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Cashflow rules

Rule 1: Proportionality. If an investment strategy follows the amount function A(t), t > 0, an investment of $k made at time 0 with the previous investment strategy, has a value of $ kA(t) A(0) at time t.

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Cashflow rules

Rule 1: Proportionality. If an investment strategy follows the amount function A(t), t > 0, an investment of $k made at time 0 with the previous investment strategy, has a value of $ kA(t) A(0) at time t. Using the amount function A(·) and proportionality:

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Cashflow rules

Rule 1: Proportionality. If an investment strategy follows the amount function A(t), t > 0, an investment of $k made at time 0 with the previous investment strategy, has a value of $ kA(t) A(0) at time t. Using the amount function A(·) and proportionality: I

Investing A(0) at time zero, we get A(t) at time t.

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Cashflow rules



I

Investing 1 at time zero, we get

A(t) A(0)

at time t.

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Cashflow rules



I

Investing 1 at time zero, we get

A(t) A(0)

I

Investing k at time zero, we get

kA(t) A(0)

at time t. at time t.

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Present value Definition 1 The present value at certain time of a cashflow is the amount of the money which need to invest at certain time in other to get the same balance as that obtained from a cashflow.

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Present value Definition 1 The present value at certain time of a cashflow is the amount of the money which need to invest at certain time in other to get the same balance as that obtained from a cashflow. Since investing k at time zero, we get kA(t) A(0) at time t, we have that: the present value at time t of a deposit of k made at time zero is kA(t) A(0) .

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Present value Definition 1 The present value at certain time of a cashflow is the amount of the money which need to invest at certain time in other to get the same balance as that obtained from a cashflow. Since investing k at time zero, we get kA(t) A(0) at time t, we have that: the present value at time t of a deposit of k made at time zero is kA(t) A(0) . Let x be the amount which need to invest at time zero to get a kA(0) balance of k at time t. We have that k = xA(t) A(0) . So, x = A(t) . Hence, the present value at time 0 of a balance of k had at time t is kA(0) A(t) . 24/52

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To know how the value of money changes over time we need to see how the value of $1 varies over time. The accumulation function a(t), t ≥ 0, is defined as the value at time t of $1 invested at time 0.

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To know how the value of money changes over time we need to see how the value of $1 varies over time. The accumulation function a(t), t ≥ 0, is defined as the value at time t of $1 invested at time 0. A(t) By proportionality, a(t) = A(0) . Observe that a(0) = 1.

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To know how the value of money changes over time we need to see how the value of $1 varies over time. The accumulation function a(t), t ≥ 0, is defined as the value at time t of $1 invested at time 0. A(t) By proportionality, a(t) = A(0) . Observe that a(0) = 1. Knowing the value function a(t) and the principal A(0), we can find the amount function A(t) using the formula A(t) = A(0)a(t).

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To know how the value of money changes over time we need to see how the value of $1 varies over time. The accumulation function a(t), t ≥ 0, is defined as the value at time t of $1 invested at time 0. A(t) By proportionality, a(t) = A(0) . Observe that a(0) = 1. Knowing the value function a(t) and the principal A(0), we can find the amount function A(t) using the formula A(t) = A(0)a(t). Using the accumulation function a(t), t ≥ 0, we have: I

The present value at time t of a deposit of k made at time zero is ka(t) (= kA(t) A(0) ).

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To know how the value of money changes over time we need to see how the value of $1 varies over time. The accumulation function a(t), t ≥ 0, is defined as the value at time t of $1 invested at time 0. A(t) By proportionality, a(t) = A(0) . Observe that a(0) = 1. Knowing the value function a(t) and the principal A(0), we can find the amount function A(t) using the formula A(t) = A(0)a(t). Using the accumulation function a(t), t ≥ 0, we have: I

The present value at time t of a deposit of k made at time zero is ka(t) (= kA(t) A(0) ).

I

The present value at time 0 of a balance of k had at time t is kA(0) k a(t) (= A(t) ).

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Example 1 The accumulation function of a fund is a(t) = (1.03)2t , t ≥ 0. (i) Amanda invests $5000 at time zero in this fund. Find the balance into Amanda’s fund at time 2.5 years. (ii) How much money does Kevin need to invest into the fund at time 0 to accumulate $10000 at time 3?

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Example 1 The accumulation function of a fund is a(t) = (1.03)2t , t ≥ 0. (i) Amanda invests $5000 at time zero in this fund. Find the balance into Amanda’s fund at time 2.5 years. (ii) How much money does Kevin need to invest into the fund at time 0 to accumulate $10000 at time 3? Solution: (i) The balance into Amanda’s fund at time 2.5 years is ka(2.5) = (5000)(1.03)2(2.5) = 5796.370371.

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Example 1 The accumulation function of a fund is a(t) = (1.03)2t , t ≥ 0. (i) Amanda invests $5000 at time zero in this fund. Find the balance into Amanda’s fund at time 2.5 years. (ii) How much money does Kevin need to invest into the fund at time 0 to accumulate $10000 at time 3? Solution: (i) The balance into Amanda’s fund at time 2.5 years is ka(2.5) = (5000)(1.03)2(2.5) = 5796.370371.

(ii) The amount which Kevin needs to invest at time 0 to accumulate $10000 at time 3 is 10000 10000 = = 8374.842567. a(3) (1.03)2(3) 32/52

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Cashflow rules Rule 2. Grows–depends–on–balance rule. If an investment follows the amount function A(t), t ≥ 0, the growth during certain period where no deposits/withdrawals are made depends on the balance on the account at the beginning of the period.

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Cashflow rules Rule 2. Grows–depends–on–balance rule. If an investment follows the amount function A(t), t ≥ 0, the growth during certain period where no deposits/withdrawals are made depends on the balance on the account at the beginning of the period. If an account has a balance of k at time t and no deposits/withdrawals are made in the future, then the future balance in this account does not depend on how the balance of k at time t was attained.

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Cashflow rules Rule 2. Grows–depends–on–balance rule. If an investment follows the amount function A(t), t ≥ 0, the growth during certain period where no deposits/withdrawals are made depends on the balance on the account at the beginning of the period. If an account has a balance of k at time t and no deposits/withdrawals are made in the future, then the future balance in this account does not depend on how the balance of k at time t was attained. In particular, the following two accounts have the same balance for times bigger than t: 1. An account where a unique deposit of k is made at time t. k 2. An account where a unique deposit of A(t) is made at time zero.

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Theorem 1 If an investment follows the amount function A(·), the present ka(t) value at time t of a deposit of $k made at time s is $ kA(t) A(s) = a(s) .

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Theorem 1 If an investment follows the amount function A(·), the present ka(t) value at time t of a deposit of $k made at time s is $ kA(t) A(s) = a(s) . Proof. We need to invest

k A(s)

at time 0 to get a balance of k at

time s. So, investing k at time s is equivalent to investing time 0. The future value at time t of an investment of 0 is kA(t) A(s) . Hence, kA(t) A(s) at time t.

k A(s)

k A(s)

at

at time

investing k at time s is equivalent to investing

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Another way to see the previous theorem is as follows. The following three accounts have the same balance at any time bigger than t: 1. An account where a unique deposit of A(0) is made at time zero. 2. An account where a unique deposit of A(s) is made at time s. 3. An account where a unique deposit of A(t) is made at time t.

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Another way to see the previous theorem is as follows. The following three accounts have the same balance at any time bigger than t: 1. An account where a unique deposit of A(0) is made at time zero. 2. An account where a unique deposit of A(s) is made at time s. 3. An account where a unique deposit of A(t) is made at time t. Hence, The present value at time t of an investment of A(s) made at time s is A(t).

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Another way to see the previous theorem is as follows. The following three accounts have the same balance at any time bigger than t: 1. An account where a unique deposit of A(0) is made at time zero. 2. An account where a unique deposit of A(s) is made at time s. 3. An account where a unique deposit of A(t) is made at time t. Hence, The present value at time t of an investment of A(s) made at time s is A(t). This means that: I

If t > s, an investment of A(s) made at time s has an accumulation value of A(t) at time t.

I

If t < s, to get an accumulation of A(s) at time s, we need to invest A(t) at time t. 40/52

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We know that: The present value at time t of an investment of A(s) made at time s is A(t), i.e. A(s) at time s is equivalent to A(t) at time t.

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We know that: The present value at time t of an investment of A(s) made at time s is A(t), i.e. A(s) at time s is equivalent to A(t) at time t. By proportionality, I

The present value at time t of an investment of 1 made at A(t) time s is A(s) , i.e. 1 at time s is equivalent to

A(t) A(s)

at time t.

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We know that: The present value at time t of an investment of A(s) made at time s is A(t), i.e. A(s) at time s is equivalent to A(t) at time t. By proportionality, I

The present value at time t of an investment of 1 made at A(t) time s is A(s) , i.e. 1 at time s is equivalent to

I

A(t) A(s)

at time t.

The present value at time t of an investment of k made at time s is kA(t) A(s) , i.e. k at time s is equivalent to

kA(t) A(s)

at time t.

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Example 2 The accumulation function of a fund follows the function t a(t) = 1 + 20 , t > 0. (i) Michael invests $3500 into the fund at time 1. Find the value of Michael’s fund account at time 4. (ii) How much money needs Jason to invest at time 2 to accumulate $700 at time 4.

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Example 2 The accumulation function of a fund follows the function t a(t) = 1 + 20 , t > 0. (i) Michael invests $3500 into the fund at time 1. Find the value of Michael’s fund account at time 4. (ii) How much money needs Jason to invest at time 2 to accumulate $700 at time 4. Solution: (i) The value of Michael’s account at time 4 is 1+

4

1.20 20 3500 a(4) a(1) = (3500) 1+ 1 = (3500) 1.05 = 4000. 20

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Example 2 The accumulation function of a fund follows the function t a(t) = 1 + 20 , t > 0. (i) Michael invests $3500 into the fund at time 1. Find the value of Michael’s fund account at time 4. (ii) How much money needs Jason to invest at time 2 to accumulate $700 at time 4. Solution: (i) The value of Michael’s account at time 4 is 1+

4

1.20 20 3500 a(4) a(1) = (3500) 1+ 1 = (3500) 1.05 = 4000. 20

(ii) To accumulate $700 at time 4, Jason needs to invest at time 2, 1.1 700 a(2) a(4) = 700 1.2 = 641.67.

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Theorem 3 Present value of a cashflow. If an investment account follows the amount function A(t), t > 0, the (equation of value) present value at time t of the cashflow Deposits Time

C1 t1

C2 t2

··· ···

Cj

A(t) . A(tj )

Cn tn

where 0 ≤ t1 < t2 < · · · < tn , is V (t) =

n X j=1

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Proof. Let s > tn . Time t1 t2 t3 t4 ··· tn

Balance before deposit 0 P1 a(t2 ) 2) C1 a(t j=1 Cj a(tj ) a(t1 ) = P2 a(t3 ) j=1 Cj a(tj ) P3 a(t4 ) j=1 Cj a(tj ) ··· Pn−1 a(tn ) j=1 Cj a(tj )

Balance after deposit C1 P2 a(t2 ) 2) C1 a(t j=1 Cj a(tj ) a(t1 ) + C2 = P3 a(t3 ) j=1 Cj a(tj ) P4 a(t4 ) j=1 Cj a(tj ) ··· Pn a(tn ) j=1 Cj a(tj )

Since the balance after deposit at time t1 is C1 , the balance before 2) deposit at time t2 is a(t a(t1 ) C1 . P 2) Since the balance after deposit at time t2 is 2j=1 Cj a(t a(tj ) , the balance before deposit time t3 is Pat a(t3 ) P2 a(t2 ) a(t3 ) 2 C = j=1 j a(tj ) j=1 Cj a(tj ) . a(t2 ) 48/52

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Hence, the balance at time s is n

n

j=1

j=1

a(s) X a(tn ) X a(s) Cj = Cj . a(tn ) a(tj ) a(tj ) The future value of the cashflow at time t is n

n

j=1

j=1

X a(t) a(t) X a(s) = . Cj Cj a(s) a(tj ) a(tj ) end of proof.

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Notice that the present value at time t of the cashflow Deposits Time

C1 t1

C2 t2

··· ···

Cn tn

is the same as the sum of the present values at time t of n separated investment accounts each following the amount function A, with the j–the account having a unique deposit of Cj at time tj .

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Example 4 The accumulation function of a fund follows the function t a(t) = 1 + 20 , t > 0. Jared invests $1000 into the fund at time 1 and he withdraws $500 at time 3. Find the value of Jared’s fund account at time 5.

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Example 4 The accumulation function of a fund follows the function t a(t) = 1 + 20 , t > 0. Jared invests $1000 into the fund at time 1 and he withdraws $500 at time 3. Find the value of Jared’s fund account at time 5. Solution: The cashflow is deposit/withdrawal Time (in years)

1000 1

−500 . 3

The value of Jared’s account at time 5 is 5 5 1 + 20 1 + 20 a(5) a(5) − 500 = 1000 − 500 1 3 a(1) a(3) 1 + 20 1 + 20 1.25 1.25 =(1000) − (500) = 1190.48 − 543.48 = 647.00. 1.05 1.15

1000

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Manual for SOA Exam FM/CAS Exam 2. Chapter 1. Basic Interest Theory. Section 1.2. Simple interest. c



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Section 1.2. Simple interest.

Simple interest Under simple interest: I

the interest paid over certain period of time is proportional to the length of this period of time and the principal.

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I

if i is the effective annual rate of simple interest, the amount of interest earned by a deposit of k held for t years is kit. The balance at time t years is k + kit = k(1 + it).

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I


I

interest is found using the principal not the earned interest. To find the earned interest, we need to know the amount of principal, not the balance.

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I


I

interest is found using the principal not the earned interest. To find the earned interest, we need to know the amount of principal, not the balance.

I

balances under simple interest follow the proportionality rule and rule about the addition of several deposits/withdrawals. However, the rule ”grows–depends–on–balance” does not hold. 5/21

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Suppose that an account earns simple interest with annual effective rate of i. I

If an investment of 1 is made at time zero, then the balance in this account at time t years is a(t) = 1 + it .

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I

If an investment of k is made at time zero, then the balance in this account at time t years is k(1 + it).

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I

If an investment of k is made at time zero, then the balance in this account at time t years is k(1 + it).

I

If an investment of k is made at time s years, then the balance in this account at time t years, t > s, is k(1 + i(t − s)). Notice that the investment is held for t − s years, and the earned interest is ki(t − s).

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Notice that the amount k(1 + i(t − s)) is not Making an investment of k (1+is) (1

k (1+is)

ka(t) a(s)

=

k(1+it) (1+is) .

at time zero, we have a balance of

+ is) = k at time s. Making an investment of

k (1+is)

at

k (1+is) (1

time zero, we have a balance of + it) at time t. This is not the balance at time t years in an account with an investment of k made at time s years. I

Making an investment of

k (1+is)

at time zero, we have a

k balance of (1+is) (1 + is) = k at time s. But since interest does not earn interest, the amount of interest earned in the k i(t − s). Hence, the balance at time t is period [s, t] is (1+is)

k+ I

k k(1 + is) + ki(t − s) k(1 + it) i(t − s) = = . (1 + is) (1 + is) (1 + is)

Making an investment of k at time s years, we have a balance of k(1 + i(t − s)) at time t. 9/21

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Present value for simple interest

I

A deposit of k made at time s has a future value of k(1 + i(t − s)) at time t, if t > s.

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Present value for simple interest

I

A deposit of k made at time s has a future value of k(1 + i(t − s)) at time t, if t > s.

I

To get a balance of k time s, we need to make a deposit of 1 k 1+i(s−t) at time t, if t < s.

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Present value for simple interest Theorem 1 If deposits/withdrawals are make according with the table, Deposits Time

C1 t1

C2 t2

··· ···

Cn tn

where 0 ≤ t1 < t2 < · · · < tn to an account earning simple interest with annual effective rate of i, then the balance at time t years, where t > tn , is given by B=

n X

Cj (1 + i(t − tj )) =

j=1

n X j=1

Cj +

n X

Cj i(t − tj ).

j=1

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Proof. Time

t1 t2 t3 ··· tk ··· tn t

Deposit/withdr. at that time

Principal after the deposit

C1 C2 C3 ··· Ck ··· Cn 0

C1 C1 + C2 P3 j=1 Cj ··· Pk j=1 Cj ··· Pn Cj Pj=1 n j=1 Cj

Amount of interest earned up to that time 0 C1 i(t2 − t1 ) P2 j=1 Cj i(t3 − tj ) ··· Pk−1 j=1 Cj i(tk − tj ) ··· Pn−1 Cj i(tn − tj ) Pj=1 n j=1 Cj i(t − tj )

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The amount of interest earned up to time t3 is C1 i(t2 − t1 ) + (C1 + C2 )i(t3 − t2 ) = C1 i(t3 − t1 ) + C2 i(t3 − t2 ) =

2 X

Cj i(t3 − tj ).

j=1

The amount of interest earned up to time tk is the amount of interest earned up to time tk−1 plus the amount of interest earned in the period [tk−1 , tk ], which is k−2 X

=

Cj i(tk−1 − tj ) +

j=1

j=1

k−1 X

k−1 X

Cj i(tk−1 − tj ) +

j=1

=

k−1 X

k−1 X

Cj i(tk − tk−1 )

Cj i(tk − tk−1 )

j=1

Cj i(tk − tj ).

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Theorem 2 If deposits/withdrawals are make according with the table, Deposits Time

C1 t1

C2 t2

··· ···

Cn tn

where 0 ≤ t1 < t2 < · · · < tn to an account earning simple interest and the balance at time t years, where t > tn , is B, then the annual effective rate of i P B − nj=1 Cj . i = Pn j=1 Cj (t − tj ) Proof. Solving for i in B = value of i.

Pn

j=1 Cj

+

Pn

j=1 Cj i(t

− tj ), we get the

15/21

c




In the formula, B− i = Pn


Pn

j=1 Cj

j=1 Cj (t

B P− n

− tj )

,

Pn

j=1 Cj

is the total amount of interest earned, j=1 Cj (t − tj ) is the sum of the balances times the amount balances are in the account.

16/21

c





Example 3 Jeremy invests $1000 into a bank account which pays simple interest with an annual rate of 7%. Nine months later, Jeremy withdraws $600 from the account. Find the balance in Jeremy’s account one year after the first deposit was made.

17/21

c





Example 3 Jeremy invests $1000 into a bank account which pays simple interest with an annual rate of 7%. Nine months later, Jeremy withdraws $600 from the account. Find the balance in Jeremy’s account one year after the first deposit was made. Solution: The cashflow of deposits is deposit/withdrawal Time (in years)

1000 0

−600 . 0.75

The balance one year after the first deposit was made is n X

Cj (1 + i(t − tj ))

j=1

=(1000)(1 + (1 − 0)(0.07)) + (−600)(1 + (1 − 0.75)(0.07)) =459.5. 18/21

c





Time

Deposit made at this time

Principal after deposit

0 0.75 1

1000 −600 0

1000 400 400

Amount of interest earned in the last period 0 (1000)(0.07)(0.75) = 52.5 (400)(0.07)(1 − 0.75) = 7

The balance one year after the first deposit was made is 400 + 52.5 + 7 = 459.5.

19/21

c





Example 4 On September 1, 2006, John invested $25000 into a bank account which pays simple interest. On March 1, 2007, John’s wife made a withdrawal of 5000. The accumulated value of the bank account on July 1, 2007 was $20575. Calculate the annual effective rate of interest earned by this account.

20/21

c





Example 4 On September 1, 2006, John invested $25000 into a bank account which pays simple interest. On March 1, 2007, John’s wife made a withdrawal of 5000. The accumulated value of the bank account on July 1, 2007 was $20575. Calculate the annual effective rate of interest earned by this account. Solution: Let September 1, 2006 be time 0. Then, March 1, 2007 6 is time 12 years; and July 1, 2007 is time 10 12 years. The annual effective rate of interest earned by this account is P B − nj=1 Cj 20575 − 25000 + 5000 i = Pn = 10 6 25000 10 j=1 Cj (t − tj ) 12 − 5000 12 − 12 575 = 3%. = 19166.66667

21/21

c




Manual for SOA Exam FM/CAS Exam 2. Chapter 1. Basic Interest Theory. Section 1.3. Compounded interest. c



1/27

c




Section 1.3. Compounded interest.

Compound interest Under compound interest the amount function is A(t) = A(0)(1 + i)t , t ≥ 0, where i is the effective annual rate of interest.

2/27

c





Compound interest Under compound interest the amount function is A(t) = A(0)(1 + i)t , t ≥ 0, where i is the effective annual rate of interest. Under compound interest, the effective rate of interest over a certain period of time depends only on the length of this period, i.e. for each 0 ≤ s < t,

A(t − s) − A(0) A(t) − A(s) = . A(s) A(0)

Notice that A(t) − A(s) A(0)(1 + i)t − A(0)(1 + i)s = = (1 + i)t−s − 1. A(s) A(0)(1 + i)s The effective rate of interest earned in the n–th year is in =

A(n) − A(n − 1) A(0)(1 + i)n − A(0)(1 + i)n−1 = = i. A(n − 1) A(0)(1 + i)n−1 3/27

c





Under compound interest, the present value at time t of a deposit of k made at time s is kA(t) kA(0)(1 + i)t = = k(1 + i)t−s . A(s) A(0)(1 + i)s

4/27

c





Under compound interest, the present value at time t of a deposit of k made at time s is kA(t) kA(0)(1 + i)t = = k(1 + i)t−s . A(s) A(0)(1 + i)s If deposits/withdrawals are made according with the table Deposits Time (in years)

C1 t1

C2 t2

··· ···

Cn tn

where 0 ≤ t1 < t2 < · · · < tn , into an account earning compound interest with an annual effective rate of interest of i, then the present value at time t of the cashflow is V (t) =

n X

Cj (1 + i)t−tj .

j=1

In particular, value of the considered cashflow at time Pn the present −t j zero is j=1 Cj (1 + i) . c



5/27



Example 1 A loan with an effective annual interest rate of 5.5% is to be repaid with the following payments: (i) 1000 at the end of the first year. (ii) 2000 at the end of the second year. (iii) 5000 at the end of the third year. Calculate the loaned amount at time 0.

6/27

c





Example 1 A loan with an effective annual interest rate of 5.5% is to be repaid with the following payments: (i) 1000 at the end of the first year. (ii) 2000 at the end of the second year. (iii) 5000 at the end of the third year. Calculate the loaned amount at time 0. Solution: The cashflow of payments to the loan is Payments Time

1000 1

2000 2

5000 3

The loaned amount at time zero is the present value at time zero of the cashflow of payments, which is (1000)(1.055)−1 + (2000)(1.055)−2 + (5000)(1.055)−3 =947.8672986 + 1796.904831 + 4258.068321 = 7002.840451. 7/27

c





The accumulation function for simple interest is a(t) = 1 + it, which is a linear function. The accumulation function for compound interest is a(t) = (1 + i)t , which is an increasing convex function. We have that (i) If 0 < t < 1, then (1 + i)t < 1 + it. (ii) If 1 < t, then 1 + it < (1 + i)t .

8/27

c





Figure 1: comparison of simple and compound accumulation functions

9/27

c





Usually, we solve for variables in the formula, A(t) = A(0)(1 + i)t , using the TI–BA–II-Plus calculator. To turn on the calculator press ON/OFF . To clear errors press CE/C . It clears the current displays (including error messages) and tentative operations. When entering a number, you realized that you make a mistake you can clear the whole display by pressing CE/C . When entering numbers, if you would like to save some of the entered digits, you can press → as many times as digits you would like to remove. Digits are deleted starting from the last entered digit. It is recommended to set–up the TI-BA–II–Plus calculator to 9 decimals. You can do that doing 2nd , FORMAT , 9 , ENTER , 2nd , QUIT . 10/27

c





We often will use the time value of the money worksheet of the calculator. There are 5 main financial variables in this worksheet: I

The number of periods N .

I

The nominal interest for year I/Y .

I

The present value PV .

I

The payment per period PMT .

I

The future value FV .

You can use the calculator to find one of these financial variables, by entering the rest of the variables in the memory of the calculator and then pressing CPT financial key , where financial key is either N , % i , PV , PMT or FV .

11/27

c





Here, financial key is either N , % i , PV , PMT or FV . I

You can recall the entries in the time value of the money worksheet, by pressing RCL financial key .

I

To enter a variable in the entry financial key , type the entry and press financial key . The entry of variables can be done in any order.

I

To find the value of any of the five variables (after entering the rest of the variables in the memory) press CPT financial key .

I

When computing a variable, a formula using all five variables and two auxiliary variables is used

12/27

c





To set–up C/Y =1 and P/Y =1, do 2nd , P/Y , 1 , ENTER , ↓ , 1 , ENTER , 2nd , QUIT . To check that this is so, do 2nd P/Y ↓ 2nd QUIT . If PMT equals zero, C/Y =1 and P/Y =1, you have the formula,  − N I/Y  PV + FV 1 + = 0. 100

(1)

You can use this to solve for any element of the four elements in the formula A(t) = A(0)(1 + i)t . Unless it is said otherwise, we will assume that the entries for C/Y and P/Y are both 1 and PMT is 0. c


13/27




Example 2 Mary invested $12000 on January 1, 1995. Assuming composite interest at 5 % per year, find the accumulated value on January 1, 2002.

14/27

c





Example 2 Mary invested $12000 on January 1, 1995. Assuming composite interest at 5 % per year, find the accumulated value on January 1, 2002. Solution: A(t) = 12000(1 + 0.05)7 = 16885.21. You can do this in the calculator by entering: 0 PMT 7 N 5 I/Y 12000 PV CPT FV . Note that since the calculator, uses the formula  PV + FV 1 +

I/Y 100

− N 

= 0.

the display in your calculator is negative. 15/27

c





Example 3 At what annual rate of compound interest will $200 grow to $275 in 5 years?

16/27

c





Example 3 At what annual rate of compound interest will $200 grow to $275 in 5 years? Solution: We solve for i in 275 = 200(1 + i)5 and get i = 6.5763%. In the calculator, you do −275 FV 5 N 200 PV CPT I/Y . Since the calculator, uses the formula (1), either the present value or the future value has to be entered as negative number (and the other one as a positive number). If you enter both the present value and the future value as positive values, you get the error message Error 5 . To clear this error message press CE/C .

17/27

c





Example 4 How many years does it take $200 grow to $275 at an effective annual rate of 5%?

18/27

c





Example 4 How many years does it take $200 grow to $275 at an effective annual rate of 5%? Solution: We solve for t in 275 = 200(1 + 0.05)t and get that t = 6.5270 years. In the calculator, you do −275 FV 5 I/Y 200 PV CPT N .

19/27

c





Example 5 At an annual effective rate of interest of 8% how long would it take to triple your money?

20/27

c





Example 5 At an annual effective rate of interest of 8% how long would it take to triple your money? Solution: We solve for t in 3 = (1 + 0.08)t and get t = 14.2749 years. In the calculator, you do −3 FV 8 I/Y 1 PV CPT N .

21/27

c





Example 6 How much money was needed to invest 10 years in the past to accumulate $ 10000 at an effective annual rate of 5%?

22/27

c





Example 6 How much money was needed to invest 10 years in the past to accumulate $ 10000 at an effective annual rate of 5%? Solution: We solve for A(0) in 10000 = A(0)(1 + 0.05)10 and get that A(0) = 6139.13. In the calculator, you do 10000 FV 5 I/Y 10 N CPT PV .

23/27

c





The calculator has a memory worksheet with values in the memory, which stores ten numbers. These ten numbers are called: M0 , · · · , M9 .To enter the number in the display into the i–th entry of the memory, press STO i , where i is an integer from 0 to 9. To recall the number in the memory entry i, press RCL i , where i is an integer from 0 to 9. The command STO + i adds the value in display to the entry i in the memory. You can see all the numbers in the memory by accessing the memory worksheet. To enter this worksheet press 2nd MEM . Use the arrows ↑ , ↓ to move from entry to another. To entry a new value in one entry, type the number and press ENTER .

24/27

c






25/27

c






1000 1

2000 2

5000 3

The loaned amount at time zero is the present value at time zero of the cashflow of payments, which is (1000)(1.055)−1 + (2000)(1.055)−2 + (5000)(1.055)−3 =947.8672986 + 1796.904831 + 4258.068321 = 7002.840451. 26/27

c





Using the calculator, you do −1000 FV 1 N 5.5 I/Y CPT PV and get (1000)(1.055)−1 = 947.8672986. You enter this number in the memory of the calculator doing STO 1 Next doing −2000 FV 2 N CPT PV you find (2000)(1.055)−2 = 1796.904831. Notice that you do not have to reenter the percentage interest rate. You enter this number in the memory of the calculator doing STO 2 Next doing −5000 FV 3 N CPT PV you get (5000)(1.055)−3 = 4258.068321. You enter this number in the memory of the calculator doing STO 3 . You can recall and add the three numbers doing CRCL 1 + CRCL 2 + CRCL 3 = and get 7002.840451. 27/27

c




Manual for SOA Exam FM/CAS Exam 2. Chapter 1. Basic Interest Theory. Section 1.4. Present value and discount. c


Extract from: ”Arcones’ Manual for the SOA Exam FM/CAS Exam 2, Financial Mathematics. Spring 2009 Edition”, available at http://www.actexmadriver.com/

1/19

c




Section 1.4. Present value and discount.

Present value and discount

Suppose that we make an investment of $k in an account earning compound interest with effective annual rate of interest i. t years later the balance in this account is k(1 + i)t . Here, k(1 + i)t is the future value of the investment t years in the future. Under compound interest, balances multiply by (1 + i)t every t years. $k at time s is worth $k(1 + i)t at time s + t.

2/19

c






Suppose that we make an investment of $k in an account earning compound interest with effective annual rate of interest i. t years later the balance in this account is k(1 + i)t . Here, k(1 + i)t is the future value of the investment t years in the future. Under compound interest, balances multiply by (1 + i)t every t years. $k at time s is worth $k(1 + i)t at time s + t. The quantity (1 + i)t is called the t–year interest factor.

3/19

c






Suppose that we make an investment of $k in an account earning compound interest with effective annual rate of interest i. t years later the balance in this account is k(1 + i)t . Here, k(1 + i)t is the future value of the investment t years in the future. Under compound interest, balances multiply by (1 + i)t every t years. $k at time s is worth $k(1 + i)t at time s + t. The quantity (1 + i)t is called the t–year interest factor. The quantity (1 + i) is called the interest factor. $k at time s is worth $k(1 + i) at time s + 1.

4/19

c





Often, we need to find the amount of money t years in the past needed to accumulate certain principal. The present value t years in the past is the amount of money which will accumulate to the principal over t years. In the case of compound interest with effective annual rate of 1 interest i, the present value of $1 t years in the past is (1+i) t . If we invested

1 (1+i)t

t years ago in account earning compound interest,

1 then the current balance is $1. The quantity (1+i) t is called the t t year discount. $k at time s is worth $kν at time s − t. 1 The quantity ν = 1+i is called the discount factor. In order to accumulate $1, we need $ν one year in the past.

5/19

c





Under the accumulation function a(t), a(n) a(n−1) .

I

The n–th year interest factor is

I

The effective rate of interest in the n–th year is . in = a(n)−a(n−1) a(n−1)

I

The n year discount factor is νn =

I

The effective rate of discount in the n–th year is . dn = a(n)−a(n−1) a(n)

a(n−1) a(n) .

in and dn are both proportions of interest over amount values, but in uses the amount value in the past and dn uses the amount value in the future. Since the amount value in the future is bigger than the amount value in the past, dn < in .

6/19

c





Notice that I

the n–th year interest factor is equal to 1 + in .

I

νn = 1 − dn .

I

1 = (1 + in )νn = (1 + in )(1 − dn )

I

{1 unit at time n − 1} ≡ {1 + in units at time n}. So, in dn = 1+i n

I

{1 − dn unit at time n − 1} ≡ {1 units at time n}. So, dn in = 1−d . n

7/19

c





Under compound interest, the effective rate of discount dn is n −(1+i)n−1 i constant dn = a(n)−a(n−1) = (1+i)(1+i) = 1+i . Under n a(n) compound interest, ν=

1 i , d = 1 − ν, d = and (1 − d)(1 + i) = 1. 1+i i +1

8/19

c





Example 1 Peter invests $738 in a bank account. One year later, his bank account is $765. (i) Find the effective annual interest rate earned by Peter in that year. (ii) Find the effective annual discount rate earned by Peter in that year.

9/19

c





Example 1 Peter invests $738 in a bank account. One year later, his bank account is $765. (i) Find the effective annual interest rate earned by Peter in that year. (ii) Find the effective annual discount rate earned by Peter in that year. Solution: (i)Peter earns an interest amount of 765 − 738 = 27. The effective annual interest rate earned by Peter is 27 738 = 3.658537%.

10/19

c





Example 1 Peter invests $738 in a bank account. One year later, his bank account is $765. (i) Find the effective annual interest rate earned by Peter in that year. (ii) Find the effective annual discount rate earned by Peter in that year. Solution: (i)Peter earns an interest amount of 765 − 738 = 27. The effective annual interest rate earned by Peter is 27 738 = 3.658537%. (ii) The effective annual discount rate earned by Peter is 27 765 = 3.529412%.

11/19

c





Example 2 If i = 7%, what are d and ν?

12/19

c





Example 2 If i = 7%, what are d and ν? i Solution: We have that d = 1+i = 1 1 = 1+0.07 = 0.934579. ν = 1+i

0.07 1+0.07

= 6.5421% and

13/19

c





Example 3 If ν = 0.95, what are d and i?

14/19

c





Example 3 If ν = 0.95, what are d and i? Solution: We have that d = 1 − ν = 1 − 0.95 = 0.05 and i = ν1 − 1 = 1−0.95 0.95 = 5.2632%.

15/19

c





Example 4 What is the present value of $5,000 to be received in 7 years at an annual effective rate of discount of 7%?

16/19

c





Example 4 What is the present value of $5,000 to be received in 7 years at an annual effective rate of discount of 7%? Solution: The value is (5000)(1 − 0.07)7 = 3008.504353.

17/19

c





Example 5 At time t = 0, Paul deposits $3500 into a fund crediting interest with an annual discount factor of 0.96. Find the fund value at time 2.5.

18/19

c





Example 5 At time t = 0, Paul deposits $3500 into a fund crediting interest with an annual discount factor of 0.96. Find the fund value at time 2.5. Solution: (3500)(0.96)−2.5 = 3876.055.

19/19

c




Manual for SOA Exam FM/CAS Exam 2. Chapter 1. Basic Interest Theory. Section 1.5. Nominal rates of interest and discount. c



1/24

c




Section 1.5. Nominal rates of interest and discount.

Nominal rate of interest When dealing with compound interest, often we will rates different from the annual effective interest rate. Suppose that an account follows compound interest with an annual nominal rate of interest compounded m times a year of i (m) , then I

$1 at time zero accrues to $(1 +

i (m) m )

at time

1 m

years.

2/24

c





Nominal rate of interest When dealing with compound interest, often we will rates different from the annual effective interest rate. Suppose that an account follows compound interest with an annual nominal rate of interest compounded m times a year of i (m) , then i (m) m ) at (m) + i m ).

I


I

The

1 m –year

interest factor is (1

time

1 m

years.

3/24

c






I


I

The

I

The

1 m –year interest factor is (1 ( m1 -year ) m–thly effective interest

time

rate is

1 m

years.

i (m) m .

4/24

c






i (m) m ) at (m) + i m ).


time

1 m

years.

1 m –year interest factor is (1 (m) I The ( 1 -year ) m–thly effective interest rate is i m m . m (m) I $1 at time zero grows to $ 1 + i in one year. m I

The

5/24

c






i (m) m ) at (m) + i m ).


time

1 m

years.

1 m –year interest factor is (1 (m) I The ( 1 -year ) m–thly effective interest rate is i m m . m (m) I $1 at time zero grows to $ 1 + i in one year. m (m) mt I $1 at time zero grows to $ 1 + i in t years. m I

The

6/24

c






I


I

The

I I I I

time

1 m

years.

1 m –year interest factor is (1 (m) The ( m1 -year ) m–thly effective interest rate is i m . (m) m $1 at time zero grows to $ 1 + i m in one year. (m) mt $1 at time zero grows to $ 1 + i m in t years. (m) mt . The accumulation function is a(t) = 1 + i m 7/24

c





Example 1 Paul takes a loan of $569. Interest in the loan is charged using compound interest. One month after a loan is taken the balance in this loan is $581. (i) Find the monthly effective interest rate, which Paul is charged in his loan. (ii) Find the annual nominal interest rate compounded monthly, which Paul is charged in his loan.

8/24

c





Example 1 Paul takes a loan of $569. Interest in the loan is charged using compound interest. One month after a loan is taken the balance in this loan is $581. (i) Find the monthly effective interest rate, which Paul is charged in his loan. (ii) Find the annual nominal interest rate compounded monthly, which Paul is charged in his loan. Solution: (i) The monthly effective interest rate, which Paul is charged in his loan is i (12) 581 − 569 = = 2.108963093%. 12 569

9/24

c





Example 1 Paul takes a loan of $569. Interest in the loan is charged using compound interest. One month after a loan is taken the balance in this loan is $581. (i) Find the monthly effective interest rate, which Paul is charged in his loan. (ii) Find the annual nominal interest rate compounded monthly, which Paul is charged in his loan. Solution: (i) The monthly effective interest rate, which Paul is charged in his loan is i (12) 581 − 569 = = 2.108963093%. 12 569 (ii) The annual nominal interest rate compounded monthly, which Paul is charged in his loan is i (12) = (12)(0.02108963093) = 25.30755712%. 10/24

c





Two rates of interest or discount are said to be equivalent if they give rise to same accumulation function. Since, the accumulation function under an annual effective rate of interest i is a(t) = (1 + i)t , we have that a nominal annual rate of interest i (m) compounded m times a year is equivalent to an annual effective rate of interest i, if the rates !mt i (m) a(t) = 1 + m and a(t) = (1 + i)t agree. This happens if and only if !m i (m) 1+ = 1 + i. m 11/24

c





Example 1 John takes a loan of 8,000 at a nominal annual rate of interest of 10% per year convertible quarterly. How much does he owe after 30 months?

12/24

c





Example 1 John takes a loan of 8,000 at a nominal annual rate of interest of 10% per year convertible quarterly. How much does he owe after 30 months? Solution: We find

0.10 8000 1 + 4

30 ·4 12

= 8000 (1 + 0.025)10 = 10240.68.

In the calculator, we do: 8000 PV 2.5 I/Y 10 N CPT FV .

13/24

c





The calculator TI–BA–II–Plus has a worksheet to convert nominal rates of interest into effective rates of interest and vice versa. To enter this worksheet press 2nd ICONV . There are 3 entries in this worksheet: NOM , EFF and C/Y . C/Y is the number of times the nominal interest is converted in a year. The relation between these variables is   C/Y EFF NOM  = 1 + 1+ . 100 100 C/Y You can enter a value in any of these entries by moving to that entry using the arrows: ↑ and ↓ . To enter a value in one entry, type the value and press ENTER . You can compute the corresponding nominal (effective) rate by moving to the entry NOM ( EFF ) and pressing the key CPT . It is possible to enter negative values in the entries NOM and EFF . However, the value in the entry C/Y has to be positive. 14/24

c





Example 2 If i (4) = 5% find the equivalent effective annual rate of interest.

15/24

c





Example 2 If i (4) = 5% find the equivalent effective annual rate of interest. 4 Solution: We solve 1 + i = 1 + 0.05 and get i = 5.0945%. In 4 the calculator, you enter the worksheet ICONV and enter: NOM equal to 5 and C/Y equal to 4. Then, go to EFF and press CPT . To quit, press 2nd , QUIT .

16/24

c





Example 3 If i = 5%, what is the equivalent i (4) ?

17/24

c





Example 3 If i = 5%, what is the equivalent i (4) ? (4) 4 Solution: We solve 1 + i 4 = 1 + 0.05 we get that (4) 1/4 i = 4 (1 + 0.05) − 1 = 4.9089%. In the calculator, you enter the worksheet ICONV and enter: EFF equal to 5 and C/Y equal to 4. Then, go to NOM and press CPT . To quit, press 2nd , QUIT .

18/24

c





d (m)

The nominal rate of discount is defined as the value such (m) that 1 unit at the present is equivalent to 1 − dm units invested m1 years ago, i.e. 1 d (m) units at time 0} ≡ 1 unit at time . {1 − m m This implies that ( {1 unit at time 0} ≡

1 1−

d (m) m

1 units at time m

) .

The accumulation function for compound interest under a the nominalrate of discount d (m) convertible m times a year is a(t) = 1 −

d (m) m

1+i =

−mt

. We have that

i (m) 1+ m

!m −1

= (1 − d)

=

d (m) 1− m

!−m . 19/24

c





In the calculator TI–BA–II–Plus, you may: I

given i (m) , find i, by entering i (m) → NOM and m → C/Y , then in EFF press CPT .

I

given i, find i (m) , by entering i → EFF and m → C/Y , then in NOM press CPT .

I

given d (m) , find d, by entering −d (m) → NOM and m → C/Y , then in EFF press CPT . d appears with a negative sign.

I

given d, find d (m) , by entering −d → EFF and m → C/Y , then in NOM press CPT . d (m) appears with a negative sign. 1 1−d

I

given i, find d, by using the formula i =

I

given d, find i, by using the formula d = 1 −

− 1. 1 1+i . 20/24

c





Example 4 If d (4) = 5% find i.

21/24

c





Example 4 If d (4) = 5% find i. Solution: We solve 1 −

d (4) 4

−4

= 1 + i to get d = 4.9070% and

i = 5.1602%. In the calculator, in the ICONV worksheet, we enter −5 in NOM , 4 in C/Y and we find that EFF is −4.9070%, then we do −4.9070 % + 1 = 1/x − 1 = to get i = 5.1602%.

22/24

c





Example 5 If i = 3% find d (2) .

23/24

c





Example 5 If i = 3% find d (2) . (2) −2 Solution: We solve for d (2) in 1 − d 2 = 1 + i. First we find that d = 2.9126% doing 3 % + 1 = 1/x − 1 = Then, using the ICONV worksheet, we get that d (2) = 2.9341%.

24/24

c




Manual for SOA Exam FM/CAS Exam 2. Chapter 1. Basic Interest Theory. Section 1.6. Force of interest. c



1/12

c




Section 1.6. Force of interest.

Force of interest The force of interest δt of an amount function A(t) is defined by 0 (t) d δt = dt ln A(t) = AA(t) .

2/12

c





Force of interest The force of interest δt of an amount function A(t) is defined by 0 (t) d δt = dt ln A(t) = AA(t) . The force of interest is the fraction of the instantaneous rate of change of the accumulation function and the accumulation function.

3/12

c





Force of interest The force of interest δt of an amount function A(t) is defined by 0 (t) d δt = dt ln A(t) = AA(t) . The force of interest is the fraction of the instantaneous rate of change of the accumulation function and the accumulation function. To find the force of interest, we may use the accumulation function, d d d d ln A(t) = ln(A(0)a(t)) = ln(A(0)) + ln(a(t)) dt dt dt dt d = ln(a(t)). dt

4/12

c





Example 1 Consider the amount function A(t) = 25 1 + the force of interest equal of 0.5.

t 3 4 .

At what time is

5/12

c





Example 1 Consider the amount function A(t) = 25 1 + the force of interest equal of 0.5.

t 3 4 .

At what time is

Solution: We have that t 3 t ln(A(t)) = ln 25 1 + = ln 25 + 3 ln 1 + . 4 4 The force of interest is δt =

1 t d d ln 25 + 3 ln 1 + =3 4 ln(A(t)) = dt dt 4 1+

From the equation,

3 4+t

t 4

=

3 . 4+t

= 21 , we get that t = 2.

6/12

c





The force of interest is also called the rate of interest continuously compounded and the continuous interest rate. We have that A(t + h) − A(t) h→0 A(t) · h interest earned over the next h years = lim . h→0 investment at time t · h δt = lim

The nominal annual rate earned in the next m1 years compounded m times a year at time t is m(a t + m1 − a(t)) a t + m1 − a(t) = . a(t) a(t) m1 We have that lim

m→∞

a t+

1 m

− a(t)

a(t) m1

= δt . 7/12

c





Under compound interest, a(t) = (1 + i)t and δt =

d d d ln a(t) = ln(1 + i)t = t ln(1 + i) = ln(1 + i) dt dt dt

Under compound interest, the force of interest is a constant δ, such that δ = ln(1 + i) = − ln ν. Under compound interest, lim i (m) = lim d (m) = δ.

m→∞

m→∞

In the case of simple interest, a(t) = 1 + it and d i δt = dt ln(1 + it) = 1+it . The force of interest is decreasing with t.

8/12

c





From the force of interest δt , we may find the accumulation function a(t), using

Theorem 2 For each t ≥ 0, a(t) = e

Rt 0

δs ds

.

9/12

c





From the force of interest δt , we may find the accumulation function a(t), using

Theorem 2 For each t ≥ 0, a(t) = e

Rt 0

δs ds

.

Proof. Since δs = Z

d ds

ln a(s) and a(0) = 1,

t

Z δs ds =

0

So, a(t) = e

0 Rt 0

δs ds

t

t d ln a(s) ds = ln a(s) = ln a(t). ds 0

.

10/12

c





Example 3 2

A bank account credits interest using a force of interest δt = t3t 3 +2 . A deposit of 100 is made in the account at time t = 0. Find the amount of interest earned by the account from the end of the 4–th year until the end of the 8–th year.

11/12

c





Example 3 2

A bank account credits interest using a force of interest δt = t3t 3 +2 . A deposit of 100 is made in the account at time t = 0. Find the amount of interest earned by the account from the end of the 4–th year until the end of the 8–th year. Rt

Solution: First, we find a(t) = e 0 δs ds . t Z t Z t 3s 2 3 δs ds = ds = ln(s + 2) 3 0 0 s +2 0 3 t + 2 = ln(t 3 + 2) − ln 2 = ln 2 and

t3 + 2 t3 =1+ . 2 2 The amount of interest earned in the considered period is 83 43 100(a(8) − a(4)) = (100) 1 + − 1+ = 22400. 2 2 a(t) = e

Rt 0

δs ds

=e

ln

“

t 3 +2 2

”

=

12/12

c



Chapter 2. Cashflows.

Manual for SOA Exam FM/CAS Exam 2. Chapter 2. Cashflows. Section 2.1. Cashflows. c



1/18

c




Section 2.1. Cashflows.

Net present value of cash flows Recall that a cashflow is a series of payments made at different times. We can represent a cashflow in a table: Investments Time (in periods)

C1 t1

C2 t2

··· ···

Cn tn

Assuming compound interest, the present value of a cashflow at time t is n n X X tj −t V (t) = Cj ν = Cj (1 + i)t−tj , j=1

j=1

where i is the effective interest per period and ν is the discount factor per period. The previous equation is the equation of value. Under the accumulation function a(·), the equation of value is V (t) =

n X j=1

Cj

a(t) . a(tj ) 2/18

c





Example 1 An investor can invest in a project which requires an investment of $37400 at time 0. The investment pays $25000 at time 1 and $15000 at time 2. The investor’s capital is currently earning an effective annual rate of interest of 4.5%. Should the investor invest in the project?

3/18

c





Example 1 An investor can invest in a project which requires an investment of $37400 at time 0. The investment pays $25000 at time 1 and $15000 at time 2. The investor’s capital is currently earning an effective annual rate of interest of 4.5%. Should the investor invest in the project? Solution: The net present value of the investment in the project is 25000(1.045)−1 + 15000(1.045)−2 − 37400 =23923.445 + 13735.9493 − 37400 = 259.3943. Yes, the investor should invest in the project.

4/18

c





A way to analyze the profitability of an investment is to find the present value at time of the cashflow derived from the investment. The net present value of an investment is the present value of the inflows minus the present value of the outflows. Suppose that a company can select between taking two investments. Which investment has the biggest net present value depends on the used interest rate. To valuate investments, companies use their cost of capital. The cost of capital of a company is an estimation of how much the company has to pay for every dollar it borrows. This cost of capital is found using the whole capital components of the company.

5/18

c





Example 2 A company has cost of capital of 7.5% as an annual effective rate of interest. Two investment projects have the following forecasted cash flows: Project A Project B Time in years

−$20,000 −$20,000 0

0 0 1

0 0 2

$25,000 $10,000 3

$10,000 $26,000 4

(i) Find the profit made under each investment project. (ii) Which project has the highest profit? (iii) Compute the net present value for each project using the company’s cost of capital. (iv) Which project has the highest net present value?

6/18

c





Solution:(i) The profit for project A is −20000 + (25000) + (10000) = 15000. The profit for project B is −20000 + (10000) + (26000) = 16000. (ii) Project B has the highest profit. (iii) The present value of project A is −20000 + (25000)(1.075)−3 + (10000)(1.075)−4 = 7612.02. The present value of project B is −20000 + (10000)(1.075)−3 + (26000)(1.075)−4 = 7518.419. (iv) Project A has the highest net present value. 7/18

c





The calculator TI–BA–II–Plus has a cashflow worksheet, which allows to work with cashflows when the deposit times are nonnegative numbers. After entering the cashflow in the calculator, you can find either the present value of the cashflow or the internal rate of return. The internal rate of return is the effective periodic rate of interest. There are 3 keys to enter different parts of this worksheet. I

Pressing the key CF , you can enter the cash flow data.

I

Pressing the key NPV opens a worksheet with two variables NPV and I . Using this worksheet, you can compute the net present value.

I

Pressing the IRR you compute the internal rate of return.

8/18

c





When you press the key CF , you can change the entries: CFo , C01 , F01 , C02 , F02 , . . . ., C24 , F24 . CFo is the initial contribution. C01 is the amount of the first round of contributions. F01 is the (number of payments) frequency of the first round of contributions. C02 is the amount of the second round of contributions. F02 is the frequency of the second round of contributions. C03 , F03 , . . . , C24 , F24 are defined similarly. To move from one entry to the next, use the arrows ↓ and ↑ . To enter a number in one entry, type the number, and then press ENTER . To clear the values in the worksheet, press 2nd CLR WORK while in the worksheet. To exit the worksheet type 2nd QUIT . The cashflow, which we have is: Payments Time

CFo 0

C01 1 to F01

C02 F01+1 to F01+F02

··· ··· 9/18

c





After you have entered the date you can calculate either the net present value or the internal rate of return. I

To calculate the net present value, press the key NPV , enter the periodic interest rate in the entry I , then go the entry NPV using one of the arrows ↓ and ↑ . Finally press CPT .

I

To calculate the internal rate of return, press the keys IRR and CPT . If the equation does not have a solution, you get ”Error 5”. If the equation has several solutions, you get the one with smallest absolute value.

10/18

c





Example 3 Joel wishes to borrow a sum of money. In return, he is prepared to pay as follows: $100 after 1 year, $200 after 2 years, $300 after 3 years, and $400 after 4 years. If i = 12%, how much can he borrow?

11/18

c





Example 3 Joel wishes to borrow a sum of money. In return, he is prepared to pay as follows: $100 after 1 year, $200 after 2 years, $300 after 3 years, and $400 after 4 years. If i = 12%, how much can he borrow? Solution: The cashflow is contributions Time

100 1

200 2

300 3

400 4

He can borrow: (100)(1 + 0.12)−1 + (200)(1 + 0.12)−2 +(300)(1 + 0.12)−3 + (400)(1 + 0.12)−4 = 716.4657955 To do this problem using the CF worksheet. Press CF , and enter CFo =0, C01 =100, F01 =1, C02 =200, F02 =1, C03 =300, F03 =1, C04 =400, F04 =1, 2nd QUIT . Go to NPV , enter I =12, and compute NPV and get 716.4657955. c



12/18




13/18

c






1000 1

2000 2

5000 3

The loaned amount at time zero is the present value at time zero of the cashflow of payments, which is (1000)(1.055)−1 + (2000)(1.055)−2 + (5000)(1.055)−3 = 7002.840451. Press CF , and enter CFo =0, C01 =1000, F01 =1, C02 =2000, F02 =1, C03 =5000, F03 =1, 2nd QUIT . Go to NPV , enter I =5.5, and compute NPV and get 7002.840451. c



14/18



Example 5 Helen borrows $5000 from her credit card account at a nominal annual interest rate of 20% per year convertible monthly. Two months later, she pays $1000 back. Four months after the payment she borrows $2000. How much does she owe one year after the loan is taken out? Solution: The cashflow is inflow/outflow Time (in months)

5000 0

0 1

−1000 2

0 3

0 4

0 5

2000 6

15/18

c






5000 0

0 1

−1000 2

0 3

0 4

0 5

2000 6

16/18

c






5000 0

0 1

−1000 2

0 3

0 4

0 5

2000 6

The equation of value at time 1 year is

0.20 5000 1 + 12 =7125.737519

12

0.20 − 1000 1 + 12

10

0.20 + 2000 1 + 12

6

17/18

c






5000 0

0 1

−1000 2

0 3

0 4

0 5

2000 6

Press CF , and enter CFo =5000, C01 =0, F01 =1, C02 =−1000, F02 =1, C03 =0, F03 =3, C04 =2000, F04 =1, 2nd QUIT . Go to NPV , enter I =1.66666(=20/12), and compute NPV = 5843.69. This is the present value at time 0 of the loan. The future value of the loan at time 1 year is 5843.69(1.01666)12 = 7125.737519. 18/18

c




Manual for SOA Exam FM/CAS Exam 2. Chapter 2. Cashflows. Section 2.2. Method of equated time. c



1/7

c




Section 2.2. Method of equated time.

Given the cashflow Investments Time

C1 t1

C2 t2

··· ···

Cn tn

P we would like to find a time ˜t such that a lump sum C = nj=1 Cj invested at time ˜t is equivalent to the previous cashflow. To find ˜t , we solve the equation, ˜

C νt =

n X

Cj ν tj ,

(1)

j=1

and get ˜t =

P ln( nj=1 Cj ν tj /C ) ln ν

(2)

2/7

c





Method of equated time

The method of equated time consists on approximating ˜t by Pn Pn j=1 Cj tj j=1 Cj tj ¯t = . (3) = Pn C j=1 Cj This is the average time of all the times tj with the weight at tj .

C Pn j

k=1

Ck

3/7

c





The first order Taylor expansion of ν t = (1 + i)−t on i is 1 − ti. So, Equation (1) is approximately C (1 − ti) =

n X

Cj (1 − tj i),

j=1

whose solution is Pn ¯t =

j=1 Cj tj

C

Pn

j=1 = Pn

Cj tj

j=1 Cj

.

If the interest were simple, the future value at time tn of the considered cashflow would be n X C (1 + (tn − t)i) = Cj (1 + (tn − tj )i), j=1 Pn

Cj tj

. whose solution is t = j=1C The approximation to t using the method of equating time is the solution to the considered problem when the accumulation function follows simple interest. c



4/7



Example 1 Payments of $300, $100 and $200 are due at the ends of years 1, 3, and 5, respectively. Assume an annual effective rate of interest of 5% per year. (i) Find the point in time at which a payment of $600 would be equivalent. (ii) Find the approximation to this point using the method of equated time.

5/7

c





Example 1 Payments of $300, $100 and $200 are due at the ends of years 1, 3, and 5, respectively. Assume an annual effective rate of interest of 5% per year. (i) Find the point in time at which a payment of $600 would be equivalent. (ii) Find the approximation to this point using the method of equated time. Solution: (i) The time ˜t solves the equation ˜

600ν t = 300ν + 100ν 3 + 200ν 5 =285.71429 + 86.38376 + 156.70523 = 528.80328, where ν = (1.05)−1 . Hence, ˜t =

ln(600/528.80328) ln(1.05)

= 2.58891.

6/7

c





Example 1 Payments of $300, $100 and $200 are due at the ends of years 1, 3, and 5, respectively. Assume an annual effective rate of interest of 5% per year. (i) Find the point in time at which a payment of $600 would be equivalent. (ii) Find the approximation to this point using the method of equated time. Solution: (i) The time ˜t solves the equation ˜

600ν t = 300ν + 100ν 3 + 200ν 5 =285.71429 + 86.38376 + 156.70523 = 528.80328, = 2.58891. where ν = (1.05)−1 . Hence, ˜t = ln(600/528.80328) ln(1.05) (ii) The equated time approximation to this point is ¯t =

(300)(1) + (100)(3) + (200)(5) = 2.666667. 600 7/7

c




Manual for SOA Exam FM/CAS Exam 2. Chapter 2. Cashflows. Section 2.3. Yield rates. c



1/16

c




Section 2.3. Yield rates.

Yield of return Suppose that the future value at time t of the cashflow: Investments Time

V0 0

C1 t1

C2 t2

··· ···

Cn tn

is FV . Then, the rate of return i of the investment satisfies the equation, FV = V0 ν −t +

n X

Cj ν tj −t = V0 (1 + i)t +

j=1

n X

Cj (1 + i)t−tj .

j=1

The rate of return i, i > −1, solving this equation is called the yield rate of return or internal rate of return. This equation can have either no solutions, or one solution, or several solutions. We are interested in values of i with i > −1. If i < −1, then (1 + i)n > 0 is n is even and (1 + i)n < 0 is n is odd. Values of i with i ≤ −1 do not make any sense. c



2/16



Example 1 Suppose that John invest $3000 in a business. One year later, John sells half of this business to a partner for $6000. Two years after the beginning, the business is in red and John has to pay $4000 to close this business. What is the rate of interest John’s got in his investment?

3/16

c





Example 1 Suppose that John invest $3000 in a business. One year later, John sells half of this business to a partner for $6000. Two years after the beginning, the business is in red and John has to pay $4000 to close this business. What is the rate of interest John’s got in his investment? Solution: The cashflow is: Inflow Time

−3000 0

6000 1

−4000 2

Since John lost money, one expect that i is negative. However, there is no solution. We have to solve −3000(1 + i)2 + 6000(1 + i) − 4000 = 0, or 3(1 + i)2 − 6(1 + i) + 4 = 0. Using the quadratic formula, √ √ 6 ± 62 − 4 · 4 · 3 6 ± −12 1+i = = . 2 2 There is no solution. c


4/16




Example 2 What is the yield rate on a transaction in which a person makes payments of $100 immediately and $100 at the end of two years, in exchange for a payment of $201 at the end of one year? Find all possible solutions.

5/16

c





Example 2 What is the yield rate on a transaction in which a person makes payments of $100 immediately and $100 at the end of two years, in exchange for a payment of $201 at the end of one year? Find all possible solutions. Solution: The cashflow is: Inflow Time

−100 0

201 1

−100 2

We have to solve −100 + 201(1 + i)−1 − 100(1 + i)−2 = 0, or 100(1 + i)2 − 201(1 + i)1 + 100 = 0. Using the quadratic formula, √ √ 201 ± 2012 − 4 · 100 · 100 201 ± 201 1+i = = . 200 200 The two solutions are i = 10.5124922% and i = −9.512492197%. 6/16

c





Since the internal rate of return could either do not exist or have several solutions, it is not a good indication of the performance of general investment strategy. However there exists a unique rate of return i with i > −1 if either all outflows happen before all the inflows, or all inflows happen before all the outflows.

7/16

c





Suppose that you an investment strategy consisting of investing (positive) payments of C1 , . . . , Cm at times t1 < · · · < tm . At times s1 < · · · < sn , we get respective (positive) returns P1 , . . . , Pn , where s1 > tm . The cashflow is −C1 t1

Inflows Time

−C2 t2

··· ···

−Cm tm

P1 s1

P2 s2

··· ···

Pn sn

In this case, there exists a unique solution to the equation n X

Pk (1 + i)−sk −

Pk Pk=1 n I Pk Pk=1 n I k=1 Pk I

Cj (1 + i)−tj = 0, i > −1.

j=1

k=1

Besides, Pn

m X

>

Pm

Cj , then i > 0.

<

Pj=1 m

Cj , then i < 0.

=

Pj=1 m

j=1 Cj ,

then i = 0. 8/16

c





Example 3 As the budgeting officer for Road Kill Motors Inc., you are evaluating the purchase of a new car factory. The cost of the factory is $4 million today. It will provide inflows of $1.4 million at the end of each of the first three years. Find the effective rate of interest which this investment will provide your company.

9/16

c





Example 3 As the budgeting officer for Road Kill Motors Inc., you are evaluating the purchase of a new car factory. The cost of the factory is $4 million today. It will provide inflows of $1.4 million at the end of each of the first three years. Find the effective rate of interest which this investment will provide your company. Solution: The cashflow is Contributions −4 1.4 1.4 1.4 Time 0 1 2 3 An equation of value for the cashflow is 0 = 4 − (1.4)(1 + i)−1 − (1.4)(1 + i)−2 − (1.4)(1 + i)−3 . In the TI–BA–II–Plus calculator, we can find i, by going to CF , and enter CFo =−4, C01 =1.4, F01 =3, 2nd , QUIT . We can move between different entries using the arrows ↓ and ↑ . Press IRR CPT and get IRR = i = 2.47974%. c



10/16



Example 4 Find the internal rate of return such that a payment of 400 at the present, 200 at the end of one year, and 300 at the end of two years, accumulate to 1000 at the end of 3 years.

11/16

c





Example 4 Find the internal rate of return such that a payment of 400 at the present, 200 at the end of one year, and 300 at the end of two years, accumulate to 1000 at the end of 3 years. Solution: The cashflow is Contributions Time

−400 0

−200 1

−300 2

1000 3

An equation of value for the cashflow is 0 = −400(1 + i)3 − 200(1 + i)2 − 300(1 + i) + 1000. In the TI–BA–II–Plus calculator, we can find i, by going to CF , and enter CFo =−400, C01 =−200, F01 =1, C02 =−300, F02 =1, C03 =1000, F03 =1, 2nd , QUIT . We can move between different entries using the arrows ↓ and ↑ . Press IRR CPT and get IRR = 5.0709%. c


12/16




Example 5 An investment fund is established at time 0 with a deposit of $5000. $6000 is added at the end of 6 months. The fund value, including interest, is $11500 at the end of 1 year. Find the internal rate of return as a annual nominal rate convertible semiannually.

13/16

c





Example 5 An investment fund is established at time 0 with a deposit of $5000. $6000 is added at the end of 6 months. The fund value, including interest, is $11500 at the end of 1 year. Find the internal rate of return as a annual nominal rate convertible semiannually. Solution: The cashflow is Investments 5000 6000 11500 Time (in half years) 0 1 2 An equation of value for the cashflow is !−1 !−2 i (2) i (2) 0 = (5000) + (6000) 1 + − (11500) 1 + . 2 2 In the TI–BA–II–Plus calculator, press CF , and enter CFo =5000, C01 =6000, F01 =1, C02 =−11500, F02 =1. (2) Press IRR , CPT and get IRR = i 2 = 3.095064303% and (2) i (2) = 6.190128606%. The six–month effective interest rate is i 2 . c



14/16



Example 6 An investment fund is established at time 0 with a deposit of $5000. $6000 is added at the end of 6 months. The fund value, including interest, is $11500 at the end of 1 year. Find the internal rate of return as a annual nominal rate convertible monthly.

15/16

c





Example 6 An investment fund is established at time 0 with a deposit of $5000. $6000 is added at the end of 6 months. The fund value, including interest, is $11500 at the end of 1 year. Find the internal rate of return as a annual nominal rate convertible monthly. Solution: The cashflow is Investments 5000 6000 11500 Time (in months) 0 6 12 An equation of value for the cashflow is !−6 !−12 i (12) i (12) 0 = (5000) + (6000) 1 + − (11500) 1 + . 12 12 In the TI–BA–II–Plus calculator, press CF , and enter CFo =5000, C01 =0, F01 =5, C02 =6000, F02 =1, C03 =0, F03 =5, C04 =−11500, F04 =1. Press IRR , CPT and get (12) IRR = i 12 = 0.509314804% and i (12) = 6.111777648%. c



16/16


Manual for SOA Exam FM/CAS Exam 2. Chapter 2. Cashflows. Section 2.4. Dollar–weighted and time–weighted rates of return. c



1/9

c




Section 2.4. Dollar–weighted and time–weighted rates of return.

Dollar–weighted and time–weighted rates of return If the cashflow Investments Time

V0 0

C1 t1

C2 t2

··· ···

Cn tn

has future value FV at time t, then its equation of value is FV = V0 (1 + i)t +

n X

Cj (1 + i)t−tj .

j=1

Using the first order Taylor expansion 1 + it of (1 + i)t , the previous equation of value is approximately, FV = V0 (1 + it) +

n X

Cj (1 + i(t − tj )).

(1)

j=1

Observe Equation (1) represents the future value of the cashflow when simple interest is used. c



2/9



The interest rate i which solves FFV = V0 (1 + it) +

n X

Cj (1 + i(t − tj )).

j=1

is called the dollar weighted rate of return, which is P FV − V0 − nj=1 Cj P . i= V0 t + nj=1 (t − tj )Cj

(2)

V0 can be interpreted as the initial balance in an account. Cj is the deposit at time tj . FV is the Pnfinal balance in the account at time t. Hence, I = FV − V − j=1 Cj is the interest earned in the Pn 0 account. V0 t + j=1 (t − tj )Cj is the sum of the deposits multiplied by the time which the deposits are in the account. 3/9

c





Example 1 On January 1, 2000, the balance in account is $25200. On April 1, 2000, $500 are deposited in this account and on July 1, 2001, a withdraw of $1000 is made. The balance in the account on October 1, 2001 is $25900. What is the annual rate of interest in this account according with the dollar–weighted method?

4/9

c





Example 1 On January 1, 2000, the balance in account is $25200. On April 1, 2000, $500 are deposited in this account and on July 1, 2001, a withdraw of $1000 is made. The balance in the account on October 1, 2001 is $25900. What is the annual rate of interest in this account according with the dollar–weighted method? Solution: The cashflow Investments Time in years

25200 0

500 3/12

−1000 18/12

has a FV at time 21/12 of $25900. So, the annual dollar–weighted rate of interest is 25900 − 25200 − 500 + 1000 (25200)(21/12) + (500)(18/12) − 1000(3/12) 1200 1200 = = = 2.6906% 44100 + 750 − 250 44600 5/9

c





Suppose that we make investments in a fund over time and we know the outstanding balance before each deposit or withdrawal occurs. Let B0 be the initial balance in the fund. Let Bj be the balance in the fund immediately before time tj , for 1 ≤ j ≤ n. Let Wj be the amount of each deposit or withdrawal at time tj . Wj > 0 for deposits and Wj < 0 for withdrawal. In a table, we have: Time Balance before depos./withdr. Depos./Withdr. Balance after depos./withdr.

0

t1

t2

···

tn−1

tn

−

B1

B2

···

Bn−1

Bn

−

W1

W2

···

Wn−1

−

B0

B1 + W1

B2 + W2

···

Bn−1 + Wn−1

−

The time–weighted annual rate of return i is the solution of (1 + i)tn =

B1 B2 B3 Bn · · ··· . B0 B1 + W1 B2 + W2 Bn−1 + Wn−1 6/9

c





In the j–th period of time, the balance of the fund has changed from Bj−1 + Wj−1 to Bj . So, the interest factor rate in the j–th Bj period of time is 1 + ij = Bj−1 +W , where ij is the effective rate j−1 of return in the period [tj−1 , tj ]. Observe that if the investment followed an annual effective rate of interest of i, the interest factor from time tj−1 to time tj would be (1 + i)tj −tj−1 . Assuming that 1 + ij = (1 + i)tj −tj−1 , we get that (1+i1 )(1+i2 ) · · · (1+in ) = (1+i)t1 (1+i)t2 −t1 · · · (1+i)tn −tn−1 = (1+i)tn . The time–weighted annual rate of return i is the solution of (1 + i)tn =

B2 B3 Bn B1 · · ··· . B0 B1 + W1 B2 + W2 Bn−1 + Wn−1

Usually, the account balance does not follow compound interest with a fixed effective rate i. Usually, 1 + ij and (1 + i)tj −tj−1 may be different. 7/9

c





Example 2 For an investment account, you are given: Date Account Balance (before deposit or withdrawal) Deposit Withdrawal

11/1/04

3/1/05

8/1/05

2/1/06

4/1/06

14,516

14,547

18,351

16,969

18,542

– –

3,000 –

– 2,000

2500 –

– –

Calculate the annual effective yield rate by the time weighted method.

8/9

c





Example 2 For an investment account, you are given: Date Account Balance (before deposit or withdrawal) Deposit Withdrawal

11/1/04

3/1/05

8/1/05

2/1/06

4/1/06

14,516

14,547

18,351

16,969

18,542

– –

3,000 –

– 2,000

2500 –

– –

Calculate the annual effective yield rate by the time weighted method. Solution: The annual effective yield rate i by the time weighted method satisfies B1 2 3 n (1 + i)17/12 = B · B1B+W · B2B+W · s Bn−1B+W 0 1 2 n−1 18351 16969 18542 = 14547 = 1.035877 14516 14547+3000 18351−2000 16969+2500 and i = 2.5193371%. c


9/9



Manual for SOA Exam FM/CAS Exam 2. Chapter 2. Cashflows. Section 2.5. Investment and portfolio methods. c



1/8

c




Section 2.5. Investment and portfolio methods.

Investment and portfolio methods Suppose that an investment fund pools money from several identities (individuals or corporations) and makes investments on behalf of them. Then, the fund faces the question: how to allocate the returns between the different identities? There are two main ways to allocate interest to the various accounts: the portfolio method and the investment year method. The portfolio method is an accounting method that credits all funds one specified current rate of interest, regardless of when the money was placed in the account. Usually this rate of interest changes from year to year. Let i y denote the annual interest rate credited in year y . If x is invested at the beginning of the year y , then the balance in the account in the year y + t is x

t−1 Y

(1 + i y +j ) = x(1 + i y )(1 + i y +1 ) · · · (1 + i y +t−1 ).

j=0 2/8

c





Example 1 Suppose that an investment account credits investors using the portfolio method with the annual rates in the following table: Calendar year of portfolio rate y 1999 2000 2001 2002

Portfolio rates iy 4.50% 5.50% 4.00% 6.50%

Suppose that 100 was invested on January 1, 1999. (i) Find the balance on January 1, 2000. (ii) The balance on January 1, 2001. (iii) The balance on July 1, 2001. 3/8

c





Solution: (i) The balance on January 1, 2000, is (100)(1.045) = 104.5. (ii) The balance on January 1, 2001, is (100)(1.045)(1.055) = 110.2475. (iii) The balance on July 1, 2001, is (100)(1.045)(1.055)(1.04)1/2 = 112.4308.

4/8

c





The investment year method is an accounting method in which an investment fund keeps records of the interest rates it earns annually on funds assigned each year to accounts within the general account. The investment year method is also called the new money method. We will assume that accounts are made according with the year at which the money was invested. For example, suppose that an investment account credits investors according with the investment year method using the following table: Calendar year of original investment y 1999 2000 2001 2002 2003

Investment year rates i1y 4.25% 4.56% 4.05% 4.45% 4.25%

i2y 4.35% 4.73% 4.04% 4.15% 4.35%

i3y 4.47% 4.75% 4.13% 4.23% 4.55%

i4y 4.57% 4.98% 4.17% 4.36% 9.55%

i5y 4.70% 4.04% 4.24% 4.44% 5.65% 5/8

c




Calendar year of original investment y 1999 2000 2001 2002 2003


Investment year rates i1y 4.25% 4.56% 4.05% 4.45% 4.25%

i2y 4.35% 4.73% 4.04% 4.15% 4.35%

i3y 4.47% 4.75% 4.13% 4.23% 4.55%

i4y 4.57% 4.98% 4.17% 4.36% 9.55%

i5y 4.70% 4.04% 4.24% 4.44% 5.65%

This means that money invested during 2000 earns an effective rate of interest of 4.56% during 2000, it earns an effective rate of interest of 4.73% during 2001, and so on. For example, if an account is open with an investment of x invested on January 1, 2000, then: the balance on January 1, 2001 is (100)(1.0456); the balance on January 1, 2002 is (100)(1.0456)(1.0473); and so on. 6/8

c





Example 2 An investment fund applies the investment year method for the first two years, after which a portfolio rate is used. The following table of interest rates is used: Calendar year of original investment y 2000 2001 2002 2003 2004 2005

Investment year rates i1y i2y 5.25% 5.25% 5.35% 5.35% 5.45% 5.45% 5.45% 5.45% 5.50% 5.35% 5.50% 5.55%

Portfolio rates i y +2 5.40% 5.65% 5.10% 5.34% 5.55% 5.65%

7/8

c





(i) Ashley invests $1000 into the fund on January 1, 2000. The investment year method is applicable for the first two years, after which a portfolio rate is used. Calculate Ashley’s account accumulation on January 1, 2006. (ii) Elizabeth invests $1000 into the fund on January 1, 2000. But, she redeemed her investment from the fund at the end of every year and reinvested the money at the new money rate. Calculate Elizabeth’s accumulation account on January 1, 2006. Solution: (i) Ashley’s account accumulation on January 1, 2006 is (1000)(1.0525)(1.0525)(1.0540)(1.0565)(1.0510)(1.0534) = 1365.684. (ii) Elizabeth’s investment value on January 1, 2006 is (1000)(1.0525)(1.0535)(1.0545)(1.0545)(1.055)(1.050) = 1365.814.

8/8

c




Manual for SOA Exam FM/CAS Exam 2. Chapter 2. Cashflows. Section 2.6. Continuous payments. c



1/10

c




Section 2.6. Continuous payments.

Continuous payments

Suppose that the payments are made very often. Then by approximation, instead of a sum, we have an integral. It is like the payments are made continuously. Let V (t) be the outstanding fund balance at time t of the cashflow. Assume that contributions are made continuously at an instantaneous rate C (t), then the equation of value is Z t t V (t) = V (0)(1 + i) + C (s)(1 + i)t−s ds. (1) 0

2/10

c





(1) appears as the limit of the equation of value for the cashflow: Inflow Time

V (0) 0

C (t1 )(t1 − 0) t1

C (t2 )(t2 − t1 ) t2

··· ···

C (tn )(tn − tn−1 ) tn

as max1≤j≤n (tj − tj−1 ) → 0, where 0 = t0 < t1 < t2 < · · · < tm = t. The equation of value at time t for this cashflow is n X V (t) = V (0)(1 + i)t + C (tj )(tj − tj−1 )(1 + i)t−tj , j=1

which tends to V (t) = V (0)(1 + i)t +

Z

t

C (s)(1 + i)t−s ds

0

as max1≤j≤n (tj − tj−1 ) → 0. Recall that the Riemann integral of a function f is defined as Z t n X f (s) ds = lim f (tj )(tj − tj−1 ). 0

max1≤j≤n (tj −tj−1 )→0

c


j=1


3/10



Example 1 A continuous–year annuity pays a constant rate 1 at time t where 0 ≤ t ≤ n. Interest is compounded with an annual rate of interest of i. (i) Find the present value of the annuity at time 0. (ii) Find the future value of the annuity at time n. Solution: (i) The present value of this continuous annuity is Z PV =

−t

(1 + i) 0

=

n

Z dt =

n

e 0

−t ln(1+i)

−e −t ln(1+i) n dt = ln(1 + i) 0

e −n ln(1+i) 1 − (1 + i)−n 1 − = . ln(1 + i) ln(1 + i) ln(1 + i)

4/10

c





Example 1 A continuous–year annuity pays a constant rate 1 at time t where 0 ≤ t ≤ n. Interest is compounded with an annual rate of interest of i. (i) Find the present value of the annuity at time 0. (ii) Find the future value of the annuity at time n. Solution: (i) The present value of this continuous annuity is Z PV =

−t

(1 + i) 0

=

n

Z dt =

n

e 0

−t ln(1+i)

−e −t ln(1+i) n dt = ln(1 + i) 0

e −n ln(1+i) 1 − (1 + i)−n 1 − = . ln(1 + i) ln(1 + i) ln(1 + i)

(ii) The future value of the continuous annuity at time n is Z n (1 + i)n − 1 FV = (1 + i)n−t dt = (1 + i)n PV = . ln(1 + i) 0 5/10

c





If instead of compound interest, the time value of money follows the accumulation function a(t), then the future value at time t of an initial outstanding balance V (0) and continuous payments C (s), in the interval 0 ≤ s ≤ t is Z t a(t) V (t) = V (0)a(t) + C (s) ds. a(s) 0

6/10

c





Example 2 3

t The force of interest at time t is δt = 10 . Find the present value of a four–year continuous annuity which has a rate of payments at time t of 5t 3 .

Solution: The accumulation function is Z t a(t) = exp δs ds 0

7/10

c





Example 2 3


Solution: The accumulation function is Z t Z a(t) = exp δs ds = exp 0

0

t

s3 ds 10

8/10

c





Example 2 3



0

t

s3 ds 10

t4

= e 40 .

9/10

c





Example 2 3



0

t

s3 ds 10

t4

= e 40 .

The present value of the four-year continuous annuity is 4 Z 4 Z 4 4 −s 4 C (s) 3 − s40 ds = 5s e ds = −50e 40 = 50 − 50e −6.4 a(s) 0 0 0 =49.91692.

10/10

c



Chapter 3. Annuities.

Manual for SOA Exam FM/CAS Exam 2. Chapter 3. Annuities. Section 3.1. Geometric series. c



1/??

c




We use the summation notation

Section 3.1. Geometric series.

Pn

i=m xi

to mean

xm + xm+1 + · · · + xn−1 + xn . Usually arithmetic rules hold. In particular: Pn Pn Pn I i=m xi + i=m yi i=m (xi + yi ) =

2/??

c






Pn

i=m xi

to mean

xm + xm+1 + · · · + xn−1 + xn . Usually arithmetic rules hold. In particular: Pn Pn Pn I i=m xi + i=m yi i=m (xi + yi ) = Pn Pn I i=m xi . i=m axi = a

3/??

c






Pn

i=m xi

to mean

xm + xm+1 + · · · + xn−1 + xn . Usually arithmetic rules hold. In particular: Pn Pn Pn I i=m xi + i=m yi i=m (xi + yi ) = Pn Pn I i=m xi . i=m axi = a I

If m ≤ k ≤ n, n X

xi =

i=m

k X

xi +

i=m

n X

xi .

i=k+1

4/??

c






Pn

i=m xi

to mean

xm + xm+1 + · · · + xn−1 + xn . Usually arithmetic rules hold. In particular: Pn Pn Pn I i=m xi + i=m yi i=m (xi + yi ) = Pn Pn I i=m xi . i=m axi = a I

If m ≤ k ≤ n, n X

xi =

i=m I

Pn

i=m

k X

xi +

i=m

n X

xi .

i=k+1

1 = n − m + 1.

5/??

c





Definition 1 The sequence of real numbers arithmetic sequence.

∞ n=0

= {a + nd}∞ n=0 is called an

6/??

c






∞ n=0


∞ The sequence {xn }∞ n=0 = {a + nd}n=0 satisfies that for each n ≥ 1, xn−1 + d = xn . Notice that

xn−1 + d = a + (n − 1)d + d = a + nd = xn .

7/??

c






∞ n=0



xn−1 + d = a + (n − 1)d + d = a + nd = xn .

Theorem 1 If a sequence {xn }∞ n=0 of real numbers satisfies xn = xn−1 + d, for each n ≥ 1, then xn = x0 + nd for each n ≥ 1.

8/??

c






∞ n=0



xn−1 + d = a + (n − 1)d + d = a + nd = xn .

Theorem 1 If a sequence {xn }∞ n=0 of real numbers satisfies xn = xn−1 + d, for each n ≥ 1, then xn = x0 + nd for each n ≥ 1.

Proof. The proof is by induction on n. The case n = 0 is obvious. Assume that the case n holds. Then, xn+1 = xn + d = x0 + nd + d = x0 + (n + 1)d. 9/??

c





Theorem 2 P n(n+1) n . j=1 j = 2

10/??

c





Theorem 2 P n(n+1) n . j=1 j = 2 Proof.

2

n X

j = (1 + 2 + · · · + (n − 1) + n) + (1 + 2 + · · · + (n − 1) + n)

j=1

=(1 + 2 + · · · + (n − 1) + n) + (n + (n − 1) + · · · + 2 + 1) =(1 + n) + (2 + n − 1) + (3 + n − 2) · · · + (n + 1) =(n + 1) + (n + 1) + (n + 1) · · · + (n + 1) = n(n + 1).

11/??

c






2

n X

j = (1 + 2 + · · · + (n − 1) + n) + (1 + 2 + · · · + (n − 1) + n)

j=1

=(1 + 2 + · · · + (n − 1) + n) + (n + (n − 1) + · · · + 2 + 1) =(1 + n) + (2 + n − 1) + (3 + n − 2) · · · + (n + 1) =(n + 1) + (n + 1) + (n + 1) · · · + (n + 1) = n(n + 1).

Previous theorem can be proved by induction.

12/??

c






2

n X

j = (1 + 2 + · · · + (n − 1) + n) + (1 + 2 + · · · + (n − 1) + n)

j=1

=(1 + 2 + · · · + (n − 1) + n) + (n + (n − 1) + · · · + 2 + 1) =(1 + n) + (2 + n − 1) + (3 + n − 2) · · · + (n + 1) =(n + 1) + (n + 1) + (n + 1) · · · + (n + 1) = n(n + 1).

Previous theorem can be proved P by induction. Note that in the summation nj=1 j, there are n numbers and the Pn n(n+1) . average of these numbers is n+1 j=1 j = 2 . Hence, 2 13/??

c





For an arithmetic sequence, n X j=0

(a + jd) =

n X j=0

a+d

n X j=0

j = (n + 1)a + d

n(n + 1) . 2

14/??

c





Example 1 Find

P100

k=10 k.

15/??

c





Example 1 Find

P100

k=10 k.

Solution 1:P P P9 100 100 k=10 k = k=1 k − k=1 k =

(100)(101) 2

−

(9)(10) 2

= 5005.

16/??

c





Example 1 Find

P100

k=10 k.

Solution 1:P P P9 100 100 k=10 k = k=1 k − k=1 k =

(100)(101) 2

−

(9)(10) 2

= 5005.

Solution 2: By the change of variables k = j + 9, 100 X k=10

k=

91 X j=1

(j + 9) =

(91)(92) + (9)(91) = 5005. 2

Notice that if k = 10, then j = 9; and if k = 100, then j = 91.

17/??

c





Definition 2 The sequence {ar n }∞ n=0 is called a geometric sequence.

18/??

c





Definition 2 The sequence {ar n }∞ n=0 is called a geometric sequence. The geometric sequence {ar n }∞ n=0 satisfies that for each n ≥ 1, rxn−1 = rar n−1 = ar n = xn , where xn = ar n .

19/??

c





Theorem 3 If a sequence satisfies xn = rxn−1 , for each n ≥ 1, then xn = x0 r n for each n ≥ 1.

20/??

c





Theorem 3 If a sequence satisfies xn = rxn−1 , for each n ≥ 1, then xn = x0 r n for each n ≥ 1.

Proof. The proof is by induction on n. The case n = 0 is obvious. Assume that the case n holds. Then, xn+1 = rxn = rx0 r n = x0 r n+1 = xn+1 .

21/??

c





Theorem 4 For any r ∈ R, r

n+1

− 1 = (r − 1)

n X

rj.

j=0

Proof. Pn

j (r

= (r +

r2

j=0 r

− 1) = (1 + r + r 2 + · · · + r n )(r − 1) + · · · + r n+1 ) − (1 + r + r 2 + · · · + r n ) = r n+1 − 1.

22/??

c





In particular, we that (x 2 − 1) = (x − 1)(1 + x), (x 3 − 1) = (x − 1)(1 + x + x 2 ), (x 4 − 1) = (x − 1)(1 + x + x 2 + x 3 ), (x 5 − 1) = (x − 1)(1 + x + x 2 + x 3 + x 4 ).

23/??

c





Corollary 1 P n+1 (i) If r 6= 1, nj=0 r j = r r −1−1 . Pn (ii) If r = 1, j=0 r j = n + 1.

24/??

c





Corollary 1 P n+1 (i) If r 6= 1, nj=0 r j = r r −1−1 . Pn (ii) If r = 1, j=0 r j = n + 1.

Proof. (i) If r 6= 1, from r n+1 − 1 = (r − 1) n X

rj =

j=0

Pn

j=0 r

j,

we get that

r n+1 − 1 . r −1

(ii) If r = 1, n X j=0

rj =

n X

1 = n + 1.

j=0

25/??

c





Example 2 Find

P20

k=5 2

k.

26/??

c





Example 2 Find

P20

k=5 2

k.

Solution: 20 X k=5

k

2 =

20 X k=0

k

2 −

4 X k=0

2k =

221 − 1 25 − 1 − = 221 −25 = 2097120. 2−1 2−1

27/??

c





We also have that for 1 ≤ m ≤ n and r 6= 1, n X j=m

j

r =

n X j=0

j

r −

m−1 X j=0

rj =

r n+1 − 1 r m − 1 r n+1 − r m − = . r −1 r −1 r −1

28/??

c





We also have that for 1 ≤ m ≤ n and r 6= 1, n X

j

r =

j=m

n X j=0

j

r −

m−1 X j=0

rj =

r n+1 − 1 r m − 1 r n+1 − r m − = . r −1 r −1 r −1

Example 3 Find

P20

k=5 2

k.

29/??

c





We also have that for 1 ≤ m ≤ n and r 6= 1, n X

j

r =

j=m

n X

j

r −

j=0

m−1 X

rj =

j=0

r n+1 − 1 r m − 1 r n+1 − r m − = . r −1 r −1 r −1

Example 3 Find

P20

k=5 2

Solution:

k.

P20

k k=5 2

=

221 −25 2−1

= 221 − 25 = 2097120.

30/??

c




Theorem 5 P ∞

For |r | < 1,

j=0 r

j

=


1 1−r .

31/??

c




Theorem 5 P ∞

For |r | < 1,

j=0 r

j

=


1 1−r .

Proof. If |r | < 1, then ln(|r |) < 0 and |r n+1 | ≤ |r |n+1 = e (n+1) ln(|r |) → 0, as n → ∞. Hence, ∞ X j=0

r j = lim

n→∞

n X

1 r n+1 − 1 = . n→∞ r − 1 1−r

r j = lim

j=0

32/??

c





Corollary 2 For |r | < 1, ∞ X

rj =

j=n

rn . 1−r

33/??

c





Corollary 2 For |r | < 1, ∞ X

rj =

j=n

rn . 1−r

Proof. By the change of variables j = k + n, ∞ X j=n

rj =

∞ X

r k+n = r n

k=0

∞ X k=0

rk =

rn . 1−r

34/??

c





Example 4 Find

P∞

k=9 3

−k .

35/??

c





Example 4 Find

P∞

k=9 3

−k .

Solution: ∞ X k=9

3−k =

3−9 = 0.0000762079. 1 − (1/3)

36/??

c





Theorem 6 For r 6= 1, n X j=1

jr j =

nr n+2 − (n + 1)r n+1 + r . (r − 1)2

37/??

c





Theorem 6 For r 6= 1, n X

jr j =

j=1

nr n+2 − (n + 1)r n+1 + r . (r − 1)2

Proof. Taking derivatives with respect to r in the inequality Pn j = r n+1 −1 , we get that r j=0 r −1 Pn

j=1 jr

j−1

=

(n+1)r n (r −1)−(r n+1 −1) (r −1)2

=

nr n+1 −(n+1)r n +1 (r −1)2

38/??

c





Corollary 3 For |r | < 1,

P∞

j=1 jr

j

=

r . (r −1)2

39/??

c





Corollary 3 For |r | < 1,

P∞

j=1 jr

j

=

r . (r −1)2

Proof. If |r | < 1, then ∞ X j=1

j

jr = lim

n→∞

n X j=1

nr n+2 − (n + 1)r n+1 + r r = . 2 n→∞ (r − 1) (r − 1)2

jr j = lim

40/??

c





Example 5 Find

P∞

−k k=1 k4 .

41/??

c





Example 5 Find

P∞

−k k=1 k4 .

Solution: ∞ X k=1

k4−k =

4−1 4 = = 0.4444444444. (1 − 4−1 )2 9

42/??

c




Manual for SOA Exam FM/CAS Exam 2. Chapter 3. Annuities. Section 3.2. Level payment annuities. c



1/82

c




Section 3.2. Level payment annuities.

An annuity is a sequence of payments made at equal intervals of time. We have n periods of times [0, t], [t, 2t], [2t, 3t], . . . [(n − 1)t, nt] with the same length. By a change of units, we will assume that intervals have unit length. So, the intervals are [0, 1], [1, 2], [2, 3], . . . [n − 1, n]. We order the periods as follows:

Time interval

Name

[0, 1] [1, 2] [2, 3] ··· [n − 2, n − 1] [n − 1, n]

1st period 2nd period 3rd period ··· (n − 1)–th period n–th period

Beginning of the period time 0 time 1 time 2 ··· time n − 2 time n − 1

End of the period time 1 time 2 time 3 ··· time n − 1 time n 2/82

c





An annuity is said to have level payments if all payments Cj are equal.

3/82

c





An annuity is said to have level payments if all payments Cj are equal. An annuity has non–level payments if some payments Cj are different from other ones.

4/82

c





An annuity is said to have level payments if all payments Cj are equal. An annuity has non–level payments if some payments Cj are different from other ones. The payments can be made either at the beginning or at the end of intervals of time.

5/82

c





An annuity is said to have level payments if all payments Cj are equal. An annuity has non–level payments if some payments Cj are different from other ones. The payments can be made either at the beginning or at the end of intervals of time. For an annuity–immediate payments are made at the end of the intervals of time. An annuity–immediate is a cashflow of the type: Contributions Time

0 0

C1 1

C2 2

··· ···

Cn n

6/82

c





An annuity is said to have level payments if all payments Cj are equal. An annuity has non–level payments if some payments Cj are different from other ones. The payments can be made either at the beginning or at the end of intervals of time. For an annuity–immediate payments are made at the end of the intervals of time. An annuity–immediate is a cashflow of the type: Contributions Time

0 0

C1 1

C2 2

··· ···

Cn n

For an annuity–due the payments are made at the beginning of the intervals of time. An annuity–due is a cashflow of the type: Contributions Time

C0 0

C1 1

··· ···

Cn−1 n−1

0 n 7/82

c





The cashflow of an annuity–immediate with level payments of one is Contributions Time

0 0

1 1

1 2

··· ···

1 n

If the time value of the money follows an accumulation function a(t), then the present value of an annuity–immediate with level annual payments of one is n

an| =

X 1 1 1 1 + + ··· + = . a(1) a(2) a(n) a(j) j=1

The accumulated value of an annuity–immediate with level annual payments of one is n

sn| =

a(n) X a(n) a(n) a(n) + + ··· + = . a(1) a(2) a(n) a(j) j=1

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Example 1 You are given that δt =

1 8+t ,

t ≥ 0. Find an| and sn| .

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Example 1 You are given that δt =

1 8+t ,

t ≥ 0. Find an| and sn| .

Solution: We have that Z t Z t t 8+t 1 ds = ln(8 + s) 0 = ln . δs ds = 8 0 0 8+s So, a(t) = e

Rt 0

δs

=

8+t 8 ,

an| =

n n X X 1 8 = a(j) 8+j j=1

and sn| =

j=1

n X a(n) j=1

a(j)

=

n X 8+n j=1

8+j

.

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The cashflow of an annuity–due with n level payments of one is Contributions Time

1 0

1 1

1 2

··· ···

1 n−1

0 n

The present value of an annuity–due with n level annual payments of one is n−1

X 1 1 1 1 än| = 1 + + + ··· + = . a(1) a(2) a(n − 1) a(j) j=0

The future value at time n of an annuity–due with level annual payments of one is n−1

¨sn| = a(n) +

X a(n) a(n) a(n) a(n) + + ··· + = . a(1) a(2) a(n − 1) a(j) j=0

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Example 2 Suppose that the annual effective interest rate for year n is 2 . Find än| and ¨sn| . in = n+4

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Example 2 Suppose that the annual effective interest rate for year n is 2 . Find än| and ¨sn| . in = n+4 Solution: Since 1 + in = 1 +

2 n+4

=

n+6 n+4 ,

a(n) = (1 + i1 )(1 + i1 ) · · · (1 + in ) =

1+6 2+6 3+6 n+6 · · ··· 1+4 2+4 3+4 n+4

(n + 6)(n + 5) , 30 n n−1 n−1 X X X 1 30 1 1 än| = = = (30) − a(j) (j + 6)(j + 5) j +5 j +6 j=1 j=0 j=0 1 1 1 1 =30 + ··· + − 30 + ··· + 0+5 n−1+5 1+5 n+5 1 1 6n n(n + 6) =(30) − = , ¨sn| = a(n)än| = . 5 n+5 n+5 5

=

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Remember: The cashflow of an annuity–immediate with n level payments of one is Contributions Time

0 0

1 1

1 2

··· ···

1 n

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Remember: The cashflow of an annuity–immediate with n level payments of one is Contributions Time

0 0

1 1

1 2

··· ···

1 n

The cashflow of an annuity–due with n level payments of one is Contributions Time

1 0

1 1

1 2

··· ···

1 n−1

0 n

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Theorem 1 If i 6= 0, the present value of an annuity–immediate with level payments of one is an|i = an| = ν + ν 2 + · · · + ν n = =

ν(1 − ν n ) 1 − νn 1 − νn = 1 = 1−ν i ν −1

1 − (1 + i)−n , i

where we have used that ν(1 + i) = 1.

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ν(1 − ν n ) 1 − νn 1 − νn = 1 = 1−ν i ν −1

1 − (1 + i)−n , i

where we have used that ν(1 + i) = 1. If i 6= 0, the future value of an annuity–immediate with level payments of one at time n is sn|i = sn| = (1+i)n−1 +(1+i)n−2 +· · ·+(1+i)+1 =

(1 + i)n − 1 . i

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ν(1 − ν n ) 1 − νn 1 − νn = 1 = 1−ν i ν −1

1 − (1 + i)−n , i

where we have used that ν(1 + i) = 1. If i 6= 0, the future value of an annuity–immediate with level payments of one at time n is sn|i = sn| = (1+i)n−1 +(1+i)n−2 +· · ·+(1+i)+1 =

(1 + i)n − 1 . i

If i = 0, an|i = sn|i = n. 18/82

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For example, a1|i = ν, a2|i = ν + ν 2 , a3|i = ν + ν 2 + ν 3 , a4|i = ν + ν 2 + ν 3 + ν 4 . s1|i = 1, s2|i = 1 + 1 + i, s3|i = 1 + 1 + i + (1 + i)2 , s4|i = 1 + 1 + i + (1 + i)2 + (1 + i)3 .

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Example 3 Calculate the present value of $5000 paid at the end of each year for 15 years using an annual effective interest rate of 7.5%.

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Example 3 Calculate the present value of $5000 paid at the end of each year for 15 years using an annual effective interest rate of 7.5%. Solution: The present value is (5000)a15|0.075 = (5000)

1 − (1.075)−15 = 44135.59873. 0.075

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If every period is exactly one year, then i in the formulas above is the annual effective rate of interest. If the length of a period is not a year, i in the formulas above is the effective rate of interest per period. If each period lasts t years, then the t–year interest factor is (1 + i)t and the t–year effective rate of interest is (1 + i)t − 1. So, if each period lasts t years, we use the previous formulas with i replaced by (1 + i)t − 1, where the last i is the annual effective rate of interest.

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For example, suppose that the payments are made each 1/m years. Then, the interest factor for 1/m years is (1 + i)1/m and the 1/m (m) year interest rate is (1 + i)1/m − 1 = i m , where i (m) is the nominal annual rate of interest convertible m times a year. For example, the present value at time 0 of the annuity Contributions Time (in years)

0 0

1 1/m

1 2/m

··· ···

1 n/m

is an|(1+i)1/m −1 = an|i (m) /m . The future value at time n/m years of the previous cashflow is sn|(1+i)1/m −1 = sn|i (m) /m .

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Example 4 John invest $500 into an account at the end of each month for 5 years. The annual effective interest rate is 4.5%. Calculate the balance of this account at the end of 5 years.

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Example 4 John invest $500 into an account at the end of each month for 5 years. The annual effective interest rate is 4.5%. Calculate the balance of this account at the end of 5 years. Solution: The number of payments made is (5)(12) = 60. The cashflow is Contributions Time (in months)

500 1

500 2

500 2

··· ···

500 60

The one–month effective interest rate is (1.045)1/12 − 1. Hence, the balance of this account at the end of 5 years is (500)s60|(1.045)1/12 −1 = (500) =(500)

((1.045)1/12 − 1 + 1)60 − 1 (1.045)1/12 − 1

(1.045)5 − 1 = 33495.8784. (1.045)1/12 − 1 25/82

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If payments are made every m years, the interest factor per period is (1 + i)m − 1. For example, the future value at time nm years of the annuity Contributions Time (in years)

0 0

1 m

1 2m

··· ···

1 nm

is an|(1+i)m −1 , where i is the annual effective rate of interest. The future value at time nm years of the previous annuity is sn|(1+i)m −1 .

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Example 5 A cashflow pays $8000 at the end of every other year for 16 years. The first payment is made in two years. The annual effective interest rate is 6.5%. Calculate the present value of this cashflow.

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Example 5 A cashflow pays $8000 at the end of every other year for 16 years. The first payment is made in two years. The annual effective interest rate is 6.5%. Calculate the present value of this cashflow. Solution: The cashflow is Contributions Time (in years)

8000 2

8000 4

8000 6

··· ···

8000 16

Notice that eight payments are made. The two–year effective interest rate is (1.065)2 − 1. Hence, the present value of the cashflow is (8000)a8|(1.065)2 −1 = (8000) =(8000)

1 − ((1.065)2 − 1 + 1)−8 (1.065)2 − 1

1 − (1.065)−16 = 37841.21717. (1.065)2 − 1 28/82

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Theorem 2 If i 6= 0, the present value of an annuity–due with level payments of one is än|i = än| = 1 + ν + ν 2 + · · · + ν n−1 =

1 − νn 1 − νn = , 1−ν d

where we have used that ν = 1 − d. If i 6= 0, the future value at time n of an annuity–due with level payments of one is ¨sn|i = ¨sn|i = (1 + i)n + (1 + i)n−1 + · · · + (1 + i) =

(1 + i)n+1 − (1 + i) (1 + i)n − 1 = , i d

where we have used that 1 − d = If i = 0, än|i = ¨sn|i = n.

1 1+i

and d =

i 1+i .

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The annuities factors which have introduced give the present value of the cashflow

at different times. The present value The present value The present value The present value

Contributions Time

1 1

of of of of

at at at at

the the the the

cashflow cashflow cashflow cashflow

1 2 time time time time

··· ···

1 n

0 is an|i . 1 is än|i . n is sn|i . n + 1 is ¨sn|i .

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1

1

1

0

1

2

3

6

6


1

1

n−1 n 6

an|i än|i

n+1 6

sn|i ¨sn|i

Figure 1: Present value of an annuity at different times

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Often, the contributions do not start at time 0. But, an|i is always the present value of a level unity annuity one period before the first payment. än|i is the present value of a level unity annuity at the time of the first payment. sn|i is the future value of a level unity annuity at the time of the last payment. ¨sn|i is the future value of a level unity annuity one period after the last payment. For example, for the cashflow Contributions Time The The The The

present present present present

value value value value

of of of of

the the the the

1 t +1

cashflow cashflow cashflow cashflow

1 t +2 at at at at

time time time time

··· ···

1 t +n

t is an|i . t + 1 is än|i . t + n is sn|i . t + n + 1 is ¨sn|i .

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1

1

1

t t +1t +2t +3 6 6


1

1

t+n−1 t+n t+n+1

6

an|i än|i

6

sn|i ¨sn|i

Figure 2: Present value of an annuity at different times

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Example 6 Investment contributions of $4500 are made at the beginning of the year for 20 years into an account. This account pays an annual effective interest rate is 7%. Calculate the accumulated value at the end of 25 years. Solution: The cashflow of payments is Contributions Time

4500 0

4500 1

··· ···

4500 19

We can find the accumulated value at the end of 25 years doing

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4500 0

4500 1

··· ···

4500 19

We can find the accumulated value at the end of 25 years doing either 1 − (1 + 0.07)−20 (1 + 0.07)26 0.07 (1 + 0.07)26 − (1 + 0.07)6 = = 276854.31, 0.07 a20|0.07 (1 + 0.07)26 =

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4500 0

4500 1

··· ···

4500 19

We can find the accumulated value at the end of 25 years doing or 1 − (1 + 0.07)−20 (1 + 0.07)25 1 − (1 + 0.07)−1 (1 + 0.07)26 − (1 + 0.07)6 = = 276854.31, 0.07 ä20|0.07 (1 + 0.07)25 =

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4500 0

4500 1

··· ···

4500 19

We can find the accumulated value at the end of 25 years doing or (1 + 0.07)20 − 1 (1 + 0.07)6 0.07 (1 + 0.07)26 − (1 + 0.07)6 = = 276854.31, 0.07 s20|0.07 (1 + 0.07)6 =

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4500 0

4500 1

··· ···

4500 19

We can find the accumulated value at the end of 25 years doing or (1 + 0.07)20 − 1 (1 + 0.07)5 1 − (1 + 0.07)−1 (1 + 0.07)26 − (1 + 0.07)6 = = 276854.31. 0.07 ¨s20|0.07 (1 + 0.07)5 =

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Theorem 3 (Interest factor relations for annuities) än|i = (1 + i)an|i , sn|i = (1 + i)n an|i , ¨sn|i = (1 + i)n+1 an|i

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Theorem 3 (Interest factor relations for annuities) än|i = (1 + i)an|i , sn|i = (1 + i)n an|i , ¨sn|i = (1 + i)n+1 an|i

Proof. Consider the cashflow: Contributions Time The The The The

present present present present

value value value value

at at at at

time time time time

0 0

1 1

1 2

··· ···

1 n

0 of the cashflow is an|i . 1 of the cashflow is än|i . n of the cashflow is sn|i . n + 1 of the cashflow is ¨sn|i . 40/82

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An application of än|i = (1 + i)an|i .

Example 7 If an|i = 11.5174109 and än|i = 11.9205203, find n.

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An application of än|i = (1 + i)an|i .

Example 7 If an|i = 11.5174109 and än|i = 11.9205203, find n. Solution: We have that än|i = (1 + i)an|i . So, 1 + i = 11.9205203 11.5174109 = 1.035 and i = 3.5%. Solving −n

11.5174109 = an|3.5% = 1−(1.035) , we get 0.035 (1.035)−n = 1 − (11.5174109)(0.035) = 0.5968906185 and 0.5968906185 n = − log log = 15. 1.035

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An application of ¨sn|i = (1 + i)sn|i .

Example 8 If ¨sn|i = 21 and sn|i = 20, find i.

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An application of ¨sn|i = (1 + i)sn|i .

Example 8 If ¨sn|i = 21 and sn|i = 20, find i. Solution: We have that 21 = ¨sn|i = (1 + i)sn|i = (1 + i)(20). So, i =

21 20

− 1 = 5%.

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Theorem 4 (Induction relations) än|i = 1 + an−1|i and sn|i = 1 + ¨sn−1|i .

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Theorem 4 (Induction relations) än|i = 1 + an−1|i and sn|i = 1 + ¨sn−1|i .

Proof. Consider the cashflows Cashflow 1 Cashflow 2 Cashflow 3

Contributions Contributions Contributions Time

1 0 1 0

0 1 1 1

0 1 1 2

··· ··· ··· ···

··· ··· ··· ···

0 1 1 n−1

0 0 0 n

We have that the third cashflow is the sum of the first two. The present value of the previous cashflows are 1 and an−1|i and än|i , respectively. This implies the first relation. The second relation follows similarly. 46/82

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An application of än|i = 1 + an−1|i .

Example 9 If ä10|i = 8, find a9|i .

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An application of än|i = 1 + an−1|i .

Example 9 If ä10|i = 8, find a9|i . Solution: We have that a9|i = ä10|i − 1 = 7.

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An application of sn|i = 1 + ¨sn−1|i

Example 10 If ¨s10|i = 15, find s11|i .

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An application of sn|i = 1 + ¨sn−1|i

Example 10 If ¨s10|i = 15, find s11|i . Solution: We have that s11|i = 1 + ¨s10|i = 16.

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Theorem 5 (Amortization relations) 1 1 1 1 = + i and = + d. an|i sn|i än|i ¨sn|i

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Proof of a1 = s 1 + i: Suppose that a loan of $1 is paid in n n|i n|i payments made at the end of the each period. Then, each payment should be a1 . Suppose that we pay the loan as follows, n|i we pay i at the end of each period and put x in an extra account paying an effective rate of interest i. Since the initial loan is $1 the interest accrued at the end of the first period is i. But, we pay i at the end of the first period. Hence, immediately after this payment we owe $1. Proceeding in this way, we deduce that we owe $1 immediately after each payment. The money in the extra account accumulates to xsn|i at time n. In order to pay the loan, we need x = s 1 . Our total payments at the end of the each period are n|i

1 sn|i

+ i. Since this series of payments repays the loan of $1, we

must have that a1 = s 1 + i. The proof of the second formula is n|i n|i similar and it is omitted. 52/82

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An application of 1 1 = + i. an|i sn|i

Example 11 If sn|i = 15.9171265 and an|i = 8.86325164, calculate n.

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An application of 1 1 = + i. an|i sn|i

Example 11 If sn|i = 15.9171265 and an|i = 8.86325164, calculate n. Solution: Using that 1 8.86325164

i= get n = 12.

−

1 an|i

1 15.9171265

=

1 sn|i

+ i ,we get that

= 5%. From an|5% = 8.86325164, we

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In the calculator TI–BA–II–Plus we can use the time value of money worksheet to solve problems with annuities. There are 5 main financial variables in this worksheet: I The number of periods N . I The nominal interest for year I/Y . The present value PV . I The payment per period PMT . I The future value FV . Recall that how to use the money worksheet was explained in Section 1.3. It is recommended that you set–up P/Y =1 and I

C/Y =1, by pressing: 2nd , P/Y , 1 , ENTER , ↓ , 1 , ENTER , 2nd , QUIT . Unless it is said otherwise, we will assume that the entries for C/Y and P/Y are both 1. To check that this is so, do 2nd P/Y ↓ 2nd QUIT . 55/82

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When BGN is set–up at END (and C/Y and P/Y have value 1), the value time of money formula in the calculator is PV + PMT ·

1 − (1 + i)−N + FV (1 + i)−N = 0. i

Using the calculator, we can solve for any variable in the equation: L + Pan|i + F (1 + i)−n = 0. This equation is equivalent to L(1 + i)n + Psn|i + F = 0. In the calculator, we input the variables we know using: L as the PV , P as the PMT , F as the FV , i as the I/Y , n as the N . L, P and F can take negative values. 56/82

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To solve for any variable from the equation L + Pan|i = 0 we input in the calculator: L as PV , P as PMT , 0 as FV , i as the I/Y , n as N . If we are solving for either I/Y or N , PV and PMT must have different signs. To solve for any variable from the equation Psn|i + F = 0. we input: F as the FV , P as the PMT , i as the I/Y , n as the N and 0 as the PV . If we are solving for either I/Y or N , FV and PMT must have different signs. 57/82

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Example 12 What must you deposit at the end of each of the next 10 years in order to accumulate 20,000 at the end of the 10 years assuming i = 5%.

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Example 12 What must you deposit at the end of each of the next 10 years in order to accumulate 20,000 at the end of the 10 years assuming i = 5%. Solution: We solve for P in 20000 = Ps10|0.05 and get P = 1590.091499. In the calculator TI–BA–II–Plus, press: 10 N 20000 FV 5 I/Y 0 PV CPT PMT Note the display in the calculator is negative.

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Example 13 If i (12) = 9% and $300 is deposited at the end of the each month for 1 year, what will the accumulated value be in 1 year?

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Example 13 If i (12) = 9% and $300 is deposited at the end of the each month for 1 year, what will the accumulated value be in 1 year? Solution: We solve for FV in FV = 300s12|0.09/12 and get FV = 3752.275907. In the calculator, press: 12 N 300 PMT 0.75 I/Y 0 PV CPT FV where we used that 9/12 = 0.75. Since the length of a period is a month: I

The monthly effective interest rate is

I

The number of periods is 12 months.

i (12) 12

=

0.09 12

= 0.75%.

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When BGN is set–up at BGN, the value time of money formula in this calculator is 1 − (1 + i)−N PV + PMT · (1 + i) · + FV (1 + i)−N = 0. i To solve for a variable from the equation: L + Pän|i + F (1 + i)−n = 0, we proceed as before, with payments set–up at beginning. Previous equation is equivalent to L(1 + i)n + P¨sn|i + F = 0, To change the setting of the payments (either at the beginning of the period, or at the end of the period), press: 2nd , BGN , 2nd , SET , 2nd , Quit . If the calculator is set–up with payments at the end of the periods, there is no indicator in the screen. If the calculator is set–up with payments at the beginning of the periods, the indicator ”BGN” appears in the screen. c



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Example 14 A company purchases 100 acres of land for $200,000 and agrees to remit 10 equal annual installments of $27,598 each at the beginning of the year. What is the annual interest rate on this loan?

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Example 14 A company purchases 100 acres of land for $200,000 and agrees to remit 10 equal annual installments of $27,598 each at the beginning of the year. What is the annual interest rate on this loan? Solution: We solve for i in the equation 200000 = 27598ä10|i to get i = 8%. In the calculator, set payments at the beginning of the period and press: 10 N −27598 PMT 200000 PV 0 FV CPT I/Y

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Example 15 An annuity pays $7000 at the end of the year for 7 years with the first payment made 5 years from now. The effective annual rate of interest is 6.5%. Find the present value of the this annuity.

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Example 15 An annuity pays $7000 at the end of the year for 7 years with the first payment made 5 years from now. The effective annual rate of interest is 6.5%. Find the present value of the this annuity. Solution 1: The cashflow of payments is Payments Time

7000 5

7000 6

7000 7

7000 8

7000 9

7000 10

7000 11

Using an immediate annuity, (7000)a7|0.06 is the present value of the annuity, one period before the first payment, i.e. (7000)a7|0.06 is the present value of the annuity at time 4. So, the present value of the annuity is (1.06)−4 (7000)a7|0.06 = (0.7920937)(7000)(5.582381) =30952.38. 66/82

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Example 15 An annuity pays $7000 at the end of the year for 7 years with the first payment made 5 years from now. The effective annual rate of interest is 6.5%. Find the present value of the this annuity. Solution 2: The cashflow of payments is Payments Time

7000 5

7000 6

7000 7

7000 8

7000 9

7000 10

7000 11

Using a due annuity, (7000)ä7|0.06 is the present value of the annuity, at the time of the first payment, i.e. (7000)ä7|0.06 is the present value of the annuity at time 5. So, the present value of the annuity is (1.06)−5 (7000)ä7|0.06 = (0.7472582)(7000)(5.917324326) =30952.38. 67/82

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Example 16 An annuity–immediate pays $7000 at the end of the year for 7 years. The current annual effective rate of interest is 4.5% for the first three years and 5.5% thereafter. Find the present value of this annuity.

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Example 16 An annuity–immediate pays $7000 at the end of the year for 7 years. The current annual effective rate of interest is 4.5% for the first three years and 5.5% thereafter. Find the present value of this annuity. Solution: The cashflow is Payments 7000 7000 7000 7000 7000 7000 7000 Time 1 2 3 4 5 6 7 Consider two cashflows: one with the first three payments and another one with the last four payments. The present value at time 0 of the first three payments is (7000)a3|0.045 . The present value at time 3 of the last four payments is (7000)a4|0.055 . The present value at time 0 of the last four payments is (7000)(1.045)−3 a4|0.055 . The present value of the whole annuity is (7000)a3|0.045 + (7000)(1.045)−3 a4|0.055 =19242.75048 + 21500.85804 = 40743.60852. 69/82

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Theorem 6 Consider the cashflow Contributions Time (in years)

0 0

1 m 1 m

1 m 2 m

1 m m m

··· ···

1 m m+1 m

··· ···

··· ···

1 m nm m

The present value of this cashflow is (m)

an|i =

1 − νn . i (m)

The future value at time n of this cashflow is (m)

sn|i =

(1 + i)n − 1 . i (m)

Proof: The present value of the cashflow is 1 m anm|i (m) /m

=

1 m

„ «−mn (m) 1− 1+ i m

c


i (m) m

=

1−ν n . i (m)


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Example 17 Suppose that Arthur takes a mortgage for L at an annual nominal rate of interest of 7.5% compounded monthly. The loan is paid at the end of the month with level payments of $1200 for n years. Suppose at the last minute, Arthur changes the conditions of his loan so that the payments will biweekly. The duration of the loan and the effective annual rate remain unchanged. Calculate the amount of the biweekly payment. Assume that there are 365/7 weeks in a year.

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Solution: Let P be the biweekly payment. We have that (12)

L = (12)(1200)an|i = (12)(1200)

1 − νn i (12)

and (365/14)

L = (365/14)(P)an|i

= (365/14)(P)

1 − νn i (365/14)

Hence, (12)(1200) (365/14)(P) = . i (12) i (365/14) and P=

(12)(1200)i (365/14) (12)(1200)(0.07485856348) = = 551.2871743, (12) (0.075)(365/14) i (365/14)

using that i

(365/14)

0.075 = (365/14) 1 + 12

12(14/365) = 7.485856348%. 72/82

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Theorem 7 Consider the cashflow 1 m

Contributions Time (in years)

1 m 1 m

0

1 m 2 m

··· ···

1 m nm−1 m

0 nm m


än|i =

1 − νn . d (m)

The future value at time n of this cashflow is (m)

¨sn|i =

(1 + i)n − 1 . d (m)

Proof: The present value of the cashflow is −mn i (m) 1 − 1 + m 1 1 1 − νn änm|i (m) /m = = (m) . (m) d m m d m

c



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Theorem 8 For integers k, n ≥ 1, ank|i = an|i äk|(1+i)n −1 .

74/82

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Theorem 8 For integers k, n ≥ 1, ank|i = an|i äk|(1+i)n −1 .

Proof. We have that ank|i an|i

=

1−ν nk i 1−ν n i

= 1 + ν n + ν 2n + · · · + ν n(k−1) = äk|(1+i)n −1 .

75/82

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An application of ank|i = an|i äk|(1+i)n −1 .

Example 18 Carrie receives 200,000 from a life insurance policy. She uses the fund to purchase to two different annuities, each costing 100,000. The first annuity is a 24–year annuity–immediate paying k per year to herself. The second annuity is a 8 year annuity–immediate paying 2k per year to her boyfriend. Both annuities are based upon an annual effective interest rate of i, i > 0. Determine i.

76/82

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An application of ank|i = an|i äk|(1+i)n −1 .

Example 18 Carrie receives 200,000 from a life insurance policy. She uses the fund to purchase to two different annuities, each costing 100,000. The first annuity is a 24–year annuity–immediate paying k per year to herself. The second annuity is a 8 year annuity–immediate paying 2k per year to her boyfriend. Both annuities are based upon an annual effective interest rate of i, i > 0. Determine i. Solution: We have that 100000 = ka24|i = 2ka8|i . So, a24|i a8|i + i)8

2=

= ä3|(1+i)8 −1 . Using the calculator, we get that

(1

− 1 = 0.618034 and i = 6.1997%.

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Theorem 9 For integers k, n ≥ 1,

ank|i sn|i


= ak|(1+i)n −1 .

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c




Theorem 9 For integers k, n ≥ 1,

ank|i sn|i


= ak|(1+i)n −1 .

Proof. By the previous theorem, ank|i = an|i äk|(1+i)n −1 . The claim follows noticing that an|i = (1 + i)−n sn|i and äk|(1+i)n −1 = (1 + i)n ak|(1+i)n −1 .

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An application of ank|i = sn|i ak|(1+i)n −1 .

Example 19 The present value of a 4n–year annuity–immediate of 1 at the end of every year is 16.663. The present value of a 4n–year annuity–immediate of 1 at the end of every fourth year is 3.924. Find n and i.

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An application of ank|i = sn|i ak|(1+i)n −1 .

Example 19 The present value of a 4n–year annuity–immediate of 1 at the end of every year is 16.663. The present value of a 4n–year annuity–immediate of 1 at the end of every fourth year is 3.924. Find n and i. Solution: We know that 16.663 = a4n|i and 3.924 = an|(1+i)4 −1 . Dividing the first equation over the second one, we get that 4.246432 =

a4n|i 16.663 = = s4|i 3.924 an|(1+i)4 −1

and i = 4%. From the equation 16.663 = a4n|4% , we get n = 7.

81/82

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Pn

Since an|i = j=1 (1 + i)−j as a function of n, an|i increases with n. As a function of i, i ≥ 0, an|i decreases with i. We have that an|0 = n and an|∞ = 0. So, n > an|i > 0 for i > 0; and an|i > n for 0 > i > −1. It is proved in the manual that an|i =

∞ X (n + j − 1)! (−i)j−1 . (n − 1)!j! j=1

The first order Taylor expansion of an|i on i is an|i ≈ n −

n(n + 1) i. 2

The second order Taylor expansion of an|i on i is an|i ≈ n −

n(n + 1) n(n + 1)(n + 2) 2 i+ i . 2 6 82/82

c




Manual for SOA Exam FM/CAS Exam 2. Chapter 3. Annuities. Section 3.3. Level payment perpetuities. c



1/16

c




Section 3.3. Level payment perpetuities.

A perpetuity is a series of payments made forever along equal intervals of time. By a change of units, we will assume that intervals have unit length. The payments can be made either at the beginning or at the end of the intervals. A perpetuity has level payments if all payments Cj , j ≥ 0, are equal. A perpetuity has non–level payments if some payments Cj are different from other ones.

2/16

c





For a perpetuity–immediate the payments are made at the end of the periods of time, i.e. at the times 1, 2, . . . . So, a perpetuity–immediate is a cashflow of the type: Payments Time

C1 1

C2 2

C3 3

··· ···

For a perpetuity–due the payments are made at the beginning of the intervals of time, i.e. at the times 0, 1, 2, . . . . So, a perpetuity immediate is a cashflow of the type: Payments Time

C0 0

C1 1

C2 2

··· ···

3/16

c





Theorem 1 The cashflow value of a perpetuity–immediate with level payments of one is Payments Time

1 1

1 2

1 3

··· ···

The present value of a perpetuity–immediate with level payments of one is ν 1 a∞|i = ν + ν 2 + · · · = = . 1−ν i

4/16

c





Example 1 The present value of a series of payments of 3 at the end of every eight years, forever, is equal to 9.5. Calculate the effective annual rate of interest.

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c





Example 1 The present value of a series of payments of 3 at the end of every eight years, forever, is equal to 9.5. Calculate the effective annual rate of interest. Solution: The 8–year interest factor is (1 + i)8 . So, the 8–year effective interest rate is (1 + i)8 − 1. We have that 9.5 = (3)a∞|(1+i)8 −1 = (1+i)3 8 −1 and 3 1/8 − 1 = 3.489979511%. i = 1 + 9.5

6/16

c





Theorem 2 The cashflow value of a perpetuity–due with level payments of one is Payments Time

1 0

1 1

1 2

··· ···

The present value of an perpetuity–due with level payments of one is 1 1 1+i ä∞|i = 1 + ν + ν 2 + · · · = = = . 1−ν d i

7/16

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Example 2 John uses his retirement fund to buy a perpetuity–due of 20,000 per year based on an annual nominal yield of interest i = 8% compounded monthly. Find John’s retirement fund.

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Example 2 John uses his retirement fund to buy a perpetuity–due of 20,000 per year based on an annual nominal yield of interest i = 8% compounded monthly. Find John’s retirement fund. Solution: Since i (12) = 8%, i = 8.299950681%. John’s retirement fund is worth (20000)ä∞|i = (20000)

1+i 1.08299950681 = (20000) = 260963.8554. i 0.08299950681

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Example 3 A perpetuity pays $1 at the end of every year plus an additional $1 at the end of every second year. The effective rate of interest is i = 5%. Find the present value of the perpetuity at time 0.

10/16

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Solution: The cashflow of the perpetuity is Payments Time

1 1

2 2

1 3

2 4

1 5

2 6

··· ···

This cashflow can be decomposed into the cashflows: Payments Time

1 1

1 3

1 5

··· ···

Payments Time

2 2

2 4

2 6

··· ···

and

The present value at time 0 of the first part of the cashflow is 1+i . The present value at time 0 of the second part of the (1+i)2 −1 cashflow is (1+i)2 2 −1 . Hence, the present value at time of the total cashflow is 1+i 2 3+i 3 + 0.05 + = = = 29.7561. 2 2 (1 + i) − 1 (1 + i) − 1 i(2 + i) 0.05(2 + 0.05) 11/16

c





Theorem 3 The cashflow value of the perpetuity–immediate with level payments of one per year and m payments per year is Payments Time

0 0

1 m 1 m

1 m 2 m

··· ···

The present value of the previous perpetuity–immediate is 1 1 1 1/m · ν + ν 2/m + · · · = · ν 1/m · m m 1 − ν 1/m 1 1 = = (m) . −1/m m(ν − 1) i (m)

a∞|i =

12/16

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Example 4 A perpetuity pays x at the end of each month. The nominal annual rate of interest compounded monthly is i (12) . Calculate the percentage of increase in the value of this perpetuity if the nominal annual rate of interest compounded monthly decreases by 10%.

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Example 4 A perpetuity pays x at the end of each month. The nominal annual rate of interest compounded monthly is i (12) . Calculate the percentage of increase in the value of this perpetuity if the nominal annual rate of interest compounded monthly decreases by 10%. Solution: At the rate i (12) , the present value of the perpetuity is x . At the rate i (12) (0.9), the present value of the perpetuity is i (12) x . The percentage of increase in the value of this perpetuity i (12) (0.9) is x x − i (12) 1 i (12) (0.9) = − 1 = 11.11111111%. x 0.9 i (12)

14/16

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Theorem 4 The cashflow value of the perpetuity–due with level payments of one per year and m payments per year is Payments Time

1 m

0

1 m 1 m

1 m 2 m

··· ···

The present value of an perpetuity–due with level payments of one per year and m payments per year is (m)

ä∞|i =

1 1 1 1 · 1 + ν 1/m + ν 2/m + · · · = · = (m) . m m 1 − ν 1/m d

15/16

c





Example 5 The present value of a series of payments of $500 at the beginning of every month years, forever, is equal to $10000. Calculate the nominal annual discount rate compounded monthly. Solution: We have that 10000 = 500 and d (12) =

500 10000

1 d (12)

.

= 0.05.

16/16

c




Manual for SOA Exam FM/CAS Exam 2. Chapter 3. Annuities. Section 3.4. Non-level payment annuities and perpetuities. c



1/37

c




Section 3.4. Non-level payment annuities and perpetuities.

Theorem 1 (Geometric Annuity) Let i, r > −1. The present value of the annuity Payments Time

1 1

1+r 2

is (Ga)n|i,r =

(1 + r )2 3

··· ···

(1 + r )n−1 n

1 1 an| i−r = ä i−r . 1 + r 1+r 1 + i n| 1+r

2/37

c




i−r Proof: Let i 0 = 1+r . Then, cashflow at time 0 is n X

j−1

(1 + r )

1+i 1+r


= 1 + i 0 . The present value of the

−j

(1 + i)

j=1

j=1

=

1 1+r

n 1 X 1 + i −j = 1+r 1+r

n X

1 + i0

−j

=

j=1

1 a 0. 1 + r n|i

Using that än|i = (1 + i)an|i , we get that 1 1 1 1 an|i 0 = än|i 0 = ä 0 . 0 1+r 1+r 1+i 1 + i n|i

3/37

c





Example 1 An annuity provides for 10 annuals payments, the first payment a year hence being $2600. The payments increase in such a way that each payment is 3% greater than the previous one. The annual effective rate of interest is 4%. Find the present value of this annuity.

4/37

c





Example 1 An annuity provides for 10 annuals payments, the first payment a year hence being $2600. The payments increase in such a way that each payment is 3% greater than the previous one. The annual effective rate of interest is 4%. Find the present value of this annuity. Solution: The cashflow is Payments Time

2600 1

2600(1.03) 2

2600(1.03)2 3

··· ···

2600(1.03)9 10

The present value at time 0 of the annuity is (2600) (Ga)n|i,r =

2600 a −− 0.04−0.03 = 2524.271845a10|0.9708737864% 1.03 10 | 1.03

=23945.54454. 5/37

c





Example 2 An annuity provides for 20 annuals payments, the first payment a year hence being $4500. The payments increase in such a way that each payment is 4.5% greater than the previous one. The annual effective rate of interest is 4.5%. Find the present value of this annuity.

6/37

c





Example 2 An annuity provides for 20 annuals payments, the first payment a year hence being $4500. The payments increase in such a way that each payment is 4.5% greater than the previous one. The annual effective rate of interest is 4.5%. Find the present value of this annuity. Solution: The cashflow is Payments Time

4500 1

4500(1.045) 2

4500(1.045)2 3

··· ···

4500(1.045)19 20

The present value at time 0 of the annuity is (4500) (Ga)n|i,r = (4500)

1 20 a20 = 86124.40191. −− 0.045−0.045 = 4500 | 1.045 1.045 1.045 7/37

c





Example 3 Chris makes annual deposits into a bank account at the beginning of each year for 10 years. Chris initial deposit is equal to 100, with each subsequent deposit k% greater than the previous year deposit. The bank credits interest at an annual effective rate of 4.5%. At the end of 10 years, the accumulated amount in Chris account is equal to 1657.22. Calculate k.

8/37

c





Example 3 Chris makes annual deposits into a bank account at the beginning of each year for 10 years. Chris initial deposit is equal to 100, with each subsequent deposit k% greater than the previous year deposit. The bank credits interest at an annual effective rate of 4.5%. At the end of 10 years, the accumulated amount in Chris account is equal to 1657.22. Calculate k. Solution: The cashflow is Payments Time

100 0

100(1 + r ) 1

100(1 + r )2 2

··· ···

100(1 + r )9 9

The accumulated amount at the end of 10 years is 1 1657.22 = (100) 1+i än| i−r (1 + i)11 . Hence, 1+r

ä10| 0.045−r = 1+r

0.045−r 1+r

(1657.22)(1.045)−10 100

= 10.67129833,

= −0.0141511755, r =

0.045+0.0141511755 1−0.0141511755

= 6% and k = 6. 9/37

c





Corollary 1 The present value of the perpetuity Payments Time

1 1

1+r 2

(1 + r )2 3

is

( (Ga)∞|i,r =

1 i−r

∞

··· ···

(1 + r )n−1 n

··· ···

if i > r , if i ≤ r .

10/37

c





Corollary 1 The present value of the perpetuity Payments Time

1 1

1+r 2

(1 + r )2 3

is

( (Ga)∞|i,r =

1 i−r

∞

··· ···

(1 + r )n−1 n

··· ···

if i > r , if i ≤ r .

Proof. If i > r , (Ga)∞|i,r = If i ≤ r , (Ga)∞|i,r =

1 i−r 1+r a∞| 1+r 1 i−r 1+r a∞| 1+r

=

1 1 1+r i−r

1+r

=

1 i−r .

= ∞.

11/37

c





Example 4 An perpetuity–immediate provides annual payments. The first payment of 13000 is one year from now. Each subsequent payment is 3.5% more than the one preceding it. The annual effective rate of interest is i = 6%. Find the present value of this perpetuity.

12/37

c





Example 4 An perpetuity–immediate provides annual payments. The first payment of 13000 is one year from now. Each subsequent payment is 3.5% more than the one preceding it. The annual effective rate of interest is i = 6%. Find the present value of this perpetuity. Solution: The present value is 1 1 (13000) (Ga)∞|i,r = (13000) i−r = (13000) 0.06−0.035 = 520000.

13/37

c





Theorem 2 (Increasing Annuity) The present value of the annuity Payments Time

1 1

2 2

3 3

··· ···

n n

is

an|i (1 + i) − nν n än|i − nν n = . i i The accumulated value of this cashflow at time n is (Ia)n|i =

(Is)n|i =

¨sn|i − n . i

14/37

c





Proof: The cashflow is the sum of the cashflows: cashflow cashflow cashflow ··· ··· cashflow Time

1 2 3

n

1 0 0 ··· ··· 0 1

1 1 0 ··· ··· 0 2

1 1 1 ··· ··· 0 3

··· ··· ··· ··· ··· ··· ···

1 1 1 ··· ··· 0 n−1

1 1 1 ··· ··· 1 n

The future value at time n of all these cashflows is (Is)n|i =

n X j=1

sj|i =

n X ¨sn|i − n (1 + i)j − 1 = . i i j=1

15/37

c





In the calculator, it is possible to find än|i − nν n in one computation. With payments set–up at the beginning, we enter n in N , i in I/Y , 1 in PMT and −n in FV . We ask the calculator to compute PV .

16/37

c





Example 5 Find the present value at time 0 of an annuity–immediate such that the payments start at 1, each payment thereafter increases by 1 until reaching 10, and they remain at that level until 25 payments in total are made. The effective annual rate of interest is 4%.

17/37

c





Example 5 Find the present value at time 0 of an annuity–immediate such that the payments start at 1, each payment thereafter increases by 1 until reaching 10, and they remain at that level until 25 payments in total are made. The effective annual rate of interest is 4%. Solution: The cashflow is Payments Time

1 1

2 2

3 3

··· ···

10 10

10 11

··· ···

10 25

The present value at time 0 of the perpetuity is (Ia)10|0.04 + (1 + 0.04)−10 10a15|0.04 än|4% − 10(1 + 0.04)−10 + (1 + 0.04)−10 (10)a15|0.04 0.04 =41.99224806 + 75.11184164 = 117.1040897. =

18/37

c





Theorem 3 (Decreasing Annuity) The present value of the annuity Payments Time

n 1

n−1 2

n−2 3

··· ···

1 n

is

n − an|i . i The accumulated value of this cashflow at time n is (Da)n|i =

(Ds)n|i =

n(1 + i)n − sn|i . i

19/37

c





Proof. The cashflow is the sum of the cashflows: cashflow cashflow cashflow ··· ··· cashflow Time

1 2 3

n

1 1 1 ··· ··· 1 1

1 1 1 ··· ··· 0 2

1 1 1 ··· ··· 0 3

··· ··· ··· ··· ··· ··· ···

1 1 0 ··· ··· 0 n−1

1 0 0 ··· ··· 0 n

The present value at time 0 of all these cashflows is n X j=1

aj|i =

n X 1 − (1 + i)−j j=1

i

=

n − an|i . i

20/37

c





In the calculator, it is possible to find n(1 + i)n − sn|i in one computation. We enter n in N , i in I/Y , 1 in PMT and −n in PV . We ask the calculator to compute FV .

21/37

c





Example 6 Find the present value of a 15–year decreasing annuity–immediate paying 150000 the first year and decreasing by 10000 each year thereafter. The effective annual interest rate of 4.5%.

22/37

c





Example 6 Find the present value of a 15–year decreasing annuity–immediate paying 150000 the first year and decreasing by 10000 each year thereafter. The effective annual interest rate of 4.5%. Solution: The cashflow of payments is Payments Time

(15)(10000) 1

(14)(10000) 2

··· ···

(1)(10000) 15

The present value of this cashflow is (10000) (Da)15|4.5% = (10000)

15 − a15|4.5% 0.045

= 946767.616.

23/37

c





Theorem 4 (Increasing Perpetuity) The present value of the perpetuity Payments Time is (Ia)∞|i =

1 1

2 2

3 3

··· ···

1+i . i2

24/37

c





Proof: The cashflow is the sum of the infinitely many cashflows: cashflow 1 cashflow 2 cashflow 3 ··· ··· Time

1 0 0 ··· ··· 1

1 1 0 ··· ··· 2

1 1 1 ··· ··· 3

··· ··· ··· ··· ··· ···

1 1 1 ··· ··· n−1

1 1 1 ··· ··· n

··· ··· ··· ··· ··· ···

··· ··· ··· ··· ··· ···

The present value at time 0 of all these cashflows is ∞ X 1 j=1

i

(1 + i)−(j−1) =

1 1 1+i = 2 . 1 i 1 − 1+i i

25/37

c





Example 7 An investor is considering the purchase of 500 ordinary shares in a company. This company pays dividends at the end of each year. The next payment is one year from now and it is $3 per share. The investor believes that each subsequent payment per share will increase by $1 each year forever. Calculate the present value of this dividend stream at a rate of interest of 6.5% per annum effective.

26/37

c





Solution: The cashflow of payments is Payments Time

3 1

4 2

5 3

··· ···

The cashflow is the sum of the following cashflows Payments Payments Time

2 1 1

2 2 2

2 3 3

··· ··· ···

Hence, the present value of this dividend stream is (500)(2)a∞|6.5% + (500)(1) (Ia)∞|6.5% 1 1.065 =(500)(2) + (500)(1) 0.065 0.0652 =15384.6154 + 126035.503 = 141420.118. 27/37

c





Theorem 5 (Rainbow Immediate) The present value of the annuity Payments Time

1 1

2 2

··· ···

n−1 n−1

n n

n−1 n+1

··· ···

2 2n − 2

1 2n − 1

is än|i an|i .

28/37

c





Proof: The value of the annuity is (Ia)n|i + ν n (Da)n−1|i = =

(1+i)an|i −ν n (1+i)an|i i

=

än|i −nν n

+

i (1+i)an|i (1−ν n ) i

ν n (n−1−an−1|i ) i

=

än|i −ν n (1+an−1|i ) i

= (1 + i)a−n−|i an|i = än|i an|i

29/37

c





Example 8 A 15 year annuity pays 1000 at the end of year 1 and increases by 1000 each year until the payment is 8000 at the end of year 8. Payments then decrease by 1000 each year until a payment of 1000 is paid at the end of year 15. The annual effective interest rate of 6.5%. Compute the present value of this annuity.

30/37

c





Example 8 A 15 year annuity pays 1000 at the end of year 1 and increases by 1000 each year until the payment is 8000 at the end of year 8. Payments then decrease by 1000 each year until a payment of 1000 is paid at the end of year 15. The annual effective interest rate of 6.5%. Compute the present value of this annuity. Solution: The cashflow is Paym. Time

1000 1

2000 2

··· ···

7000 7

8000 8

7000 9

··· ···

2000 14

1000 15

The present value is 1000ä8|i a8|i = (1000)(6.08875096)(6.48451977) = 39482.626.

31/37

c





Theorem 6 (Paused Rainbow Immediate) The present value of the annuity Paym. Time

1 1

2 2

··· ···

n−1 n−1

n n

n n+1

n−1 n+2

··· ···

2 2n − 1

1 2n

is än+1|i an+1|i .

32/37

c





Theorem 6 (Paused Rainbow Immediate) The present value of the annuity Paym. Time

1 1

2 2

··· ···

n−1 n−1

n n

n−1 n+2

n n+1

··· ···

2 2n − 1

1 2n

is än+1|i an+1|i . Proof: The present value of the annuity is (Ia)n|i + ν n (Da)−n−|i = (1+i)a

−(1−ia

¨ an|i −nν n i

+

ν n (n−an|i ) i

=

¨ an|i −ν n an|i i

)a

n|i n|i n|i = i = an|i (1 + an|i ) = an|i än+1|i .

33/37

c





Example 9 A 20 year annuity pays 5000 at the end of year 1 and increases by 5000 each year until the payment is 50000 at the end of year 10. The payment remains constant for one year. Payments then decrease by 5000 each year until a payment of 5000 is paid at the end of year 20. The annual effective interest rate of 4%. Compute the present value of this annuity.

34/37

c





Example 9 A 20 year annuity pays 5000 at the end of year 1 and increases by 5000 each year until the payment is 50000 at the end of year 10. The payment remains constant for one year. Payments then decrease by 5000 each year until a payment of 5000 is paid at the end of year 20. The annual effective interest rate of 4%. Compute the present value of this annuity. Solution: The cashflow is Paym. Time

(1)(5000) 1

··· ···

(10)5000 10

(10)5000 11

··· ···

(1)(5000) 20

The present value is 5000ä11|i a10|i = (5000)(8.11089578)(9.11089578) = 369487.631 35/37

c





Theorem 7 The present value of the annuity Payments Time

0 0

1 m 1 m

1 m 2 m

··· ···

is (m)

(Ia)n|i =

1 m

2 m

1

1+

1 m

än|i − nν n i (m)

··· ···

2 m

2

··· ···

··· ···

n m

n

.

For the proof, see the manual.

36/37

c






0 0

1 m2 1 m

2 m2 2 m

is

I

(m)

a

(m) n|i

3 m2 3 m

··· ···

··· ···

n m2

n

(m)

=


.

For the proof, see the manual.

37/37

c




Manual for SOA Exam FM/CAS Exam 2. Chapter 3. Annuities. Section 3.5. Continuous annuities. c



1/15

c




Section 3.5. Continuous annuities.

Continuous annuities Annuities with length of period very small are approximately continuous annuities. For example, the cashflows Inflow Time

0 0

1 m 1 m

1 m 2 m

··· ···

1 m m m

1 m m+1 m

··· ···

··· ···

1 m nm m

and Inflow Time

1 m

0

1 m 1 m

1 m 2 m

··· ···

1 m m m

1 m m+1 m

··· ···

··· ···

1 m nm−1 m

0 nm m

tend to a continuous cashflow with rate C (t) = 1, 0 ≤ t ≤ n, as m → ∞. 2/15

c





Theorem 1 The present value of a continuous annuity with rate C (t) = 1, 0 ≤ t ≤ n, is 1 − νn ¯a−n−|i = . δ The future value at time n of a continuous annuity with rate of one is (1 + i)n − 1 . ¯sn|i = δ Proof: We have that Z n e t ln ν ν t dt = ln ν 0

n n −nδ 1 − νn = 1−ν = 1−e = . − ln ν δ δ 0

3/15

c





Recall

Theorem 2 Consider the cashflow Inflow Time

0 0

1 m 1 m

1 m 2 m

··· ···

1 m m m

1 m m+1 m

··· ···

··· ···

1 m nm m

Then, the present value of this cashflow is (m)

an|i =

1 − νn , i (m)

where i (m) is the nominal annual rate of interest convertible m times at year. The future value at time n of this cashflow is (m)

sn|i =

(1 + i)n − 1 . i (m) 4/15

c





Recall

Theorem 3 Consider the cashflow Inflow Time

1 m

0

1 m 1 m

1 m 2 m

··· ···

1 m m m

1 m m+1 m

··· ···

··· ···

1 m nm−1 m

0 nm m


än|i =

1 − νn , d (m)

where d (m) is the nominal annual rate of discount convertible m times at year. The future value at time n of this cashflow is (m)

¨sn|i =

(1 + i)n − 1 . d (m) 5/15

c





Theorem 4 (m)

(m)

¯an|i = lim an|i = lim än|i m→∞

m→∞

and (m)

(m)

¯sn|i = lim sn|i = lim ¨sn|i . m→∞

Proof: (m)

m→∞

1 − νn 1 − νn = = ¯an|i . m→∞ i (m) δ 1 − νn 1 − νn = = ¯an|i . = lim m→∞ d (m) δ

lim an|i = lim

m→∞

(m) lim ä m→∞ n|i

6/15

c





Given a real number x, the integer part of x is the largest integer smaller than or equal to x, i.e. the integer k satisfying k ≤ x < k + 1. The integer part of x is noted by [x]. Next theorem considers the continuous annuity with rate equal to the integer part.

Theorem 5 The present value of a continuous annuity with C (t) = [t], 0 ≤ t ≤ n, is än|i − nν n (I ¯a)n|i = . δ

7/15

c





Proof. The present value of the continuous cashflow is Z (I ¯a)n|i = =

n X

=

C (s)ν s ds =

0

j(e −jδ

j=1

n

e −(j−1)δ )

− −δ

=

n Z X j=1 n X

j

jν s ds =

j−1

n X j(ν j − ν j−1 ) ln ν j=1

j(e −(j−1)δ

−

e −jδ )

δ

j=1

1 + e −δ + · · · + e −(n−1)δ − ne −nδ . δ

Now, e −δ = ν and 1 + e −δ + · · · + e −(n−1)δ = 1 + ν + · · · + ν n−1 = So, (I ¯a)n|i =

1 − νn = än|i . 1−ν

¨ an|i −nν n . δ 8/15

c





Recall


0 0

1 m 1 m

1 m 2 m

··· ···

is (m)

(Ia)n|i =

1 m

2 m

1

1+


··· ···

1 m

2 m

2

··· ···

··· ···

n m

n

.

9/15

c





Theorem 7 (m) n |i

(I ¯a)n|i = lim (Ia)−− . m→∞

10/15

c





Theorem 7 (m) n |i

(I ¯a)n|i = lim (Ia)−− . m→∞

Proof. We have that (m)

lim (Ia)n|i = lim

m→∞

m→∞


=

än|i − nν n = (I ¯a)n|i . δ

11/15

c





Theorem 8 The present value of a continuous annuity with C (t) = t, 0 ≤ t ≤ n, is ¯an|i − nν n ¯I ¯a = . n|i δ

Proof. By the change of variables x = δs, Z n Z n Z n s s ¯I ¯a = C (s)ν ds = sν ds = se −sδ ds n|i 0 0 0 nδ Z nδ −2 −x −2 −x =δ xe dx = δ (−1 − x)e 0

0

¯an|i − nν n ne −nδ 1 − e −nδ − = . =δ −2 − δ −2 e −nδ (1 + nδ) = δ2 δ δ 12/15

c





Recall


0 0

1 m2 1 m

2 m2 2 m

is

I

(m)

a

(m) n|i

3 m2 3 m

··· ···

··· ···

n m2

n

(m)

=


.

13/15

c





Theorem 10 (m) (m) ¯I ¯a = lim I a . n|i m→∞

n|i

14/15

c





Theorem 10 (m) (m) ¯I ¯a = lim I a . n|i m→∞

n|i

Proof.

lim

m→∞

=

I

(m)

a

(m) n|i

(m)

= lim

än|i − nν n

m→∞

¯an|i − 1 − ν n nν n − = δ2 δ δ

nν n

i (m)

1 − νn nν n − m→∞ i (m) d (m) i (m)

= lim

= ¯I ¯a n|i .

15/15

c



Chapter 4. Amortization and sinking bonds.

Manual for SOA Exam FM/CAS Exam 2. Chapter 4. Amortization and sinking bonds. Section 4.1. Amortization schedules. c



1/52

c




Section 4.1. Amortization schedules.

In this chapter, we study different problems related with the payment of a loan. Suppose that a borrower (also called debtor) takes a loan from a lender. The borrower will make payments which eventually will repay the loan. Payments made by the borrower can be applied to the outstanding balance or not. According with the amortization method, all the payments made by the borrower reduce the outstanding balance of the loan. When a loan is paid usually, the total amount of loan payments exceed the loan amount. The finance charge is the total amount of interest paid (the total payments minus the loan payments).

2/52

c





The simplest way to pay a loan is by unique payment. Suppose that a borrower takes a loan with amount L at time zero and the lender charges an annual effective rate of interest of i. If the borrower pays the loan with a lump sum P at time n, then P = L(1 + i)n . The finance charge in this situation is L(1 + i)n − L.

3/52

c





Example 1 Juan borrows $35,000 for four years at an annual nominal interest rate of 7.5% convertible monthly. Juan will pay the loan with a unique payment at the end of four years. (i) Find the amount of this payment. (ii) Find the finance charge which Juan is charged in this loan.

4/52

c





Example 1 Juan borrows $35,000 for four years at an annual nominal interest rate of 7.5% convertible monthly. Juan will pay the loan with a unique payment at the end of four years. (i) Find the amount of this payment. (ii) Find the finance charge which Juan is charged in this loan. Solution: (i) The amount of the loan payment is (12)(4) = 47200.97. (35000) 1 + 0.075 12 (ii) The finance charge which Juan is charged in this loan is 47200.97 − 35000 = 12200.97.

5/52

c





Suppose that a borrower takes a loan of L at time 0 and repays the loan in a series of payments C1 , . . . , Cn at times t1 , . . . , tn , where 0 < t1 < t2 < · · · < tn . The debtor cashflow is Inflows Time

L 0

−C1 t1

−C2 t2

−Cn t3

··· ···

−Cn tn

Assume that the loan increases with a certain accumulation function a(t), t ≥ 0. Since the loan will be repaid, the present value a time zero (or any other time) of this cashflow is zero. Hence n X Cj . L= a(tj ) j=1

The finance charge for this loan is n X j=1

Cj − L =

n X j=1

n n X X Cj 1 = Cj − Cj 1 − . a(tj ) a(tj ) j=1

j=1

6/52

c





According with the retrospective method, the outstanding balance at certain point is the present value of the loan at that time minus the present value of the payments made at that time. For the cashflow Inflows Time

L 0

−C1 t1

−C2 t2

−Cn t3

··· ···

−Cn tn

the outstanding balance immediately after the k–th payment, is Bk = La(tk ) −

k X a(tk )Cj j=1

a(tj )

.

Of course, we have that B0 = L, Bn = 0.

7/52

c





According to the prospective method, the outstanding balance after the k–th payment is equal to the present value of the remaining payments. For the cashflow Inflows Time

−C1 t1

L 0

−C2 t2

−Cn t3

··· ···

−Cn tn

the outstanding balance immediately after the k–th payment, is Bk =

n X a(tk )Cj a(tj )

j=k+1

Of course, we have that La(tk ) −

n X a(tk )Cj = . a(tj ) a(tj )

k X a(tk )Cj j=1

j=k+1

8/52

c





An inductive relation for the outstanding balance is Bk = Bk−1

a(tk ) − Ck . a(tk−1 )

Previous relation says that the outstanding balance after the k–th payment is the accumulation of the previous outstanding balance minus the amount of the payment made.

9/52

c





During the period [tk−1 , tk ], the amount of interest accrued is a(tk ) −1 . Ik = Bk−1 a(tk−1 ) Immediately before the k–th payment, the outstanding balance is a(tk ) Bk−1 + Ik = Bk−1 a(t . Immediately after the k–th payment, the k−1 ) outstanding balance is Bk = Bk−1 + Ik − Ck . The k–th payment Ck can be split as Ik plus Ck − Ik . Ik is called the interest portion of the k–th payment. Ck − Ik is called the principal portion of the k–the payment. If Ck − Ik < 0, then the outstanding balance increases during the k–th period. Notice that a(tk ) − 1 = Bk−1 − Bk Ck − Ik = Ck − Bk−1 a(tk−1 ) is the reduction on principal made during the the k–period. The total Pn amount of reduction on principal is equal to the loan amount: k=1 (Bk − Bk−1 ) = Bn − B0 = L. c



10/52



Under compound interest, L=

n X

Cj (1 + i)−tj .

j=1

The outstanding balance immediately after the k–th payment is tk

Bk = L(1 + i) −

k X

tk −tj

Cj (1 + i)

j=1

=

n X

Cj (1 + i)tk −tj .

j=k+1

The inductive relation for outstanding balances is Bk = Bk−1 (1 + i)tk −tk−1 − Ck . The amount of interest accrued during the period [tk−1 , tk ] is Ik = Bk−1 (1 + i)tk −tk−1 − 1 . The principal portion of the k–the payment is Ck − Ik = Ck − Bk−1 (1 + i)tk −tk−1 − 1 = Bk−1 − Bk . P P The finance charge is nj=1 Cj − L = nj=1 Cj (1 − (1 + i)−tj ) . c



11/52



Usually, we consider payments made at equally spaced intervals of time and compound interest. Suppose that a borrower takes a loan L at time 0 and repays the loan in a series of level payments C1 , . . . , Cn at times t0 , 2t0 , . . . , nt0 . By a change of units, we may assume that t0 = 1. Hence, the debtor cashflow is Inflows Time

L 0

−C1 1

−C2 2

−Cn 3

··· ···

−Cn n

Let i be the effective rate of interest per period. Then, we have that n X L= Cj (1 + i)−j . j=1

12/52

c





The outstanding balance immediately after the k–th payment, is k

Bk = L(1 + i) −

k X

k−j

Cj (1 + i)

=

j=1

n X

Cj (1 + i)k−j .

j=k+1

The amount of interest accrued during the k–th year is iBk−1 . The principal portion of the k–th payment is Ck − iBk−1 = Bk − Bk−1 . Hence, the outstanding balance after the k–th payment is Bk = Bk−1 − (Ck − iBk−1 ) = (1 + i)Bk−1 − Ck . The finance charge is n X j=1

Cj − L =

n X

−j

Cj (1 − (1 + i) ) =

j=1

n X

Cj (1 − ν j ).

j=1

13/52

c





Example 2 Roger buys a car for $25,000 by making level payments at the end of the month for three years. Roger is charged an annual nominal interest rate of 8.5% compounded monthly in his loan. (i) Find the amount of each monthly payment. (ii) Find the total amount of payments made by Roger. (iii) Find the total interest paid by Roger during the duration of the loan. (iv) Calculate the outstanding loan balance immediately after the 12–th payment has been made using the retrospective method. (v) Calculate the outstanding loan balance immediately after the 12–th payment has been made using the prospective method.

14/52

c





Example 2 Roger buys a car for $25,000 by making level payments at the end of the month for three years. Roger is charged an annual nominal interest rate of 8.5% compounded monthly in his loan. (i) Find the amount of each monthly payment. (ii) Find the total amount of payments made by Roger. (iii) Find the total interest paid by Roger during the duration of the loan. (iv) Calculate the outstanding loan balance immediately after the 12–th payment has been made using the retrospective method. (v) Calculate the outstanding loan balance immediately after the 12–th payment has been made using the prospective method. Solution: (i) Let P be the monthly payment. Pa36 and P = 789.1884356. −− |0.085/12

We have that 25000 =

15/52

c





Example 2 Roger buys a car for $25,000 by making level payments at the end of the month for three years. Roger is charged an annual nominal interest rate of 8.5% compounded monthly in his loan. (i) Find the amount of each monthly payment. (ii) Find the total amount of payments made by Roger. (iii) Find the total interest paid by Roger during the duration of the loan. (iv) Calculate the outstanding loan balance immediately after the 12–th payment has been made using the retrospective method. (v) Calculate the outstanding loan balance immediately after the 12–th payment has been made using the prospective method. Solution: (ii) The total amount of payments (36)(789.1884356) = 28410.78368.

made

by

Roger

is

16/52

c





Example 2 Roger buys a car for $25,000 by making level payments at the end of the month for three years. Roger is charged an annual nominal interest rate of 8.5% compounded monthly in his loan. (i) Find the amount of each monthly payment. (ii) Find the total amount of payments made by Roger. (iii) Find the total interest paid by Roger during the duration of the loan. (iv) Calculate the outstanding loan balance immediately after the 12–th payment has been made using the retrospective method. (v) Calculate the outstanding loan balance immediately after the 12–th payment has been made using the prospective method. Solution: (iii) The total interest paid by Roger during the duration of the loan is 28410.78368 − 25000 = 3410.78368.

17/52

c





Example 2 Roger buys a car for $25,000 by making level payments at the end of the month for three years. Roger is charged an annual nominal interest rate of 8.5% compounded monthly in his loan. (i) Find the amount of each monthly payment. (ii) Find the total amount of payments made by Roger. (iii) Find the total interest paid by Roger during the duration of the loan. (iv) Calculate the outstanding loan balance immediately after the 12–th payment has been made using the retrospective method. (v) Calculate the outstanding loan balance immediately after the 12–th payment has been made using the prospective method. Solution: (iv) The outstanding loan balance immediately after the 12–th payment has been made using the retrospective method. is (25000)(1+0.085/12)12 −(789.1884356)s12 = 17361.71419. −− |0.085/12 c



18/52



Example 2 Roger buys a car for $25,000 by making level payments at the end of the month for three years. Roger is charged an annual nominal interest rate of 8.5% compounded monthly in his loan. (i) Find the amount of each monthly payment. (ii) Find the total amount of payments made by Roger. (iii) Find the total interest paid by Roger during the duration of the loan. (iv) Calculate the outstanding loan balance immediately after the 12–th payment has been made using the retrospective method. (v) Calculate the outstanding loan balance immediately after the 12–th payment has been made using the prospective method. Solution: (v) The outstanding loan balance immediately after the 12–th payment has been made using the prospective method is (789.1884356)a24 = 17361.71419. −− |0.085/12 c



19/52



Example 3 A loan is being repaid with 10 payments of $3000 followed by 20 payments of $5000 at the end of each year. The effective annual rate of interest is 4.5%. (i) Calculate the amount of the loan. (ii) Calculate the outstanding loan balance immediately after the 15–th payment has been made by both the prospective and the retrospective method. (iii) Calculate the amounts of interest and principal paid in the 16–th payment.

20/52

c





Example 3 A loan is being repaid with 10 payments of $3000 followed by 20 payments of $5000 at the end of each year. The effective annual rate of interest is 4.5%. (i) Calculate the amount of the loan. (ii) Calculate the outstanding loan balance immediately after the 15–th payment has been made by both the prospective and the retrospective method. (iii) Calculate the amounts of interest and principal paid in the 16–th payment. Solution: (i) The cashflow of payments is Inflows Time

3000 1

3000 2

··· ···

3000 10

5000 11

5000 12

··· ···

5000 30

The loan amount is 3000a10 + (1.045)−10 5000a20 = −− −− |4.5% |4.5% 23738.1545 + 41880.8518 = 65619.0063. c



21/52



Example 3 A loan is being repaid with 10 payments of $3000 followed by 20 payments of $5000 at the end of each year. The effective annual rate of interest is 4.5%. (i) Calculate the amount of the loan. (ii) Calculate the outstanding loan balance immediately after the 15–th payment has been made by both the prospective and the retrospective method. (iii) Calculate the amounts of interest and principal paid in the 16–th payment. Solution: (ii) The outstanding loan balance immediately after the 15–th payment using the retrospective method is 65619.0063(1.045)15 − 3000(1.045)5 s10 − 5000s−5−|4.5% −− |4.5% = 126991.311 − 45940.0337 − 27353.5486 = 53697.7287. 22/52

c





Example 3 A loan is being repaid with 10 payments of $3000 followed by 20 payments of $5000 at the end of each year. The effective annual rate of interest is 4.5%. (i) Calculate the amount of the loan. (ii) Calculate the outstanding loan balance immediately after the 15–th payment has been made by both the prospective and the retrospective method. (iii) Calculate the amounts of interest and principal paid in the 16–th payment. Solution: The outstanding loan balance immediately after the 15–th payment using the prospective method is 5000a15 = 53697.7286. −− |4.5%

23/52

c





Example 3 A loan is being repaid with 10 payments of $3000 followed by 20 payments of $5000 at the end of each year. The effective annual rate of interest is 4.5%. (i) Calculate the amount of the loan. (ii) Calculate the outstanding loan balance immediately after the 15–th payment has been made by both the prospective and the retrospective method. (iii) Calculate the amounts of interest and principal paid in the 16–th payment. Solution: (iii) The amount of interest paid in the 16–th payment is (53697.7286)(0.045) = 2416.39779. The amount of interest paid in the 16–th payment is 5000 − 2416.39779 = 2583.60221.

24/52

c





Next, we consider the amortization method of repaying a loan with level payments made at the end of periods of the same length. Let L be the amount borrowed. Let P be the level payment. Let n be the number of payments. Let i be the effective rate of interest per payment period. The cashflow of payments is Inflows Time

P 1

P 2

P 3

··· ···

P n

We have that L = Pa−n−|i . The outstanding principal after the k–th payment is Bk = L(1 + i)k − Ps−k−|i = P(a−n−|i (1 + i)k − s−k−|i ) = Pan−k−−|i . The interest portion of the k–th payment is iBk−1 = iPan+1−k−−|i = P(1 − ν n+1−k ). The principal reduction of the k–th payment is Bk−1 − Bk = P − iBk−1 = Pν n+1−k . 25/52

c





Using that Bk−1 = Pan+1−k−−|i and Bk−1 − Bk = Pν n+1−k , we get that Bk = P(an+1−k−−|i − ν n+1−k ). The outstanding principal after the k–th payment can be found using all these formulas Bk = L(1 + i)k − Ps−k−|i = P(a−n−|i (1 + i)k − s−k−|i ) =Pan−k−−|i = P(an+1−k−−|i − ν n+1−k ).

26/52

c





The following is the amortization schedule for a loan of L = Pa−n−|i with level payments of P. Period 0 1 2 3 ··· ··· k ··· ··· n−1 n

Payment − P P P ··· ··· P ··· ··· P P

Interest paid − P(1 − ν n ) P(1 − ν n−1 ) P(1 − ν n−2 ) ··· ··· P(1 − ν n+1−k ) ··· ··· P(1 − ν 2 ) P(1 − ν)

Principal repaid − Pν n Pν n−1 Pν n−2 ··· ··· Pν n+1−k ··· ··· Pν 2 Pν

Outstanding balance L = Pa−n−|i −− Pan−1 |i −− Pan−2 |i −− Pan−3 |i ··· ··· Pan−k−−|i ··· ··· Pa−1−|i 0

27/52

c





Example 4 The following is the amortization schedule of a loan of $20,000.00 at an effective interest rate of 8% for 12 years. Time 0 1 2 3 4 5 6 7 8 9 10 11 12

Payment amount − 2653.90 2653.90 2653.90 2653.90 2653.90 2653.90 2653.90 2653.90 2653.90 2653.90 2653.90 2653.91

Interest paid − 1600.00 1515.69 1424.63 1326.29 1220.08 1105.38 981.49 847.70 703.20 547.15 378.61 196.59

Principal reduction − 1053.90 1138.21 1229.27 1327.61 1433.82 1548.52 1672.41 1806.20 1950.70 2106.75 2275.29 2457.32

Balance 2000.00 18946.10 17807.89 16578.62 15251.01 13817.19 12268.67 10596.26 8790.06 6839.36 4732.61 2457.32 0.00 28/52

c





Example 5 A loan of 100,000 is being repaid by 15 equal annual installments made at the end of each year at 6% interest effective annually. (i) Find the amount of each annual installment. (ii) Find the finance charge of this loan. (iii) Find how much interest is accrued in the first year. (iv) Find how principal is repaid in the first payment. (v) Find the balance in the loan immediately after the first payment.

29/52

c





Example 5 A loan of 100,000 is being repaid by 15 equal annual installments made at the end of each year at 6% interest effective annually. (i) Find the amount of each annual installment. (ii) Find the finance charge of this loan. (iii) Find how much interest is accrued in the first year. (iv) Find how principal is repaid in the first payment. (v) Find the balance in the loan immediately after the first payment. Solution: (i)We solve 100000 = Pa15 and get P = 10296.2764. −− |6%

30/52

c





Example 5 A loan of 100,000 is being repaid by 15 equal annual installments made at the end of each year at 6% interest effective annually. (i) Find the amount of each annual installment. (ii) Find the finance charge of this loan. (iii) Find how much interest is accrued in the first year. (iv) Find how principal is repaid in the first payment. (v) Find the balance in the loan immediately after the first payment. Solution: (ii) The finance charge is (15)(10296.2764) − 100000 = 54444.146.

31/52

c





Example 5 A loan of 100,000 is being repaid by 15 equal annual installments made at the end of each year at 6% interest effective annually. (i) Find the amount of each annual installment. (ii) Find the finance charge of this loan. (iii) Find how much interest is accrued in the first year. (iv) Find how principal is repaid in the first payment. (v) Find the balance in the loan immediately after the first payment. Solution: (iii) The amount of interest accrued in the first year is (100000)(0.06) = 6000.

32/52

c





Example 5 A loan of 100,000 is being repaid by 15 equal annual installments made at the end of each year at 6% interest effective annually. (i) Find the amount of each annual installment. (ii) Find the finance charge of this loan. (iii) Find how much interest is accrued in the first year. (iv) Find how principal is repaid in the first payment. (v) Find the balance in the loan immediately after the first payment. Solution: (iv) The amount of principal repaid in the first year is 10296.2764 − 6000 = 4296.2764.

33/52

c





Example 5 A loan of 100,000 is being repaid by 15 equal annual installments made at the end of each year at 6% interest effective annually. (i) Find the amount of each annual installment. (ii) Find the finance charge of this loan. (iii) Find how much interest is accrued in the first year. (iv) Find how principal is repaid in the first payment. (v) Find the balance in the loan immediately after the first payment. Solution: (v) The balance in the loan immediately after the first payment is 100000 − 4296.2764 = 95703.7236.

34/52

c





Example 6 A loan L is being paid with 20 equal annual payments at the end of each year. The principal portion of the 8–th payment is 827.65 and the interest portion is 873.81. Find L.

35/52

c





Example 6 A loan L is being paid with 20 equal annual payments at the end of each year. The principal portion of the 8–th payment is 827.65 and the interest portion is 873.81. Find L. Solution: We know that 827.65 = Pν n+1−k = Pν 13 , 873.81 = P(1 − ν n+1−k ) = P(1 − ν 13 ). Adding the two equations, we get that P = 827.65 + 873.81 = 1701.46. From the equation 827.65 = 1701.46(1 + i)−13 , we get that i = 5.7%. Hence, L = 1701.46a20 = 20000. −− |5.7%

36/52

c





A way to pay a loan is to pay interest as it accrues and to pay the principal in level installments. Suppose that a loan of amount L is paid at the end of each year for n years. At the end of each year two payments are made: one paying the interest accrued and another one making a principal payment of Ln . At the end of j years the outstanding balance is L(n−j) n . Hence, the interest payment at L(n+1−j) the end of j years is i . The total payment made at the end n L(n+1−j) L of the j–th year is n + i = Ln (1 + i(n + 1 − j)). n

37/52

c





Example 7 Samuel takes a loan of $175000. He will pay the loan in 15 years by paying the interest accrued at the end of each year, and paying level payments of the principal at the end of each year. The annual effective rate of interest of the loan is 8.5%.

38/52

c





Example 7 Samuel takes a loan of $175000. He will pay the loan in 15 years by paying the interest accrued at the end of each year, and paying level payments of the principal at the end of each year. The annual effective rate of interest of the loan is 8.5%. (i) Find the amount of each payment of principal.

39/52

c





Example 7 Samuel takes a loan of $175000. He will pay the loan in 15 years by paying the interest accrued at the end of each year, and paying level payments of the principal at the end of each year. The annual effective rate of interest of the loan is 8.5%. (i) Find the amount of each payment of principal. Solution: (i) The annual payment of principal is 175000 = 11666.67. 15

40/52

c





Example 7 Samuel takes a loan of $175000. He will pay the loan in 15 years by paying the interest accrued at the end of each year, and paying level payments of the principal at the end of each year. The annual effective rate of interest of the loan is 8.5%. (ii) Find the outstanding principal owed at the end of the ninth year.

41/52

c





Example 7 Samuel takes a loan of $175000. He will pay the loan in 15 years by paying the interest accrued at the end of each year, and paying level payments of the principal at the end of each year. The annual effective rate of interest of the loan is 8.5%. (ii) Find the outstanding principal owed at the end of the ninth year. Solution: (ii) The outstanding principal owed at the end of the ninth year is (175000)(15−9) = 70000. 15

42/52

c





Example 7 Samuel takes a loan of $175000. He will pay the loan in 15 years by paying the interest accrued at the end of each year, and paying level payments of the principal at the end of each year. The annual effective rate of interest of the loan is 8.5%. (iii) Find the interest accrued during the tenth year.

43/52

c





Example 7 Samuel takes a loan of $175000. He will pay the loan in 15 years by paying the interest accrued at the end of each year, and paying level payments of the principal at the end of each year. The annual effective rate of interest of the loan is 8.5%. (iii) Find the interest accrued during the tenth year. Solution: (iii) The amount of interest paid at the end of the tenth year is (0.085)70000 = 5950.

44/52

c





Example 7 Samuel takes a loan of $175000. He will pay the loan in 15 years by paying the interest accrued at the end of each year, and paying level payments of the principal at the end of each year. The annual effective rate of interest of the loan is 8.5%. (iv) Find the total amount of payments made at the end of the tenth year.

45/52

c





Example 7 Samuel takes a loan of $175000. He will pay the loan in 15 years by paying the interest accrued at the end of each year, and paying level payments of the principal at the end of each year. The annual effective rate of interest of the loan is 8.5%. (iv) Find the total amount of payments made at the end of the tenth year. Solution: (iv) The total amount of payments made at the end of the tenth year is 11666.67 + 5950 = 17616.67.

46/52

c





Example 7 Samuel takes a loan of $175000. He will pay the loan in 15 years by paying the interest accrued at the end of each year, and paying level payments of the principal at the end of each year. The annual effective rate of interest of the loan is 8.5%. (v) Find the total amount of payments which Samuel makes.

47/52

c





Example 7 Samuel takes a loan of $175000. He will pay the loan in 15 years by paying the interest accrued at the end of each year, and paying level payments of the principal at the end of each year. The annual effective rate of interest of the loan is 8.5%. (v) Find the total amount of payments which Samuel makes. Solution: (v) The interest payment at the end of j years is i L(n+1−j) = (0.085) (175000)(16−j) . Hence, the total interest payn 15 ments are (0.085)

15 X (175000)(16 − j)

15

j=1

175000 =(0.085) 15

(16)(15) (16)(15) − 2

= 119000.

The total amount of payments which Samuel makes is 119000 + 175000 = 294000. 48/52

c





Example 8 A loan of $150000 is going to be paid over 20 years with monthly payments. The first payment is one month from now. During each year, the payments are constant. But, they increase by 3% each year. The annual effective rate of interest is 6%. Calculate the total amount of the payments made during the first year. Calculate the outstanding loan balance on the loan ten years from now after the payment is made.

49/52

c





Solution: Let P be the monthly payment during the first year. During the k–th year, 12 payments of P(1.03)k−1 are made. The value of these payments at the end of the k–th year is P(1.03)k−1 s12|i (12) /12 = P(1.03)k−1 12.32652834, where we have used that i (12) = 5.84106068%. So, the cashflow of payments is equivalent to Payments Time

P12.3265 1

P(1.03)12.3265 2

··· ···

P(1.03)19 12.3265 20

The present value of this cashflow is the loan amount: 12.3265P 12.3265P an| i−r = a20| 0.06−0.03 = 179.493145P 150000 = 1+0.03 1+r 1+r 1.03 and P = 835.6865105. The outstanding loan balance on the loan ten years from now after the payment is made is (835.6865105)(1.03)9 (12.3265)a10| 0.06−0.03 = 201586.9934. 1+0.03

50/52

c





Example 9 Mary takes on a loan of $135,000. The loan is being repaid by a 10–year increasing annuity–immediate. The initial payment is 10000, and each subsequent payment is x larger than the preceding payment. The annual effective interest rate is 6.5%. Determine the principal outstanding immediately after the 5–th payment.

51/52

c





Solution: The cashflow is Contributions Time

10000 1

10000 + x 2

10000 + 2x 3

··· ···

10000 + 9x 10

The present value of the payments is 135000 = (10000 − x)a10|6.5% + x(Ia)10|6.5% =(10000 − x)(7.188830223) + 35.82836665x. So, x = 135000−(7.188830223)(10000) 35.82836665−7.188830223 = 2203.656401. The payments to be made after the 5-th payment are Contributions Time

10000 + (5)(2203.656401) 6

··· ···

10000 + (9)(2203.656 10

Its present value at time 5 is (10000 + (4)(2203.656401))a5|6.5% + (2203.656401)(Ia)5|6.5% =78187.55276 + 26321.63894 = 104509.1917. 52/52

c




Manual for SOA Exam FM/CAS Exam 2. Chapter 4. Amortization and sinking bonds. Section 4.2. Sinking funds. c



1/17

c




Section 4.2. Sinking funds.

An alternate way to repay a loan is to make two payments, one directly to the lender and another to an auxiliary fund. The auxiliary fund is called a sinking fund. The payments made directly to the lender apply to the principal. The deposits made into the sinking fund do not. Usually, the sinking fund accumulates with a different interest rate than the rate charged by the lender. At the end of the duration of the loan, the borrower withdraws the total accumulated in the sinking fund and uses this money to pay the loan to the lender.

2/17

c





Usually, we consider the case of payments made at the end of each of n periods. The simplest case is the one when all the payments are equal. Let i be the periodic effective rate charged by the lender on the loan. At the end of each period, the borrower pays P directly to lender. The borrower deposits Q into a sinking fund earning a rate of interest j. Usually, j < i. The cashflow of payments to the principal is Contributions Time

0 0

P 1

P 2

··· ···

P n−1

P +R n

where R is the lump–sum payment obtained by withdrawing the total accumulated in the sinking fund at the end of n periods. The cashflow of deposits in the sinking fund is Contributions Time

0 0

Q 1

Q 2

Q 3

··· ···

Q n

Hence, the accumulated value in the sinking fund at time n is R = Qs−n−|j . c



3/17



The borrower’s total cashflow is Contributions Time

L 0

−P − Q 1

−P − Q 2

−P − Q 3

··· ···

−P − Q n

The lender cashflow is Contributions Time

−L 0

P 1

P 2

P 3

··· ···

P +R n

In order to the loan to be repaid: L = Pa−n−|i + R(1 + i)−n = Pa−n−|i + Qs−n−|j (1 + i)−n .

4/17

c





Example 1 Dave borrows 150,000 from a trust company at an annual effective rate of interest of 9.5%. He agrees to pay the interest annually at the end of the each year, and build up a sinking fund which will repay the loan at the end of 15 years. The sinking fund accumulates at annual effective rate of interest of 4.5%. Calculate:

5/17

c





Example 1 Dave borrows 150,000 from a trust company at an annual effective rate of interest of 9.5%. He agrees to pay the interest annually at the end of the each year, and build up a sinking fund which will repay the loan at the end of 15 years. The sinking fund accumulates at annual effective rate of interest of 4.5%. Calculate: (i) The annual interest payments.

6/17

c





Example 1 Dave borrows 150,000 from a trust company at an annual effective rate of interest of 9.5%. He agrees to pay the interest annually at the end of the each year, and build up a sinking fund which will repay the loan at the end of 15 years. The sinking fund accumulates at annual effective rate of interest of 4.5%. Calculate: (i) The annual interest payments. Solution: (i) The annual interest payment is 150000(0.095) = 14250.

7/17

c





Example 1 Dave borrows 150,000 from a trust company at an annual effective rate of interest of 9.5%. He agrees to pay the interest annually at the end of the each year, and build up a sinking fund which will repay the loan at the end of 15 years. The sinking fund accumulates at annual effective rate of interest of 4.5%. Calculate: (ii) The annual sinking fund payment.

8/17

c





Example 1 Dave borrows 150,000 from a trust company at an annual effective rate of interest of 9.5%. He agrees to pay the interest annually at the end of the each year, and build up a sinking fund which will repay the loan at the end of 15 years. The sinking fund accumulates at annual effective rate of interest of 4.5%. Calculate: (ii) The annual sinking fund payment. Solution: (ii) 150000 = Qs15 and Q = 7217.071217. −− |4.5%

9/17

c





Example 1 Dave borrows 150,000 from a trust company at an annual effective rate of interest of 9.5%. He agrees to pay the interest annually at the end of the each year, and build up a sinking fund which will repay the loan at the end of 15 years. The sinking fund accumulates at annual effective rate of interest of 4.5%. Calculate: (iii) Dave’s total annual outlay.

10/17

c





Example 1 Dave borrows 150,000 from a trust company at an annual effective rate of interest of 9.5%. He agrees to pay the interest annually at the end of the each year, and build up a sinking fund which will repay the loan at the end of 15 years. The sinking fund accumulates at annual effective rate of interest of 4.5%. Calculate: (iii) Dave’s total annual outlay. Solution: (iii) Dave’s total annual outlay is 14250 +7217.071217 = 21467.07122.

11/17

c





Example 1 Dave borrows 150,000 from a trust company at an annual effective rate of interest of 9.5%. He agrees to pay the interest annually at the end of the each year, and build up a sinking fund which will repay the loan at the end of 15 years. The sinking fund accumulates at annual effective rate of interest of 4.5%. Calculate: (iv) The annual effective rate of interest i 0 for which the payments made at the end of the year will be equal to Dave’s total annual outlay.

12/17

c





Example 1 Dave borrows 150,000 from a trust company at an annual effective rate of interest of 9.5%. He agrees to pay the interest annually at the end of the each year, and build up a sinking fund which will repay the loan at the end of 15 years. The sinking fund accumulates at annual effective rate of interest of 4.5%. Calculate: (iv) The annual effective rate of interest i 0 for which the payments made at the end of the year will be equal to Dave’s total annual outlay. Solution: (iv) 150000 = (21467.07122)a15 and i 0 = −− 0 |i 11.52440895%. Notice that although Dave borrows at a rate 9.5%, by getting only 4.5% in his sinking fund, the actual rate of interest Dave which is paying is higher.

13/17

c





Example 1 Dave borrows 150,000 from a trust company at an annual effective rate of interest of 9.5%. He agrees to pay the interest annually at the end of the each year, and build up a sinking fund which will repay the loan at the end of 15 years. The sinking fund accumulates at annual effective rate of interest of 4.5%. Calculate: (v) Find i + n−1 n+1 (i − j), where i = 9.5%, j = 4.5% and n = 15, and compare with i 0 .

14/17

c





Example 1 Dave borrows 150,000 from a trust company at an annual effective rate of interest of 9.5%. He agrees to pay the interest annually at the end of the each year, and build up a sinking fund which will repay the loan at the end of 15 years. The sinking fund accumulates at annual effective rate of interest of 4.5%. Calculate: (v) Find i + n−1 n+1 (i − j), where i = 9.5%, j = 4.5% and n = 15, and compare with i 0 . (15−1)(0.95−0.045) Solution: (v) i + n−1 = 0.13875. n+1 (i − j) = 0.095 + 15+1 which is sort of close to 0.1152440895.

15/17

c





Example 2 Steve repays a loan of $18,000 by making interest payment at the end of the year for 15 years and equal deposits at the end of each year into a sinking fund for 15 years. At the end of the 15 years, Steve withdraws the balance from the sinking fund and pays the loan. The sinking fund earns 6% effective annually. Immediately after the fourth payment, the yield on the sinking fund increases to 7% effective annually. At that time, Steve adjusts his sinking fund payment to x so that the sinking fund will accumulate to $18,000, 15 years after the original loan date. Find x.

16/17

c





Example 2 Steve repays a loan of $18,000 by making interest payment at the end of the year for 15 years and equal deposits at the end of each year into a sinking fund for 15 years. At the end of the 15 years, Steve withdraws the balance from the sinking fund and pays the loan. The sinking fund earns 6% effective annually. Immediately after the fourth payment, the yield on the sinking fund increases to 7% effective annually. At that time, Steve adjusts his sinking fund payment to x so that the sinking fund will accumulate to $18,000, 15 years after the original loan date. Find x. Solution: Let Q be the initial payment Joe makes to the sinking fund. Then, 18000 = Qs15 . Hence, Q = 773.3297512. The −− |6% balance in the sinking fund immediately after the fourth payment is 773.3297512s−4−|6% = 3383.020703. The final accumulation in the sinking fund is 18000. So, 18000 = 3383.020703(1.07)11 + xs11 −− |7% and x = 689.2751, which can be found doing: 11 N 7 I/Y 3383.020703 PV −18000 FV CPT PMT c



17/17


Manual for SOA Exam FM/CAS Exam 2. Chapter 4. Amortization and sinking bonds. Section 4.3. Reinvestment. c



1/6

c




Section 4.3. Reinvestment.

Suppose that a bank account pays interest in the original deposit, but not in the obtained interest. Then, it will wise to withdraw the interest and invest it in another account. In other situations, it makes sense to reinvest the earned interest in a different investment. For example, a stock pays dividends and/or capital gains, which can be invested somewhere else. We obtain a flow of interest payments which are invested at different rate from the one in the initial investment. Suppose that a mortgage company makes a loan to a customer. To know the mortgage company’s return in its investment we need to take in account the interest rate charged to the customer in the loan and the interest rate which the mortgage company gets in the monthly payments it receives.

2/6

c





Example 1 Payments of $2500 are invested at the end of each quarter for 5 years. The payments earn interest at an annual nominal interest rate converted quarterly of 14% and the interest payments are reinvested at an annual nominal interest rate converted quarterly of 10%. Find the total accumulation for both accounts at the end of 5 years.

3/6

c





Solution: The balance in the first account is Balance in the 1st account Time (in quarters)

2500 1

2(2500) 2

··· ···

··· ···

(20)(2500) 20

Since the rate of interest per quarter is 14%/4 = 3.5%, the interest payments are Cashflow of interest Time (in quarters)

(1)(0.035)(2500) 2

··· ···

19(0.035)(2500) 20

The total accumulation for both accounts at the end of the 5–th year is (20)(2500) + (0.035)(2500) (Is)19 −− |0.025 (1.025)s −− −19 19 |0.025 = 50000 + (87.5) 0.025 = 50000 + (87.5) 24.54465761−19 = 69406.30164. 0.025 4/6

c





Example 2 John invests 1000 at the beginning of each year for 5 years at an annual effective interest rate of 10% and reinvests the interest at an annual effective interest rate of 8%. Calculate the total value of his investment at the end of 5 years.

5/6

c





Solution: The balance in John’s account is Balance in John’s account Time

1000 0

2000 1

3000 2

4000 3

5000 4

400 4

500 5

5000 5

The interest payments which John gets are John’s interest payments Time

0 0

100 1

200 2

300 3

The total accumulation of John’s investments at time 5 is 5000 + 100 (Is)−5−|0.08 = 5000 + 100 =5000 + 100

6.335929037 − 5 0.08

(1.08)s−5−|0.08 − 5

!

0.08

= 6669.911296.

6/6

c



Chapter 5. Bonds.

Manual for SOA Exam FM/CAS Exam 2. Chapter 5. Bonds. Section 5.1. Securities. c



1/??

c



Chapter 5. Bonds.

Section 5.1. Securities.

Securities

When a corporation or a public institution needs to raise money, it arranges contracts with investors. These contracts are called financial assets or securities. From the investor’s view point, securities are investment instruments, endorsed by a corporation, government, or other organization. The borrower offers either debt or ownership (equity). In the case of debt, the borrower agrees to make a series of payments to the investor. In the case of equity securities, the borrower gives a part of the ownership of the corporation to the investor.

2/??

c



Chapter 5. Bonds.


Stock For a corporation, claims to its ownership are determined by shares of stock. The proportion of a company which an investor owns is the fraction of its shares owned over the total number of shares outstanding. There are several types of stocks. Some stocks provide voting rights to the holder. Some stocks entitle the holder to receive payments, such as dividends and/or capital appreciation. Different types of stocks have different order of preference to the company’s assets in the case of liquidation (the company is forced to sells its assets, pay outstanding debts, and distribute the remainder to shareholders). Common stock entitles the holder to payments of dividends and capital appreciation. Common stock gives voting rights. A preferred stock is a stock which provides a fixed dividend that does not fluctuate. Preferred stock shareholders do not enjoy voting rights. c



3/??

Chapter 5. Bonds.


Bonds

A way for a government or corporation to raise money is to issue bonds. A bond is a certificate issued from a borrower to a lender agreeing to make payment(s) in a loan. The price of a bond P is the amount that the lender pays (loans) to the government or corporation for the bond. Types of available bonds are: U.S. government securities, municipal bonds, corporate bonds, mortgage and asset–backed securities, federal agency securities and foreign government bonds.

4/??

c



Chapter 5. Bonds.

Manual for SOA Exam FM/CAS Exam 2. Chapter 5. Bonds. Section 5.2. Price of a bond. c



1/17

c



Chapter 5. Bonds.

Section 5.2. Price of a bond.

Bonds There are two kind of bonds: accumulation bonds and bonds with coupons. The time at which the loan is repaid is called the maturity date (or redemption date). I

In the case of accumulation bonds, the borrower agrees to pay the loan plus interest at a unique date, called the redemption time. An accumulation bond is also called a zero coupon bond.

I

The most common bonds are bonds with coupons. For bonds with coupons, the borrower agrees to make period payments (coupons) plus a balloon payment (the redemption value) C at the maturity date.

2/17

c



Chapter 5. Bonds.


Cashflow of a bond with coupons.

Every bond has a face value (or par value) F . The coupon payment is Fr . Here, r is the coupon rate per interest period. Often, the payments are semiannually and 2r is the annual nominal rate of interest convertibly semiannually. A bond is called redeemable at par if C = F . Unless said otherwise we assume that a bond is redeemable at par. Let n be the number of interest periods until the redemption date. Let i be the yield rate per interest period. The cashflow for the borrower is Contributions Time

P 0

−Fr 1

−Fr 2

−Fr 3

··· ···

−Fr − C n

3/17

c



Chapter 5. Bonds.


Variables for a bond. I I I I I I I I I I I I I I

P = the price of a bond. F = the par or face value of a bond. C = the redemption value of a bond. r = the coupon rate of a bond. Fr = the amount of a coupon. i = the yield rate of the bond per coupon period. 1 ν = 1+i = the discount factor per coupon period. n = the number of coupon payment periods. g = Fr C = the modified coupon rate of a bond G = Fri = the base amount of a bond. K = C ν n = the present value, compounded at the yield rate, of the redemption value of a bond P − C = the premium (if P > C ). C − P = the discount (if C > P). k = P−C premium as a fraction of redemption value. C

c



4/17

Chapter 5. Bonds.


Price of a bond The basic formula for the price of a bond is P = Fra−n−|i + C (1 + i)−n = Fra−n−|i + K . The premium/discount formula for the price of a bond is P = Fra−n−|i + C ν n = Fra−n−|i + C (1 − ia−n−|i ) =C + (Fr − Ci)a−n−|i = C + C (g − i)a−n−|i . The base amount formula for the price of a bond is P = Fra−n−|i +C ν n = Gia−n−|i +C ν n = G (1−ν n )+C ν n = G +(C −G )ν n . The Makehan formula for the price of a bond is 1 − νn P = Fra−n−|i + C ν n = Cg + C νn i g g = (C − C ν n ) + C ν n = (C − K ) + K . i i 5/17

c



Chapter 5. Bonds.


Example 1 Find the price of a 10–year bond, redeemable at par, with face value of $10,000 and coupon rate of 10%, convertible quarterly, that will yield 8%, convertible quarterly.

6/17

c



Chapter 5. Bonds.


Example 1 Find the price of a 10–year bond, redeemable at par, with face value of $10,000 and coupon rate of 10%, convertible quarterly, that will yield 8%, convertible quarterly. Solution: We know that F = C = 10000, n = (10)(4) = 40, r = 0.10 4 = 0.025, Fr = (10000)(0.025) = 250. and 0.08 i = 4 = 0.02. So, the price of the bond is P = Fra−n−|i + C (1 + i)−n = 250a40 + 10000(1.02)−40 −− |0.02 =11367.77396. In the calculator, do: 40 N 2 I/Y 250 PMT 10000 PMT CPT PV .

7/17

c



Chapter 5. Bonds.


Example 2 A 30 year bond matures at its face value of 10,000. It pays semiannual coupons of 600. Calculate the price of the bond if the annual nominal interest rate convertible semiannually is 7.5%.

8/17

c



Chapter 5. Bonds.


Example 2 A 30 year bond matures at its face value of 10,000. It pays semiannual coupons of 600. Calculate the price of the bond if the annual nominal interest rate convertible semiannually is 7.5%. Solution: We know that F = C = 10000, n = (30)(2) = 60, Fr = 600, and i = 7.5% 2 = 3.75%. The price of the bond is (600)a60 + (10000)(1 + 0.0375)−60 = 15341.03109. −− |0.0375

9/17

c



Chapter 5. Bonds.


Remember that unless said otherwise a bond is redeemable at par.

Example 3 What is the price of a 5–year 100 par–value bond having quarterly coupons at a quarter rate of 1.5% that is bought to yield a nominal annual rate of 12% convertible monthly?

10/17

c



Chapter 5. Bonds.


Remember that unless said otherwise a bond is redeemable at par.

Example 3 What is the price of a 5–year 100 par–value bond having quarterly coupons at a quarter rate of 1.5% that is bought to yield a nominal annual rate of 12% convertible monthly? Solution: Solution: We know that F = C = 100, n = (5)(4) = 20, r = 0.015 and Fr = (100)(0.015) = 1.5. Let j be the effective yield rate per quarter. We have that i (12) = 12%, i = 12.68250301% and i (4) = 12.1204%, j = i (4) /4 = 3.0301%. Hence, P = 1.5a20 + 100(1.030301)−20 = 77.29919664. −− |3.0301%

11/17

c



Chapter 5. Bonds.


Example 4 The price of a zero coupon 1000 face value bond is 599.4584. The yield rate convertible semi–annually is 6.5%. Calculate the maturity date.

12/17

c



Chapter 5. Bonds.


Example 4 The price of a zero coupon 1000 face value bond is 599.4584. The yield rate convertible semi–annually is 6.5%. Calculate the maturity date. Solution: Let n be the maturity date in years. We have that 2n 599.4584 = (1000) 1 + 0.065 and n = 8 years. 2

13/17

c



Chapter 5. Bonds.


Example 5 What is the yield as an annual effective rate of interest on a 100 par–value 10–year bond with coupon rate 6%, convertible monthly, that is selling for 90?

14/17

c



Chapter 5. Bonds.


Example 5 What is the yield as an annual effective rate of interest on a 100 par–value 10–year bond with coupon rate 6%, convertible monthly, that is selling for 90? Solution: We know that P = 90, C = F = 100, n = (10)(12) = 120, r = 0.06 12 = 0.005 and Fr = (100)(0.005) = 0.5. Let j = i (12) /12. Then, −120 . 90 = P = Fra−n−|j + C (1 + j)−12n = (0.5)a120 −− + 100(1 + j) |j

Hence, j = 0.618181404%, i (12) = 12j = 7.419376846% and i = 7.676949087%.

15/17

c



Chapter 5. Bonds.


Example 6 A 1000 par value 10–year bond with semiannual coupons and redeemable at 1200 is purchased to yield 8% convertible semiannually. The first coupon is 50. Each subsequent coupon is 3% greater than the preceding coupon. Find the price of the bond.

16/17

c



Chapter 5. Bonds.


Example 6 A 1000 par value 10–year bond with semiannual coupons and redeemable at 1200 is purchased to yield 8% convertible semiannually. The first coupon is 50. Each subsequent coupon is 3% greater than the preceding coupon. Find the price of the bond. Solution: The cashflow of coupons is Coupons Time (in half–years)

50 1

(50)(1.03) 2

··· ···

(50)(1.03)19 20

The present value of the payments is 1 a −− 0.04−0.03 + (1200)(1.04)−20 1.03 20 | 1.03 =878.5721 + 547.6643 = 1426.2364. (50)

17/17

c



Chapter 5. Bonds.

Manual for SOA Exam FM/CAS Exam 2. Chapter 5. Bonds. Section 5.3. Book value and amortization schedules. c



1/??

c



Chapter 5. Bonds.

Section 5.3. Book value and amortization schedules.

Book value The book value Bk of a bond at time k of a bond is the present value of the payments to be made, i.e. the present value at time k of the remaining n − k coupons and the redemption value C . This is also the outstanding balance of the loan at that time. So, Bk can be found using any of the following expressions: Bk = Fran−k−−|i +C ν n−k = C +(Fr −Ci)an−k−−|i = C +C (g −i)an−k−−|i . Notice that Fran−k−−|i + C ν n−k = Fran−k−−|i + C (1 − ian−k−−|i ) =C + (Fr − Ci)an−k−−|i = C + C (g − i)an−k−−|i .

2/??

c



Chapter 5. Bonds.


Of course, we have that B0 = P and Bn = C . The previous formula for Bk is that of the outstanding balance of a loan using the prospective method. Using the retrospective method, the book value of a bond is Bk = P(1 + i)k − Frs−k−|i . The previous formula is equivalent to P = Bk (1 + i)−k + Fra−k−|i . A way to interpret the previous formula is as follow. Bk is the balance in the loan after the first k payments are made. The present value of Bk and the payments made until that moment equals the initial balance.

3/??

c



Chapter 5. Bonds.


Example 1 Zack buys a 20 year bond with a par value of 4000 and 10% semiannual coupons. He attains an annual yield of 5% convertible semiannually. The redemption value of the bond is 1200. Find the book value of the bond at the end of the 12–th year.

4/??

c



Chapter 5. Bonds.


Example 1 Zack buys a 20 year bond with a par value of 4000 and 10% semiannual coupons. He attains an annual yield of 5% convertible semiannually. The redemption value of the bond is 1200. Find the book value of the bond at the end of the 12–th year. Solution: We have that F = 4000, r = 5%, Fr = (4000)(0.05) = 200, n = 40, k = 24 and C = 1200. Hence, B24 = Fran−k−−|i + C ν n−k = (200)a16 + (1200)(1.05)−16 −− |5% =3419.350452.

5/??

c



Chapter 5. Bonds.


Example 2 An n–year 5000 par value bond pays 6% annual coupons. At annual yield of 3%, the book value of the bond at the end of year 7 is 5520. Calculate the price of the bond.

6/??

c



Chapter 5. Bonds.


Example 2 An n–year 5000 par value bond pays 6% annual coupons. At annual yield of 3%, the book value of the bond at the end of year 7 is 5520. Calculate the price of the bond. Solution: We have that F = C = 5000, r = 6%, Fr = (5000)(0.06) = 300, B7 = 5520. Hence, P = Bk (1 + i)−k + Fra−k−|i = 5520(1.03)−7 + (300)a−7−|3% = 6357.35.

7/??

c



Chapter 5. Bonds.


Inductive relation for book value Theorem 1 Bk+1 = Bk (1 + i) − Fr .

8/??

c



Chapter 5. Bonds.


Inductive relation for book value Theorem 1 Bk+1 = Bk (1 + i) − Fr . Proof: Bk (1 + i) − Fr = (Fran−k−−|i + C ν n−k )(1 + i) − Fr =Fr ((1 + i)an−k−−|i − 1) + C ν n−k−1 n−k−1 =Fran−k−1 = Bk+1 . −− + C ν |i

9/??

c



Chapter 5. Bonds.


Inductive relation for book value Theorem 1 Bk+1 = Bk (1 + i) − Fr . Proof: Bk (1 + i) − Fr = (Fran−k−−|i + C ν n−k )(1 + i) − Fr =Fr ((1 + i)an−k−−|i − 1) + C ν n−k−1 n−k−1 =Fran−k−1 = Bk+1 . −− + C ν |i

Bk is the outstanding balance at time k. One period later, the principal has increased to Bk (1 + i), i.e. Bk (1 + i) is the outstanding balance immediately before the (k + 1)–th payment is made. Immediately after the (k + 1)–th payment to principal of Fr is made, the outstanding balance is Bk+1 = Bk (1 + i) − Fr . c



10/??

Chapter 5. Bonds.


Example 3 Consider a 30–year $50,000 par–value bond with semiannual coupons, with r = 0.03, and yield rate 10%, convertible semiannually. (i) Find the book value of the bond immediately after the 25–th coupon payment. (ii) Find book price immediately before the 26–th coupon payment. (iii) Find the book value of the bond immediately after the 26–th coupon payment.

11/??

c



Chapter 5. Bonds.


Example 3 Consider a 30–year $50,000 par–value bond with semiannual coupons, with r = 0.03, and yield rate 10%, convertible semiannually. (i) Find the book value of the bond immediately after the 25–th coupon payment. (ii) Find book price immediately before the 26–th coupon payment. (iii) Find the book value of the bond immediately after the 26–th coupon payment. Solution: (i) We have that F = C = 50000, i = 0.05, Fr = 50000(0.03) = 1500 and n = 60. So, B25 = Fran−k−−|i +C ν n−k = 1500a35 +50000(1.05)−35 = 33625.80571. −− |0.05

12/??

c



Chapter 5. Bonds.


Example 3 Consider a 30–year $50,000 par–value bond with semiannual coupons, with r = 0.03, and yield rate 10%, convertible semiannually. (i) Find the book value of the bond immediately after the 25–th coupon payment. (ii) Find book price immediately before the 26–th coupon payment. (iii) Find the book value of the bond immediately after the 26–th coupon payment. Solution: (i) We have that F = C = 50000, i = 0.05, Fr = 50000(0.03) = 1500 and n = 60. So, B25 = Fran−k−−|i +C ν n−k = 1500a35 +50000(1.05)−35 = 33625.80571. −− |0.05 (ii) Just before the next coupon payment the book value is 33625.80571(1.05) = 35307.096. 13/??

c



Chapter 5. Bonds.


Example 3 Consider a 30–year $50,000 par–value bond with semiannual coupons, with r = 0.03, and yield rate 10%, convertible semiannually. (i) Find the book value of the bond immediately after the 25–th coupon payment. (ii) Find book price immediately before the 26–th coupon payment. (iii) Find the book value of the bond immediately after the 26–th coupon payment. Solution: (i) We have that F = C = 50000, i = 0.05, Fr = 50000(0.03) = 1500 and n = 60. So, B25 = Fran−k−−|i +C ν n−k = 1500a35 +50000(1.05)−35 = 33625.80571. −− |0.05 (ii) Just before the next coupon payment the book value is 33625.80571(1.05) = 35307.096. (iii) The book value of the bond immediately after the 26–th coupon payment is 35307.096 − 1500 = 33807.096. c



14/??

Chapter 5. Bonds.


General inductive relation for book value By induction from the previous formula, we get that Bk+m = Bk (1 + i)m − Frsm −− . |i Notice that: I

Bk+m is the outstanding balance immediately after the k + m payment.

I

Bk (1 + i)m is the accrued balance at time k + m of the outstanding balance immediately after the k payment.

I

Frsm −− is the future value at time k + m of the coupon |i payments k + 1, k + 2, . . . , k + m, i.e. the coupons payments from when the outstanding was Bk until when the outstanding is Bk+m . 15/??

c



Chapter 5. Bonds.


Example 4 An n–year 4000 par value bond with 9% semiannual coupons has an annual nominal yield of i, i > 0, convertible semiannually. The book value of the bond at the end of year 4 is 3812.13 and the book value at the end of year 7 is 3884.27. Calculate i.

16/??

c



Chapter 5. Bonds.


Example 4 An n–year 4000 par value bond with 9% semiannual coupons has an annual nominal yield of i, i > 0, convertible semiannually. The book value of the bond at the end of year 4 is 3812.13 and the book value at the end of year 7 is 3884.27. Calculate i. Solution: We have that F = C = 4000, r = 4.5% and Fr = 4000(0.045) = 180. The end of year 4 is the end of the 8–th period. The end of year 7 is the end of the 14–th period. Hence, B8 = 3812.13 and B14 = 3884.27 and 3884.27 = 3812.13(1 + i)6 − (180)s−6−|i . using the calculator with 6 N −3812.13 PV 180 PMT 3884.27 FV CPT I/Y we get that i = 5% (this is the effective interest rate per period). The annual nominal rate of interest convertible semiannually is i = 10%. 17/??

c



Chapter 5. Bonds.


Amount of interest contained in the k–th coupon. The amount of interest contained in the k–th coupon is Ik = iBk−1 , which can obtained using any of the following formulas: Ik = iBk−1 = Fr − (Fr − Ci)ν n−k+1 = Cg − C (g − i)ν n−k+1 . Notice that iBk−1 = iFran+1−k−−|i + iC ν n−k+1 = Fr (1 − ν n−k+1 ) + iC ν n−k+1 =Fr − (Fr − Ci)ν n−k+1 = Cg − (Cg − Ci)ν n−k+1 =Cg − C (g − i)ν n−k+1 .

18/??

c



Chapter 5. Bonds.


Principal portion in the the k–th coupon.

The principal portion in the k–th coupon is Pk = Bk−1 − Bk = Fr − Bk−1 i = Fr − Ik and it can be obtained using any of the following formulas: Pk = (Fr − Ci)ν n−k+1 = C (g − i)ν n−k+1 . Pk is the change in the book value of the bond (principal adjustment) between times k − 1 and k. Pk could be either negative, or zero or positive. Pk is the amortization in the k–th payment.

19/??

c



Chapter 5. Bonds.


Example 5 Kendal buys a 5000 par–value 10 year bond with 8% semiannual coupons to yield 4% converted semiannually. Find the amount of interest and principal in the 5–th coupon.

20/??

c



Chapter 5. Bonds.


Example 5 Kendal buys a 5000 par–value 10 year bond with 8% semiannual coupons to yield 4% converted semiannually. Find the amount of interest and principal in the 5–th coupon. Solution: We have that F = C = 5000, r = g = 0.04, Fr = 200, n = 20 and i = 2%. Using that Ik = Fr − (Fr − Ci)ν n+1−k and Pk = (Fr − Ci)ν n+1−k , we get that I5 = 200 − (200 − (5000)(0.02))(1.02)−(20+1−5) = 127.1554 and P5 = (200 − (5000)(0.02))(1.02)−(20+1−5) = 72.8446.

21/??

c



Chapter 5. Bonds.


Example 6 For a bond pays annual coupons. The principal amortized in the first coupon is half that amortized in the 10–th coupon. Calculate the annual yield rate.

22/??

c



Chapter 5. Bonds.


Example 6 For a bond pays annual coupons. The principal amortized in the first coupon is half that amortized in the 10–th coupon. Calculate the annual yield rate. Solution: We know that (Fr − Ci)ν n = 12 (Fr − Ci)ν n−9 . Hence, (1 + i)9 = 2 and i = 8.005973889%.

23/??

c



Chapter 5. Bonds.


The amortization schedule of a bond is Time 0 1 2 ··· k ··· n−1 n

Payment − Fr Fr ··· Fr ··· Fr Fr + C

Interest paid (iBk−1 ) − Fr − (Fr − Ci)ν n Fr − (Fr − Ci)ν n−1 ··· Fr − (Fr − Ci)ν n−k+1 ··· Fr − (Fr − Ci)ν 2 Fr − (Fr − Ci)ν

Principal repaid − (Fr − Ci)ν n (Fr − Ci)ν n−1 ··· (Fr − Ci)ν n+1−k ··· (Fr − Ci)ν 2 (Fr − Ci)ν + C

Book value (Bk ) C + (Fr − Ci)a−n−|i C + (Fr − Ci)an−1 −− |i C + (Fr − Ci)an−2 −− |i ··· C + (Fr − Ci)an−k −− |i ··· C + (Fr − Ci)a−1−|i 0

The total payments in a bond are nFr + C . The total coupon interest (sum of the column of interest payments) is nFr + C − P. The total coupon principal (sum of the column of payments to principal) is P. 24/??

c



Chapter 5. Bonds.


Notice that for each time k the interest paid plus the principal paid equal to the total payment. We also have that Ik = iBk−1 = i C + (Fr − Ci)an−k+1 = iC + i(Fr − Ci)an−k+1 −− −− |i |i =Fr + (Fr − Ci)(1 − ν n−k+1 ) = Fr − (Fr − Ci)ν n−k+1 . and Bk = Bk−1 − (Fr − Ci)ν n+1−k n+1−k =C + (Fr − Ci)an−k+1 −− − (Fr − Ci)ν |i n+1−k =C + (Fr − Ci)(an−k+1 ) = C + (Fr − Ci)an−k−−|i . −− − ν |i

25/??

c



Chapter 5. Bonds.


Example 7 A 1000 par–value 3–year bond pays 6%, convertible semiannually, and has a yield rate of 8%, convertible semiannually. (i) What is the interest paid in the 3rd coupon? (ii) What is the change in book value contained in the 3rd coupon? (iii) Construct a bond amortization schedule.

26/??

c



Chapter 5. Bonds.


Example 7 A 1000 par–value 3–year bond pays 6%, convertible semiannually, and has a yield rate of 8%, convertible semiannually. (i) What is the interest paid in the 3rd coupon? (ii) What is the change in book value contained in the 3rd coupon? (iii) Construct a bond amortization schedule. Solution: (i) We know that F = C = 1000, r = 0.03, Fr = 30, n = 6, i = 0.04 and Ci = 40. The interest paid in the 3rd coupon is I3 = Fr −(Fr −Ci)ν n−k+1 = 30−(30−40)(1.04)−4 = 38.54804191. We also can do B2 = 30a−4−|0.04 + 1000(1.04)−4 = 963.7010478 and I3 = B2 (0.04) = 38.54804191. 27/??

c



Chapter 5. Bonds.


Example 7 A 1000 par–value 3–year bond pays 6%, convertible semiannually, and has a yield rate of 8%, convertible semiannually. (i) What is the interest paid in the 3rd coupon? (ii) What is the change in book value contained in the 3rd coupon? (iii) Construct a bond amortization schedule. Solution: (ii) Since I3 > Fr , the book value increases in the 3rd coupon. The increase in the book value in the 3rd coupon is I3 − Fr = 38.54804191 − 30 = 8.54804191. We also can do B3 = 30a−3−|0.04 + 1000(1.04)−3 = 972.2490897 and B3 − B2 = 972.2490897 − 963.7010478 = 8.5480419.

28/??

c



Chapter 5. Bonds.


(iii) Here is the bond amortization schedule: Time 0 1 2 3 4 5 6

Coupon

Interest

Principal Adjustment

30 30 30 30 30 30

37.90 38.22 38.59 38.89 39.25 39.62

7.90 8.22 8.59 8.89 9.25 9.62

Book Value 947.58 955.48 963.70 972.25 981.14 990.37 1000.00

29/??

c



Chapter 5. Bonds.


The price of a typical bond changes in the opposite direction from a change in interest rates. As interest rates rise, the price of a bond falls. A bond assures a determined number of payments in the future, if interest rates rise, the present value of these payments decreases. We make an unrealized capital loss, if the market value is less than the book value. Reciprocally, if interest rates decline, the price of a bond rises. The market value of a bond is the price at which a bond is bought/sold. When rates of interest change, the market value of a bond changes. We make an unrealized capital gain, if the market value is bigger than the book value.

30/??

c



Chapter 5. Bonds.


Example 8 Oliver buys a ten–year 5000 face value bond with semiannual coupons at annual rate of 6%. He buys his bond to yield 8% compounded semiannually and immediately sell them to an investor to yield 4% compounded semiannually. What is Oliver’s profit in this investment?

31/??

c



Chapter 5. Bonds.


Example 8 Oliver buys a ten–year 5000 face value bond with semiannual coupons at annual rate of 6%. He buys his bond to yield 8% compounded semiannually and immediately sell them to an investor to yield 4% compounded semiannually. What is Oliver’s profit in this investment? Solution: We have that F = C = 5000, r = 0.03, Fr = 150 and n = 20. Oliver buys his bond for (150)a20 + 5000(1.04)−20 = 4320.484 −− |0.04 Oliver sells his bond for (150)a20 + 5000(1.02)−20 = 5817.572 −− |0.02 Oliver’s profit is 5817.572 − 4320.484 = 1497.088. 32/??

c



Chapter 5. Bonds.


Example 9 On January 1, 2000, Maxwell bought a 10–year $5000 non–callable bond with coupons at 7% convertible semiannually. Maxwell bought the bond to yield 7%, compounded semiannually. On July 1, 2005, the market value of bonds is based on a 5% interest rate, compounded semiannually. Calculate the unrealized capital gain on July 1, 2005.

33/??

c



Chapter 5. Bonds.


Example 9 On January 1, 2000, Maxwell bought a 10–year $5000 non–callable bond with coupons at 7% convertible semiannually. Maxwell bought the bond to yield 7%, compounded semiannually. On July 1, 2005, the market value of bonds is based on a 5% interest rate, compounded semiannually. Calculate the unrealized capital gain on July 1, 2005. Solution: We have that F = 5000, r = 0.035 and Fr = 175. On July 1, 2005, there are nine remaining coupons. On July 1, 2005, the book value of the bond is 175a−9−|3.5% + 5000(1.035)−9 = 5000. On July 1, 2005, the market value of the bond is 175a−9−|2.5% + 5000(1.025)−9 = 5398.543276. The unrealized capital gain is 5398.543276 − 5000 = 398.543276. c



34/??

Chapter 5. Bonds.


Premium. If the investor is paying more than the redemption value, i.e. if P > C , we say that the bond has being bought at premium. The premium is P − C . We have that P − C = (Fr − Ci)a−n−|i = C (g − i)a−n−|i . A bond has been bought at premium if and only if g > i. For a bond bought at premium Fr > Ci and P = B0 > B1 > B2 > · · · > Bn = C . Notice that Bk − C = (Fr − Ci)an−k−−|i . Pk = Bk−1 − Bk = (Fr − Ci)ν n+1−k is the write–up Pn in premium in the k–th coupon. The premium is P − C = j=1 (Bj−1 − Bj ).

35/??

c



Chapter 5. Bonds.


Discount.

If the investor is paying less than the redemption value, i.e. if P < C , we say that the bond has being bought at discount. The discount is C − P. We have that C − P = (Ci − Fr )a−n−|i = C (i − g )a−n−|i . A bond has been bought at discount if and only if g < i. For a bond bought at discount Fr < Ci and P = B0 < B1 < B2 < · · · < Bn = C . |Pk | = Bk − Bk−1 = (Ci − Fr )ν n+1−k is the write–up in discount Pn in the k–th coupon. The discount is C − P = j=1 (Bj − Bj−1 ).

36/??

c



Chapter 5. Bonds.


Example 10 A 10 year 50000 face value bond pays semi–annual coupons of 3000. The bond is bought to yield a nominal annual interest rate of 6% convertible semi–annually. Calculate the premium paid for the bond.

37/??

c



Chapter 5. Bonds.


Example 10 A 10 year 50000 face value bond pays semi–annual coupons of 3000. The bond is bought to yield a nominal annual interest rate of 6% convertible semi–annually. Calculate the premium paid for the bond. Solution: We know that F = C = 50000, Fr = 3000, n = 20, and i = 3%. The price of the bond is P = Fra−n−|i + C (1 + i)−n = (3000)a20 + (50000)(1.03)−20 −− |3% =72316.21229. The premium of the bond is P − C = 72316.21229 − 50000 = 22316.21229.

38/??

c



Chapter 5. Bonds.


Example 11 Oprah buys a 10000 par–value 15 year bond with 9% semiannual coupons to yield 5% converted semiannually. (i) Find the premium in the bond. (ii) Find the write up in premium in the 8–th coupon.

39/??

c



Chapter 5. Bonds.


Example 11 Oprah buys a 10000 par–value 15 year bond with 9% semiannual coupons to yield 5% converted semiannually. (i) Find the premium in the bond. (ii) Find the write up in premium in the 8–th coupon. Solution: (i) We have that F = C = 10000, r = 0.045, Fr = 450 and n = 30. Oprah buys her bond for (450)a30|2.5% + 10000(1.025)−30 = 14186.06 The premium of the bond is P − C = 14186.06 − 10000 = 4186.06.

40/??

c



Chapter 5. Bonds.


Example 11 Oprah buys a 10000 par–value 15 year bond with 9% semiannual coupons to yield 5% converted semiannually. (i) Find the premium in the bond. (ii) Find the write up in premium in the 8–th coupon. Solution: (i) We have that F = C = 10000, r = 0.045, Fr = 450 and n = 30. Oprah buys her bond for (450)a30|2.5% + 10000(1.025)−30 = 14186.06 The premium of the bond is P − C = 14186.06 − 10000 = 4186.06. (ii) The write–up in premium in the 8–th coupon is (Fr − Ci)ν n+1−k = (10000)(0.045 − 0.025)(1.025)−23 = 113.34. We also can do the problem in the following way: B7 = (450)a23 + 10000(1.025)−23 = 13466.42 −− |2.5% B8 = (450)a22 + 10000(1.025)−22 = 13353.08 −− |2.5% and B7 − B8 = 13466.42 − 13353.08 = 113.34. c



41/??

Chapter 5. Bonds.

Manual for SOA Exam FM/CAS Exam 2. Chapter 5. Bonds. Section 5.4. Book value between coupon payments dates. c



1/10

c



Chapter 5. Bonds.

Section 5.4. Book value between coupon payments dates.

Book value between coupon payments dates We know that the book value of a bond immediately after a coupon payment is given by Bk = Fran−k|i +C ν n−k = C +(Fr −Ci)an−k|i = C +C (g −i)an−k|i . We want to determine the book value of a bond between successive coupon dates. We use as a unit of time coupon periods. We want to determine the value of a bond at time k + t periods, where k is an integer and 0 ≤ t < 1. The present value at time k + t of the remaining payments is Bk (1 + i)t . Immediately before the payment of the (k + 1)–th coupon, the price of the bond is Bk (1 + i). Immediately after the payment of the (k + 1)–th coupon, the price of the bond is Bk (1 + i) − Fr . Hence, the price of a bond is a discontinuous function (see Figure 1). 2/10

c



Chapter 5. Bonds.


Figure 1: Theoretical flat and market values 3/10

c



Chapter 5. Bonds.


During the time period (k, k + 1) a bond accrues a coupon payment. The accrued value at time t ∈ [0, 1) of the next coupon is denoted by Frt . We have that Fr0 = 0, limt→1− Frt = Fr . The flat price of a bond is the money that actually changes hands at m is the price of a bond the date of sale. The market price Bk+t excluding the accrued value of the next coupon. Hence, we have f m + Fr . that Bk+t = Bk+t t The flat price (also known as the dirty price) is the book value of a bond. It is the price that an investor pays for a bond. The market price (also known as the clean price) is the price of bond quoted in a newspaper.

4/10

c



Chapter 5. Bonds.


There are three methods to determine the flat value, accrued coupon and market value of a bond: I

Theoretical method: f Bk+t m Bk+t

I

(1 + i)t − 1 = Bk (1 + i) , Frt = Fr i t −1 (1 + i) t = Bk (1 + i) − Fr . i t

Practical method: f m Bk+t = Bk (1 + ti), Frt = tFr , Bk+t = Bk (1 + ti) − tFr

I

Semi–theoretical method: f m Bk+t = Bk (1 + i)t , Frt = tFr , Bk+t = Bk (1 + i)t − tFr

5/10

c



Chapter 5. Bonds.


The practical method assumes that the flat price accrues under simple interest. It assumes that the accrued coupon is proportional to the time since the last coupon payment. As to the theoretical method, Bk (1 + i)t is the present value of the payments to be made. This is actual outstanding balance in the loan at time k + t. We have that (1 + i)t − 1 m t Bk+t = Bk (1 + i) − Fr i −t n−k 1−ν ν −1 n−k −t = Fr + Cν ν − Fr i i n−k−t 1−ν =Fr + C ν n−k−t = Fran−k−t|i + C ν n−k−t , i s

where as|i = 1−ν i , s > 0 and s is not necessarily a positive integer. The market price according with the theoretical method has a continuous function. 6/10

c



Chapter 5. Bonds.


Example 1 Find the flat price, the accrued interest and the market price of a 1000 10–year bond with 4% annual coupons, bought to yield 3%, four months after the second coupon has been issued. Use all three methods.

7/10

c



Chapter 5. Bonds.


Example 1 Find the flat price, the accrued interest and the market price of a 1000 10–year bond with 4% annual coupons, bought to yield 3%, four months after the second coupon has been issued. Use all three methods. Solution: We have that F = 1000, n = 10, r = 0.04, Fr = 40, i = 3%, k = 2 and t = 13 . So, B2 = Fran−k−−|i + C ν n−k = 40a−8−|3% + (1000)(1.03)−8 = 1070.20. Using the theoretical method, f B2+(1/3) = B2 (1 + i)1/3 = (1070.20)(1.03)1/3 = 1080.80, " # " # (1 + i)1/3 − 1 (1 + 0.03)1/3 − 1 Fr(1/3) = Fr = (40) = 13.20 i 0.03 m f B2+(1/3) = B2+(1/3) − Fr(1/3) = 1080.80 − 13.20 = 1067.60. 8/10

c



Chapter 5. Bonds.


Example 1 Find the flat price, the accrued interest and the market price of a 1000 10–year bond with 4% annual coupons, bought to yield 3%, four months after the second coupon has been issued. Use all three methods. Solution: We have that F = 1000, n = 10, r = 0.04, Fr = 40, i = 3%, k = 2 and t = 13 . So, B2 = Fran−k−−|i + C ν n−k = 40a−8−|3% + (1000)(1.03)−8 = 1070.20. Using the practical method, f B2+(1/3) = B2 (1 + (1/3)i) = (1070.20)(1 + (0.03)/3)) = 1080.90,

Fr(1/3) = (1/3)Fr = (1/3)(40) = 13.33 m f B2+(1/3) = B2+(1/3) − Fr(1/3) = 1080.90 − 13.33 = 1067.57

9/10

c



Chapter 5. Bonds.


Example 1 Find the flat price, the accrued interest and the market price of a 1000 10–year bond with 4% annual coupons, bought to yield 3%, four months after the second coupon has been issued. Use all three methods. Solution: We have that F = 1000, n = 10, r = 0.04, Fr = 40, i = 3%, k = 2 and t = 13 . So, B2 = Fran−k−−|i + C ν n−k = 40a−8−|3% + (1000)(1.03)−8 = 1070.20. Using the semi–theoretical method: f B2+(1/3) = B2 (1 + i)1/3 = (1070.20)(1.03)1/3 = 1080.80,

Fr(1/3) = (1/3)Fr = (1/3)(40) = 13.33, m f B2+(1/3) = B2+(1/3) − Fr(1/3) = 1080.79 − 13.33 = 1067.46.

10/10

c



Chapter 5. Bonds.

Manual for SOA Exam FM/CAS Exam 2. Chapter 5. Bonds. Section 5.5. Callable bonds. c



1/10

c



Chapter 5. Bonds.

Section 5.5. Callable bonds.

Callable bonds A callable bond is a bond which gives the issuer (not the investor) the right to redeem prior to its maturity date, under certain conditions. When issued, the call provisions explain when the bond can be redeemed and what the price will be. In most cases, there is some period of time during which the bond cannot be called. This period of time is named the call protection period. The earliest time to call the bond is named the call date. The call price is the amount of money the insurer must pay to buy the bond back. Usually, bonds can be only called immediately after the payment of a coupon. We will study the computation of the yield rate of return for the investor in this situation. Since the investor does not know the cashflow obtained from his investment, he will assume that the issuer calls the bond under the worst possible situation (in the sense of lowest possible interest rates). 2/10

c



Chapter 5. Bonds.


If the bond is called immediately after the payment of k–th coupon, the present value of the obtained payments is Pk = Fra−k−|i + C ν k = C + (Fr − Ci)a−k−|i = C + C (g − i)a−k−|i . Pk is the price which the investor would pay for the bond assuming that the bond is called immediately after the k coupon. As smaller as Pk is, as worst for the lender is. Between all possible choices to recall a bond, the borrower will choose the option with the smallest price. Assuming that the redemption value is a constant and that a bond can be called after any coupon payment: (i) if Fr > Ci (bond sells at a premium), Pk increases with k, and we assume the redemption date is the earliest possible. (ii) if Fr < Ci (bond sells at a discount), Pk decreases with k, and we assume that the redemption date is the latest possible. If one investor wants to get an effective rate of interest of i per period, then the maximum price which the investor should pay is the lowest possible Pk under that particular rate i. 3/10

c



Chapter 5. Bonds.


Example 1 Consider a 100 par–value 8% bond with semiannual coupons callable at 120 on any coupon date starting 5 years after issue for the next 5 years, at 110 starting 10 years after issue for the next 5 years and maturing at 105 at the end of 15 years. What is the highest price which an investor can pay and still be certain of a yield of 9% converted semiannually.

4/10

c



Chapter 5. Bonds.


Example 1 Consider a 100 par–value 8% bond with semiannual coupons callable at 120 on any coupon date starting 5 years after issue for the next 5 years, at 110 starting 10 years after issue for the next 5 years and maturing at 105 at the end of 15 years. What is the highest price which an investor can pay and still be certain of a yield of 9% converted semiannually. Solution: We have that F = 100, r = 4%, Fr = 4, n = 30 and i = 4.5%. Let k be the number of the half year when the bond is called. If 10 ≤ k ≤ 19, then C = 120 and Ci = (120)(0.045) = 5.4. So, the bond sells at a discount. We have that lowest price which the investor can get is P19 = 4a19 + 120(1.045)−19 = 102.37. −− |4.5% 5/10

c



Chapter 5. Bonds.


Example 1 Consider a 100 par–value 8% bond with semiannual coupons callable at 120 on any coupon date starting 5 years after issue for the next 5 years, at 110 starting 10 years after issue for the next 5 years and maturing at 105 at the end of 15 years. What is the highest price which an investor can pay and still be certain of a yield of 9% converted semiannually. Solution: We have that F = 100, r = 4%, Fr = 4, n = 30 and i = 4.5%. Let k be the number of the half year when the bond is called. If 20 ≤ k ≤ 29, then C = 110 and Ci = (110)(0.045) = 4.95. So, the bond sells at a discount. We have that lowest price which the investor can get is P29 = 4a29 + (110)(1.045)−29 = 94.78. −− |4.5% 6/10

c



Chapter 5. Bonds.


Example 1 Consider a 100 par–value 8% bond with semiannual coupons callable at 120 on any coupon date starting 5 years after issue for the next 5 years, at 110 starting 10 years after issue for the next 5 years and maturing at 105 at the end of 15 years. What is the highest price which an investor can pay and still be certain of a yield of 9% converted semiannually. Solution: We have that F = 100, r = 4%, Fr = 4, n = 30 and i = 4.5%. Let k be the number of the half year when the bond is called. If k = 30, then the price is P30 = 4a30 + 105(1.045)−30 = 93.19. −− |4.5%

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c



Chapter 5. Bonds.


Example 1 Consider a 100 par–value 8% bond with semiannual coupons callable at 120 on any coupon date starting 5 years after issue for the next 5 years, at 110 starting 10 years after issue for the next 5 years and maturing at 105 at the end of 15 years. What is the highest price which an investor can pay and still be certain of a yield of 9% converted semiannually. Solution: We have that F = 100, r = 4%, Fr = 4, n = 30 and i = 4.5%. Let k be the number of the half year when the bond is called. We conclude that the highest price which an investor can pay and still be certain of a yield of 9% converted semiannually is 93.19.

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Chapter 5. Bonds.


Example 2 Joshua paid 800 for a 15-year 1000 par value bond with semiannual coupons at a nominal annual rate of 4% convertible semiannually. The bond can be called at 1300 on any coupon date starting at the end of year 7. What is the minimum annual nominal rate convertible semiannually yield that Joshua could receive? Answer: 7.3521%.

9/10

c



Chapter 5. Bonds.


Example 2 Joshua paid 800 for a 15-year 1000 par value bond with semiannual coupons at a nominal annual rate of 4% convertible semiannually. The bond can be called at 1300 on any coupon date starting at the end of year 7. What is the minimum annual nominal rate convertible semiannually yield that Joshua could receive? Answer: 7.3521%. Solution: We have that F = 1000, P = 800, C = 1300, r = 0.02 and n = 30. Since P < C , the bond was bought at discount. We assume that the redemption value is as late as possible. From the equation −30 800 = 20a30 −− + 1300(1 + i) |i we get that i = 3.6760 and i (2) = 7.3521%. 10/10

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More securities

Manual for SOA Exam FM/CAS Exam 2. Chapter 5. Bonds. Section 5.6. More securities. c



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More securities

Serial bonds A serial bond is a collection of bonds issued at the same time but with different redemption dates. The price of a serial bond is the sum of the prices of the individual bonds. Let Pk , Ck , Kk be the price value, the redemption value and the present value of redemption value of the k–th bond. Let P 0 , C 0 , K 0 be the price value, redemption value and present value of redemption value of the serial bond. For each k we have that g (Ck − Kk ). i P P P Let P 0 = kj=1 Pj , C 0 = kj=1 Cj and K 0 = kj=1 Kj . Hence, Pk = Kk +

P0 = K 0 +

g 0 (C − K 0 ). i 2/11

c



More securities

Example 1 A 12% serial bond with semiannual coupons and par value of 1000 will be redeemed by the following schedule: (i) 100 at the end of years 10 through 14; and (ii) 500 at the end of year 15. Calculate the price of the bond on the issue date to yield 10% per annum convertible semiannually.

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More securities

Example 1 A 12% serial bond with semiannual coupons and par value of 1000 will be redeemed by the following schedule: (i) 100 at the end of years 10 through 14; and (ii) 500 at the end of year 15. Calculate the price of the bond on the issue date to yield 10% per annum convertible semiannually. Solution: We have that g = Fr C = 6% and i = 5%. Since the annual nominal rate compounded semiannually is 10%, the annual effective rate of interest is 10.25%. The redemption values and times of redemption are given by the following table: Redemption value Time in years

100 10

100 11

100 12

100 13

100 14

500 15

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More securities

Solution: We have that g = Fr C = 6% and i = 5%. Since the annual nominal rate compounded semiannually is 10%, the annual effective rate of interest is 10.25%. The redemption values and times of redemption are given by the following table: Redemption value Time in years

100 10

100 11

100 12

100 13

100 14

500 15

Hence, K0 =

14 X

(100)(1 + 0.1025)−k + (500)(1 + 0.1025)−15

k=10

=(100)(1.1025)−9 a−5−|10.25% + (500)(1.1025)−15 =156.51 + 115.69 = 272.20. Hence, 0.06 g (1000 − 272.20) = 1145.56. P 0 = K 0 + (C 0 − K 0 ) = 272.20 + i 0.05 5/11

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More securities

Preferred stock

Preferred stock is like a perpetual bond. This stock pays dividends forever. The price of the stock is the present value of future dividends. If a preferred stock pays an annual dividend D, then the price of this stock is P = Da∞ −− = |i

D , i

where i the annual effective rate of interest.

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More securities

Example 2 You have acquired some Microsot preferred stock that pays $3000 per year forever. What is the present value of this investment? Assume that i = 6%.

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More securities

Example 2 You have acquired some Microsot preferred stock that pays $3000 per year forever. What is the present value of this investment? Assume that i = 6%. Solution:

3000 0.06

= 50000.

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More securities

Common stock For common stock the dividends are not known in advance. So one has to project what these dividends will be in the future. For example, let D be the dividend at the end of the current period and assume that the next dividends change geometrically with common ratio 1 + k with −1 < k < i, then the cashflow of dividends is: Contributions Time, in years

D 1

D(1 + k) 2

D(1 + k)2 3

··· ···

The price of the common stock is P=

D 1 a∞ . −− i−k = D | 1+k 1+k i −k

Recall that P=D

1 1+k (1 + k)2 D 1 1 +D +D +. . . = · =D . 2 3 1+k 1+i (1 + i) (1 + i) 1 + i 1 − 1+i i −k 9/11

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More securities

Example 3 Each quarter the corporation plans to pay 45% of its earnings as a stock dividend. The earnings of a corporation increase at 1% per quarter indefinitely. At the start of a quarter, an investor purchases the stock to yield a nominal rate of 5% compounded quarterly. The first stock dividend is 2.4 payable at the end of the quarter. Calculate the theoretical price of the stock.

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More securities

Example 3 Each quarter the corporation plans to pay 45% of its earnings as a stock dividend. The earnings of a corporation increase at 1% per quarter indefinitely. At the start of a quarter, an investor purchases the stock to yield a nominal rate of 5% compounded quarterly. The first stock dividend is 2.4 payable at the end of the quarter. Calculate the theoretical price of the stock. Solution: Since the earnings of a corporation increase at 1% per quarter, the dividends also increase at 1% per quarter. The cashflow of dividends is Dividends Time

2.4 1

(2.4)(1.01) 2

(2.4)(1.01)2 3

··· ···

The present value of the cashflow of dividends is 2.4 P = (0.05/4)−0.01 = 960. (i (4) /4)−k 11/11

c



Chapter 6. Variable interest rates and portfolio insurance.

Manual for SOA Exam FM/CAS Exam 2. Chapter 6. Variable interest rates and portfolio insurance. Section 6.1. Inflation. c



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Section 6.1. Inflation.

Inflation

Inflation is the fall in the purchasing power of money over time. It is usually measured with reference to an index representing the cost of certain goods and services. One the most frequently used index is the consumer index price. The consumer index price is released by the Bureau of Labor Statistics monthly. The Bureau of Labor Statistics finds the consumer index price by averaging the changes of prices of a market basket of goods and services. Sometimes, it is convenient for all the calculations related with an investment to find the units of real purchasing power rather than units of ordinary currency.

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Example 1 Peter uses his 450000 in his retirement fund to buy a perpetuity–immediate. The perpetuity is expected to pay dividends at the end of each year forever. The next payment (payable one year from now) is x, and is expected to increase at a rate of 3% per year. This increase is made to take in account for the inflation. The current annual effective rate of interest is 4.5%. Calculate x.

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c





Example 1 Peter uses his 450000 in his retirement fund to buy a perpetuity–immediate. The perpetuity is expected to pay dividends at the end of each year forever. The next payment (payable one year from now) is x, and is expected to increase at a rate of 3% per year. This increase is made to take in account for the inflation. The current annual effective rate of interest is 4.5%. Calculate x. Solution: The cashflow of payments is Payments Time

x 1

x(1.03)1 2

x(1.03)2 3

··· ···

Using the formula for the present value of a geometric perpetuity, x 450000 = 0.045−0.03 and x = (450000)(0.045 − 0.03) = 6750.

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Example 2 Richard invests $10,000 at the end of each year for 10 years into an account earning an effective rate of interest is 6.5%. The annual rate of inflation is 3.5% over the 10 year period. Calculate the value at the end of 10 years of Richard’s investment in today’s dollars.

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Example 2 Richard invests $10,000 at the end of each year for 10 years into an account earning an effective rate of interest is 6.5%. The annual rate of inflation is 3.5% over the 10 year period. Calculate the value at the end of 10 years of Richard’s investment in today’s dollars. Solution: The balance at the end of ten years is (10000)s10|6.5% = 134944.2254. The value at the end of 10 years of Richard’s investment in today’s dollars is 134944.2254(1.035)−10 = 95664.50019.

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Manual for SOA Exam FM/CAS Exam 2. Chapter 6. Variable interest rates and portfolio insurance. Section 6.2. Arbitrage. c



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Section 6.2. Arbitrage.

Arbitrage

In economics, arbitrage is the practice of taking advantage of a state of imbalance between two (or possibly more) markets by a combination of matching deals to make a profit. A simple case of arbitrage consists in buying something in one place and selling it in another place at the same time. Suppose that the exchange rates (after taking out the fees for making the exchange) in London are £5 = U10 and the exchange rates in Tokyo are £6 = U10. Converting U10 to £6 in Tokyo and converting that £6 into U12 in London, for a profit of U2, would be arbitrage.

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Manual for SOA Exam FM/CAS Exam 2. Chapter 6. Variable interest rates and portfolio insurance. Section 6.3. Term structure of interest rates. c



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Section 6.3. Term structure of interest rates.

Term structure of interest rates The relationship between yield and time to mature is called the term structure of interest rates. As larger as money is tied up in an investment as more likely a default is. Usually, interest rates increase with maturity date. For US Treasury zero–coupons bonds, different interest rates are given according with the maturity date.

Definition 1 A yield curve is a graph that shows interest rates (vertical axis) versus (maturity date) duration of a investment/loan (horizontal axis). Yield curves are studied to predict of changes in economic activity (economic growth, inflation, etc.). 2/25

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Example 1 A bank offers CD’s with the following interest rates length of the investment (in years) Interest rate

1 year 7%

2 years 8%

3 years 8.5%

4 years 9%

5 years 9.25%

Graph the yield curve for these interest rates.

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c





Example 1 A bank offers CD’s with the following interest rates length of the investment (in years) Interest rate

1 year 7%

2 years 8%

3 years 8.5%

4 years 9%

5 years 9.25%

Graph the yield curve for these interest rates. Solution: Assuming that the interest rate is a linear function between the given points, the yield curve is given by the graph in Figure 1.

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Figure 1: Yield curve for Example 1 5/25

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The interest rates appearing in the yield curve are called the spot rates. Thus, for Example 1, the spot rates are 7%, 8%, 8.5%, 9% and 9.25%.

Definition 2 The j year spot rate sj is the rate of interest charged in a loan paid with a unique payment at the end of j years. Note that is the j year spot rate sj is as an effective annual rate of interest, the current j year interest factor is (1 + sj )j . Money invested now multiply by (1 + sj )j in j years. The price of a zero coupon j–year bond with face value F is P = F (1 + sj )−j .

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Spot rates refer to a fixed maturity date. Usually, bonds have coupon payments over time. But, often strip bonds are traded. Strip or zero coupon bonds are bonds that have being ”separated” into their component parts (each coupon payment and the face value). Often strip bonds are obtained from US Treasury bonds. A financial trader (strips) ”separates” the coupons from a US Treasury bond, by accumulating a large number of US Treasury bonds and selling the rights of obtaining a particular payment to an investor. In this way, the investor can buy a strip bond as an individual security. The strip bond market consists of coupons and residuals, with coupons representing the interest portion of the original bond and the residual representing the principal portion. An investor will get a unique payment from a strip bond. In this situation, interest rates of a strip bond depend on the maturity date. The yield rate of a zero–coupon bond is called its spot rate. 7/25

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The following table consists of the Daily Treasury Yield Curve Rates, which can be found at http://www.treas.gov/offices/domestic-finance/debtmanagement/interest-rate/yield.html Date 07/01/04 07/02/04 07/06/04 07/07/04 07/08/04 07/09/04

1 mo 1.01 1.07 1.11 1.16 1.14 1.14

3 mo 1.22 1.30 1.34 1.30 1.27 1.28

6 mo 1.64 1.61 1.68 1.64 1.63 1.63

1 yr 2.07 2.02 2.15 2.00 1.99 2.00

2 yr 2.64 2.54 2.56 2.56 2.55 2.55

3 yr 3.08 2.96 2.99 2.99 2.97 2.96

5 yr 3.74 3.62 3.65 3.67 3.65 3.64

7 yr 4.18 4.08 4.10 4.10 4.09 4.08

10 yr 4.57 4.48 4.49 4.50 4.49 4.49

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Suppose that a zero–coupon bond with a face value of F and maturity j years has a price of Pj . Then, Pj (1 + sj )j = F , where sj is the j year spot rate. Note that a payment of Pj now is exchanged by a payment of F in j years. (1 + sj )j is the interest factor from year zero to year j. If the current interest rates follow the accumulation function is a(t), t ≥ 0, then a(j) = (1 + sj )j , i.e. sj = (a(j))1/j − 1.

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Example 2 The following table lists prices of zero-coupon $1000 bonds with their respective maturities: Number of years to maturity 1 2 5 10

Price $980.39 $957.41 $888.18 $781.20

Calculate the 1–year, 2–year, 5–year, and 10–year spot rates of interest.

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Price $980.39 $957.41 $888.18 $781.20

Calculate the 1–year, 2–year, 5–year, and 10–year spot rates of interest. Solution: Since (1000)(1 + s1 )−1 = 980.39, (1000)(1 + s2 )−2 = 957.41, (1000)(1 + s5 )−5 = 888.17, and (1000)(1 + s10 )−10 = 781.1984, we get s1 = 2.00%, s2 = 2.20% , s5 = 2.40%, s10 = 2.50%. 11/25

c





The present value at time zero of a cashflow Contributions Time

0 0

C1 1

C2 2

··· ···

Cn n

following the spot rates spot rate maturity time

s1 1

s2 2

··· ···

sn n

is given by the formula PV =

n X

(1 + sj )−j Cj .

j=1

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Example 3 (i) Find the price of a 2–year 1000 par value 6% bond with semi–annual coupons using the spot rates: nominal annual interest rate convertible semiannually maturity time (in half years)

4%

5%

6%

7%

1

2

3

4

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4%

5%

6%

7%

1

2

3

4

Solution: (i) The cashflow of payments is Payments Time (in half years)

30 1

30 2

30 3

1030 4

The present value of these payments is −1 −2 0.04 0.05 PV = (30) 1 + + (30) 1 + 2 2 −3 −4 0.06 0.07 + (30) 1 + + (1030) 1 + 2 2 =29.41176 + 28.55443 + 27.45425 + 897.5855 = 983.0059 c



14/25




4%

5%

6%

7%

1

2

3

4

(ii) Find the annual effective yield rate of the previous bond, if bought at the price in (i).

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4%

5%

6%

7%

1

2

3

4

(ii) Find the annual effective yield rate of the previous bond, if bought at the price in (i). Solution: (ii) To find the yield rate, we solve for i (2) in 983.0059 = 30a−4−|i (2) /2 +1000(1+i (2) /2)−4 , to get i (2) = 6.92450%. The annual effective yield rate is i = 7.0443%. Note that i (2) = 6.92450% is a sort of average of the spot rates used to find the price of the bond. Since the biggest payment is at time t = 4 half years, i (2) = 6.92450% is close to 7%.

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The one year forward rate for the j–th year fj is defined as fj =

(1 + sj )j − 1. (1 + sj−1 )j−1

fj is also called the 1 year forward rate from time j − 1 to time j. fj is also called the 1 year forward rate from the j–th year. fj is also called the (j − 1)–year forward rate. fj is also called the (j − 1)–year deferred 1–year forward rate. fj is also called the (j − 1)–year forward rate, 1–year interest rate. 1 + fj is the interest factor from year j − 1 to year j.

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(1 + sj−1 )j−1 is the interest factor from year zero to year j − 1. (1 + sj )j is the interest factor from year zero to year j. Hence, we have that (1 + sj−1 )j−1 (1 + fj ) = (1 + sj )j . Notice that by the definition of the one year forward rate (1 + s2 )2 = (1 + s1 )(1 + f2 ), (1 + s3 )3 = (1 + s2 )2 (1 + f3 ) = (1 + s1 )(1 + f2 )(1 + f3 ), (1 + s4 )4 = (1 + s3 )3 (1 + f4 ) = (1 + s1 )(1 + f2 )(1 + f3 )(1 + f4 ). In general, (1 + sn )n = (1 + s1 )(1 + f2 )(1 + f3 ) · · · (1 + fn ).

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Example 4 Suppose that the following spot rates are given: maturity time (in years) Interest rate

1

2

3

4

5

12.00%

11.75%

11.25%

10.00%

9.25%

Calculate the one–year forward rates for years 2 through 5.

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Example 4 Suppose that the following spot rates are given: maturity time (in years) Interest rate

1

2

3

4

5

12.00%

11.75%

11.25%

10.00%

9.25%

Calculate the one–year forward rates for years 2 through 5. Solution: f2 = f3 = f4 = f5 = The The The The

one–year one–year one–year one–year

forward forward forward forward

(1.1175)2 1.12 (1.1125)3 (1.1175)2 (1.1)4 (1.1125)3 (1.0925)5 (1.1)4

rate rate rate rate

for for for for

− 1 = 0.115006 − 1 = 0.102567 − 1 = 0.063336 − 1 = 0.063008

year year year year

c


2 3 4 5

is is is is

11.5006%. 10.2567%. 6.3336%. 6.3008%.


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Price $96.15 $92.10 $87.63 $82.27

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Price $96.15 $92.10 $87.63 $82.27

(i) Calculate the 1–year, 2–year, 3–year, and 4–year spot rates of interest.

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Price $96.15 $92.10 $87.63 $82.27

(i) Calculate the 1–year, 2–year, 3–year, and 4–year spot rates of interest. Solution: (i) Note that the price of j–th bond is Pj = (100)(1 + sj )−j . To get the j year spot rate sj , we solve (100)(1 + s1 )−1 = 96.15, (100)(1 + s2 )−2 = 92.10, (100)(1 + s3 )−3 = 87.63, and (100)(1 + s4 )−4 = 82.27, to get s1 = 4.00%, s2 = 4.20% , s3 = 4.50%, s4 = 5.00%. c



23/25




Price $96.15 $92.10 $87.63 $82.27

(ii) Calculate the 1–year, 2–year, and 3–year forward rates of interest.

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Price $96.15 $92.10 $87.63 $82.27

(ii) Calculate the 1–year, 2–year, and 3–year forward rates of interest. Solution: (ii) To get the j − 1 year forward rate fj , we do fj = (1+sj )j (1+sj−1 )j−1

−1 = We get that:

Pj−1 Pj

− 1, where Pj is the price of the j–th bond.

f2 = f3 = f4 =

96.15 92.10 92.10 87.63 87.63 82.27

− 1 = 4.397394%, − 1 = 5.100993%, − 1 = 6.515133%.

c



25/25


Manual for SOA Exam FM/CAS Exam 2. Chapter 6. Variable interest rates and portfolio insurance. Section 6.4. Duration, convexity. c



1/90

c




Section 6.4. Duration, convexity.

Duration Next we will assume that the rate of interest is constant over maturity.

Definition 1 The duration (or Macaulay’s duration) of a cashflow Contributions Time in years

C1 1

C2 2

··· ···

Cn n

with Cj ≥ 0 for each 1 ≤ j ≤ m, is defined as Pn Pn n j −j X Cj ν j j=1 jCj ν j=1 jCj (1 + i) ¯ P P d = Pn = = j . n n j −j k j=1 Cj ν j=1 Cj (1 + i) k=1 Ck ν j=1

2/90

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Main Properties of volatility I

The duration is an average of the times when the payments of the cashflow are made: d¯ =

n X

n

j=1

where wj = I

I

Cj ν j k k=1 Ck ν

Pn

X Cj ν j = jwj , k k=1 Ck ν

j Pn

j=1

satisfy wj ≥ 0 and

Pn

j=1 wj

= 1.

wj is the fraction of the present value of contribution at time t over the present value of the whole cashflow. If Cj0 > 0 and Cj = 0, for each j 6= j0 , then d¯ = j0 , for each rate of interest i.

I

The units of the duration are years.

I

The Macaulay duration is a measure of the price sensitivity of a cashflow to interest rate changes. 3/90

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Example 1 An investment pays 1000 at the end of year two and 1000 at the end of year 12. The annual effective rate of interest is 8%. Calculate the Macaulay duration for this investment.

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Example 1 An investment pays 1000 at the end of year two and 1000 at the end of year 12. The annual effective rate of interest is 8%. Calculate the Macaulay duration for this investment. Solution: Pn j (2)(1000)(1.08)−2 + (12)(1000)(1.08)−12 j=1 jCj ν ¯ d = Pn = j (1000)(1.08)−2 + (1000)(1.08)−12 j=1 Cj ν =5.165633881 years.

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Theorem 1 Let r > 0. If the Macaulay duration of the cashflow Contributions Time in years

C1 1

C2 2

··· ···

Cn n

¯ then the Macaulay duration of the cashflow is d, Contributions Time in years

rC1 1

rC2 2

··· ···

rCn n

¯ is d.

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Theorem 1 Let r > 0. If the Macaulay duration of the cashflow Contributions Time in years

C1 1

C2 2

··· ···

Cn n


rC1 1

rC2 2

··· ···

rCn n

¯ is d. Proof: The duration of the modified cashflow is Pn Pn j j j=1 jrCj ν j=1 jCj ν ¯ Pn P = = d. n j j rC ν C ν j j j=1 j=1 7/90

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Example 2 The Macaulay duration of a 10–year annuity–immediate with annual payments of $1000 is 5.6 years. Calculate the Macaulay duration of a 10–year annuity–immediate with annual payments of $50000.

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Example 2 The Macaulay duration of a 10–year annuity–immediate with annual payments of $1000 is 5.6 years. Calculate the Macaulay duration of a 10–year annuity–immediate with annual payments of $50000. Solution: Both cashflows have duration 5.6 years.

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Theorem 2 If the Macaulay duration of the cashflow Contributions Time in years

C1 1

C2 2

··· ···

Cn n


C1 t +1

C2 t +2

··· ···

Cn t +n

is d¯ + t.

10/90

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Theorem 2 If the Macaulay duration of the cashflow Contributions Time in years

C1 1

C2 2

··· ···

Cn n


C1 t +1

C2 t +2

··· ···

Cn t +n

is d¯ + t. Proof: Pn P P Pn j j t nj=1 Cj ν j + nj=1 jCj ν j j=1 (t + j)Cj ν j=1 jCj ν Pn P P = = t + . n n j j j j=1 Cj ν j=1 Cj ν j=1 Cj ν 11/90

c





Example 3 The Macaulay duration of a 10–year annuity–immediate with annual payments of $1000 is 5.6 years. Calculate the Macaulay duration of a 10–year annuity–due with annual payments of $5000.

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Example 3 The Macaulay duration of a 10–year annuity–immediate with annual payments of $1000 is 5.6 years. Calculate the Macaulay duration of a 10–year annuity–due with annual payments of $5000. Solution: The Macaulay duration of the two annuities does not dependent on the amount of the payment. So, we may assume that the two annual payments agree. Since the cashflow of an annuity–due is obtained from the cashflow of an annuity–immediate by translating payments 1 year, the answer is 5.6 − 1 = 4.6 years.

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Theorem 3 Suppose that two cashflows have durations d¯1 and d¯2 , respectively, present values P1 and P2 , respectively. Then, the duration of the combined cashflow is P1 d¯1 + P2 d¯2 d¯ = . P1 + P2 By induction the previous formula holds for a combination of finitely many cashflows. Suppose that we have n cashflows. The j–the cashflow has present value Pj and duration d¯j . Then, the duration of the combined cashflow is Pn ¯ j=1 Pj (i)dj ¯ . d = Pn j=1 Pj (i) 14/90

c





Proof: Suppose that the considered cashflows are Contributions Time

0 0

C1 1

C2 2

··· ···

Cn n

Contributions Time

0 0

D1 1

D2 2

··· ···

Dn n

and

Then, the combined cashflow is Contributions Time

0 0

C1 + D1 1

C2 + D2 2

··· ···

Cn + Dn n

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We have that P1 = of duration,

Pn

j=1 Cj ν

j


and P2 =

Pn

j=1 Dj ν

j.

By definition

Pn Pn j j j=1 jCj ν j=1 jCj ν ¯ = d1 = Pn j P1 j=1 Cj ν

and

Pn d¯2 =

jDj ν j Pj=1 n j j=1 Dj ν

Pn =

j=1 jDj ν

P2

j

.

Hence, Pn Pn P j(Cj + Dj )ν j jCj ν j + nj=1 jDj ν j d¯1 P1 + d¯2 P2 j=1 j=1 P P d¯ = Pn = = . n n j j j P1 + P2 j=1 (Cj + Dj )ν j=1 Cj ν + j=1 Dj ν

16/90

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Example 4 An insurance has the following portfolio of investments: (i) Bonds with a value of $1,520,000 and duration 4.5 years. (ii) Stock dividends payments with a value of $1,600,000 and duration 14.5 years. (iii) Certificate of deposits payments with a value of $2,350,000 and duration 2 years. Calculate the duration of the portfolio of investments.

17/90

c





Example 4 An insurance has the following portfolio of investments: (i) Bonds with a value of $1,520,000 and duration 4.5 years. (ii) Stock dividends payments with a value of $1,600,000 and duration 14.5 years. (iii) Certificate of deposits payments with a value of $2,350,000 and duration 2 years. Calculate the duration of the portfolio of investments. Solution: The duration of the portfolio is Pn ¯ j=1 Pj (i)dj d¯ = Pn j=1 Pj (i) (4.5)(1520000) + (14.5)(1600000) + (2)(2350000) 1520000 + 1600000 + 2350000 =6.351005484 years. =

18/90

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Theorem 4 The Macaulay duration of a level payments annuity–immediate is d¯ =

(Ia)−n−|i a−n−|i

.

19/90

c





Theorem 4 The Macaulay duration of a level payments annuity–immediate is d¯ =


.

Proof. We have that

Pn

j=1 d¯ = Pn

jPν j

j=1 Pν

j

=


.

20/90

c





Example 5 Calculate Macaulay the duration of a 15–year annuity immediate with level payments if the current effective interest rate per annum is 5%.

21/90

c





Example 5 Calculate Macaulay the duration of a 15–year annuity immediate with level payments if the current effective interest rate per annum is 5%. Solution: The Macaulay the duration is d¯ =


=

(Ia)15 −− |5% a15 −− |5%

=

73.66768937 = 7.097313716. 10.37965804

22/90

c





Theorem 5 The duration of a level payments perpetuity–immediate is 1+i d¯ = . i

Proof. We have that P∞ j (Ia)∞ −− |i j=1 jPν ¯ = = d = P∞ j Pν a −− j=1 ∞ |i

1+i i2 1 i

=

1+i . i

23/90

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Example 6 Suppose that the Macaulay duration of a perpetuity immediate with level payments of 1000 at the end of each year is 21. Find the current effective rate of interest.

24/90

c





Example 6 Suppose that the Macaulay duration of a perpetuity immediate with level payments of 1000 at the end of each year is 21. Find the current effective rate of interest. Solution: We have that 21 = d¯ = 1+i . So, i = 1 = 5%. i

20

25/90

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Theorem 6 The duration of n year bond with r % annual coupons, face value F and redemption value C is d¯ =

Fr (Ia)−n−|i + Cnν n Fra−n−|i + C ν n

.

Proof. Since the cashflow Contributions Time

Fr 1

Fr 2

··· ···

Fr n−1

Fr + C n

the duration is d¯ =

Fr

Pn

Fr

jν j Pj=1 n j j=1 ν

+ Cnν n + C νn

=


. 26/90

c





Example 7 Megan buys a 10–year 1000–face–value bond with a redemption value of 1200 which pay annual coupons at rate 7.5%. Calculate the Macaulay duration if the effective rate of interest per annum is 8%.

27/90

c





Example 7 Megan buys a 10–year 1000–face–value bond with a redemption value of 1200 which pay annual coupons at rate 7.5%. Calculate the Macaulay duration if the effective rate of interest per annum is 8%. Solution: We have that d¯ =

= =


(1000)(0.075)(Ia)10 + (1200)(10)(1.08)−10 −− |8% (1000)(0.075)a10 + (1200)(1.08)−10 −− |8% (75)(32.68691288) + 5558.321857 = 7.562958059. (75)(6.710081399) + 555.8321857

28/90

c





We have the following table: Duration d¯ d¯ = n

Cashflow zero–coupon bond level payments annuity–immediate

d¯ =

level payments perpetuity–immediate regular bond

d¯ =

(Ia)−−

n |i

a−− n |i d¯ = 1+i i Fr (Ia)−− +Cnν n n |i

Fra−− +C ν n n |i

29/90

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Volatility Consider a cashflow Contributions Time in years

C1 1

Definition 2

··· ···

C2 2

Cn n

0

(i) The quantity ν¯ = − d lndiP(i) = − PP(i) is called the volatility or modified duration. P P Notice that P(i) = nj=1 Cj ν j = nj=1 Cj (1 + i)−j and P P 0 (i) = nj=1 Cj (−j)(1 + i)−j−1 . So,

Pn ν¯ =

j=1 P n

jCj ν j+1

j=1 Cj ν

j

.

30/90

c




Since


Pn

j=1 d¯ = Pn

jCj ν j

j=1 Cj ν

and

Pn ν¯ =

j=1 P n

j

jCj ν j+1

j=1 Cj ν

j

,

¯ ν¯ = ν d.

31/90

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Main Properties of volatility 0

I I

(i) ν¯ = − PP(i) .

The volatility measures the loss of present value of the cashflow as i increases relative to the PV of the cashflow.

I

Pn ν¯ =

j+1 j=1 jCj ν Pn . j j=1 Cj ν

I

If Cj ≥ 0, for each 1 ≤ j ≤ n, ν¯ > 0. ¯ ν¯ = ν d.

I

Volatility is measured in years.

I

32/90

c





The present value P(i) of the above cashflow as a function on i, i.e. n n X X P(i) = Cj ν j = Cj (1 + i)−j . j=1

j=1

It is easy to see that 0

P (i) =

n X

−j−1

Cj (−j)(1+i)

00

, and P (i) =

j=1

n X

Cj j(j+1)(1+i)−j−2 .

j=1

If Cj > 0, for each 1 ≤ j ≤ n, then P 0 (i) < 0 and P 00 (i) > 0, for each i ≥ 0. This implies that P(i) is a decreasing convex function on i.

33/90

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Since P(i), i ≥ 0, is a decreasing function of i, so is ln P(i). Hence, Pn −j−1 d ln P(i) P 0 (i) j=1 Cj (−j)(1 + i) Pn 0<− =− =− −j di P(i) j=1 Cj (1 + i) Pn j+1 j=1 jCj ν = Pn . j j=1 Cj ν Hence, ν¯ > 0.

34/90

c





¯ we have that the volatility satisfies some of the Since ν¯ = ν d, properties of the duration. Suppose that we have n cashflows. The j–the cashflow has present value Pj and duration ν¯j . Then, the duration of the combined cashflow is Pn νj j=1 Pj (i)¯ ν¯ = Pn . j=1 Pj (i)

35/90

c





Example 8 A perpetuity pays 100 immediately. Each subsequent payment in increased by inflation. The current annual effective rate of interest is 6.5%. Calculate the modified duration of the perpetuity assuming that inflation will be 5% annually.

36/90

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Example 8 A perpetuity pays 100 immediately. Each subsequent payment in increased by inflation. The current annual effective rate of interest is 6.5%. Calculate the modified duration of the perpetuity assuming that inflation will be 5% annually. Solution: The present value of the perpetuity is P(i) = −100 i > 0.05. Hence, P 0 (i) = (i−0.05) 2, 0

(0.065) ν¯ = − PP(0.065) =

1 0.065−0.05

100 i−0.05 ,

if

= 66.66666667.

37/90

c





Example 9 A portfolio consists of four bonds. The prices and modified durations of the four bonds are given by the table: Bond Bond A Bond B Bond C Bond D

Present value $15050 $10350 $67080 $16750

Modified duration in years 4.3 10.4 7.6 6.5

Find the volatility of the whole portfolio.

38/90

c





Example 9 A portfolio consists of four bonds. The prices and modified durations of the four bonds are given by the table: Bond Bond A Bond B Bond C Bond D

Present value $15050 $10350 $67080 $16750

Modified duration in years 4.3 10.4 7.6 6.5

Find the volatility of the whole portfolio. Solution: We have that Pn νj j=1 Pj (i)¯ ν¯ = Pn j=1 Pj (i) (15050)(4.3) + (10350)(10.4) + (67080)(7.6) + (16750)(6.5) 15050 + 10350 + 67080 + 16750 =7.241948183 years.

=

39/90

c





Let P(i) be the present value of a portfolio, when i is the effective rate of interest. By a Taylor expansion, for h close to zero, ¯ = P(i) (1 − ν¯h) . P(i + h) ≈ P(i) + P 0 (i)h = P(i) 1 − ν dh

40/90

c






Example 10 A portfolio of bonds is worth 535000 at the current rate of interest of 4.75%. Its Macaulay duration is 6.375. Estimate the value of the portfolio if interest rates decrease by 0.10%.

41/90

c






Example 10 A portfolio of bonds is worth 535000 at the current rate of interest of 4.75%. Its Macaulay duration is 6.375. Estimate the value of the portfolio if interest rates decrease by 0.10%. ¯ . In our case, Solution: We have that P(i + h) ≈ P(i) 1 − ν dh P(0.0475 − 0.0010) ≈ 535000(1 − (1.0475)−1 (6.375)(−0.001)) =538255.9666.

42/90

c





If interest rates change from i into i + h, the percentage of change in the present value of the portfolio is P(i + h) − P(i) P(i) + P 0 (i)h − P(i) ¯ = −¯ ≈ = −ν dh ν h. P(i) P(i)

43/90

c






Example 11 A bond has a volatility of 4.5 years, at the current annual interest rate of 5%. Calculate the percentage of loss of value of the bond if the annual effective interest rate increase 250 basis points.

44/90

c






Example 11 A bond has a volatility of 4.5 years, at the current annual interest rate of 5%. Calculate the percentage of loss of value of the bond if the annual effective interest rate increase 250 basis points. Solution: The percentage of change is −¯ ν h = −(4.5)(0.025) = −0.1125 = −11.25%. The bond loses 11.25% of its value.

45/90

c





Duration is a measurement of how long in years it takes for the payments to be made. Mainly, we will consider applications to the bond market. Duration is an important measure for investors to consider, as bonds with higher durations are riskier and have a higher price volatility than bonds with lower durations. We have the following rules of thumb: I

Higher coupon rates lead to lower duration.

I

Longer terms to maturity usually lead to longer duration.

I

Higher yields lead to lower duration.

46/90

c





The price of a bond decreases as the rate of interest increases. Suppose that you believe that interest rates will drop soon. You want to make a benefit by buying a bond today and selling it later for a higher price. The profit you make is P(i + h) − P(i), where i is the interest you buy the bond and i + h is the interest rate when you sell the bond. Notice that you make a benefit if h < 0. The rate of return in your investment is P(i + h) − P(i) ≈ −¯ ν h. P(i) So, between all possible bonds, you will make a biggest profit investing in the bond with the highest possible volatility.

47/90

c





Example 12 Suppose that you are comparing two five–year bonds with a face value of 1000, and are expecting a drop in yields of 1% almost immediately. The current yield is 8%. Bond 1 has 6% annual coupons and bond 2 has annual 12% coupons. You would like to invest 100,000 in the bond giving you the biggest return. (i) Which would provide you with the highest potential gain if your outlook for rates actually occurs? (ii) Find the duration of each bond.

48/90

c





Solution: (i) The price of the bond 1 is (60)a−5−|8% + 1000(1.08)−5 = 920.15. Its price after the change of interest rates is (60)a−5−|7% + 1000(1.07)−5 = 959.00. The gain is 38.85. The percentage of change is 959.00−920.15 = 4.22% 920.15

49/90

c





Solution: (i) The price of the bond 1 is (60)a−5−|8% + 1000(1.08)−5 = 920.15. Its price after the change of interest rates is (60)a−5−|7% + 1000(1.07)−5 = 959.00. The gain is 38.85. The percentage of change is 959.00−920.15 = 4.22% 920.15 The price of the bond 2 is (120)a−5−|8% + 1000(1.08)−5 = 1159.71. Its price after the change of interest rates is (120)a−5−|7% + 1000(1.07)−5 = 1205.01. The gain is 45.30. The percentage of change is 1205.01−1159.71 = 3.91%. 1159.71 Bond 1 is a better investment than bond 2 (if we believe that the interest rates are going to fall to 1%). c



50/90



(ii) For the first bond, F = C = 1000, Fr = 60, i = 8% and n = 5. Its price is Fra−n−|i + F ν n = 60a−5−|8% + 1000(1.08)−5 = 920.15. and its Macaulay duration is d¯ = =

Fr (Ia)−n−|i + Fnν n Fra−n−|i + F ν n

=

60(Ia)−5−|0.08 + (1000)(5)(1.08)−5 920.15

60(11.3651) + 3402.92 = 4.4393. 920.15

51/90

c





(ii) For the first bond, F = C = 1000, Fr = 60, i = 8% and n = 5. Its price is Fra−n−|i + F ν n = 60a−5−|8% + 1000(1.08)−5 = 920.15. and its Macaulay duration is d¯ =


=

60(Ia)−5−|0.08 + (1000)(5)(1.08)−5 920.15

60(11.3651) + 3402.92 = 4.4393. 920.15 For the second bond, F = 1000, Fr = 120, i = 8% and n = 5. Its price is =

Fra−n−|i + F ν n = 120a−5−|8% + 1000(1.08)−5 = 1159.71. and its Macaulay duration is d¯ =

120(Ia)−5−|0.08 + (1000)(5)(1.08)−5 1159.71

=

120(11.3651) + 3402.92 1159.71

=4.1103. c


52/90




Convexity

Definition 3 The convexity of the cashflow Contributions Time in years

C1 1

C2 2

··· ···

Cn n

is defined as P 00 (i) c¯ = = P(i)

Pn

−j−2 j=1 Cj j(j + 1)(1 + i) Pn −j j=1 Cj (1 + i)

Pn =

j=1 Cj j(j + 1)ν Pn −j j=1 Cj ν

−j−2

.

Convexity is measured in years2 .

53/90

c






Convexity measures the rate of change of the volatility: d d P 0 (i) P 00 (i)P(i) − P 0 (i)P 0 (i) ν¯ = = = c¯ − (¯ ν )2 . di di P(i) (P(i))2

I

The second order Taylor expansion of the present value with respect to the yield is: h2 00 h2 0 P(i + h) ≈ P(i) + P (i)h + P (i) = P(i) 1 − ν¯h + c¯ . 2 2

I

Convexity is a measure of the curvature of the price–yield curve for a bond. Convexity is related with the second term in the Taylor expansion of the PV.

I

Using duration and convexity, we measure of how sensitive the present value of a cashflow is to interest rate changes. 54/90

c






Using duration and convexity, we have the following Taylor expansion: h2 P(i + h) ≈ P(i) 1 − ν¯h + c¯ . 2

I

The percentage change in the PV of a cashflow is P(i + h) − P(i) h2 ≈ −¯ ν h + c¯. P(i) 2

I

Convexity can be used to compare bonds. If two bonds offer the same duration and yield but one exhibits greater convexity, the bond with greater convexity is more affected by interest rates. 55/90

c





Example 13 A portfolio of bonds is worth 350000 at the current rate of interest of 5.2%. Its modified duration is 7.22. Its convexity is 370. Estimate the value of the portfolio if interest rates increase by 0.2%.

56/90

c





Example 13 A portfolio of bonds is worth 350000 at the current rate of interest of 5.2%. Its modified duration is 7.22. Its convexity is 370. Estimate the value of the portfolio if interest rates increase by 0.2%. 2

Solution: We have that P(i + h) ≈ P(i) 1 − ν¯h + h2 c¯ . In our case, (0.002)2 P(0.052 + 0.002) = 350000 1 − (7.22)(0.002) + (370) 2 =345205.

57/90

c





Example 14 Calculate the duration, the modified duration and the convexity of a $5000 face value 15–year zero–coupon bond if the current effective annual rate of interest is 7.5%.

58/90

c





Example 14 Calculate the duration, the modified duration and the convexity of a $5000 face value 15–year zero–coupon bond if the current effective annual rate of interest is 7.5%. Solution: Since P(i) = (5000)(1 + i)−15 , P 0 (i) = (5000)(−15)(1 + i)−16 , P 00 (i) = (5000)(−15)(−16)(1 + i)−17 , we have that 0 (0.75) −1 = 13.95348837 years, ν¯ = −P P(0.75) = (15)(1 + 0.075) d¯ = (1 + i)¯ ν = (1.075)(13.95348837) = 15 years and c¯ = (−15)(−16)(1 + 0.75)−2 = 78.36734694 years2 .

59/90

c





Example 15 Calculate the duration, the modified duration and the convexity of a level payments perpetuity–immediate with payments at the end of the year if the current effective annual rate of interest is 5%.

60/90

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Example 15 Calculate the duration, the modified duration and the convexity of a level payments perpetuity–immediate with payments at the end of the year if the current effective annual rate of interest is 5%. P 0 (i) = −C , P 00 (i) = 2C , we have that i2 i3 ¯ = 20 years, d = (1 + i)¯ ν = (1.05)(20) = 21

Solution: Since P(i) = ν¯ =

−P 0 (0.05) P(0.05)

ears and c¯ =

=

1 0.05

2 i2

=

2 (0.05)2

C i ,

= 800 years2 .

61/90

c





Example 16 A 100 par value 3 year bond pays annual coupons at a rate 7% coupon rate (with annual coupon payments). The current annual effective interest rate is 7%.

62/90

c





Example 16 A 100 par value 3 year bond pays annual coupons at a rate 7% coupon rate (with annual coupon payments). The current annual effective interest rate is 7%. (i) Calculate the duration, the modified duration and the convexity of the bond.

63/90

c





Example 16 A 100 par value 3 year bond pays annual coupons at a rate 7% coupon rate (with annual coupon payments). The current annual effective interest rate is 7%. (i) Calculate the duration, the modified duration and the convexity of the bond. Contributions 7 7 107 Solution: (i) The cashflow is Time 1 2 3 The duration is (7)(1.07)−1 + 2(7)(1.07)−2 + 3(107)(1.07)−3 = 2.808018. d¯ = 100 The modified duration is ν¯ =

2.808018 1.07

= 2.6243. The convexity is

(7)(1)(2)(1.07)−3 + (7)(2)(3)(1.07)−4 + (107)(3)(4)(1.07)−5 100 =9.58944. c¯ =

64/90

c





Example 16 A 100 par value 3 year bond pays annual coupons at a rate 7% coupon rate (with annual coupon payments). The current annual effective interest rate is 7%. (ii) If the interest rate change from 7% to 8%, what is the percentage change in the price of the bond?

65/90

c





Example 16 A 100 par value 3 year bond pays annual coupons at a rate 7% coupon rate (with annual coupon payments). The current annual effective interest rate is 7%. (ii) If the interest rate change from 7% to 8%, what is the percentage change in the price of the bond? Solution: (ii) If i = 7%, the price of the bond is 7a−3−|7% + 100(1.07)3 = 100. If i = 8%, the price of the bond is 7a−3−|8% + 100(1.08)3 = 97.4229. The change in percentage is

97.4229 100

− 1 = −2.5771%.

66/90

c





Example 16 A 100 par value 3 year bond pays annual coupons at a rate 7% coupon rate (with annual coupon payments). The current annual effective interest rate is 7%. (iii) Using the duration rule, including convexity, what is the percentage change in the bond price?

67/90

c





Example 16 A 100 par value 3 year bond pays annual coupons at a rate 7% coupon rate (with annual coupon payments). The current annual effective interest rate is 7%. (iii) Using the duration rule, including convexity, what is the percentage change in the bond price? Solution: (iii) The estimation in the change in percentage is h2 (0.01)2 c¯ = −(2.6243)(0.01) + (9.58944) 2 2 = − 0.025760528 = −2.576353%. − ν¯h +

68/90

c





Theorem 7 Suppose that we have n different investments. The j–th investment has present value Pj and convexity c¯j . Then, the convexity of the combined investments is Pn cj j=1 Pj (i)¯ . c¯ = Pn j=1 Pj (i)

69/90

c





Theorem 7 Suppose that we have n different investments. The j–th investment has present value Pj and convexity c¯j . Then, the convexity of the combined investments is Pn cj j=1 Pj (i)¯ . c¯ = Pn j=1 Pj (i)

Proof. Notice that Pn c¯ =

00 j=1 Pj (i)

Pj (i)

Pn =

cj j=1 Pj (i)¯ Pj (i)

.

70/90

c





Example 17 A company has issued debt using the following bonds: Bond Bond A Bond B Bond C Bond D

Present value 100000 50000 120000 80000

Macaulay’s duration 5.3 3.4 12.2 2.3

convexity 1.2 3.2 6.2 3.6

Find the Macaulay’s duration and the convexity for the entire portfolio.

71/90

c





Solution: Let Pj , d¯j and c¯j be the present value, Macaulay’s duration and convexity, respectively, of the j–th bond, 1 ≤ j ≤ 4. Then, the Macaulay’s duration of the whole portfolio is Pn ¯ j=1 Pj dj ¯ d = Pn j=1 Pj 100000(5.3) + 50000(3.4) + 120000(12.2) + 80000(2.3) 100000 + 50000 + 120000 + 80000 =6.708571429. =

The convexity of the whole portfolio is Pn ¯j j=1 Pj c c¯ = Pn j=1 Pj 100000(1.2) + 50000(3.2) + 120000(6.2) + 80000(3.6) 100000 + 50000 + 120000 + 80000 =3.748571429. =

72/90

c





In the case of payments made every m1 years, it is usual to use the nominal rate of interest i (m) as the variable. The present value of the cashflow Contributions Time (in years)

C1

C2

1 m

2 m

··· ···

is P(i

(m)

)=

n X

Cj

j=1

i (m) 1+ m

Cn n m

!−j .

73/90

c





The duration (or Macaulay’s duration) of the cashflow is −j −j Pn Pn j i (m) i (m) C jC 1 + 1 + j j j=1 m m m 1 j=1 d¯ = −j = −j years. Pn P m n i (m) i (m) C C 1 + 1 + j=1 j j=1 j m m The volatility is P 0 (i (m) ) ν¯ = − =− P(i (m) )

i (m) −j−1 1 j=1 Cj (−j)(1 + m ) m Pn i (m) −j j=1 Cj (1 + m )

Pn

=

i (m) 1+ m

!−1 ¯ d.

The convexity is Pn i (m) −j−2 1 P 00 (i (m) ) j=1 Cj (−j)(−j − 1)(1 + m ) m2 c¯ = = Pn i (m) −j P(i (m) ) j=1 Cj (1 + m ) Pn (m) i −j−2 1 j=1 Cj j(j + 1)(1 + m ) . = 2 Pn i (m) −j m C (1 + ) j j=1 m 74/90

c





Example 18 You are given the following information about a bond: I

The term–to–maturity is 2 years.

I

The bond has a 9% annual coupon rate, paid semiannually.

I

The annual bond–equivalent yield–to–maturity is 8%.

I

The par value is $100.

75/90

c







I


I


I


(i) Calculate the current price of the bond.

76/90

c







I


I


I


(i) Calculate the current price of the bond. Solution: (i) Since F = 100, Fr = 4.5, i = 4% and n = 4, the price is is Fra−n−|i + Pν n = (4.5)a−4−|4% + 100(1.04)−4 = 101.8149.

77/90

c







I


I


I


(ii) Calculate the Macaulay duration of the bond.

78/90

c







I


I


I


(ii) Calculate the Macaulay duration of the bond. Solution: (ii) The Macaulay duration is (in years) d¯ =

1 2

Fr (Ia)−− +nF ν n n |i

Fra−− +F ν n n |i

=

1 (4.5)(8.896856)+(4)(100)(0.854804) 2 101.8149

= 1.875744.

79/90

c







I


I


I


(iii) Calculate the convexity of the bond.

80/90

c







I


I


I


(iii) Calculate the convexity of the bond. Solution: (iii) The convexity is (in years2 ) 1 × 4 (4.5)(1)(2)(1.04)−3+(4.5)(2)(3)(1.04)−4+(4.5)(3)(4)(1.04)−5+(104.5)(4)(5)(1.04)−6 101.8149 1 8.0009 + 23.0797 + 44.3841 + 1651.7574 = = 4.241083 4 101.8149

c¯ =

where we have used

1 4

because the time is in half years.

c



81/90





I


I


I


(iv) For a 200 basis point increase in yield, determine the amount of error in using duration and convexity to estimate the price change.

82/90

c





(iv) For a 200 basis point increase in yield, determine the amount of error in using duration and convexity to estimate the price change. Solution: (iv) We need to find h2 h2 0 00 . P(i+h)−P(i)−P (i)h−P (i) = P(i+h)−P(i) 1 − ν¯h + c¯ 2 2 The price of the bond after the change in interest rates is P(i +h) = Fra−n−|i+h +P(1+i +h)−n = (4.5)a−4−|6% +100(1.06)−4 = 94.8023.

So, the error is h2 P(i + h) − P(i) 1 − ν¯h + c¯ 2 (0.02)2 −1 =94.8023 − 101.8149 1 − (1.875744)(1.04) (0.02) + 4.241083 2 =94.8023 − 101.8149(0.9647762) = −3.426292. 83/90

c





Example 18 Find the price and Macaulay duration of the following fixed–income securities, given the annual effective rate of interest 4.75% and par value of each bond is $1,000.

84/90

c





Example 18 Find the price and Macaulay duration of the following fixed–income securities, given the annual effective rate of interest 4.75% and par value of each bond is $1,000. (i) 3–year bond with 5.00% annual coupons

85/90

c





Example 18 Find the price and Macaulay duration of the following fixed–income securities, given the annual effective rate of interest 4.75% and par value of each bond is $1,000. (i) 3–year bond with 5.00% annual coupons Solution: (i) We have F = 1000, r = 0.05, i = 4.75%, Fr = 50 and n = 3. The price of the bond is Fra−n−|i + F ν n = 50a−3−|4.75% + 1000(1.0475)−3 = 1006.84. The Macaulay duration is (in years) d¯ = =


=

50(Ia)−3−|0.0475 + (1000)(3)(1.0475)−3 1006.84

50(5.3875) + 2610.11 = 2.8599. 1006.84 86/90

c





Example 18 Find the price and Macaulay duration of the following fixed–income securities, given the annual effective rate of interest 4.75% and par value of each bond is $1,000. (ii) 3–year bond with 5.00% semiannual coupons

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c





Example 18 Find the price and Macaulay duration of the following fixed–income securities, given the annual effective rate of interest 4.75% and par value of each bond is $1,000. (ii) 3–year bond with 5.00% semiannual coupons (2) Solution: (ii) We have i = 0.0475, i (2) = 0.046949, i 2 = 0.0234745, F = 1000, r = 0.025, Fr = 25 and n = 6. The price of the bond is 25a−6−|0.0234745 + 1000(1.0234745)−6 = 1008.45. The Macaulay duration is (in years) n −6 1 25(Ia)−6−|0.0234745 + (1000)(6)(1.0234745) 1 Fr (Ia)−n−|i + Fnν = d¯ = 2 Fra−n−|i + F ν n 2 1008.45

=

(12.5)(19.0026) + 2610.11 = 2.823782. 1008.45 88/90

c





Example 18 Find the price and Macaulay duration of the following fixed–income securities, given the annual effective rate of interest 4.75% and par value of each bond is $1,000. (iii) 3–year bond with 5.00% quarter coupons.

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c





Example 18 Find the price and Macaulay duration of the following fixed–income securities, given the annual effective rate of interest 4.75% and par value of each bond is $1,000. (iii) 3–year bond with 5.00% quarter coupons. (4) Solution: (iii) We have i = 0.0475, i (4) = 0.046677, i 4 = 0.0116692, F = 1000, Fr = 12.5 and n = 6. The price of the bond is 12.5a12 + 1000(1.0116692)−12 = 1009.25. −− |0.0116692 The Macaulay duration is (in years) n −12 −− 1 12.5(Ia)12 1 Fr (Ia)−n−|i + Fnν |0.0116692 + (1000)(12)(1.0116692) = d¯ = 4 Fra−n−|i + F ν n 4 1009.25

=

(3.125)(70.8531) + 2610.11 = 2.8056. 1009.25 90/90

c




Manual for SOA Exam FM/CAS Exam 2. Chapter 6. Variable interest rates and portfolio insurance. Section 6.5. Asset–liability management. c



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c




Section 6.5. Asset–liability management.

Hedging

The money collected by insurance companies is invested and subject to risk (possible losses). For example, bonds can drop in value if the rate of interest changes. Several methods have being developed to minimize the risk of investing. A method, often sophisticated, employed to minimize investment risk is called hedging. In this section, we study several hedging methods.

2/20

c





Match assets and liabilities

If possible we would like to match assets and liabilities. i.e. the total amount of contributions in assets equals the total amount of contributions in liabilities. I

If an insurance company has more liabilities than assets, it may fail to meet its commitments to its policyholders.

I

If an insurance company has more assets than liabilities, it will not the make the appropriate profit for the capital available.

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c





Example 1 A company has liabilities of 2000, 5000 and 10000 payable at the end of years 1, 2 and 5 respectively. The investments available to the company are the following zero–coupon 1000 par value bonds: Bond Bond A Bond B Bond C Bond D Bond E

Maturity (years) 1 year 2 years 3 years 4 years 5 years

Effective Annual Yield 4.5% 5.0 5.5 6.0 6.5

Determine the cost for matching these liabilities exactly.

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c





Example 1 A company has liabilities of 2000, 5000 and 10000 payable at the end of years 1, 2 and 5 respectively. The investments available to the company are the following zero–coupon 1000 par value bonds: Bond Bond A Bond B Bond C Bond D Bond E

Maturity (years) 1 year 2 years 3 years 4 years 5 years

Effective Annual Yield 4.5% 5.0 5.5 6.0 6.5

Determine the cost for matching these liabilities exactly. Solution: We need to buy 2 Bonds A, 5 Bonds B and 10 Bonds E. The cost is 2000(1.045)−1 + 5000(1.05)−2 + 10000(1.065)−5 = 13747.83136. 5/20

c





Example 2 A bond portfolio manager in a pension fund is designing a bond portfolio. His company has an obligation to pay 50000 at the end of each year for 3 years. He can purchase a combination of the following three bonds in order to exactly match its obligation: (i) 1–year 5% annual coupon bond with a yield rate of 6%. (ii) 2–year 7% annual coupon bond with a yield rate of 7%. (ii) 3–year 9% annual coupon bond with a yield rate of 8%. (i) How much of each bond should you purchase in order to exactly match the liabilities? (ii) Find the cost of such a combination of bonds.

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c





Solution: (i) Suppose that the face values of the bonds we buy are x, y , z, in each of the bonds, respectively. Then, the cashflows are Liabilities Bond 1 Bond 2 Bond 3 Time

50000 (1.05)x (0.07)y (0.09)z 1

50000 0 (1.07)y (0.09)z 2

50000 0 0 (1.09)z 3

In order to match assets and liabilities, 50000 = (1.05)x + (0.07)y + (0.09)z, 50000 = (1.07)y + (0.09)z, 50000 = (1.09)z. Hence, z= y= x=

50000 1.09 = 45871.56, 50000−(0.09)(45871.56) = 42870.61645, 1.07 50000−(0.07)(42870.61645)−(0.09)(45871.56) 1.05

= 40829.15852. 7/20

c





(ii) Find the cost of such a combination of bonds. Solution: (ii) The total price of the bonds is (40829.15852)(1.05)(1.06)−1 + (42870.61645)((0.07)a−2−|7% + (1.07)−2 ) + (45871.56)((0.09)a−3−|8% + (1.08)−3 ) =40035.97426 + 42870.61645 + 47053.71459 = 129960.3053.

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Theory of immunization

Fluctuation in interest rates can cause losses to a financial institution. Suppose that a financial institution has a cashflow of assets and a cashflow of liabilities. The present values of these cashflows is sensitive to changes of interest rates. If interest rates fall, the present value of the cashflow of liabilities increases. If interest rates increase, the present value of the cashflow of assets decreases. To mitigate the risk associated with a change in the interest rates, Redington (1954) introduced the theory of immunization. Immunization is a hedging method against the risk associated with changes in interest rates.

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According with the traditional immunization theory, a portfolio is immunized against fluctuations in interest rates if the 3 criteria are satisfied: 1. The present value of the assets must equal the present value of the liabilities. 2. The duration of the assets must equal the duration of the liabilities. 3. The convexity of the assets must be greater than that of the liabilities. The first of the previous 3 conditions is an efficiency condition. The last conditions are imposed so that interest rate risk for the assets offsets the interest rate risk for the liabilities.

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If the the immunization conditions are satisfied, then PA (i) = PL (i), νA = νL , c¯A > c¯L . The approximations to the present value of assets and liabilities are: h2 PA (i + h) ≈ PA (i) 1 − νA h + c¯A 2 and h2 PL (i + h) ≈ PL (i) 1 − νL h + c¯L . 2

Hence h2 (¯ c − c¯L ) > 0. 2 For small variation in interest rates the previous approximations are accurate. Since high variations in interest rates in short periods of time are unlikely, it is possible to hedge against interest rate variations by immunizing periodically. PA (i + h) − PL (i + h) = PA (i)

c



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Example 3 An actuarial department needs to set–up an investment program to pay for a loan of $20000 due in 2 years. The only available investments are: (i) a money market fund paying the current rate of interest. (ii) 5–year zero–coupon bonds earning 4%. Assume that the current rate of interest is 4%. Develop an investment program satisfying the theory of immunization. Graph the present value of asset minus liabilities versus interest rates.

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c





Solution: The investment program invest x in the money maker fund, and y in the zero coupon bond. The PV of the cashflow of assets and liabilities is P(i) = x + y (1.04)5 (1 + i)−5 − 20000(1 + i)−2 . We solve for x and y such that P(0.04) = 0 and P 0 (0.04) = 0. Since P 0 (i) = −(5)y (1.04)5 (1 + i)−6 + (2)(20000)(1 + i)−3 , we need to solve x+y = 20000(1.04)−2 , and −(5)y (1.04)−1 +(2)(20000)(1.04)−3 = 0. −2

= 7396.45 and We get y = (2)(20000)(1.04) 5 −2 x = 20000(1.04) − y = 11094.67. Since P 00 (i) = (5)(6)y (1.04)5 (1 + i)−7 − (2)(3)(20000)(1 + i)−4 , P 00 (0.04) = (5)(6)7396.45(1.04)−2 −(2)(3)(20000)(1.04)−4 = 102576.5. The convexity of the cashflow is positive. The investment strategy consisting in allocate 11094.67 in the money market account and 7396.45 in bonds satisfies the immunization requirements. c



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The graph of the present value of asset minus liabilities versus interest rates is Figure 1.

Figure 1: Present value of assets minus liabilities

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c





Example 4 An actuarial department has determined that the company has a liability of $10,000 that will be payable in seven years. The company has two choices of assets to invest in: a 5–year zero-coupon bond and a 10–year zero coupon bond. The interest rate is 5%. How can the actuarial department immunize its portfolio?

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c





Solution: The investment program invest x in the 5–year bond, and y in the 10–year bond. The PV of the cashflow of assets and liabilities is P(i) = x(1.05)5 (1 + i)−5 + y (1.05)10 (1 + i)−10 − 10000(1 + i)−7 . So, P 0 (i) = −(5)x(1.05)5 (1 + i)−6 − (10)y (1.05)10 (1 + i)−11 + (7)10000(1 + i)−8 and P 00 (i) = (5)(6)x(1.05)5 (1 + i)−7 + (10)(11)y (1.05)10 (1 + i)−12 − (7)(8)10000(1 + i)−9 . We solve for x and y in the equations P(0.05) = 0 and P 0 (0.05) = 0, i.e. x + y = (10000)(1.05)−7 , and 5x + 10y = (70000)(1.05)−7 . We get x = (6000)(1.05)−7 = 4264.08 and y = (4000)(1.05)−7 = 2842.72. c



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We have that P 00 (0.05) = (5)(6)(6000)(1.05)−7 (1.05)5 (1.05)−7 + (10)(11)(4000)(1.05)−7 (1.05)10 (1.05)−12 − (7)(8)(10000)(1.05)−9 =(60000)(1.05)−9 > 0. So, the investment program satisfies the immunization requirements.

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c





Example 5 An actuarial department has determined that the company has a liability of $10,000 that will be payable in two years. The company has two choices of assets to invest in: a one–year zero–coupon bond and a three–year zero–coupon bond. The interest rate is 6%. (i) Find an investment portfolio which immunizes this portfolio. (ii) Find the interval of interest rates at which the present value of assets is bigger than the present value of liabilities.

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c





Solution: (i) The investment program invest x in the one–year zero–coupon bond, and y in the three–year zero–coupon bond. The PV of the cashflow of assets and liabilities is P(i) = x(1.06)(1 + i)−1 + y (1.06)3 (1 + i)−3 − 10000(1 + i)−2 . So, P 0 (i) = x(1.06)(−1)(1 + i)−2 + y (1.06)3 (−3)(1 + i)−4 − 10000(−2)(1 + i)−3 and P 00 (i) = x(1.06)(−1)(−2)(1 + i)−3 + y (1.06)3 (−3)(−4)(1 + i)−5 − 10000(−2)(−3)(1 + i)−4 . We solve for x and y in the equations P(0.06) = 0 and P 0 (0.06) = 0, i.e. x + y = (10000)(1.06)−2 , and x + 3y = (20000)(1.06)−2 . We get x = y = (5000)(1.06)−2 = 4449.98222. c



19/20



We have that P 00 (0.06) = (10000)(1.06)−2 > 0. So, the investment program satisfies the immunization requirements. (ii) The present value of assets is bigger than the present value of liabilities if 0 ≤(5000)(1.06)−1 (1 + i)−1 + (5000)(1.06)(1 + i)−3 − 10000(1 + i)−2 which is equivalent to (5000)(1.06)−1 (1 + i)2 + (5000)(1.06) − 10000(1 + i) =(5000)(1.06)−1 (1 + i)2 − (2)(1.06) + (1.06)2 =(5000)(1.06)−1 (i − 0.06)2 ≥ 0. The present value of assets is bigger than the present value of liabilities for any interest rate. c



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Chapter 7. Derivatives markets.

Manual for SOA Exam FM/CAS Exam 2. Chapter 7. Derivatives markets. Section 7.1. Derivatives. c



1/52

c




Section 7.1. Derivatives.

Risk sharing I

A risk is a contingent financial loss. Changes in commodity prices, currency exchange rates and interest rates are potential risks for a business. A farmer faces the possible fall of the price of his/her crop. Surging oil prices can wipe out airlines’ profits. Manufacturing companies face high rising prices of commodities. These changes in prices could hurt the viability of a business.

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c





Risk sharing I


I

Many of the risks faced by business are diversifiable. A risk is diversifiable if it is unrelated to another risk. Markets permit diversifiable risks to be widely shared.

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c





Risk sharing I


I


I

Risk is nondiversifiable when it does vanish when spread across many investors.

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c





Risk sharing I


I


I

Risk is nondiversifiable when it does vanish when spread across many investors.

I

A way to do risk sharing for companies is to do contracts to avoid risks. 5/52

c





Derivatives Definition 1 A derivative is a contract which specifies the right or obligation to receive or deliver certain asset for a certain price. The value of a derivative contract depends on the value of another asset.

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c





Derivatives Definition 1 A derivative is a contract which specifies the right or obligation to receive or deliver certain asset for a certain price. The value of a derivative contract depends on the value of another asset. They are several possible reasons to enter into a derivative market:

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c





Derivatives Definition 1 A derivative is a contract which specifies the right or obligation to receive or deliver certain asset for a certain price. The value of a derivative contract depends on the value of another asset. They are several possible reasons to enter into a derivative market: I

Risk management. Parties enter derivatives to avoid risks.

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c





Derivatives Definition 1 A derivative is a contract which specifies the right or obligation to receive or deliver certain asset for a certain price. The value of a derivative contract depends on the value of another asset. They are several possible reasons to enter into a derivative market: I I

Risk management. Parties enter derivatives to avoid risks. Speculation. Parties enter derivatives to make money. A market–maker enters into derivatives to make money.

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c





Derivatives Definition 1 A derivative is a contract which specifies the right or obligation to receive or deliver certain asset for a certain price. The value of a derivative contract depends on the value of another asset. They are several possible reasons to enter into a derivative market: I I I

Risk management. Parties enter derivatives to avoid risks. Speculation. Parties enter derivatives to make money. A market–maker enters into derivatives to make money. Reduce transaction costs. Derivatives can be used to reduce commodity costs, borrowing costs, etc.

10/52

c





Derivatives Definition 1 A derivative is a contract which specifies the right or obligation to receive or deliver certain asset for a certain price. The value of a derivative contract depends on the value of another asset. They are several possible reasons to enter into a derivative market: I I I I

Risk management. Parties enter derivatives to avoid risks. Speculation. Parties enter derivatives to make money. A market–maker enters into derivatives to make money. Reduce transaction costs. Derivatives can be used to reduce commodity costs, borrowing costs, etc. Arbitrage. When derivatives are miss priced, investors can make a profit.

11/52

c





Derivatives Definition 1 A derivative is a contract which specifies the right or obligation to receive or deliver certain asset for a certain price. The value of a derivative contract depends on the value of another asset. They are several possible reasons to enter into a derivative market: I I I I I

Risk management. Parties enter derivatives to avoid risks. Speculation. Parties enter derivatives to make money. A market–maker enters into derivatives to make money. Reduce transaction costs. Derivatives can be used to reduce commodity costs, borrowing costs, etc. Arbitrage. When derivatives are miss priced, investors can make a profit. Regulatory arbitrage. Sometimes business enter into derivatives to get around regulatory limitations, accounting regulations and taxes. 12/52

c





Example 1 Suppose that a farmer grows wheat and a baker makes bread using wheat and other ingredients. If the price of the wheat decreases, the farmer loses money. If the price of the wheat increases, the baker loses money. In order to avoid possible financial losses which may jeopardy their businesss profitability, the farmer and the baker can agree to sell/buy wheat one year from now at a certain price. The contract they enter is a derivative. It is a win–win contract for both of them. The two risks are diversifiable.

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c





Example 1 Suppose that a farmer grows wheat and a baker makes bread using wheat and other ingredients. If the price of the wheat decreases, the farmer loses money. If the price of the wheat increases, the baker loses money. In order to avoid possible financial losses which may jeopardy their businesss profitability, the farmer and the baker can agree to sell/buy wheat one year from now at a certain price. The contract they enter is a derivative. It is a win–win contract for both of them. The two risks are diversifiable. Usually, the contract is not made directly between them. A market–maker or scalper makes a contract with the farmer and another with the baker. The farmer and the baker enter into this contract to do hedging.

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c





Derivatives in Practice. I

Derivatives are often traded on commodities, stock, stock indexes, currency exchange rates and interest rates. Common commodities in derivatives are agricultural (corn, soybeans, wheat, live cattle, cattle feeder, hogs lean, sugar, coffee, orange juice), metals (gold, silver, copper, lead, aluminum, platinum) and energy (crude oil, ethanol, natural gas, gasoline).

15/52

c






I

Derivatives are often traded on commodities, stock, stock indexes, currency exchange rates and interest rates. Common commodities in derivatives are agricultural (corn, soybeans, wheat, live cattle, cattle feeder, hogs lean, sugar, coffee, orange juice), metals (gold, silver, copper, lead, aluminum, platinum) and energy (crude oil, ethanol, natural gas, gasoline). Derivative contracts for agricultural commodities have been traded in the U.S. for more than 100 years. The biggest markets in derivatives are the Chicago Board for Trade, the Chicago Mercantile Exchange, the New York Mercantile Exchange, and the Eurex (Frankfurt, Germany).

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c






I

I

Derivatives are often traded on commodities, stock, stock indexes, currency exchange rates and interest rates. Common commodities in derivatives are agricultural (corn, soybeans, wheat, live cattle, cattle feeder, hogs lean, sugar, coffee, orange juice), metals (gold, silver, copper, lead, aluminum, platinum) and energy (crude oil, ethanol, natural gas, gasoline). Derivative contracts for agricultural commodities have been traded in the U.S. for more than 100 years. The biggest markets in derivatives are the Chicago Board for Trade, the Chicago Mercantile Exchange, the New York Mercantile Exchange, and the Eurex (Frankfurt, Germany). The market in derivatives is regulated by the (SEC) Securities and Exchange Commission and the (CFTC) Commodity Futures Trading Commission. 17/52

c





Buying an asset I

(Scalpers) Market–makers buy/sell stock and derivatives making transactions possible. Usually, scalpers are financial institutions.

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Buying an asset I


I

In order to make a living, scalpers buy low and sell high.

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Buying an asset I


I


I

The price at which the scalper buys is called the bid price. The bid price of an asset is the price at which a scalper takes bids for an asset (a bid is an offer).

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Buying an asset I


I


I


I

The price at which the scalper sells is called the offer price or ask price.

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Buying an asset I


I


I


I


I

The bid price is lower than the offer price. The difference between the bid price and the offer price is called the bid–ask spread.

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Buying an asset I


I


I


I


I

The bid price is lower than the offer price. The difference between the bid price and the offer price is called the bid–ask spread.

I

bid–ask percentage spread is the bid–ask spread divided over the ask price. Scalpers also ask for commissions. 23/52

c




bid ask or offer price


price at which the scalper buys price at which the scalper sells

The bid price is lower than the offer price.

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Example 2 An online currency exchange service’s bid rate for Japanese yens is $0.00838 and its ask rate is $0.00847. Find the bid–ask percentage spread.

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c





Example 2 An online currency exchange service’s bid rate for Japanese yens is $0.00838 and its ask rate is $0.00847. Find the bid–ask percentage spread. Solution: The ask percentage spread is 0.00847−0.00838 = 1.062574%. 0.00847

26/52

c





Example 3 $1 million face value six month T bill is traded by a government security dealer who give the following annual nominal discount yield convertible semiannually: Bid 4.96%

Ask 4.94%

(i) Find the price the dealer is asking for the T bill (ii) Find the price the dealer is buying the T bill. (iii) Find the bid–ask percentage spread.

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c






Ask 4.94%

(i) Find the price the dealer is asking for the T bill (ii) Find the price the dealer is buying the T bill. (iii) Find the bid–ask percentage spread. Solution: (i) (1000000)(1 − (0.0494/2)) = 975300

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c






Ask 4.94%

(i) Find the price the dealer is asking for the T bill (ii) Find the price the dealer is buying the T bill. (iii) Find the bid–ask percentage spread. Solution: (i) (1000000)(1 − (0.0494/2)) = 975300 (ii) (1000000)(1 − (0.0496/2)) = 975200.

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c






Ask 4.94%

(i) Find the price the dealer is asking for the T bill (ii) Find the price the dealer is buying the T bill. (iii) Find the bid–ask percentage spread. Solution: (i) (1000000)(1 − (0.0494/2)) = 975300 (ii) (1000000)(1 − (0.0496/2)) = 975200. Notice that the dealer sells the T bill for money than he buys it.

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c






Ask 4.94%

(i) Find the price the dealer is asking for the T bill (ii) Find the price the dealer is buying the T bill. (iii) Find the bid–ask percentage spread. Solution: (i) (1000000)(1 − (0.0494/2)) = 975300 (ii) (1000000)(1 − (0.0496/2)) = 975200. Notice that the dealer sells the T bill for money than he buys it. (iii) The bid–ask percentage spread is 975300−975200 = 0.01025325541%. 975300 31/52

c





Long and short positions

I

When some one owns an asset, we say to this person has a long position in this asset. Later, he may sell the asset and receive cash. Buying an asset means to make an investment. Buying an asset is like lending money.

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c





Long and short positions

I

When some one owns an asset, we say to this person has a long position in this asset. Later, he may sell the asset and receive cash. Buying an asset means to make an investment. Buying an asset is like lending money.

I

When some one needs to buy an asset in the future, it is said that this person has a short position in the asset.

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c





Short sale A short sale of an asset entails borrowing an asset and then immediately selling the asset receiving cash. Later, the short seller must buy back the asset paying cash for it and return it to the lender. The act of buying the replacement asset and return it to the lender is called to close or cover the short position.

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c





Short sale A short sale of an asset entails borrowing an asset and then immediately selling the asset receiving cash. Later, the short seller must buy back the asset paying cash for it and return it to the lender. The act of buying the replacement asset and return it to the lender is called to close or cover the short position. believe price will increase price will decrease

derivative purchase short sale

desired outcome buy low now and sell high later sell high now and buy low later

Table: An investor makes money buying low and selling high.

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There are three main reasons to short sell:

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There are three main reasons to short sell: I

Speculation. An investor enters a short sale because he believes that the price of the stock will drop and desires to profit from this price movement.

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c







I

Financing. A short sale is a way to borrow money.

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I

Financing. A short sale is a way to borrow money.

I

Hedging. A short sale can be undertaken to offset the risk of owning an asset.

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In order to a short sale to take place several conditions must be taken into account:

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In order to a short sale to take place several conditions must be taken into account: I

Availability of a lender of stock. A lender must interested in making some money and losing temporarily his voting rights on the company issuing the stock.

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I

Credit risk of the short seller. The short seller make be required to set–up a bank account with a deposit as collateral.

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I

Credit risk of the short seller. The short seller make be required to set–up a bank account with a deposit as collateral.

I

Scarcity of shares.

If the stock pays dividends, the short seller must return the paid dividend payments to the stock lender.

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c





Example 4 Mary short sells 200 shares of XYZ stock which has a bid price of $18.12 and ask price of $18.56. Her broker charges her a 0.5% commission to take on the short sale. Mary covers her position 6 months later when the bid price is $15.74 and the ask price is $15.93. The commission to close the short sale is $15. XYZ stock did not pay any dividends in those six months. How much does Mary earn in this short sale?

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c





Example 4 Mary short sells 200 shares of XYZ stock which has a bid price of $18.12 and ask price of $18.56. Her broker charges her a 0.5% commission to take on the short sale. Mary covers her position 6 months later when the bid price is $15.74 and the ask price is $15.93. The commission to close the short sale is $15. XYZ stock did not pay any dividends in those six months. How much does Mary earn in this short sale? Solution: Mary short sells the stock for (200)(18.12)(1 − 0.005) = $3605.88. Mary covers her position for (200)(15.93) + 15 = $3201. Mary earns 3605.88 − 3201 = $404.88. 45/52

c




I


Often, short sellers are required to make a deposit as collateral into an account with the lender. This account is called the margin account. This deposit is called the margin requirement. Usually, the margin requirement is a percentage of the current price of the stock. The margin requirement could be bigger than the current asset price. If this is so, the excess of the margin over the current asset price is called haircut.

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I

I


Often, short sellers are required to make a deposit as collateral into an account with the lender. This account is called the margin account. This deposit is called the margin requirement. Usually, the margin requirement is a percentage of the current price of the stock. The margin requirement could be bigger than the current asset price. If this is so, the excess of the margin over the current asset price is called haircut. This margin account generates an interest for the investor. Demand on short sales is a factor to determine the margin account interest rate. The margin interest rate is called the repo rate for bonds and the short rebate for stock.

47/52

c




I

I

I


Often, short sellers are required to make a deposit as collateral into an account with the lender. This account is called the margin account. This deposit is called the margin requirement. Usually, the margin requirement is a percentage of the current price of the stock. The margin requirement could be bigger than the current asset price. If this is so, the excess of the margin over the current asset price is called haircut. This margin account generates an interest for the investor. Demand on short sales is a factor to determine the margin account interest rate. The margin interest rate is called the repo rate for bonds and the short rebate for stock. When borrowing, the lender can require payment of certain benefits lost by lending the asset. This payment requirement is called the lease rate of the asset. Usually, the lease rate of a stock is the payment of the dividends obtained while the stock was shorted. Usually, the lease rate of a bond is the payment of the coupons obtained while the bond was shorted.

c



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Sometimes short sales are subject to a margin requirement (or deposit). The investor has to set–up an account with a percentage of the current price of the stock. This margin account generates an interest for the investor. If the stock which the investor borrows pays a dividend, the investor must pay the dividend to the brokerage firm making the loan. We have the following variables in a short sale of a stock: I

P = profit on sale=price sold−price bought.

I

M = margin requirement= deposit on the short sale.

I

D = dividend paid by the short seller to the security’s owner.

I

j = rate of interest earned in the margin account.

I

I = Mj interest earned by the short seller on the margin deposit.

I

i = yield rate on the short sale. 49/52

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We have that the net profit is

Net profit = gain on short sale+interest on margin−dividend on stock = P Hence, the yield rate earner in a short sale is i=

net profit P +I −D = . margin M

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Example 5 Jason sold short 1,000 shares of FinanTech at $75 a share on January 2, 2006. Jason is required to hold a margin account with his broker equal to 50% of the short security’s initial value. Jason’s margin account earns an annual effective interest rate of i. There is a $0.25 per share dividend paid on December 31, 2006. On January 2, 2007, Jason buys back stock to cover his position at a price of $70 per share. Jason’s annual effective yield in this investment is 17.6667%. Find i.

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Example 5 Jason sold short 1,000 shares of FinanTech at $75 a share on January 2, 2006. Jason is required to hold a margin account with his broker equal to 50% of the short security’s initial value. Jason’s margin account earns an annual effective interest rate of i. There is a $0.25 per share dividend paid on December 31, 2006. On January 2, 2007, Jason buys back stock to cover his position at a price of $70 per share. Jason’s annual effective yield in this investment is 17.6667%. Find i. Solution: We have that P = (75)(1000) − (70)(10000) = 5000, M = (75)(1000)(0.50) = 37500, I = 37500i and D = (1000)(0.25) = 250. Jason’s annual effective yield is 0.176667 = Hence, i =

5000 + 37500i − 250 P +I −D = . M 37500

(0.176667)(37500)−5000+250 37500

= 5%. 52/52

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Manual for SOA Exam FM/CAS Exam 2. Chapter 7. Derivatives markets. Section 7.2. Forwards. c



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Section 7.2. Forwards.

Forwards

Definition 1 A forward is a contract between a buyer and seller in which they agree upon the sale of an asset of a specified quality for a specified price at a specified future date. Forward contracts are privately negotiated and are not standardized. Common forwards are in commodities, currency exchange, stock shares and stock indices.

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A forward contract states the following:

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A forward contract states the following: I

forces the seller to sell and the buyer to buy.

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I

spells out the quantity, quality and exact type of asset to be sold.

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I


I

states the delivery price and the time, date and place for the transfer of ownership of the asset.

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I


I


I

specify the time, date, place for payment.

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I


I


I


Usually a forward contract has more terms.

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I


I


I


Usually a forward contract has more terms. Sometimes instead of the asset to be delivered, there exists a cash settlement between the parties engaging in a forward contract. Either physical settlement or cash settlement can be used to settle a forward contract.

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I


I


I


Usually a forward contract has more terms. Sometimes instead of the asset to be delivered, there exists a cash settlement between the parties engaging in a forward contract. Either physical settlement or cash settlement can be used to settle a forward contract. When entering in a forward contract, parties must check counterparts for credit risk (sometimes using a collateral, bank letters, real state guarantee, etc). 10/89

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I


The asset in which the forward contract is based is called the underlier or underlying asset.

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I


I

The nominal amount (also called notional amount) of a forward contract is the quantity of the asset traded in the forward contract.

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I


I


I

The price of the asset in the forward contract is called the forward price.

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I


I


I

The price of the asset in the forward contract is called the forward price.

I

The time at which the contract settles is called the expiration date.

For example, if a forward contract involves 10,000 barrels of oil to be delivered in one year, oil is the underlying asset, 10000 barrels is the notional amount and one year is the expiration date.

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The two main reasons why an investor might be interested in forward contracts are: speculation and (hedging) reduce investment risk.

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Apart from commission, a forward contract requires no initial payment. The current price of an asset is called its spot price. Besides the spot price, to price a forward contract, several factors, such as delivery cost and time of delivery must be taken into account. The difference between the spot and the forward price is called the forward premium or forward discount. If the forward price is higher than the spot price, the asset is forwarded at a premium. The premium is the forward price minus the current spot price. If the forward price is lower than the spot price, the asset is forwarded at a discount. The discount is the current spot price minus the forward price.

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The buyer of the asset in a forward contract is called the long forward. The long forward benefits when prices rise. The seller of the asset in a forward contract is called the short forward. The short forward benefits when prices decline.

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We will denote by ST the spot price of an asset at time T . S0 is the current price of the asset. S0 is a fixed quantity. For T > 0, ST is a random variable. We will denote by F0,T to the price at time zero of a forward with expiration time T paid at time T . The payoff of a derivative is the value of this position at expiration. The payoff of a long forward contract is ST − F0,T . Notice that the bearer of a long forward contract buys an asset at time T with value ST for F0,T . Assuming that there are no expenses setting the forward contract, the profit for a long forward is ST − F0,T . The payoff of a short forward is F0,T − ST . The holder of a short forward contract sells at time T an asset with value ST for F0,T . Assuming that there are no expenses setting the forward contract, the profit for a short forward contract is F0,T − ST .

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The profits of the long and the short in a forward contract are the opposite of each other. The sum of their profits is zero. A forward contract is a zero–sum game. I I I I

The minimum long forward’s profit F0,T , which is attained when ST = 0. The maximum long forward’s profit is infinity, which is attained when ST = ∞. The minimum short forward’s profit is −∞, which is attained when ST = ∞. The maximum short forward’s profit is F0,T , which is attained when ST = 0. long forward’s profit short forward’s profit

long forward short forward

= =

minimum profit −F0,T −∞

ST − F0,T F0,T − ST

maximum profit ∞ −F0,T 19/89

c





Example 1 A gold miner enters a forward contract with a jeweler to sell him 200 ounces of gold in six months for $600 per ounce. (i) Find the jeweler’s payoff in the forward contract if the spot price at expiration of a gold ounce is $590, $595, $600, $605, $610. Graph the jeweler’s payoff. (ii) Find the gold miner’s payoff in the forward contract if the spot price at expiration of a gold ounce is $590, $595, $600, $605, $610. Graph the gold miner’s payoff.

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Solution: (i) The jeweler’s payoff is (200)(ST − 600). A table of the jeweler’s payoff is ST Payoff

590 −2000

595 −1000

600 0

605 1000

610 2000

The graph of the jeweler’s payoff is given in Figure 1. (ii) The gold miner’s payoff is (200)(600 − ST ). A table of the gold miner’s payoff is ST Payoff

590 2000

595 1000

600 0

605 −1000

610 −2000

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Figure 1: Example 1. (Long forward) Jeweler’s payoff 22/89

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Figure 2: Example 1. (Short forward) Gold miner’s payoff

Notice how figures 1 and 2 are the opposite of each other. The gold miner’s payoff equals minus the jeweler’s payoff. c



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Usually, a commissions has to be paid to enter a forward contract. Suppose that the long forward has to paid CL to the market–maker at negotiation time to enter into the forward contract. Then, the profit for a long forward is Profit of a long forward = ST − F0,T − CL (1 + i)T , where i is the annual effective rate of interest. If the short forward has to paid CS at negotiation time to the market–maker to enter the forward contract, then profit for a short forward is Profit of a short forward = F0,T − ST − CS (1 + i)T .

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Sometimes, instead of using the annual effective rate of interest, we will use the annual interest rate compounded continuously. This is another name for the force of interest. This rate is also called the annual continuous interest rate. If r is the annual continuously compounded interest rate, then the future value at time T of a payment of P made at time zero is Pe rT .

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Alternative ways to buy an asset. Suppose that you want to buy an asset. Suppose that the buyer’s payment can be made at either time zero or time T . Suppose that the transfer of ownership of an asset can be made either at time zero or at time T . There are four possible ways to buy an asset (see Table 1): Pay at time Receive the asset at time 0 T

0

T

Outright purchase Prepaid forward contract

Fully leveraged purchase Forward contract

Table 1: Alternative ways to buy an asset. 26/89

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1. Outright purchase. Both the payment and the transfer of ownership are made at time zero. The price paid per share is the current spot price S0 . 2. Fully leveraged purchase. The transfer of ownership is made at time zero. The payment is made at time T . The payment is S0 e rT , where S0 is the current spot price and r is the risk–free continuously compounded annual interest rate. P is made at 3. Prepaid forward contract. A payment of F0,T time zero. The transfer of ownership is made at time T . The P is not necessarily the current spot price S . payment F0,T 0

4. Forward contract. Both the payment and the transfer of ownership happen at time T . The price of a forward contract P , where r is is denoted by F0,T . We have that F0,T = e rT F0,T the risk–free annual interest rate continuously compounded. 27/89

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Pricing a forward contract.

Whenever an asset is delivered and paid at time zero, the fair price of the asset is its (spot) market price. The market of an outright purchase is S0 . Commissions and bid–ask spreads must be taken into account. The case of a fully leverage purchase, the price of a forward contract is just the price of a loan of S0 . The price of a fully leveraged purchase is S0 e rT , which is the price of a loan of S0 taken at time zero and paid at time T .

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We are interested in determining F0,T , the price of a forward contract. Many different factors such as the cost of storing, delivering, the convenience yield and the scarcity of the asset. Some commodities like oil have high storage costs. The convenience yield measures the cost of not having the asset, but a forward contract on it. For example, if instead of having a forward on gasoline, we have the physical asset, we may use it in case of scarcity. In the case of stock paying dividends, an stock owner receives dividend payments, and a long forward holder does not.

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Pricing a prepaid forward contract.

P , we make three cases. To find F0,T

1. Price of prepaid forward contract if there are no dividends. We consider an asset with no cost/benefit in holding the asset. This applies to the the price of a stock which does pay any dividends. It is irrelevant whether the transfer of ownership happens now or later P =S , The no arbitrage price of a prepaid forward contract is F0,T 0 where S0 is the price of an asset today.

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Example 2 XYZ stock costs $55 per share. XYZ stock does not pay any dividends. The risk–free interest rate continuously compounded 8%. Calculate the price of a prepaid forward contract that expires 30 months from today.

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Example 2 XYZ stock costs $55 per share. XYZ stock does not pay any dividends. The risk–free interest rate continuously compounded 8%. Calculate the price of a prepaid forward contract that expires 30 months from today. P = S = 55. Solution: The prepaid forward price is F0,T 0

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2. Price of prepaid forward contract when there are discrete dividends. Suppose that the stock is expected to make a dividend payment of DTi at the time ti , i = 1 . . . , n. A prepaid forward contract will entitle you receive the stock at time T without receiving the interim dividends. The prepaid forward price is P F0,T = S0 −

n X

Dti e −rti .

i=1

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Example 3 XYZ stock cost $55 per share. It pays $2 in dividends every 3 months. The first dividend is paid in 3 months. The risk–free interest rate continuously compounded 8%. Calculate the price of a prepaid forward contract that expires 18 months from today, immediately after the dividend is paid.

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Example 3 XYZ stock cost $55 per share. It pays $2 in dividends every 3 months. The first dividend is paid in 3 months. The risk–free interest rate continuously compounded 8%. Calculate the price of a prepaid forward contract that expires 18 months from today, immediately after the dividend is paid. Solution: The prepaid forward price is P F0,T = S0 −

n X

Dti e −rti = 55 −

i=1

=55 −

6 X

6 X

2e −(0.08)j(1/4)

j=1

2e −(0.02)j = 55 − 2a−6−|e 0.02 −1 = 43.80474631.

j=1

Recall that a−n−|i =

Pn

j=1 (1

+ i)−j . 35/89

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3. Price of prepaid forward contract when there are continuous dividends. In the case of an index stock, dividends are given almost daily. We may model the dividend payments as a continuous flow. Let δ be the rate of dividends given per unit of time. Suppose that dividends payments are reinvested into stock. Let tj = jT n , 1 ≤ j ≤ n. If Aj is the amount of shares at time tj , then Aj+1 = Aj (1 + δT n ). Hence, the total amount of shares δT multiplies by 1 + n in each period. Hence, one share at time zero n grows to 1 + δT at time T . Letting n → ∞, we get that one n share at time zero grows to e δT shares at time T . With $K at time 0, we can buy SK0 shares in the market at time 0. These shares grow to SK0 e δT at time T . With $K at time 0, we can buy K shares to be delivered at time T using a prepaid forward. FP 0,T

Hence, is there exists no arbitrage,

K Tδ S0 e

=

K P F0,T

and

P = S0 e −δT . F0,T 36/89

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Example 4 An investor is interested in buying XYZ stock. The current price of stock is $45 per share. This stock pays dividends at an annual continuous rate of 5%. Calculate the price of a prepaid forward contract which expires in 18 months.

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Example 4 An investor is interested in buying XYZ stock. The current price of stock is $45 per share. This stock pays dividends at an annual continuous rate of 5%. Calculate the price of a prepaid forward contract which expires in 18 months. Solution: The price of the prepaid forward contract is P F0,T = S0 e −δT = 45e −(0.05)(18/12) = 41.74845688.

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Example 5 XYZ stock costs $55 per share. The annual continuous interest rate is 0.055. This stock pays dividends at an annual continuous rate of 3.5%. A one year prepaid forward has a price of $52.60. Is there any arbitrage opportunity? If so, describe the position an arbitrageur would take and his profit per share.

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Example 5 XYZ stock costs $55 per share. The annual continuous interest rate is 0.055. This stock pays dividends at an annual continuous rate of 3.5%. A one year prepaid forward has a price of $52.60. Is there any arbitrage opportunity? If so, describe the position an arbitrageur would take and his profit per share. Solution: The no arbitrage prepaid forward price is P F0,T = S0 e −δT = 55e −0.035 = 53.10829789.

An arbitrage portfolio consists of entering a prepaid long forward contract for one share of stock and shorting e −0.035 shares of stock. The return of this transaction is 55e −0.035 − 52.60 = 0.5082978942. At redemption time, the arbitrageur covers his short position after executing the prepaid forward contract. 40/89

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Example 6 XYZ stock costs $55 per share. The annual continuous interest rate is 0.035. This stock pays dividends at an annual continuous rate of 5.5%. A one year prepaid forward has a price of $52.60. Is there any arbitrage opportunity? If so, describe the position an arbitrageur would take and his profit per share.

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Example 6 XYZ stock costs $55 per share. The annual continuous interest rate is 0.035. This stock pays dividends at an annual continuous rate of 5.5%. A one year prepaid forward has a price of $52.60. Is there any arbitrage opportunity? If so, describe the position an arbitrageur would take and his profit per share. Solution: The no arbitrage prepaid forward price is P F0,T = S0 e −δT = 55e −0.055 = 52.05668314.

An arbitrage portfolio consists of entering a prepaid short forward contract for one share of stock and buying e −0.055 shares of stock. The return of this transaction is 52.60 − 55e −0.055 = 0.5433168626. At redemption time, we use the bought stock to meet the short forward. Notice that in the previous questions, we can make arbitrage without making any investment of capital. The total price of setting the portfolios at time zero is zero. c



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Pricing a forward contract.

Both the payment and the transfer of ownership happen at time T . The price of a forward contract is the future value of the P . So, prepaid forward contract, i.e. F0,T = e rT F0,T I

The price of a forward contract for a stock with no dividends is F0,T = e rT S0 .

I

The price of a forward contract P for a stock with discrete dividends is F0,T = e rT S0 − ni=1 Dti e r (T −ti ) .

I

The price of a forward contract for a stock with continuous dividends is F0,T = e (r −δ)T S0 .

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Example 7 The current price of one share of XYZ stock is 55.34. The price of a nine–month forward contract on one share of XYZ stock is 57.6. XYZ stock is not going to pay any dividends on the next 2 years. (i) Calculate the annual compounded continuously interest rate implied by this forward contract. (ii) Calculate the price of a two–year forward contract on one share of XYZ stock.

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Example 7 The current price of one share of XYZ stock is 55.34. The price of a nine–month forward contract on one share of XYZ stock is 57.6. XYZ stock is not going to pay any dividends on the next 2 years. (i) Calculate the annual compounded continuously interest rate implied by this forward contract. (ii) Calculate the price of a two–year forward contract on one share of XYZ stock. Solution: (i) Since F0,T = e rT S0 , 57.6 = e (3/4)r 55.34 and r = (4/3) ln(57.6/55.34) = 0.05336879112.

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Example 7 The current price of one share of XYZ stock is 55.34. The price of a nine–month forward contract on one share of XYZ stock is 57.6. XYZ stock is not going to pay any dividends on the next 2 years. (i) Calculate the annual compounded continuously interest rate implied by this forward contract. (ii) Calculate the price of a two–year forward contract on one share of XYZ stock. Solution: (i) Since F0,T = e rT S0 , 57.6 = e (3/4)r 55.34 and r = (4/3) ln(57.6/55.34) = 0.05336879112. (ii) We have that F0,2 = e r 2 S0 = e (0.05336879112)(2) 55.34 = 61.57362151.

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Example 8 A stock is expected to pay a dividend of $1 per share in 2 months and again in 5 months. The current stock price is $59 per share. The risk free effective annual rate of interest is 6%. (i) What is the fair price of a 6–month forward contract? (ii) Assume that 3 months from now the stock price is $57 per share, what is the fair price of the same forward contract at that time?

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Example 8 A stock is expected to pay a dividend of $1 per share in 2 months and again in 5 months. The current stock price is $59 per share. The risk free effective annual rate of interest is 6%. (i) What is the fair price of a 6–month forward contract? (ii) Assume that 3 months from now the stock price is $57 per share, what is the fair price of the same forward contract at that time? Solution: (i) The forward price is the future value of the payments associated with owning the stock in six months: F0,0.5 = (59)(1.06)0.5 − (1)(1.06)4/12 − (1)(1.06)1/12 = 58.71974.

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Example 8 A stock is expected to pay a dividend of $1 per share in 2 months and again in 5 months. The current stock price is $59 per share. The risk free effective annual rate of interest is 6%. (i) What is the fair price of a 6–month forward contract? (ii) Assume that 3 months from now the stock price is $57 per share, what is the fair price of the same forward contract at that time? Solution: (i) The forward price is the future value of the payments associated with owning the stock in six months: F0,0.5 = (59)(1.06)0.5 − (1)(1.06)4/12 − (1)(1.06)1/12 = 58.71974. (ii) (57)(1.06)3/12 − (1)(1.06)1/12 = 56.83154.

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Example 9 An investor is interested in buying XYZ stock. The current price of stock is $30 per share. The risk–free annual interest rate continuously compounded is 0.03. The price of a fourteen–month forward contract is 30.352. Calculate the continuous dividend yield δ.

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Example 9 An investor is interested in buying XYZ stock. The current price of stock is $30 per share. The risk–free annual interest rate continuously compounded is 0.03. The price of a fourteen–month forward contract is 30.352. Calculate the continuous dividend yield δ. Solution: We have that 30.352 = F0,T = S0 e (r −δ)T = 30e (0.03−δ)(14/12) . and δ = 0.03 − (12/14) ln(30.352/30) = 0.02000140155.

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Example 10 An investor is interested in buying XYZ stock. The current price of stock is $30 per share. This stock pays dividends at an annual continuous rate of 0.02. The risk–free annual effective rate of interest is 0.045. (i) What is the price of prepaid forward contract which expires in 18 months? (ii) What is the price of forward contract which expires in 18 months?

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Example 10 An investor is interested in buying XYZ stock. The current price of stock is $30 per share. This stock pays dividends at an annual continuous rate of 0.02. The risk–free annual effective rate of interest is 0.045. (i) What is the price of prepaid forward contract which expires in 18 months? (ii) What is the price of forward contract which expires in 18 months? Solution: (i) The prepaid forward price is P F0,T = S0 e −δT = 30e −(0.02)(18/12) = 29.11336601.

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Example 10 An investor is interested in buying XYZ stock. The current price of stock is $30 per share. This stock pays dividends at an annual continuous rate of 0.02. The risk–free annual effective rate of interest is 0.045. (i) What is the price of prepaid forward contract which expires in 18 months? (ii) What is the price of forward contract which expires in 18 months? Solution: (i) The prepaid forward price is P F0,T = S0 e −δT = 30e −(0.02)(18/12) = 29.11336601.

(ii) The 18–month forward price is 29.11336601(1.045)18/12 = 31.1004631. 54/89

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off–market forward contract

A forward contract where either you pay a premium or you collect a premium for entering into the deal is called an off–market forward contract.

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Example 11 Suppose that the current value of a certain amount of a commodity is $45000. The annual effective rate of interest is 4.5%. (i) You are offered a 2–year long forward contract at a forward price of $50000. How much would you need to be paid to enter into this contract? (ii) You are offered a 2–year long forward contract at a forward price of $48000. How much would you need be willing to pay to enter into this contract?

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Example 11 Suppose that the current value of a certain amount of a commodity is $45000. The annual effective rate of interest is 4.5%. (i) You are offered a 2–year long forward contract at a forward price of $50000. How much would you need to be paid to enter into this contract? (ii) You are offered a 2–year long forward contract at a forward price of $48000. How much would you need be willing to pay to enter into this contract? Solution: (i) Let x be how much you need to be paid to enter into this contract. The current value of the commodity should be equal to the present value of the expenses needed to get the commodity using the long forward contract. Hence, 50000(1.045)−2 − x = 45000. So, x = 50000(1.045)−2 − 45000 = 786.4976.

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Example 11 Suppose that the current value of a certain amount of a commodity is $45000. The annual effective rate of interest is 4.5%. (i) You are offered a 2–year long forward contract at a forward price of $50000. How much would you need to be paid to enter into this contract? (ii) You are offered a 2–year long forward contract at a forward price of $48000. How much would you need be willing to pay to enter into this contract? Solution: (ii) Let y be how much would you need be willing to pay to enter into this contract. The current value of the commodity should be equal to the present value of the expenses needed to get the commodity using the long forward contract. Hence, 48000(1.045)−2 + y = 45000. So, y = 45000 − 48000(1.045)−2 = 1044.962341.

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Arbitrage

If the price of a forward contract does not follow the previous formulas, an arbitrageur can do arbitrage.

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Example 12 XYZ stock pays no dividends and has a current price of $42.5 per share. A long position in a forward contract is available to buy 1000 shares of stock six months from now for $43 per share. A bank pays interest at the rate of 5% per annum (continuously compounded) on a 6–month certificate of deposit. Describe a strategy for creating an arbitrage profit and determine the amount of the profit.

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Example 12 XYZ stock pays no dividends and has a current price of $42.5 per share. A long position in a forward contract is available to buy 1000 shares of stock six months from now for $43 per share. A bank pays interest at the rate of 5% per annum (continuously compounded) on a 6–month certificate of deposit. Describe a strategy for creating an arbitrage profit and determine the amount of the profit. Solution: The no arbitrage price of a forward contract is S0 e rT = (42.5)e 0.05(0.5) = 43.57589262. Hence, it is possible to do arbitrage by entering into the long forward position. An arbitrageur can: sell 1000 shares of stock for (1000)(42.5) = 42500, deposit 42500 in the bank for six months, and sign up a forward contract for a long position for 1000 shares of stock. In six months, the CD returns (1000)(42.5)e 0.05(0.5) = 43575.89262. The cost of the forward is (1000)(43) = 430000. Hence, the profit is 43575.89262 − 430000 = 575.89262. c



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Example 13 Suppose that the risk–free effective rate of interest is 5% per annum. XYZ stock is currently trading for $45.34 per share. XYZ stock is expected to pay a dividend of $1.20 per share six months from now. The price of a nine–month forward contract on one share of XYZ stock is $47.56. Is there an arbitrage opportunity on the forward contract? If so, describe the strategy to realize profit and find the arbitrage profit.

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c





Example 13 Suppose that the risk–free effective rate of interest is 5% per annum. XYZ stock is currently trading for $45.34 per share. XYZ stock is expected to pay a dividend of $1.20 per share six months from now. The price of a nine–month forward contract on one share of XYZ stock is $47.56. Is there an arbitrage opportunity on the forward contract? If so, describe the strategy to realize profit and find the arbitrage profit. Solution: The no arbitrage forward price is F0,T = e

rT

S0 −

n X

Dti e r (T −ti ) = 45.34(1.05)9/12 − 1.2(1.05)3/12

i=1

=45.81511211. We can make arbitrage by buying stock and entering a short forward contract. The profit per share at expiration is 47.56 − 45.81511211 = 1.74488789. c



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Suppose that a stock pays dividends at the continuous rate δ. In the absence of arbitrage, entering a forward for one share for F0,T is equivalent to buying e −δT shares of stock for S0 e −δT and (borrowing S0 e −δT ) selling a zero–coupon bond for S0 e −δT with expiration in T years. In both cases, at time T we have one share of stock after we make a payment of F0,T . There exists no arbitrage if S0 e −δT = F0,T e −rT . If S0 e −δT 6= F0,T e −rT , we can make arbitrage.

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c





e (r −δ)T S0 ,

If F0,T < we can enter into a long forward for one share −δT of stock, and short e shares of stock. At redemption time, we cover the short position by paying F0,T for the stock. It is like we have borrowed S0 e −δT and pay the loan for F0,T . In some sense we have created a zero–coupon bond. The position is called a synthetic zero–coupon bond. Let r 0 be the continuous annual rate of interest of the synthetic bond. This rate is called the implied repo rate. We have that S0 e −δT e r

0T

= F0,T .

Hence, if F0,T < e (r −δ)T S0 , F0,T 1 1 0 < log r = log −δT T T S0 e

S0 e (r −δ)T S0 e −δT

! < r.

By doing an arbitrage, we are able to reduce the interest rate at which we borrow. Technically, this is not call arbitrage. It is called quasi–arbitrage. We benefit from this portfolio, only if we are already borrowing. c



65/89



Notice that the synthetic bond is created observing that Long forward = Buy stock + Issue a bond implies that Issue a bond = Short stock + Long forward

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c





Reciprocally, if F0,T > e (r −δ)T S0 , we can create a portfolio earning a rate of interest bigger than the risk–free interest rate. We can enter a short forward contract for one share of stock and buy e −δT shares of stock. At redemption time, we get an inflow of F0,T . Since we invested S0 e −δT , the continuous annual interest rate r 0 , which we earned in the investment satisfies S0 e −δT e r

0T

= F0,T .

Hence, if F0,T > e (r −δ)T S0 , 1 r 0 = ln T

F0,T S0 e −δT

1 > ln T

S0 e (r −δ)T S0 e −δT

! = r.

Again this rate is called the implied repo rate. 67/89

c





Example 14 XYZ stock costs $123.118 per share. This stock pays dividends at an annual continuous rate of 2.5%. A 18 month forward has a price of $130.242. You own 10000 shares of XYZ stock. Calculate the annual continuous rate of interest at which you can borrow by shorting your stock.

68/89

c





Example 14 XYZ stock costs $123.118 per share. This stock pays dividends at an annual continuous rate of 2.5%. A 18 month forward has a price of $130.242. You own 10000 shares of XYZ stock. Calculate the annual continuous rate of interest at which you can borrow by shorting your stock. Solution: We have that F0,T 1 1 130.242 r 0 = δ+ ln = 0.025+ ln = 6.250067554%. T S0 1.5 123.118

69/89

c





Example 15 XYZ stock costs $124 per share. This stock pays dividends at an annual continuous rate of 1.5%. A 2–year forward has a price of $135.7 per share. Calculate the annual continuous rate of interest which you earn by buying stock and entering into a short forward contract, both positions for the same nominal amount.

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c





Example 15 XYZ stock costs $124 per share. This stock pays dividends at an annual continuous rate of 1.5%. A 2–year forward has a price of $135.7 per share. Calculate the annual continuous rate of interest which you earn by buying stock and entering into a short forward contract, both positions for the same nominal amount. Solution: We have that F0,T 1 1 135.7 0 = 0.015 + ln r = δ + ln = 6%. T S0 2 124

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c





F

. Notice that this is not a price. The forward premium is S0,T 0 Prices of options are called premiums. But, here the nomenclature is different. If a stock index pays dividends according with a F = e T (r −δ) . continuous rate δ, then S0,T 0

72/89

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Example 16 XYZ stock cost $55 per share. A four–month forward on XYZ stock costs $57.5. XYZ stock pays dividends according a continuous rate. (i) Calculate the four–month forward premium. (ii) Calculate the eight–month forward premium. (iii) Calculate the eight–month forward price.

73/89

c





Example 16 XYZ stock cost $55 per share. A four–month forward on XYZ stock costs $57.5. XYZ stock pays dividends according a continuous rate. (i) Calculate the four–month forward premium. (ii) Calculate the eight–month forward premium. (iii) Calculate the eight–month forward price. Solution: (i) The four–month forward premium is F0,4/12 = 57.5 S0 55 = 1.045454545.

74/89

c





Example 16 XYZ stock cost $55 per share. A four–month forward on XYZ stock costs $57.5. XYZ stock pays dividends according a continuous rate. (i) Calculate the four–month forward premium. (ii) Calculate the eight–month forward premium. (iii) Calculate the eight–month forward price. Solution: (i) The four–month forward premium is F0,4/12 = 57.5 S0 55 = 1.045454545. (ii) The eight–month forward premium is 2 F0,8/12 = e (8/12)(r −δ) = e (4/12)(r −δ) = S0

57.5 55

2 = 1.092975206.

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c





Example 16 XYZ stock cost $55 per share. A four–month forward on XYZ stock costs $57.5. XYZ stock pays dividends according a continuous rate. (i) Calculate the four–month forward premium. (ii) Calculate the eight–month forward premium. (iii) Calculate the eight–month forward price. Solution: (i) The four–month forward premium is F0,4/12 = 57.5 S0 55 = 1.045454545. (ii) The eight–month forward premium is 2 F0,8/12 = e (8/12)(r −δ) = e (4/12)(r −δ) = S0

57.5 55

2 = 1.092975206.

(iii) The eight–month forward price is F0,8/12 = (55)(1.092975206) = 60.11363633. 76/89

c





F The annualized forward premium is T1 ln S0,T . Note that in 0 the case of continuous dividends, the annualized forward premium is r − δ.

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Example 17 XYZ stock cost $55 per share. A four–month forward on XYZ stock costs $57.5. (i) Calculate the annualized forward premium (ii) Calculate the twelve–month forward price.

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Example 17 XYZ stock cost $55 per share. A four–month forward on XYZ stock costs $57.5. (i) Calculate the annualized forward premium (ii) Calculate the twelve–month forward price. Solution: (i) The annualized forward premium is F0,T 1 1 57.5 r − δ = ln = ln = 0.1333552877. T S0 4/12 55

79/89

c





Example 17 XYZ stock cost $55 per share. A four–month forward on XYZ stock costs $57.5. (i) Calculate the annualized forward premium (ii) Calculate the twelve–month forward price. Solution: (i) The annualized forward premium is F0,T 1 1 57.5 r − δ = ln = ln = 0.1333552877. T S0 4/12 55 (ii) The twelve–month forward price is F0,T = S0 e r −δ = 55e 0.1333552877 = 62.84607438.

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Hedging a forward contract. A (scalper) market maker must be able to offset the risk of trading forward contracts. Assume continuous dividends. Suppose that a scalper enters into a short forward contract. The profit at expiration for a long forward position is ST − F0,T . In order to obtain this same payoff a scalper can borrow S0 e −δT and use this money to get e −δT shares of stock. At time T , he sells the stock which he owns for S0 e (r −δ)T = F0,T . Notice that by investing the dividends, e −δT shares of stock have grown to one share at time T . Borrowing S0 e −δT and buying e −δT shares of stock is called a synthetic long forward. So, if a scalper enters into a short forward contract with a client, the scalper either matches this position with another client’s long forward contract or creates a synthetic long forward 81/89

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The profit of a short forward position is F0,T − ST . A scalper can get this profit, by (lending) buying a zero–coupon bond for S0 e −δT and shorting e −δT shares receiving S0 e −δT . Buying a zero–coupon bond for S0 e −δT and shorting a tailed position for e −δT shares is called a synthetic short forward. Again, a scalper may need to create this position to match a client’s long forward position.

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Using these strategies, a market–maker can hedge his clients positions. A transaction in which you buy the asset and short the forward contract is called cash–and–carry (or cash–and–carry hedge). It is called cash–and–carry, because the cash is used to buy the asset and the asset is kept. A cash–and–carry has no risk. You have obligation to deliver the asset, but you also own the asset. An arbitrage that involves buying the asset and selling it forward is called cash–and–carry arbitrage. A (reverse cash–and–carry hedge) reverse cash–and–carry involves short–selling and asset and entering into a long forward position.

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An arbitrageur can make money if F0,T 6= S0 e (r −δ)T . But, in the real world, transaction costs have to be taken into account. Suppose that: (i) The stock bid and ask prices are S0b and S0a , where S0b < S0a . b < Fa . (ii) The forward bid and ask prices are F0,T 0,T (iii) The cost of a transaction in the stock is KS . (iv) The cost of a transaction in the forward is KF . (v) The interest rates for borrowing and lending are rb > rl , respectively.

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c





Suppose that the arbitrageur believes that the observed forward price F0,T is too high. Then, he could: b (a) contract a short forward for F0,T (b) buy a tailed position in stock for S0a e −δT . −δT (c) borrow S0ae + KF + KS . The payoff of this combined transaction is: b F0,T − ST + ST − (S0a e −δT + Kf + KS )e rb T b =F0,T − (S0a e −δT + Kf + KS )e rb T . b > (S a e −δT + K + K )e rb T . The The scalper makes money if F0,T f S 0 previous strategy is cash–and–carry arbitrage.

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c





Suppose that the scalper believes that the observed forward price F0,T is too low. Then, he could: a (a) enter a long forward for F0,T (b) short a tailed position in stock for S0b e −δT . (c) lend S0b e −δT − KF − KS . The payoff of this combined transaction is: a ST − F0,T − ST + (S0a e −δT − KF − KS )e rl T a = − F0,T + (S0a e −δT − KF − KS )e rl T . a . The arbitrageur makes money if (S0a e −δT − KF − KS )e rl T > F0,T The previous strategy is reverse cash–and–carry arbitrage.

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Example 18 Suppose that an arbitrageur would like to enter a cash–and–carry for 10000 barrels of oil for delivery in six months. Suppose that he can borrow at an annual effective rate of interest of 4.5%. The current price of a barrel of oil is $55. (i) What is the minimum forward price at which he would make a profit? (ii) What is his profit if the forward price is $57?

87/89

c





Example 18 Suppose that an arbitrageur would like to enter a cash–and–carry for 10000 barrels of oil for delivery in six months. Suppose that he can borrow at an annual effective rate of interest of 4.5%. The current price of a barrel of oil is $55. (i) What is the minimum forward price at which he would make a profit? (ii) What is his profit if the forward price is $57? Solution: (i) He would make a profit if F0,T > 55(1.045)1/2 = 56.22388283.

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c





Example 18 Suppose that an arbitrageur would like to enter a cash–and–carry for 10000 barrels of oil for delivery in six months. Suppose that he can borrow at an annual effective rate of interest of 4.5%. The current price of a barrel of oil is $55. (i) What is the minimum forward price at which he would make a profit? (ii) What is his profit if the forward price is $57? Solution: (i) He would make a profit if F0,T > 55(1.045)1/2 = 56.22388283. (ii) The profit is (10000)(57 − (55)(1.045)1/2 ) = 7761.171743.

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c




Manual for SOA Exam FM/CAS Exam 2. Chapter 7. Derivative markets. Section 7.3. Futures. c



1/15

c




Section 7.3. Futures.

Futures

A future is a standardized contract in which two counterparts agree to buy/sell an asset for a specified price (the future price) at a specified date (the delivery date). The buyer in the future contract is called the long future. The seller in the future contract is called the short future. The main reasons to enter into a future contract are hedging and speculation. At difference of futures, forward contracts are privately negotiated and are not standardized. Forward contracts are entirely flexible. Forward contracts are tailor–made contracts.

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Futures are bought and sold in organized futures exchanges. The biggest future exchanges are the Chicago Mercantile Exchange, the Chicago Board of Trade, the International Petroleum Exchange of London, the New York Mercantile Exchange, the London Metal Exchange and the Tokyo Commodity Exchange. Futures transactions in the USA are regulated by the (CFTC) Commodity Futures Trading Commission, an agency of the USA government. The CFTC also regulates option markets. A future contract is negotiated through a brokerage firm that holds a seat on the exchange. A future contract is settled by a clearinghouse owned by or associated with the exchange. The clearinghouse matches the purchases and the sales which take place during the day. By matching trades, the clearinghouse never takes market risk because it always has offsetting positions with different counterparts. By having the clearinghouse as counterpart, an individual entering a future contract does not face the possible credit risk of its counterpart. 3/15

c





Let us consider some common futures. Crude oil futures trade in units of 1,000 U.S. barrels (42,000 gallons). The underlying is a US barrel. The notional amount is 1000 barrels. The current price is $70/barrel. Hence, the current value of a future contract on crude oil is $70000. S & P 500 future contracts trade on 250 units of the index. They are cash settled. At expiration time, instead of a sale, one of the future counterpart receive a payment according with S & P 500 spot price at expiration. The current price of S & P 500 is 1500. The current value of a future contract on S & P 500 is (250)(1500) = $375000.

4/15

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Suppose that two parties agree in a future contact for crude oil for delivery in 18 months. The contract is worth $70000. Each (investor) party makes a trade with the clearinghouse. This contract has two risks: market risk and credit risk. The market risk is related with the volatility of the price of the asset. The credit risk is related with the solvency of each party. To avoid credit risk, an individual or corporation entering a future contract must make a deposit into an account called the margin account. This deposit is called the initial margin. The margin account earns interest. The amount of the initial margin is determined by the exchange. It is usually a fraction of the market value of the futures’ underlying asset. Usually future positions are settled into the margin account either every day or every week. By every day we mean every day which the market is open. Let us suppose that a clearinghouse settles accounts daily. Suppose that the annual continuously compounded interest rate is r . 5/15

c





Every day, the profit or loss is calculated on the investor’s futures position. If there exists a loss, the investor’s broker transfers that amount from the investor’s margin account to the clearinghouse. If a profit, the clearinghouse transfers that amount to investor’s broker who then deposits it into the investor’s margin account. The profit for a long position in a future contract is Mt−(1/365) (e r /365 − 1) + N(St − St−(1/365) ), where Mt−(1/365) is the yesterday’s balance in the margin account, N is the nominal amount, St is the current price, St−(1/365) is the yesterday price. Hence, after the settlement, the balance in the investor’s margin account is Mt = Mt−(1/365) e r /365 + N(St − St−(1/365) ). The profit for a short position in a future contract is Mt−(1/365) (1 − e r /365 ) + N(St−(1/365) − St ). Marking–to–market is to calculate the value of a future contract according with the current value of the asset. c



6/15



Example 1 On July 5, 2007, John enters a long future contract for 1,000 U.S. barrels of oil at $71.6 a barrel. The margin account is 50% of the market value of the futures’ underlier. The annual continuously compounded rate of return is 0.06. (i) On July 6, 2007, the price of oil is $70.3. What is the balance in John’s margin account after settlement? (ii) On July 7, 2007, the price of oil is $72.1. What is the balance in John’s margin account after settlement?

7/15

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Solution: (i) The initial balance in John’s margin account is (0.50)(1000)(71.6) = 35800. The balance in John’s margin account on July 6, 2007, after settlement, is Mt−(1/365) e r /365 + N(St − St−(1/365) ) =(35800)e 0.06/365 + (1000)(70.3 − 71.6) = 35105.89. Since the price of the oil decreases, the value of having 1000 barrels in 18 months decreases.

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Solution: (ii) The balance in John’s margin account on July 6, 2007, after settlement, is Mt−(1/365) e r /365 + n(St − St−(1/365) ) =(35105.89)e 0.06/365 + (1000)(72.1 − 70.3) = 35711.56. Notice that this balance is different from (35800)e (0.06)(2/365) + (1000)(72.1 − 71.6) = 36311.77. In the first day, John’s account balance was smaller. So, John lost interest because the drop on price on July 6, 2007.

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c





If the balance in the margin account falls the clearinghouse has less protection against default. Investors are required to keep the margin account to a minimum level. This level is a fraction of the initial margin. The maintenance margin is the fraction of the initial margin which participants are asked to hold in their accounts. If the balance in the margin account falls below this level, an investor’s broker will require the investor to deposit funds sufficient to restore the balance to the initial margin level. Such a demand is called a margin call. If an investor fail to the deposit, the investor’s broker will immediately liquidate some or all of the investor’s positions.

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Example 2 A company enters into a short futures contract to sell 100,000 pounds of frozen orange juice for $1.4 cents per pound. The initial margin is 30% and the maintenance margin is 20%. The annual effective rate of interest is 4.5%. The account is settled every week. What is the minimum next week price which would lead to a margin call?

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Solution: The initial balance in the margin account is (0.30)(100000)(1.4) = 42000. The minimum balance in the margin account is (0.20)(100000)(1.4) = 28000. After settlement next week balance is 42000(1.045)1/52 + 100000(1.4 − S1/52 ). A margin call happens if 28000 > 42000(1.045)1/52 + 100000(1.4 − S1/52 ), or S1/52 > 1.4 −

28000 − 42000(1.045)1/52 = 1.540355672. 100000

12/15

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Besides holding the contract until expiration, there are two ways to close a future contract: offset the contract and exchange for physicals. To offset the contract means to enter a reverse position with the same broker. Since future contracts are standardized, it is possible to find a reserve position on a contract. Exchange for physicals consists selling/buying the commodity.

13/15

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The two main advantages of futures versus forwards are liquidity and counter–party risk. It is much easier to cancel before expiration a future contract than a forward contract. Since the trade is made against a clearinghouse, a participant does face credit risk. At the same time, the margin and the marking to market reduces the default risk.

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Having a margin account makes the profit/losses of the investment higher for a future than for a forward. The oscillations of the price of the asset make the earnings in the margin account more variable. Usually, if there is a profit from the change of the price of the asset, there is also a profit in the balance account. Reciprocally, if there is a loss from the price change, there exists a loss in the margin account.

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Manual for SOA Exam FM/CAS Exam 2. Chapter 7. Derivatives markets. Section 7.4. Call options. c



1/112

c




Section 7.4. Call options.

Minimums and maximums

Definition 1 Given two real numbers a and b, (i) min(a, b) denotes the (minimum) smallest of the two numbers. (ii) max(a, b) denotes the (maximum) biggest of the two numbers.

Example 1 min(10, 5) = 5, max(10, 5) = 10, min(−1, 5) = −1, max(−1, 5) = 5, min(−2, −100) = −100, max(−2, −100) = −2.

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Definition 2 Given real numbers a1 , . . . , an , (i) min(a1 , . . . , an ) denotes the (minimum) smallest of these numbers. (ii) max(a1 , . . . , an ) denotes the (maximum) biggest of these numbers.

Example 2 min(−1, 5, 3, −6) = −6, max(−1, 5, 3, −6) = 5, min(−2, −100, −50) = −100 and max(−2, −100, −50) = −2.

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Theorem 1 For each a, b, c ∈ R and each λ ≥ 0, I

min(a, b) = min(b, a).

I

max(a, b) = max(b, a).

I

min(min(a, b), c) = min(a, min(b, c)) = min(a, b, c).

I

max(max(a, b), c) = max(a, max(b, c)) = max(a, b, c).

I

min(a + c, b + c) = min(a, b) + c.

I

max(a + c, b + c) = max(a, b) + c.

I

min(λa, λb) = λ min(a, b).

I

max(λa, λb) = λ max(a, b).

I

min(−a, −b) = − max(a, b).

I

max(−a, −b) = − min(a, b).

4/112

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Definition 3 Given a real number a, |a| = a, if a ≥ 0; and |a| = −a, if a ≤ 0

Example 3 |23| = 23, | − 4| = 4.

Theorem 2 For each a, b ∈ R, min(a, b) + max(a, b) = a + b.

Proof. min(a, b) and max(a, b) are a and b in some order. Hence, min(a, b) + max(a, b) = a + b.

Theorem 3 For each a ∈ R, |a| = max(a, 0) − min(a, 0).

Proof. If a ≥ 0, then max(a, 0) = a, min(a, 0) = 0, and max(a, 0) − min(a, 0) = a = |a|. If a ≤ 0, then max(a, 0) = 0, min(a, 0) = a, and max(a, 0) − min(a, 0) = −a = |a|. 5/112

c





Call options Definition 4 A call option is a financial contract which gives the owner the right, but not the obligation, to buy a specified amount of a given asset at a specified price during a specified period of time.

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Call options Definition 4 A call option is a financial contract which gives the owner the right, but not the obligation, to buy a specified amount of a given asset at a specified price during a specified period of time. The call option owner exercises the option by buying the asset at the specified call price from the call writer. A call option is executed only if the call owner decides to do so. A call option owner executes a call option only when it benefits him, i.e. when the specified call price is smaller than the current (market value) spot price. Since the owner of a call option can make money if the option is exercised, call options are sold. The owner of the call option must pay to its counterpart for holding a call option. The price of a call option is called its premium. 7/112

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I

I I I I I I I


The (owner) buyer of a call option is called the option call holder. The holder of a call option is said to have a long call position. The seller of a call option is called the option call writer. The writer of a call is said to have a short call position. Assets used in call options are in commodities, currency exchange, stock shares and stock indices. A call option needs to specify the type and quality of the underlying. The asset used in the call option is called the underlier or underlying asset. The amount of the underlying asset to which the call option applies is called the notional amount. The specified price of an asset in a call option is called the strike price, or exercise price. A forward contract forces the buyer and seller to execute the sale. A call option is executed only if the call holder decides to do so.

c



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I

For an European option, the exercise of the option must occur at a certain time (the expiration date).

I

For an American option, the exercise of the option must occur any time by the expiration date.

I

For a Bermudan option, the buyer can exercise the call option during specified periods.

Unless say otherwise, we will assume that an option is an European option. European options are simpler and easier to study.

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Example 4 Suppose that an investor buys a call option of 100 shares of XYZ stock with a strike price of $76 per share. The exercise date is one year from now. (i) If the spot price at expiration is $70 per share, the call option holder does not exercise the option. The option is worthless. The call option holder can buy stock in the market for a price smaller than the call option price. (ii) If the (the market price) spot price at expiration is $80 per share, the call option holder exercises the call option, i.e. he buys 100 shares of XYZ stock for $76 from the option seller. Since the call option holder can sell these shares for $80 per share, the call option holder gets a payoff of 100(80 − 76) = $400.

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Let K be the strike price of a call option. Let ST be the price of the asset at expiration. I

The call option holder’s payoff is ( 0 if ST < K , ST − K if ST ≥ K . We also can write this as max(0, ST − K ).

I

The payoff for the call option writer is the opposite of the holder’s payoff. The payoff for the call option writer is − max(0, ST − K ).

I

A call–option is a zero–sum game. The sum of the two payoffs is zero.

Figure 1 shows a graph of the call option payoff as a function of ST . 11/112

c





− max(ST − K , 0)

max(ST − K , 0) 6

6

-

K

ST

-

ST

K @ @

@ @ @

Payoff for the call option holder

Payoff for the call option writer

Figure 1: Payoffs of a call option 12/112

c





Recall: I The call option holder’s payoff is max(0, ST − K ). I

The call option writer’s payoff is − max(0, ST − K ).

We get from Figure 1 that: I The minimum payoff for the call option holder is 0. The maximum payoff for the call option holder is ∞. I The minimum payoff for the call option writer is −∞. The maximum payoff for the call option writer is 0.

call option holder call option writer

minimum payoff 0 −∞

maximum payoff ∞ 0 13/112

c





Example 5 Andrew buys a 45–strike call option for XYZ stock with a nominal amount of 2000 shares. The expiration date is 6 months from now. (i) Calculate Andrew’s payoff for the following spot prices per share at expiration: 35, 40, 45, 55, 60. (ii) Calculate Andrew’s minimum and maximum payoffs.

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c





Example 5 Andrew buys a 45–strike call option for XYZ stock with a nominal amount of 2000 shares. The expiration date is 6 months from now. (i) Calculate Andrew’s payoff for the following spot prices per share at expiration: 35, 40, 45, 55, 60. (ii) Calculate Andrew’s minimum and maximum payoffs. Solution: (i) Andrew’s payoff is (2000) max(ST − 45, 0). The corresponding payoffs are: if ST = 35, payoff = (2000) max(35 − 45, 0) = 0, if ST = 40, payoff = (2000) max(40 − 45, 0) = 0, if ST = 45, payoff = (2000) max(45 − 45, 0) = 0, if ST = 50, payoff = (2000) max(50 − 45, 0) = 10000, if ST = 55, payoff = (2000) max(55 − 45, 0) = 20000. (ii) Andrew’s minimum payoff is 0. Andrew’s maximum payoff is ∞. c



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Example 6 Madison sells a 45–strike call option for XYZ stock with a nominal amount of 2000 shares. The expiration date is 6 months from now. (i) Calculate Madison’s payoff for the following spot prices at expiration: 35, 40, 45, 55, 60. (ii) Calculate Madison’s minimum and maximum payoffs.

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c





Example 6 Madison sells a 45–strike call option for XYZ stock with a nominal amount of 2000 shares. The expiration date is 6 months from now. (i) Calculate Madison’s payoff for the following spot prices at expiration: 35, 40, 45, 55, 60. (ii) Calculate Madison’s minimum and maximum payoffs. Solution: (i) Madison’s payoff is −(2000) max(ST − 45, 0). The corresponding payoffs are: if ST = 35, payoff = −(2000) max(35 − 45, 0) = 0, if ST = 40, payoff = −(2000) max(40 − 45, 0) = 0, if ST = 45, payoff = −(2000) max(45 − 45, 0) = 0, if ST = 50, payoff = −(2000) max(50 − 45, 0) = −10000, if ST = 55, payoff = −(2000) max(55 − 45, 0) = −20000. (ii) Madison’s payoff is (2000) max(ST − 45, 0). Madison’s minimum payoff is −∞. Madison’s maximum payoff is 0. c



17/112



Let Call(K , T ) be the premium per unit paid by the buyer of a call option with strike price K and expiration time T years. Notice that Call(K , T ) > 0. The premium of a call option for N units is NCall(K , T ). Let i be the risk–free annual effective rate of interest. I

The call option holder’s profit per unit is max(ST − K , 0) − Call(K , T )(1 + i)T ( −Call(K , T )(1 + i)T if ST < K , = T ST − K − Call(K , T )(1 + i) if ST ≥ K .

I

The call option seller’s profit per unit is Call(K , T )(1 + i)T − max(0, ST − K ) ( Call(K , T )(1 + i)T if ST < K , = . T Call(K , T )(1 + i) − (ST − K ) if ST ≥ K . 18/112

c





The call option holder profit max(ST − K , 0) − Call(K , T )(1 + i)T as a function of ST is nondecreasing. The call option holder benefits from the increase of the spot price. I

The minimum call option holder profit is −Call(K , T )(1 + i)T .

I

The maximum call option holder profit is ∞.

I

The profit for the call option holder is positive if ST > K + Call(K , T )(1 + i)T .

I

If ST < K + Call(K , T )(1 + i)T , the call option holder profit is negative.

19/112

c





The call option writer’s profit Call(K , T )(1 + i)T − max(0, ST − K ) as a function of ST is nonincreasing. The call option writer benefits from the decrease of the spot price. I

The minimum call option writer profit is −∞. The call option writer position is riskier than his counterpart. A call option writer can assumed unbounded loses.

I

The maximum call option writer profit is Call(K , T )(1 + i)T .

I

The profit for the call option writer is positive if ST < K + Call(K , T )(1 + i)T .

I

The profit for the call option writer is negative if ST > K + Call(K , T )(1 + i)T .

20/112

c







profit max(ST − K , 0) − Call(K , T )(1 + i)T − max(ST − K , 0) + Call(K , T )(1 + i)T minimum profit −Call(K , T )(1 + i)T −∞

maximum profit ∞ Call(K , T )(1 + i)T

Figure 2 shows the graph of the profit of a call option as a function of ST .

21/112

c




max(ST − K , 0) − C (1 + i)T


C (1 + i)T − max(ST − K , 0)

6

6

K −C (1 +

i)T

C (1 + i)T -

ST

@

K @

ST

@

Profit for the call option holder

@ @

Profit for the call option writer

Figure 2: Profit of a call option c



22/112



If r is the annual interest rate compounded continuously, then the profit for the call option holder is max(0, ST − K ) − Call(K , T )e rT and the profit of the call option writer is Call(K , T )e rT − max(0, ST − K ).

23/112

c





Example 7 Ethan buys a 35–strike call option for XYZ stock for 4.337 per share. The nominal amount of this call option is 2000 shares. The expiration date of this option is 18 months. The annual effective interest rate is 5.5%.

24/112

c





Example 7 Ethan buys a 35–strike call option for XYZ stock for 4.337 per share. The nominal amount of this call option is 2000 shares. The expiration date of this option is 18 months. The annual effective interest rate is 5.5%. (i) Calculate Ethan’s profit function.

25/112

c





Example 7 Ethan buys a 35–strike call option for XYZ stock for 4.337 per share. The nominal amount of this call option is 2000 shares. The expiration date of this option is 18 months. The annual effective interest rate is 5.5%. (i) Calculate Ethan’s profit function. Solution: (i) Ethan’s profit function is (2000)(max(ST − 35, 0) − 4.337(1.055)1.5 ) =(2000) max(ST − 35, 0) − 9400.

26/112

c





Example 7 Ethan buys a 35–strike call option for XYZ stock for 4.337 per share. The nominal amount of this call option is 2000 shares. The expiration date of this option is 18 months. The annual effective interest rate is 5.5%. (ii) Calculate Ethan’s profit for the following spot prices at expiration: 25, 30, 35, 40, 45.

27/112

c





Example 7 Ethan buys a 35–strike call option for XYZ stock for 4.337 per share. The nominal amount of this call option is 2000 shares. The expiration date of this option is 18 months. The annual effective interest rate is 5.5%. (ii) Calculate Ethan’s profit for the following spot prices at expiration: 25, 30, 35, 40, 45. Solution: (ii) Since Ethan’s profit is (2000) max(ST −35, 0)−9400, Ethan’s profit for the considered spot prices is: if ST = 25, profit = (2000) max(25 − 35, 0) − 9400 = −9400, if ST = 30, profit = (2000) max(30 − 35, 0) − 9400 = −9400, if ST = 35, profit = (2000) max(35 − 35, 0) − 9400 = −9400, if ST = 40, profit = (2000) max(40 − 35, 0) − 9400 = 600, if ST = 45, profit = (2000) max(45 − 35, 0) − 9400 = 10600. 28/112

c





Example 7 Ethan buys a 35–strike call option for XYZ stock for 4.337 per share. The nominal amount of this call option is 2000 shares. The expiration date of this option is 18 months. The annual effective interest rate is 5.5%. (iii) Calculate Ethan’s minimum and maximum profits.

29/112

c





Example 7 Ethan buys a 35–strike call option for XYZ stock for 4.337 per share. The nominal amount of this call option is 2000 shares. The expiration date of this option is 18 months. The annual effective interest rate is 5.5%. (iii) Calculate Ethan’s minimum and maximum profits. Solution: (iii) Since Ethan’s profit is (2000) max(ST −35, 0)−9400, Ethan’s minimum profit is −9400 and Ethan’s maximum profit is ∞.

30/112

c





Example 7 Ethan buys a 35–strike call option for XYZ stock for 4.337 per share. The nominal amount of this call option is 2000 shares. The expiration date of this option is 18 months. The annual effective interest rate is 5.5%. (iv) Find the spot prices at which Ethan’s profit is positive.

31/112

c





Example 7 Ethan buys a 35–strike call option for XYZ stock for 4.337 per share. The nominal amount of this call option is 2000 shares. The expiration date of this option is 18 months. The annual effective interest rate is 5.5%. (iv) Find the spot prices at which Ethan’s profit is positive. Solution: (iv) Since Ethan’s profit is (2000) max(ST −35, 0)−9400, Ethan’s profit is positive if (2000) max(ST − 35, 0) − 9400 > 0, i.e. if ST > 35 + 9400 2000 = 39.7.

32/112

c





Example 7 Ethan buys a 35–strike call option for XYZ stock for 4.337 per share. The nominal amount of this call option is 2000 shares. The expiration date of this option is 18 months. The annual effective interest rate is 5.5%. (v) Calculate the spot price at expiration at which Ethan does not make or lose money on this contract.

33/112

c





Example 7 Ethan buys a 35–strike call option for XYZ stock for 4.337 per share. The nominal amount of this call option is 2000 shares. The expiration date of this option is 18 months. The annual effective interest rate is 5.5%. (v) Calculate the spot price at expiration at which Ethan does not make or lose money on this contract. Solution: (v) Since Ethan’s profit is (2000) max(ST −35, 0)−9400, Ethan breaks even if (2000)(ST − 35) − 9400 = 0, i.e. if ST = 35 + 9400 2000 = 39.7.

34/112

c





Example 7 Ethan buys a 35–strike call option for XYZ stock for 4.337 per share. The nominal amount of this call option is 2000 shares. The expiration date of this option is 18 months. The annual effective interest rate is 5.5%. (vi) Find the spot price at expiration at which Ethan makes an annual effective yield of 4.75%.

35/112

c





Example 7 Ethan buys a 35–strike call option for XYZ stock for 4.337 per share. The nominal amount of this call option is 2000 shares. The expiration date of this option is 18 months. The annual effective interest rate is 5.5%. (vi) Find the spot price at expiration at which Ethan makes an annual effective yield of 4.75%. Solution: (vi) Ethan invests (2000)(4.337) = 8674. If his yield is 4.75%, his payoff is (8674)(1.0475)18/12 = 9300 = (2000) max(ST − 35, 0) and ST = 35 +

9300 = 39.65. 2000

36/112

c





Example 7 Ethan buys a 35–strike call option for XYZ stock for 4.337 per share. The nominal amount of this call option is 2000 shares. The expiration date of this option is 18 months. The annual effective interest rate is 5.5%. (vii) Find the annual effective rate of return earned by Ethan if the spot price at expiration is 38.

37/112

c





Example 7 Ethan buys a 35–strike call option for XYZ stock for 4.337 per share. The nominal amount of this call option is 2000 shares. The expiration date of this option is 18 months. The annual effective interest rate is 5.5%. (vii) Find the annual effective rate of return earned by Ethan if the spot price at expiration is 38. Solution: (vii) Let i be Ethan’s annual effective rate of return. Ethan invests (2000)(4.337) = 8674. His payoff is (2000) max(38 − 35, 0) = 6000. Hence, 8674(1 + i)1.5 = 6000 and i = −21.78538923%.

38/112

c





Example 8 Hannah sells a 35–strike call option for XYZ stock for 4.337 per share. The nominal amount of this call option is 2000 shares. The expiration date of this option is 18 months. The annual effective interest rate is 5.5%. Hannah invests the proceeds of the sale in a zero–coupon bond.

39/112

c





Example 8 Hannah sells a 35–strike call option for XYZ stock for 4.337 per share. The nominal amount of this call option is 2000 shares. The expiration date of this option is 18 months. The annual effective interest rate is 5.5%. Hannah invests the proceeds of the sale in a zero–coupon bond. (i) Calculate Hannah’s profit function.

40/112

c





Example 8 Hannah sells a 35–strike call option for XYZ stock for 4.337 per share. The nominal amount of this call option is 2000 shares. The expiration date of this option is 18 months. The annual effective interest rate is 5.5%. Hannah invests the proceeds of the sale in a zero–coupon bond. (i) Calculate Hannah’s profit function. Solution: (i) Hannah’s profit is − (2000)(max(ST − 35, 0) − 4.337(1.055)1.5 ) =9400 − (2000) max(ST − 35, 0).

41/112

c





Example 8 Hannah sells a 35–strike call option for XYZ stock for 4.337 per share. The nominal amount of this call option is 2000 shares. The expiration date of this option is 18 months. The annual effective interest rate is 5.5%. Hannah invests the proceeds of the sale in a zero–coupon bond. (ii) Calculate Hannah’s profit for the following spot prices at expiration: 25, 30, 35, 40, 45.

42/112

c





Example 8 Hannah sells a 35–strike call option for XYZ stock for 4.337 per share. The nominal amount of this call option is 2000 shares. The expiration date of this option is 18 months. The annual effective interest rate is 5.5%. Hannah invests the proceeds of the sale in a zero–coupon bond. (ii) Calculate Hannah’s profit for the following spot prices at expiration: 25, 30, 35, 40, 45. Solution: (ii) Since Hannah’s profit is 9400 − (2000) max(ST − 35, 0), Hannah’s profit for the considered spot prices is: if ST = 25, profit = 9400 − (2000) max(25 − 35, 0) = 9400, if ST = 30, profit = 9400 − (2000) max(30 − 35, 0) = 9400, if ST = 35, profit = 9400 − (2000) max(35 − 35, 0) = 9400, if ST = 40, profit = 9400 − (2000) max(40 − 35, 0) = −600, if ST = 45, profit = 9400 − (2000) max(45 − 35, 0) = −10600. 43/112

c





Example 8 Hannah sells a 35–strike call option for XYZ stock for 4.337 per share. The nominal amount of this call option is 2000 shares. The expiration date of this option is 18 months. The annual effective interest rate is 5.5%. Hannah invests the proceeds of the sale in a zero–coupon bond. (iii) Calculate Hannah’s minimum and maximum profits.

44/112

c





Example 8 Hannah sells a 35–strike call option for XYZ stock for 4.337 per share. The nominal amount of this call option is 2000 shares. The expiration date of this option is 18 months. The annual effective interest rate is 5.5%. Hannah invests the proceeds of the sale in a zero–coupon bond. (iii) Calculate Hannah’s minimum and maximum profits. Solution: (iii) Since Hannah’s profit is 9400 − (2000) max(ST − 35, 0), Hannah’s minimum profit is −∞ and Hannah’s maximum profit is 9400.

45/112

c





Next we consider the pricing of a call option. The profit of a call option depends on ST , which is random. In the case of uncertain scenarios, an arbitrage portfolio consists of a zero investment portfolio, which shows non–negative payoffs in all scenarios. This implies that if there exists no arbitrage, the profit function of a portfolio is either constantly zero, or its minimum is negative and its maximum positive.

46/112

c





Theorem 4 If there exist no arbitrage, then max(S0 − (1 + i)−T K , 0) < Call(K , T ) < S0 .

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c





Proof: Consider the portfolio consisting of selling a call option and buying the asset. The profit per unit at expiration is ST − max(ST − K , 0) − (S0 − Call(K , T ))(1 + i)T =ST + K − max(ST , K ) − (S0 − Call(K , T ))(1 + i)T = min(ST , K ) − (S0 − Call(K , T ))(1 + i)T . The profit is nondecreasing on ST . The minimum of this portfolio is −(S0 − Call(K , T ))(1 + i)T . The maximum of this portfolio is K − (S0 − Call(K , T ))(1 + i)T . If there exists no arbitrage and the profit function is not constant, the minimum profit is negative and the maximum profit is positive. Hence, −(S0 − Call(K , T ))(1 + i)T < 0 < K − (S0 − Call(K , T ))(1 + i)T which is equivalent to S0 − (1 + i)−T K < Call(K , T ) < S0 . 48/112

c





If the bounds in Theorem 4 do not hold, we can make arbitrage. For example, if the price of the call is bigger than the spot price, we can make money by buying the asset, selling the call and investing the proceeds in a zero–coupon bond. At redemption time, we have the asset which can use to satisfy the requirements of the call.

49/112

c





Example 9 Consider an European call option on a stock worth S0 =32, with expiration date exactly one year from now, and with strike price $30. The risk–free annual rate of interest compounded continuously is r = 5%.

50/112

c





Example 9 Consider an European call option on a stock worth S0 =32, with expiration date exactly one year from now, and with strike price $30. The risk–free annual rate of interest compounded continuously is r = 5%. (i) If the call is worth $3, find an arbitrage portfolio.

51/112

c





Example 9 Consider an European call option on a stock worth S0 =32, with expiration date exactly one year from now, and with strike price $30. The risk–free annual rate of interest compounded continuously is r = 5%. (i) If the call is worth $3, find an arbitrage portfolio. Solution: (i) We have that Call(K , T )−S0 +(1+i)−T K = 3−32+e −0.05 30 = −0.463117265 < 0. We can do arbitrage by buying the call and shorting stock. If the spot price at expiration is more than 30, we buy the stock using the call option. If the spot price at expiration is less than 30, we buy the stock at market price. Any case, we buy stock for min(ST , 30). Hence, the profit is (32 − 3)e 0.05 − min(ST , 30) ≥ (32 − 3)e 0.05 − 30 = 0.4868617949. 52/112

c





Example 9 Consider an European call option on a stock worth S0 =32, with expiration date exactly one year from now, and with strike price $30. The risk–free annual rate of interest compounded continuously is r = 5%. (i) If the call is worth $35, find an arbitrage portfolio.

53/112

c





Example 9 Consider an European call option on a stock worth S0 =32, with expiration date exactly one year from now, and with strike price $30. The risk–free annual rate of interest compounded continuously is r = 5%. (i) If the call is worth $35, find an arbitrage portfolio. Solution: (ii) In this case Call(K , T ) > S0 . We can do arbitrage by selling the call and buying stock. If the spot price at expiration is more than 30, we sell the stock to the call option holder. If the spot price at expiration is less than 30, we sell the stock at the market price. In any case, we sell stock for min(ST , 30). The profit is (35 − 32)e 0.05 + min(ST , 30) ≥ (35 − 32)e 0.05 = 3.153813289.

54/112

c





A call option is a way to buy stock in the future. A long forward is another way to buy stock in the future. Buying a call option, you are guaranteed that the price you pay is not bigger than the strike price. If you buy a call option, you can buy the asset at expiration for min(ST , K ). The baker in the example in Section 7.1, instead of buying a long forward for F0,T , he can buy a call option to hedge against high wheat prices. Doing this we will be able to buy wheat at time T for min(ST , K ). The cost of this investment strategy is Call(K , T )e rT + min(ST , K ). Recall that F0,T is the price of a forward contract with delivery in T years. The profit of a long forward is ST − F0,T . The minimum profit of a long forward is −F0,T . The maximum profit of a long forward is ∞. 55/112

c





Example 10 Joseph buys a one–year long forward for 100 shares of a stock at $74 per share. Samantha buys a call option of 100 shares of XYZ stock for $76 per share. The exercise date is one year from now. The risk free effective annual interest rate is 6%. The premium of this call is $6.4133 per share.

56/112

c





Example 10 Joseph buys a one–year long forward for 100 shares of a stock at $74 per share. Samantha buys a call option of 100 shares of XYZ stock for $76 per share. The exercise date is one year from now. The risk free effective annual interest rate is 6%. The premium of this call is $6.4133 per share. (i) Make a table with Joseph’s profit and Samantha’s profit when the spot price at expiration is $50, $70, $90 and $110. Compare these profits.

57/112

c





Example 10 Joseph buys a one–year long forward for 100 shares of a stock at $74 per share. Samantha buys a call option of 100 shares of XYZ stock for $76 per share. The exercise date is one year from now. The risk free effective annual interest rate is 6%. The premium of this call is $6.4133 per share. (i) Make a table with Joseph’s profit and Samantha’s profit when the spot price at expiration is $50, $70, $90 and $110. Compare these profits. Solution: (i) Joseph’s profit is given by the formula (100)(ST − 74). Samantha’s profit is 100 max(0, ST − K ) − 100Call(K , T )(1 + i)T = 100 max(0, ST − 76) − (100)(6.4133)(1.06) = 100 max(0, ST − 76) − 679.81. 58/112

c





Example 10 Joseph buys a one–year long forward for 100 shares of a stock at $74 per share. Samantha buys a call option of 100 shares of XYZ stock for $76 per share. The exercise date is one year from now. The risk free effective annual interest rate is 6%. The premium of this call is $6.4133 per share. (i) Make a table with Joseph’s profit and Samantha’s profit when the spot price at expiration is $50, $70, $90 and $110. Compare these profits. Solution: (i) (continuation) Joseph’s profit Samantha’s profit Spot Price

−2400 −679.81 50

−400 −679.81 70

1600 720.19 90

3600 2720.19 110

For high spot prices at expiration, Samantha’s profits are smaller than John’s profits. For low prices, Samantha’s losses are smaller than Joseph’s losses. 59/112

c





Example 10 Joseph buys a one–year long forward for 100 shares of a stock at $74 per share. Samantha buys a call option of 100 shares of XYZ stock for $76 per share. The exercise date is one year from now. The risk free effective annual interest rate is 6%. The premium of this call is $6.4133 per share. (ii) Calculate Joseph’s profit and Samantha’s minimum and maximum payoffs.

60/112

c





Example 10 Joseph buys a one–year long forward for 100 shares of a stock at $74 per share. Samantha buys a call option of 100 shares of XYZ stock for $76 per share. The exercise date is one year from now. The risk free effective annual interest rate is 6%. The premium of this call is $6.4133 per share. (ii) Calculate Joseph’s profit and Samantha’s minimum and maximum payoffs. Solution: (ii) Joseph’s minimum profit is −7400. Joseph’s maximum profit is ∞. Samantha’s minimum profit is −679.81. Samantha’s maximum profit is ∞.

61/112

c





Example 10 Joseph buys a one–year long forward for 100 shares of a stock at $74 per share. Samantha buys a call option of 100 shares of XYZ stock for $76 per share. The exercise date is one year from now. The risk free effective annual interest rate is 6%. The premium of this call is $6.4133 per share. (iii) Which is the minimum spot price at expiration at which Joseph makes a profit? Which is the minimum spot price at expiration at which Samantha makes a profit?

62/112

c





Example 10 Joseph buys a one–year long forward for 100 shares of a stock at $74 per share. Samantha buys a call option of 100 shares of XYZ stock for $76 per share. The exercise date is one year from now. The risk free effective annual interest rate is 6%. The premium of this call is $6.4133 per share. (iii) Which is the minimum spot price at expiration at which Joseph makes a profit? Which is the minimum spot price at expiration at which Samantha makes a profit? Solution: (iii) Joseph is even if ST = 74. Samantha is even if 100(ST −76)−679.81 = 0, i.e. ST = 76+(679.81/100) = 82.7981.

63/112

c





Example 10 Joseph buys a one–year long forward for 100 shares of a stock at $74 per share. Samantha buys a call option of 100 shares of XYZ stock for $76 per share. The exercise date is one year from now. The risk free effective annual interest rate is 6%. The premium of this call is $6.4133 per share. (iv) Draw the graph of the profit versus the spot price at expiration for Joseph and Samantha.

64/112

c





Example 10 Joseph buys a one–year long forward for 100 shares of a stock at $74 per share. Samantha buys a call option of 100 shares of XYZ stock for $76 per share. The exercise date is one year from now. The risk free effective annual interest rate is 6%. The premium of this call is $6.4133 per share. (iv) Draw the graph of the profit versus the spot price at expiration for Joseph and Samantha. Solution: (iv) The graphs of (long forward) Joseph’s profit and (purchased call) Samantha’s profit are in Figure 3.

65/112

c





Example 10 Joseph buys a one–year long forward for 100 shares of a stock at $74 per share. Samantha buys a call option of 100 shares of XYZ stock for $76 per share. The exercise date is one year from now. The risk free effective annual interest rate is 6%. The premium of this call is $6.4133 per share. (v) Find the spot price at redemption at which both profits are equal.

66/112

c





Example 10 Joseph buys a one–year long forward for 100 shares of a stock at $74 per share. Samantha buys a call option of 100 shares of XYZ stock for $76 per share. The exercise date is one year from now. The risk free effective annual interest rate is 6%. The premium of this call is $6.4133 per share. (v) Find the spot price at redemption at which both profits are equal. Solution: (v) We solve (100)(ST − 74) = 100 max(0, ST − 76) − 679.81 for ST . There is not solution with ST ≥ 76. If ST < 76 we have the equation (100)(ST − 74) = −679.81, or ST = 74 − 6.7981 = 67.2019.

67/112

c





Figure 3: Example 10. Profit for long forward and purchased call. 68/112

c





A purchased call option reduces losses over a long forward. Notice that in Figure 3 the losses for a long forward holder can be big if the spot price at redemption is small. A call option is an insured long position in an asset. In return for not having large losses, the possible profits in a call option are smaller. The spot price needed to make money is bigger for a purchased call than for a long forward. The profit for the call option holder is positive if ST > K + Call(K , T )(1 + i)T . The profit for the long forward is positive if ST > F0,T . By Theorem 5, K + Call(K , T )(1 + i)T > F0,T . To make a positive profit, a call option holder needs a bigger increase on the spot price than a long forward holder. 69/112

c





Theorem 5 If there exists no arbitrage, then (1 + i)−T max(F0,T − K , 0) < Call(K , T ) < (1 + i)−T F0,T .

70/112

c





Proof: Suppose that you enter into a short forward contract and you buy a call option. Both contracts have the same expiration time and nominal amount. At expiration, the profit of this strategy is F0,T − ST + max(ST − K , 0) − (1 + i)T Call(K , T ) =F0,T + max(−K , −ST ) − (1 + i)T Call(K , T ) =F0,T − min(K , ST ) − (1 + i)T Call(K , T ). This profit function is increasing on ST and it not constant. The minimum profit of this portfolio is F0,T − K − (1 + i)T Call(K , T ). The maximum profit of this portfolio is F0,T − (1 + i)T Call(K , T ). If there is no arbitrage, F0,T − K − (1 + i)T Call(K , T ) < 0 < F0,T − (1 + i)T Call(K , T ). 71/112

c





Example 11 The current price of a forward contract for 1000 units of an asset with expiration date two years from now is $120000. The risk–free annual rate of interest compounded continuously is 5%. The price of a two–year 100–strike European call option for 1000 units of the asset is $15000. Find an arbitrage portfolio and its minimum profit.

72/112

c





Example 11 The current price of a forward contract for 1000 units of an asset with expiration date two years from now is $120000. The risk–free annual rate of interest compounded continuously is 5%. The price of a two–year 100–strike European call option for 1000 units of the asset is $15000. Find an arbitrage portfolio and its minimum profit. Solution: Since e −rT (F0,T − K ) = e −(2)(0.05) (120000 − (100)(1000)) =18096.74836 > 15000, the call option is under priced. Consider the portfolio consisting of buying the call and entering into a short forward. The profit is 1000 max(ST − 100, 0) − 15000e (2)(0.05) + 120000 − 1000ST =1000 max(−100, −ST ) + 103422.4362 =103422.4362 − 1000 min(100, ST ). The minimum profit is 103422.4362 − 1000(100) = 3422.4362. c



73/112



Another motive to buy call options is to speculate. Call options allow betting in the increase of the price of a particular asset for a small cash outlay. Buying a call option, a speculator achieves leverage. Call options provide price exposure without having to pay, hold and warehouse the underlying asset. If a speculator believes that an asset price is going to increase and it is right, he can get a much higher yield of return buying a call option than buying the asset.

74/112

c





Example 12 Rachel is a speculator. She anticipates XYZ stock to appreciate from its current level of $130 per share in four months. Rachel buys a four–month 1000–share call option with a strike price of $150 per share and a premium of $1.8074 per share. Luke is also a speculator. He also expects XYZ stock to appreciate and buys XYZ stock at the current market price.

75/112

c





Example 12 Rachel is a speculator. She anticipates XYZ stock to appreciate from its current level of $130 per share in four months. Rachel buys a four–month 1000–share call option with a strike price of $150 per share and a premium of $1.8074 per share. Luke is also a speculator. He also expects XYZ stock to appreciate and buys XYZ stock at the current market price. (i) Find Rachel’s annual effective rate of return in her investment for the following spot prices at expiration 130, 150, 160 and 170.

76/112

c





Example 12 Rachel is a speculator. She anticipates XYZ stock to appreciate from its current level of $130 per share in four months. Rachel buys a four–month 1000–share call option with a strike price of $150 per share and a premium of $1.8074 per share. Luke is also a speculator. He also expects XYZ stock to appreciate and buys XYZ stock at the current market price. (i) Find Rachel’s annual effective rate of return in her investment for the following spot prices at expiration 130, 150, 160 and 170. Solution: (i) Rachel invests (1000)(1.8074) = 1807.4. Four months later, she receives (1000) max(ST − 150, 0). If ST ≤ 150, Rachel loses all her money and her yield of return is −100%. If ST = 160, Rachel receives (1000)(160 − 150) = 10000 3 10000 at expiration. Rachel’s annual rate of return is 1807.4 − 1 = 168.3702647 = 16837.02647%. If ST = 170, Rachel receives (1000)(170 − 150) = 20000 at expiration. Rachel’s annual rate of 20000 3 return is 1807.4 − 1 = 1353.962117 = 135396.2117%. c



77/112



Example 12 Rachel is a speculator. She anticipates XYZ stock to appreciate from its current level of $130 per share in four months. Rachel buys a four–month 1000–share call option with a strike price of $150 per share and a premium of $1.8074 per share. Luke is also a speculator. He also expects XYZ stock to appreciate and buys XYZ stock at the current market price. (ii) Luke sells his stock at the end of four months. Find Luke’s annual effective rate of return in his investment for the spot prices in (i).

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c





Example 12 Rachel is a speculator. She anticipates XYZ stock to appreciate from its current level of $130 per share in four months. Rachel buys a four–month 1000–share call option with a strike price of $150 per share and a premium of $1.8074 per share. Luke is also a speculator. He also expects XYZ stock to appreciate and buys XYZ stock at the current market price. (ii) Luke sells his stock at the end of four months. Find Luke’s annual effective rate of return in his investment for the spot prices in (i). Solution: (ii) Luke invests 130 per share. His annual rate of return 3 ST j satisfies ST = 130(1 + j)1/3 . So, j = 130 − 1. If ST = 130, j = 0%. If ST = 150, j = 53.61857078%. If ST = 160, j = 86.43604916%. If ST = 170, j = 123.6231224%. 79/112

c





Example 12 Rachel is a speculator. She anticipates XYZ stock to appreciate from its current level of $130 per share in four months. Rachel buys a four–month 1000–share call option with a strike price of $150 per share and a premium of $1.8074 per share. Luke is also a speculator. He also expects XYZ stock to appreciate and buys XYZ stock at the current market price. (iii) Compare the rates in (i) and (ii).

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Example 12 Rachel is a speculator. She anticipates XYZ stock to appreciate from its current level of $130 per share in four months. Rachel buys a four–month 1000–share call option with a strike price of $150 per share and a premium of $1.8074 per share. Luke is also a speculator. He also expects XYZ stock to appreciate and buys XYZ stock at the current market price. (iii) Compare the rates in (i) and (ii). Solution: (iii) In the case that XYZ stock does not appreciate, Rachel loses all her money. But in the cases where XYZ stock appreciates, Rachel makes a much higher yield than Luke.

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Let jC be the rate of return which an investor makes buying a call option. Since the payoff per share is max(ST − K , 0), we have that max(ST − K , 0) = (1 + jC )T . Call(K, T) Let jB be the rate of return which an investor makes buying an asset and holding it for T years. Since the payoff per share is ST , we have that ST = (1 + jB )T S0 . We have that jC > jB if ST max(ST − K , 0) > , Call(K, T) S0 which is equivalent to S0 > Call(K, T) and ST >

K Call(K,T) 1 1 Call(K,T) − S0

=

KS0 . S0 − Call(K, T)

We conclude that if ST is large enough, investing in an option call gives a larger yield than buying an asset. c



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Next we consider call options with different strike prices. If 0 < K1 < K2 , then max(ST − K2 , 0) ≤ max(ST − K1 , 0), i.e. the payoff of a K1 –strike call option is higher than the payoff of a K2 –strike call option (see Figure 4). Hence, the price of the call is bigger for the call with smaller strike price (see Theorem 6).

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Example 13 The current price of XYZ stock is $75 per share. The annual effective rate of interest is 5%. The redemption time is one year from now. Draw the payoff and profit diagrams for the buyer of:

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Example 13 The current price of XYZ stock is $75 per share. The annual effective rate of interest is 5%. The redemption time is one year from now. Draw the payoff and profit diagrams for the buyer of: (i) a $70 strike call option with a premium of $10.755.

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Example 13 The current price of XYZ stock is $75 per share. The annual effective rate of interest is 5%. The redemption time is one year from now. Draw the payoff and profit diagrams for the buyer of: (i) a $70 strike call option with a premium of $10.755. Solution: (i) The payoff is max(ST − 70, 0). The diagram of this payoff is in Figure 4. The profit is max(ST − 70, 0) − (10.755)(1.05) = max(ST − 70, 0) − 11.29275. The diagram of this profit is in Figure 5.

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Example 13 The current price of XYZ stock is $75 per share. The annual effective rate of interest is 5%. The redemption time is one year from now. Draw the payoff and profit diagrams for the buyer of: (ii) a $80 strike call option with a premium of $5.445.

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c





Example 13 The current price of XYZ stock is $75 per share. The annual effective rate of interest is 5%. The redemption time is one year from now. Draw the payoff and profit diagrams for the buyer of: (ii) a $80 strike call option with a premium of $5.445. Solution: (ii) The payoff is max(ST − 80, 0). The diagram of this payoff is in Figure 4. The profit is max(ST − 80, 0) − (5.445)(1.05) = max(ST − 80, 0) − 5.71725. The diagram of this profit is in Figure 5.

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Example 13 The current price of XYZ stock is $75 per share. The annual effective rate of interest is 5%. The redemption time is one year from now. Draw the payoff and profit diagrams for the buyer of: (iii) Find the spot price at redemption at which both profits are equal.

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Example 13 The current price of XYZ stock is $75 per share. The annual effective rate of interest is 5%. The redemption time is one year from now. Draw the payoff and profit diagrams for the buyer of: (iii) Find the spot price at redemption at which both profits are equal. Solution: (iii) The profit amounts are equal for some ST ∈ (70, 80). So, ST − 70 − 11.29275 = max(ST − 70, 0) − 11.29275 = max(ST − 80, 0) − 5.71725 = −5.71725 and ST = 70 + 11.29275 − 5.71725 = 75.5755.

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Figure 4: Example 13. Payoff for two calls with different strikes. 91/112

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Figure 5: Example 13. Profit for two calls with different strikes. 92/112

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Theorem 6 If 0 < K1 < K2 , then Call(K2 , T ) ≤ Call(K1 , T ) ≤ Call(K2 , T ) + (K2 − K1 )e −rT .

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Proof. We have that max(ST − K2 , 0) ≤ max(ST − K1 , 0) = K2 − K1 + max(ST − K2 , K1 − K2 ) ≤K2 − K1 + max(ST − K2 , 0). In other words, (i) The payoff for a K2 –strike call is smaller than the payoff for a K1 –strike call. (ii) The payoff for a K1 –strike call is smaller than K2 − K1 plus the payoff for a K2 –strike call. Hence, if there exist no arbitrage, then Call(K2 , T ) ≤ Call(K1 , T ) ≤ Call(K2 , T ) + (K2 − K1 )e −rT .

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Example 14 Consider two European call options on a stock worth S0 =32, both with expiration date exactly two years from now and the same nominal amount. The risk–free annual rate of interest compounded continuously is 5%. One call option has strike price $30 and the other one $35. The price of the 30–strike call is 7.

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Example 14 Consider two European call options on a stock worth S0 =32, both with expiration date exactly two years from now and the same nominal amount. The risk–free annual rate of interest compounded continuously is 5%. One call option has strike price $30 and the other one $35. The price of the 30–strike call is 7. (i) Suppose that the price of the 35–strike call option is 8, find an arbitrage portfolio.

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Example 14 Consider two European call options on a stock worth S0 =32, both with expiration date exactly two years from now and the same nominal amount. The risk–free annual rate of interest compounded continuously is 5%. One call option has strike price $30 and the other one $35. The price of the 30–strike call is 7. (i) Suppose that the price of the 35–strike call option is 8, find an arbitrage portfolio. Solution: (i) Here, Call(35, T ) ≤ Call(30, T ) does not hold. We can do arbitrage by a buying a 30–strike call option and selling a 35–strike call option, both for the same nominal amount. The profit per share is max(ST − 30, 0) − max(ST − 35, 0) + (8 − 7)e 0.05 ≥(8 − 7)e 0.05 = 1.051271096.

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Example 14 Consider two European call options on a stock worth S0 =32, both with expiration date exactly two years from now and the same nominal amount. The risk–free annual rate of interest compounded continuously is 5%. One call option has strike price $30 and the other one $35. The price of the 30–strike call is 7. (ii) Suppose that the price of the 35–strike call option is 1, find an arbitrage portfolio.

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Example 14 Consider two European call options on a stock worth S0 =32, both with expiration date exactly two years from now and the same nominal amount. The risk–free annual rate of interest compounded continuously is 5%. One call option has strike price $30 and the other one $35. The price of the 30–strike call is 7. (ii) Suppose that the price of the 35–strike call option is 1, find an arbitrage portfolio. Solution: (ii) We have that Call(K2 , T ) − Call(K1 , T ) + (K2 − K1 )e −rT =1 − 7 + (35 − 30)e −0.05 = −1.243852877 < 0. We can do arbitrage by buying a 35–strike call option and selling a 30–strike call option.

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Example 14 Consider two European call options on a stock worth S0 =32, both with expiration date exactly two years from now and the same nominal amount. The risk–free annual rate of interest compounded continuously is 5%. One call option has strike price $30 and the other one $35. The price of the 30–strike call is 7. (ii) Suppose that the price of the 35–strike call option is 1, find an arbitrage portfolio. Solution: (ii) (continuation) We can do arbitrage by buying a 35– strike call option and selling a 30–strike call option. The profit per share is max(ST − 35, 0) − max(ST − 30, 0) + (7 − 1)e 0.05 = − 5 + max(ST − 30, 5) − max(ST − 30, 0) + (7 − 1)e 0.05 ≥ − 5 + (7 − 1)e 0.05 = −5 + 6.307626578 = 1.307626578.

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The premium of a call option of an asset depends on several factors, like asset price, interest rate, expiration time, strike price, and asset price variability. We have the following rules of thump for the price of a call: I

Higher asset prices lead to higher call option prices.

I

Higher strike prices lead to lower call option prices.

I

Higher interest rates lead to higher call option prices.

I

Higher expiration time leads to higher call option prices.

I

Higher variation of an asset price leads to higher call option prices.

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Since the call option buyer’s payoff decreases as the strike increases, the (price) premium of a call option decrease as the strikes increases. Hence, between two call options with different strike prices: (i) The call option with smaller strike price has a bigger premium. (ii) If the spot price is low enough, both call options substain a loss. The loss is bigger for the call option with the smaller strike price. (iii) If the spot price is high enough, both call options have a positive profit. The profit is bigger for the call option with the smaller strike price. We can check the previous assertions analytically using Theorem 6.

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The strike price is paid at the expiration time, as higher the interest rate is as higher the call option premium is. As higher the expiration time as higher the call option premium is. The greater the past variability of the price of an asset is as more likely is that the option will be exercised. So, higher variation of an asset price leads to higher call option prices.

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The common method to find the price of a call option of a stock is to use the Black–Scholes formula1 Call(K , T ) = S0 e −δT Φ(d1 ) − Ke −rT Φ(d2 ) where d1 =

log(S0 /K ) + (r − δ + σ 2 /2)T √ ; σ T √ d2 = d1 − σ T ;

S0 is the current price of the stock; K is the strike price; r is the risk free continuously compounded annual interest rate; δ is the continuous rate of dividend payments; T is the expiration time in years of the option; σ is the implied volatility for the underlying asset and Φ the cumulative distribution function of a standard normal distribution. 1

In 1973, Fischer Black and Myron Scholes published a paper presenting the pricing formula for call and put options. c



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Table 1 shows the premium of a call option for different strike prices. We have used S0 = 75, T = 1, σ = 0.20, δ = 0, r = ln(1.05). Table 1:

K Call(K , T )

65 14.31722

70 10.75552

75 7.78971

80 5.444947

85 3.680736

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Example 15 The current price of XYZ stock is $75 per share. The annual effective rate of interest is 5%. The redemption time is one year from now. The price of stock one year from now is $73.5. Calculate the profit per share at expiration for the holder of each one of the call options in Table 1.

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Solution: The profit is max(ST − K , 0) − (1.05)Call(K , T ) = max(73.5 − K , 0) − (1.05)Call(K , T ). The corresponding profits are: if K = 65, max(73.5 − 65, 0) − (1.05)(14.31722) = −6.533081, if K = 70, max(73.5 − 70, 0) − (1.05)(10.75552) = −7.793296, if K = 75, max(73.5 − 75, 0) − (1.05)(7.78971) = −8.1791955, if K = 80, max(73.5 − 80, 0) − (1.05)(5.444947) = −5.71719435, if K = 85, max(73.5 − 85, 0) − (1.05)(3.680736) = −3.8647728.

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If K is very small, the call option will almost certainly be executed. Hence, if K is very small, Call(K , T ) = F0,T , i.e. lim Call(K , T ) = F0,T . If K is very large, the call option will K →0+

almost certainly not be executed. Hence, lim Call(K , T ) = 0. As K →∞

a function on K , Call(K , T ) is a decreasing function with lim Call(K , T ) = F0,T and limK →∞ Call(K , T ) = 0. Figure 6 K →0+

shows the graph of Call(K , T ) as a function of T .

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Figure 6: Example 16. Graph of Call(K , T ) as a function of K . 109/112

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Example 16 Using the Black–Scholes formula with T = 1, S0 = 100, T = 1, σ = 0.25, r = ln(1.06) and δ = 0.0, the following table of call option premiums was obtained: Call(K , T ) K

76.4150 25

52.8366 50

30.0399 75

12.7562 100

4.1341 125

0.8417 150

0.2672 175

0.0605 200

Figure 6 shows the graph of this function. When we consider Call(K , T ) as function of T . If T is small enough, then the option will be exercised if S0 > K with a profit of S0 − K . Hence, if S0 > K , lim Call(K , T ) = S0 − K . Notice T →0+

that by buying the call option for Call(K , T ), we buy an asset worth S0 for K . If T is small enough and S0 < K , the option is not exercised and his value is zero, i.e. lim Call(K , T ) = 0. T →0+

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An option is in–the–money option if it would have a positive payoff if exercised immediately. An option is out–the–money option if it would have a negative payoff if exercised immediately. An option is at–the–money option if it would have a zero payoff if exercised immediately. The previous definition hold for both call and put options. Put options will considered shortly. For a purchased call option, we have

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I I I


The purchased call option is in–the–money, if S0 > K . The purchased call option is out–the–money, if S0 < K . The purchased call option is at–the–money, if S0 = K .

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Manual for SOA Exam FM/CAS Exam 2. Chapter 7. Derivatives markets. Section 7.5. Put options. c



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Section 7.5. Put options.

Put options Definition 1 A put option is a financial contract which gives the (holder) owner the right, but not the obligation, to sell a specified amount of a given security at a specified price at a specified time.

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Put options Definition 1 A put option is a financial contract which gives the (holder) owner the right, but not the obligation, to sell a specified amount of a given security at a specified price at a specified time. The put option owner exercises the option by selling the asset at the specified call price to the put writer. A put option is executed only if the put owner decides to do so. A put option owner executes a put option only when it benefits him, i.e. when the specified call price is bigger than the current (market value) spot price. Since the owner of a put option can make money if the option is exercised, put options are sold. The owner of the put option must pay to its counterpart for holding a put option. The price of a put option is called its premium. 3/83

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I

I I I I I I I


The (owner) buyer of a put option is called the option put holder. The holder of a put option is said to have a long put position. The seller of a put option is called the option put writer. The writer of a put is said to have a short put position. Assets used in put options are in commodities, currency exchange, stock shares and stock indices. A put option needs to specify the type and quality of the underlying. The asset used in the put option is called the underlier or underlying asset. The amount of the underlying asset to which the put option applies is called the notional amount. The specified price of an asset in a put option is called the strike price, or exercise price. A forward contract forces the buyer and seller to execute the sale. A put option is executed only if the put holder decides to do so.

c



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I

For an European option, the exercise of the option must occur at a certain time (the expiration date).

I

For an American option, the exercise of the option must occur any time by the expiration date.

I

For a Bermudan option, the buyer can exercise the call option during specified periods.

Unless say otherwise, we will assume that an option is an European option. European options are simpler and easier to study.

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Example 1 John buys a six–month put option for 150 shares with a strike price of $45 per share. (i) If the price per share six months from now is $40, John sells 150 shares to the put option writer for (150)(45) = 6750. Since the market value of these 150 shares is (150)(40) = 6000. John makes (before expenses) 6750 − 6000 = 750 on this contract. (ii) If the price per share six months from now is $50, John does not exercise the put option.

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I


The put option buyer’s payoff per share is ( K − ST if ST < K , max(K − ST , 0) = 0 if ST ≥ K , where K is the strike price and ST is the spot price at redemption.

I

The put option writer’s payoff per share is ( −(K − ST ) if ST < K , − max(K − ST , 0) = 0 if ST ≥ K ,

Figure 1 shows the graph of the payoff of a put option. A put option is a zero–sum game.

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6

6

max(0, K − ST )

− max(0, K − ST )

K @ @ @ @

-

-

ST

ST −K

Payoff for the put option holder

Payoff for the put option writer

Figure 1: Payoffs of a put option 8/83

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Recall: I The put option holder’s payoff is max(0, K − ST ). I

The put option writer’s payoff is − max(0, K − ST ).

We get from Figure 1 that: I The minimum payoff for the put option holder is 0. The maximum payoff for the put option holder is K . I The minimum payoff for the put option writer is −K . The maximum payoff for the put option writer is 0.

put option holder put option writer

minimum payoff 0 −K

maximum payoff K 0 9/83

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Example 2 Daniel buys a 55–strike put option on XYZ stock with a nominal amount of 5000 shares. The expiration date is 6 months from now. The nominal amount of the put option is 5000 shares of XYZ stock. (i) Calculate Daniel’s payoff for the following spot prices per share at expiration: 40, 45, 55, 60, 60. (ii) Calculate Daniel’s minimum and maximum payoffs.

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Example 2 Daniel buys a 55–strike put option on XYZ stock with a nominal amount of 5000 shares. The expiration date is 6 months from now. The nominal amount of the put option is 5000 shares of XYZ stock. (i) Calculate Daniel’s payoff for the following spot prices per share at expiration: 40, 45, 55, 60, 60. (ii) Calculate Daniel’s minimum and maximum payoffs. Solution: (i) Daniel’s payoff is 5000 max(55 − ST , 0). The corresponding payoffs are: if ST = 40, payoff = (5000) max(55 − 40, 0) = 75000, if ST = 45, payoff = (5000) max(55 − 45, 0) = 50000, if ST = 50, payoff = (5000) max(55 − 50, 0) = 25000, if ST = 55, payoff = (5000) max(55 − 55, 0) = 0, if ST = 60, payoff = (5000) max(55 − 60, 0) = 0. (ii) Daniel’s minimum payoff is zero. Daniel’s maximum payoff is c

2009. Miguel Arcones. All rights reserved. Manual for SOA Exam FM/CAS Exam 2. (5000)(55) =A.275000.

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Example 3 Isabella sells a 55–strike put option on XYZ stock. The expiration date is 18 months from now. The nominal amount of the put option is 10000 shares of XYZ stock. (i) Calculate Isabella’s payoff for the following spot prices per share at expiration: 40, 45, 55, 60, 60. (ii) Calculate Isabella’s minimum and maximum payoffs.

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Example 3 Isabella sells a 55–strike put option on XYZ stock. The expiration date is 18 months from now. The nominal amount of the put option is 10000 shares of XYZ stock. (i) Calculate Isabella’s payoff for the following spot prices per share at expiration: 40, 45, 55, 60, 60. (ii) Calculate Isabella’s minimum and maximum payoffs. Solution: (i) Isabella’s payoff is −10000 max(55 − ST , 0). The corresponding payoffs are: if ST = 40, payoff = −(10000) max(55 − 40, 0) = −150000, if ST = 45, payoff = −(10000) max(55 − 45, 0) = −100000, if ST = 50, payoff = −(10000) max(55 − 50, 0) = −50000, if ST = 55, payoff = −(10000) max(55 − 55, 0) = 0, if ST = 60, payoff = −(10000) max(55 − 60, 0) = 0. (ii) Isabella’s minimum payoff is −(10000)(55) = −550000. Isabella’s maximum payoff is zero. c



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Let Put(K , T ) be the premium per unit paid of a put option with strike price K and expiration time T years. Notice that Put(K , T ) > 0. Let i be the risk free annual effective rate of interest. The put option holder’s profit is max(K − ST , 0) − Put(K , T )(1 + i)T ( K − ST − Put(K , T )(1 + i)T if ST < K , = −Put(K , T )(1 + i)T if ST ≥ K . Put(K , T )(1 + i)T is the future value at time T of the purchase price. The put option writer’s profit is − max(K − ST , 0) + Put(K , T )(1 + i)T ( −K + ST + Put(K , T )(1 + i)T if ST < K , = Put(K , T )(1 + i)T if ST ≥ K . Figure 2 shows a graph of the put profit. 14/83

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put holder put writer

put holder put writer


profit max(K − ST , 0) − Put(K , T )e rT − max(K − ST , 0) + Put(K , T )e rT

minimum profit −Put(K , T )e rT −K + Put(K , T )e rT

maximum profit K − Put(K , T )e rT Put(K , T )e rT

Figure 2 shows a graph of the put profit as a function of ST .

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6 max(K − ST , 0) − P(1 + i)T


6 P(1 + i)T − max(K − ST , 0)

K − P(1 + i)T @ @

K @

ST -

@

−P(1 + i)T

P(1 + i)T K

-

ST

P(1 + i)T − K

Profit for the put option holder Profit for the put option writer

Figure 2: Profit of a put option 16/83

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Notice that the put option holder’s profit as a function of ST is nonincreasing. The put option holder benefits from a decrease on the spot price. The minimum of the put option holder’s profit is −Put(K , T )(1 + i)T . The maximum of the put option’s holder profit is K − Put(K , T )(1 + i)T . If there exists no arbitrage −Put(K , T )(1 + i)T < 0 < K − Put(K , T )(1 + i)T , which is equivalent to 0 < Put(K , T ) < K (1 + i)−T .

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Theorem 1 If there exists no arbitrage, then max((1 + i)−T K − S0 , 0) < Put(K , T ) < K (1 + i)−T .

Proof. Consider the portfolio consisting of buying an asset and a put option on this asset, both for the same notional amount. The profit at expiration is ST + max(K − ST , 0) − S0 (1 + i)T − Put(K , T )(1 + i)T = max(K , ST ) − (Put(K , T ) + S0 )(1 + i)T . The maximum profit is ∞. The minimum profit is K − (Put(K , T ) + S0 )(1 + i)T . If there exists no arbitrage K − (Put(K , T ) + S0 )(1 + i)T < 0, which is equivalent to (1 + i)−T K − S0 < Put(K , T ). From this bound and the bounds before the theorem, the claim follows. c



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Example 4 The current price of XYZ stock is 160 per share. The annual effective interest rate is 7%. The price of a one–year European 200–strike put option for XYZ stock is $20 per share. Find an arbitrage strategy and the minimum profit per share.

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Example 4 The current price of XYZ stock is 160 per share. The annual effective interest rate is 7%. The price of a one–year European 200–strike put option for XYZ stock is $20 per share. Find an arbitrage strategy and the minimum profit per share. Solution: We have that Put(K , T ) + S0 − (1 + i)−T K = 20 + 160 − (200)(1.07)−1 = − 6.91588785 < 0. The put premium is too low. Consider the portfolio consisting of buying the put and the stock, both for the same nominal amount. The profit per share is max(200−ST , 0)−20(1.07)+ST −160(1.07) = max(200, ST )−192.6. The minimum profit per share is 200 − 192.6 = 7.4. 20/83

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Example 5 The current price of XYZ stock is 160 per share. The annual effective interest rate is 7%. The price of a one–year European 200–strike put option for XYZ stock is $190 per share. Find an arbitrage strategy and the minimum profit per share.

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Example 5 The current price of XYZ stock is 160 per share. The annual effective interest rate is 7%. The price of a one–year European 200–strike put option for XYZ stock is $190 per share. Find an arbitrage strategy and the minimum profit per share. Solution: We have that (1 + i)−T K − Put(K , T ) = (200)(1.07)−1 − 190 = − 3.08411215 < 0. The put is overpriced. Consider the portfolio consisting of selling the put. The profit per share is 190(1.07) − max(200 − ST , 0). The minimum profit per share is 190(1.07) − 200 = 3.3.

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The profit of a put option holder is positive if max(K − ST , 0) − Put(K , T )(1 + i)T > 0, which is equivalent to K − Put(K , T )(1 + i)T > ST . If K − Put(K , T )(1 + i)T < ST , the put option holder’s profit is negative.

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Example 6 Ashley buys a 85–strike put option for 4.3185816 per share. The nominal amount of the put option is 2500 shares of XYZ stock. The expiration date of this option is one year. The annual interest rate compounded continuously is 5%.

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Example 6 Ashley buys a 85–strike put option for 4.3185816 per share. The nominal amount of the put option is 2500 shares of XYZ stock. The expiration date of this option is one year. The annual interest rate compounded continuously is 5%. (i) Calculate Ashley’s profit function.

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Example 6 Ashley buys a 85–strike put option for 4.3185816 per share. The nominal amount of the put option is 2500 shares of XYZ stock. The expiration date of this option is one year. The annual interest rate compounded continuously is 5%. (i) Calculate Ashley’s profit function. Solution: (i) Ashley’s profit is (2500)(max(85 − ST , 0) − 4.3185816e 0.05 ) =(2500) max(85 − ST , 0) − 11350.

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Example 6 Ashley buys a 85–strike put option for 4.3185816 per share. The nominal amount of the put option is 2500 shares of XYZ stock. The expiration date of this option is one year. The annual interest rate compounded continuously is 5%. (ii) Calculate Ashley’s profit for the following spot prices at expiration: 75, 80, 85, 90, 95.

27/83

c





Example 6 Ashley buys a 85–strike put option for 4.3185816 per share. The nominal amount of the put option is 2500 shares of XYZ stock. The expiration date of this option is one year. The annual interest rate compounded continuously is 5%. (ii) Calculate Ashley’s profit for the following spot prices at expiration: 75, 80, 85, 90, 95. Solution: (ii) The profits corresponding to the considered spot prices are: if ST = 75, profit = (2500) max(85 − 75, 0) − 11350 = 13650, if ST = 80, profit = (2500) max(85 − 80, 0) − 11350 = 1150, if ST = 85, profit = (2500) max(85 − 85, 0) − 11350 = −11350, if ST = 90, profit = (2500) max(85 − 90, 0) − 11350 = −11350, if ST = 95, profit = (2500) max(85 − 95, 0) − 11350 = −11350. 28/83

c





Example 6 Ashley buys a 85–strike put option for 4.3185816 per share. The nominal amount of the put option is 2500 shares of XYZ stock. The expiration date of this option is one year. The annual interest rate compounded continuously is 5%. (iii) Calculate Ashley’s minimum and maximum profits.

29/83

c





Example 6 Ashley buys a 85–strike put option for 4.3185816 per share. The nominal amount of the put option is 2500 shares of XYZ stock. The expiration date of this option is one year. The annual interest rate compounded continuously is 5%. (iii) Calculate Ashley’s minimum and maximum profits. Solution: (iii) Ashley’s minimum profit is −11350. Ashley’s maximum profit is (2500)(85) − 11350 = 201150.

30/83

c





Example 6 Ashley buys a 85–strike put option for 4.3185816 per share. The nominal amount of the put option is 2500 shares of XYZ stock. The expiration date of this option is one year. The annual interest rate compounded continuously is 5%. (iv) Calculate the spot prices at which Ashley’s profit is positive.

31/83

c





Example 6 Ashley buys a 85–strike put option for 4.3185816 per share. The nominal amount of the put option is 2500 shares of XYZ stock. The expiration date of this option is one year. The annual interest rate compounded continuously is 5%. (iv) Calculate the spot prices at which Ashley’s profit is positive. Solution: (iv) Ashley’s profit is positive if (2500) max(85 − ST , 0) − 11350 > 0, which is equivalent to ST < 85 − 11350 2500 = 80.46.

32/83

c





Example 6 Ashley buys a 85–strike put option for 4.3185816 per share. The nominal amount of the put option is 2500 shares of XYZ stock. The expiration date of this option is one year. The annual interest rate compounded continuously is 5%. (v) Calculate the spot price at expiration at which Ashley breaks even.

33/83

c





Example 6 Ashley buys a 85–strike put option for 4.3185816 per share. The nominal amount of the put option is 2500 shares of XYZ stock. The expiration date of this option is one year. The annual interest rate compounded continuously is 5%. (v) Calculate the spot price at expiration at which Ashley breaks even. Solution: (v) Ashley breaks even if (2500) max(85 − ST , 0) − 11350 = 0, i.e. if ST = 85 − 11350 2500 = 80.46.

34/83

c





Example 6 Ashley buys a 85–strike put option for 4.3185816 per share. The nominal amount of the put option is 2500 shares of XYZ stock. The expiration date of this option is one year. The annual interest rate compounded continuously is 5%. (vi) Calculate Ashley’s annual yield in her investment for the spot prices at expiration in (ii).

35/83

c





Example 6 Ashley buys a 85–strike put option for 4.3185816 per share. The nominal amount of the put option is 2500 shares of XYZ stock. The expiration date of this option is one year. The annual interest rate compounded continuously is 5%. (vi) Calculate Ashley’s annual yield in her investment for the spot prices at expiration in (ii). Solution: (vi) Ashley invests (2500)(4.3185816) = 10796.454. Ashley’s return is (2500) max(85 − ST , 0). Let j be Ashley’s annual yield. Then, 10796.454(1 + j)0.5 = (2500) max(85 − ST , 0). 2 max(85−ST ,0) Hence, j = (2500)10796.454 − 1. Therefore, 2 max(85−75,0) if ST = 75, j = (2500)10796.454 − 1 = 436.1888022%, 2 max(85−80,0) if ST = 80, j = (2500)10796.454 − 1 = 34.04720055%, if ST = 85, or 90, or 95, j = −100%. 36/83

c





Example 7 William sells a 85–strike put option for 4.3185816 per share. The nominal amount of the put option is 2500 shares of XYZ stock. The expiration date of the option is one year. The annual interest rate compounded continuously is 5%.

37/83

c





Example 7 William sells a 85–strike put option for 4.3185816 per share. The nominal amount of the put option is 2500 shares of XYZ stock. The expiration date of the option is one year. The annual interest rate compounded continuously is 5%. (i) Calculate William’s profit function.

38/83

c





Example 7 William sells a 85–strike put option for 4.3185816 per share. The nominal amount of the put option is 2500 shares of XYZ stock. The expiration date of the option is one year. The annual interest rate compounded continuously is 5%. (i) Calculate William’s profit function. Solution: (i) William’s profit is (2500)(4.3185816e 0.05 − max(85 − ST , 0)) =11350 − (2500) max(85 − ST , 0).

39/83

c





Example 7 William sells a 85–strike put option for 4.3185816 per share. The nominal amount of the put option is 2500 shares of XYZ stock. The expiration date of the option is one year. The annual interest rate compounded continuously is 5%. (ii) Calculate William’s profit for the following spot prices at expiration: 75, 80, 85, 90, 95.

40/83

c





Example 7 William sells a 85–strike put option for 4.3185816 per share. The nominal amount of the put option is 2500 shares of XYZ stock. The expiration date of the option is one year. The annual interest rate compounded continuously is 5%. (ii) Calculate William’s profit for the following spot prices at expiration: 75, 80, 85, 90, 95. Solution: (ii) William’s profits for the considered spot prices are if ST = 75, profit = 11350 − (2500) max(85 − 75, 0) = −13650, if ST = 80, profit = 11350 − (2500) max(85 − 80, 0) = −1150, if ST = 85, profit = 11350 − (2500) max(85 − 85, 0) = 11350, if ST = 90, profit = 11350 − (2500) max(85 − 90, 0) = 11350, if ST = 95, profit = 11350 − (2500) max(85 − 95, 0) = 11350.

41/83

c





Example 7 William sells a 85–strike put option for 4.3185816 per share. The nominal amount of the put option is 2500 shares of XYZ stock. The expiration date of the option is one year. The annual interest rate compounded continuously is 5%. (iii) Calculate William’s minimum and maximum profits.

42/83

c





Example 7 William sells a 85–strike put option for 4.3185816 per share. The nominal amount of the put option is 2500 shares of XYZ stock. The expiration date of the option is one year. The annual interest rate compounded continuously is 5%. (iii) Calculate William’s minimum and maximum profits. Solution: (iii) William’s minimum profit is 11350 − (2500)(85) = −201150. William’s maximum profit is 11350.

43/83

c





Example 7 William sells a 85–strike put option for 4.3185816 per share. The nominal amount of the put option is 2500 shares of XYZ stock. The expiration date of the option is one year. The annual interest rate compounded continuously is 5%. (iv) Calculate the spot prices at expiration at which William makes a positive profit.

44/83

c





Example 7 William sells a 85–strike put option for 4.3185816 per share. The nominal amount of the put option is 2500 shares of XYZ stock. The expiration date of the option is one year. The annual interest rate compounded continuously is 5%. (iv) Calculate the spot prices at expiration at which William makes a positive profit. Solution: (iv) William makes a positive profit if 11350 − (2500) max(85 − ST , 0) > 0, i.e. if ST > 85 − 11350 2500 = 80.46.

45/83

c





A put option is a way to sell an asset in the future. A short forward is another way to sell an asset in the future.

46/83

c





Example 8 Rachel enters into a short forward contract for 500 shares of XYZ stock for $26.35035 per share. The exercise date is three months from now. The risk free effective annual interest rate is 5.5%. The current price of XYZ stock is $26 per share. Dan buys a put option of 500 shares of XYZ stock with a strike price of $26 per share. The exercise date is three months from now. The premium of this put option is $1.377368 per share.

47/83

c





Example 8 Rachel enters into a short forward contract for 500 shares of XYZ stock for $26.35035 per share. The exercise date is three months from now. The risk free effective annual interest rate is 5.5%. The current price of XYZ stock is $26 per share. Dan buys a put option of 500 shares of XYZ stock with a strike price of $26 per share. The exercise date is three months from now. The premium of this put option is $1.377368 per share. (i) Find Rachel’s and Dan’s profits as a function of the spot price at expiration.

48/83

c





Example 8 Rachel enters into a short forward contract for 500 shares of XYZ stock for $26.35035 per share. The exercise date is three months from now. The risk free effective annual interest rate is 5.5%. The current price of XYZ stock is $26 per share. Dan buys a put option of 500 shares of XYZ stock with a strike price of $26 per share. The exercise date is three months from now. The premium of this put option is $1.377368 per share. (i) Find Rachel’s and Dan’s profits as a function of the spot price at expiration. Solution: (i) The no arbitrage price of XYZ stock is (26)(1.055)0.25 = 26.35035454. Rachel’s profit is (500)(26.35035 − ST ) = 13175.17727 − 500ST .

49/83

c





Example 8 Rachel enters into a short forward contract for 500 shares of XYZ stock for $26.35035 per share. The exercise date is three months from now. The risk free effective annual interest rate is 5.5%. The current price of XYZ stock is $26 per share. Dan buys a put option of 500 shares of XYZ stock with a strike price of $26 per share. The exercise date is three months from now. The premium of this put option is $1.377368 per share. (i) Find Rachel’s and Dan’s profits as a function of the spot price at expiration. Solution: (i) (continuation) Dan’s profit is 500 max(0, K − ST ) − 500Put(K , T )(1 + i)T = 500 max(0, 26 − ST ) − (500)(1.377368)(1.055)0.25 = 500 max(0, 26 − ST ) − 697.9641.

50/83

c





Example 8 Rachel enters into a short forward contract for 500 shares of XYZ stock for $26.35035 per share. The exercise date is three months from now. The risk free effective annual interest rate is 5.5%. The current price of XYZ stock is $26 per share. Dan buys a put option of 500 shares of XYZ stock with a strike price of $26 per share. The exercise date is three months from now. The premium of this put option is $1.377368 per share. (ii) Make a table with Rachel’s and Dan’s profits when the spot price at expiration is $18, $20, $22, $24, $26, $28, $30.

51/83

c





Example 8 Rachel enters into a short forward contract for 500 shares of XYZ stock for $26.35035 per share. The exercise date is three months from now. The risk free effective annual interest rate is 5.5%. The current price of XYZ stock is $26 per share. Dan buys a put option of 500 shares of XYZ stock with a strike price of $26 per share. The exercise date is three months from now. The premium of this put option is $1.377368 per share. (ii) Make a table with Rachel’s and Dan’s profits when the spot price at expiration is $18, $20, $22, $24, $26, $28, $30. Solution: (ii) Rachel’s profit Dan’s profit Spot Price

4175.18 3302.04 18

3175.18 2302.04 20

2175.18 1302.04 22

1175.18 302.04 24

175.18 −697.96 26

−824.82 −697.96 28

−1824.82 −697.96 30

52/83

c





Example 8 Rachel enters into a short forward contract for 500 shares of XYZ stock for $26.35035 per share. The exercise date is three months from now. The risk free effective annual interest rate is 5.5%. The current price of XYZ stock is $26 per share. Dan buys a put option of 500 shares of XYZ stock with a strike price of $26 per share. The exercise date is three months from now. The premium of this put option is $1.377368 per share. (iii) Calculate Rachel’s and Dan’s minimum and maximum payoffs.

53/83

c





Example 8 Rachel enters into a short forward contract for 500 shares of XYZ stock for $26.35035 per share. The exercise date is three months from now. The risk free effective annual interest rate is 5.5%. The current price of XYZ stock is $26 per share. Dan buys a put option of 500 shares of XYZ stock with a strike price of $26 per share. The exercise date is three months from now. The premium of this put option is $1.377368 per share. (iii) Calculate Rachel’s and Dan’s minimum and maximum payoffs. Solution: (iii) Rachel’s minimum profit is −∞. Rachel’s maximum profit is 13175.18. Dan’s minimum profit is −697.96. Dan’s maximum profit is (500)(26) − 697.9641 = 12302.04.

54/83

c





Example 8 Rachel enters into a short forward contract for 500 shares of XYZ stock for $26.35035 per share. The exercise date is three months from now. The risk free effective annual interest rate is 5.5%. The current price of XYZ stock is $26 per share. Dan buys a put option of 500 shares of XYZ stock with a strike price of $26 per share. The exercise date is three months from now. The premium of this put option is $1.377368 per share. (iv) Calculate the spot price at expiration at which Dan and Rachel make the same profit.

55/83

c





Example 8 Rachel enters into a short forward contract for 500 shares of XYZ stock for $26.35035 per share. The exercise date is three months from now. The risk free effective annual interest rate is 5.5%. The current price of XYZ stock is $26 per share. Dan buys a put option of 500 shares of XYZ stock with a strike price of $26 per share. The exercise date is three months from now. The premium of this put option is $1.377368 per share. (iv) Calculate the spot price at expiration at which Dan and Rachel make the same profit. Solution: (iv) Since the profits are equal for some ST > 26, we solve −697.9641 = (500)(26.35035−ST ) and get ST = 26.35035+ 697.9641/500 = 27.74628.

56/83

c





Example 8 Rachel enters into a short forward contract for 500 shares of XYZ stock for $26.35035 per share. The exercise date is three months from now. The risk free effective annual interest rate is 5.5%. The current price of XYZ stock is $26 per share. Dan buys a put option of 500 shares of XYZ stock with a strike price of $26 per share. The exercise date is three months from now. The premium of this put option is $1.377368 per share. (v) Draw the graphs of the profit versus the spot price at expiration for Dan and Rachel.

57/83

c





Example 8 Rachel enters into a short forward contract for 500 shares of XYZ stock for $26.35035 per share. The exercise date is three months from now. The risk free effective annual interest rate is 5.5%. The current price of XYZ stock is $26 per share. Dan buys a put option of 500 shares of XYZ stock with a strike price of $26 per share. The exercise date is three months from now. The premium of this put option is $1.377368 per share. (v) Draw the graphs of the profit versus the spot price at expiration for Dan and Rachel. Solution: (v) The graphs of (short forward) Rachel’s profit and (purchased put option) Dan’s profit are in Figure 3.

58/83

c





Figure 3: Profit for short forward and purchased put option. 59/83

c





A purchased put option reduces losses over a short forward. The profit per unit of a short forward contract is F0,T − ST . The minimum profit for a short forward contract is −∞. The maximum profit for a short forward contract is F0,T . The profit for the put option holder is max(K − ST , 0) − Put(K , T )(1 + i)T . The minimum of the put option holder’s profit is −Put(K , T )(1 + i)T . The maximum of the put option holder’s profit is K − Put(K , T )(1 + i)T . A put option is an insured position in an asset. In return for not having large losses, the possible returns for a put option are smaller than those for a short forward.

60/83

c





Theorem 2 If there exists no arbitrage, then max((1 + i)−T (K − F0,t ), 0) ≤ Put(K , T ) ≤ K (1 + i)−T .

61/83

c





Theorem 2 If there exists no arbitrage, then max((1 + i)−T (K − F0,t ), 0) ≤ Put(K , T ) ≤ K (1 + i)−T .

Proof. Consider the portfolio consisting of entering a long forward contract and buying a put option. The profit at expiration is ST − F0,T + max(K − ST , 0) − Put(K , T )(1 + i)T = max(K , ST ) − F0,T − (Put(K , T ) + S0 )(1 + i)T . The maximum profit is ∞. The minimum profit is K − F0,T − (Put(K , T ) + S0 )(1 + i)T . If there exists no arbitrage K − F0,T − Put(K , T )(1 + i)T < 0. The claim follows from this bound and the bounds in Theorem 1. 62/83

c





Example 9 The current price of a forward of corn is $3.3 per bushel. The annual effective interest rate is 7.5%. The price of a one–year European 3.5–strike put option for corn is $0.18 per bushel. Find an arbitrage strategy and its minimum profit per bushel.

63/83

c





Example 9 The current price of a forward of corn is $3.3 per bushel. The annual effective interest rate is 7.5%. The price of a one–year European 3.5–strike put option for corn is $0.18 per bushel. Find an arbitrage strategy and its minimum profit per bushel. Solution: We have that Put(K , T ) − ((1 + i)−T (K − F0,t ) =0.18 − (1.075)(3.5 − 3.3) = −0.035 < 0. The put premium is too low. Consider the portfolio consisting of entering into a long forward contract and buying a put option, both for the same nominal amount. The profit is ST − 3.3 + max(3.5 − ST , 0) − (0.18)(1.075) = max(3.5, ST ) − 3.3 − (0.18)(1.075) The minimum profit per share is 3.5 − 3.3 − (0.18)(1.075) = 0.0065. c



64/83



Example 10 The payoff of a purchase put option is similar to the one on a policy insurance of some asset. Suppose that you car is worth $20000. In the case of an accident, the insurance company pays you max(20000 − ST , 0), where ST is the price of the car after the accident. If you own stock valued at K and buy a put option with strike price K , the payoff of the put option at expiration time is max(K − ST , 0). The payoff is precisely the loss in value of the stock. There are minor differences between these two examples. In the case of the insurance of car, usually a deductible is applied. If the deductible in your car insurance is $500, the payment by the insurance company is max(20000 − 500 − ST , 0). Since cost of an accident is always positive, ST < 20000. 65/83

c





Theorem 3 If 0 < K1 < K2 , then Put(K1 , T ) ≤ Put(K2 , T ) ≤ Put(K1 , T ) + (K2 − K1 )e −rT .

66/83

c





Proof. We have that max(K1 − ST , 0) ≤ max(K2 − ST , 0) = K2 − K1 + max(K1 − ST , K1 − K2 ) ≤K2 − K1 + max(K1 − ST , 0). In other words, (i) The payoff for a K1 –strike put is smaller than or equal to the payoff for a K2 –strike put. (ii) The payoff for a K2 –strike put is smaller than or equal to (K2 − K1 ) plus the payoff for K1 –strike put. Hence, if there exist no arbitrage, then Put(K1 , T ) ≤ Put(K2 , T ) ≤ Put(K1 , T ) + (K2 − K1 )e −rT .

67/83

c





Example 11 The price of a one–year European 3.5–strike put option for corn is $0.18 per bushel. The price of a one–year European 3.75–strike put option for corn is $0.15 per bushel. The annual effective interest rate is 7.5%. Find an arbitrage strategy and it minimum profit.

68/83

c





Example 11 The price of a one–year European 3.5–strike put option for corn is $0.18 per bushel. The price of a one–year European 3.75–strike put option for corn is $0.15 per bushel. The annual effective interest rate is 7.5%. Find an arbitrage strategy and it minimum profit. Solution: In this case, Put(K1 , T ) ≤ Put(K2 , T ) does not hold. We can do arbitrage by a buying a 3.75–strike put option and selling 3.5–strike put option, both for the same nominal amount. The profit per share is max(3.75 − ST , 0) − max(3.5 − ST , 0) − (0.15 − 0.18)(1.075) ≥ max(3.75, ST ) − max(3.5, ST ) − (0.15 − 0.18)(1.075) ≥(0.18 − 0.15)(1.075) = 0.03225.

69/83

c





Example 12 Consider two European put options on a stock, both with expiration date exactly two years from now. One put option has strike price $85 and the other one $95. The price of the 85–strike put is 8. The price of the 95–strike put option is 20. The risk–free annual rate of interest compounded continuously is 5%. Find an arbitrage portfolio and its minimum profit.

70/83

c





Example 12 Consider two European put options on a stock, both with expiration date exactly two years from now. One put option has strike price $85 and the other one $95. The price of the 85–strike put is 8. The price of the 95–strike put option is 20. The risk–free annual rate of interest compounded continuously is 5%. Find an arbitrage portfolio and its minimum profit. Solution: In this case Put(K2 , T ) ≤ Put(K1 , T ) + (K2 − K1 )e −rT does not hold. Notice that Put(K1 , T )+(K2 −K1 )e −rT = 8+(95−85)e −(2)(0.05) = 17.04837418. We can do arbitrage by buying a 85–strike put option and selling a 95–strike put option. The profit per share is max(85 − ST , 0) − max(95 − ST , 0) + (20 − 8)e (2)(0.05) = max(85 − ST , 0) − 10 − max(85 − ST , −10) + 13.26205102 ≥ − 10 + 13.26205102 = 3.26205102. 71/83

c





We have the following rules of thump for price of a put option of an asset: I

Higher asset prices lead to higher put option prices.

I

Higher strike prices lead to higher put option prices.

I

Higher interest rates lead to higher put option prices.

I

Higher expiration time lead to higher put option prices.

I

Higher variation of an asset price lead to higher put option prices.

72/83

c





Since the put option buyer’s payoff increases as the strike increases, the premium of a put option increases as the strike price increases. Hence, between two put options with different strike prices: (i) The put option with smaller strike price has a smaller price. (ii) If the spot price is low enough, both put options have a positive profit. The loss is bigger for put option with the higher strike price. (iii) If the spot price is high enough, both put options have a loss. The profit is bigger for put option with the bigger strike price.

73/83

c





Since the strike price is paid at the expiration time, as higher the interest rate is as higher the put option price is. As higher the expiration time is as higher the put option price is. Usually, the price of a put option is found using the Black—Scholes formula. The Black–Scholes formula for the price of a put option is Put(K , T ) = Ke −rT Φ(−d2 ) − S0 e −δT Φ(−d1 ).

74/83

c





Table 1 shows the premium of a put option for different strike prices. We have used the Black–Scholes formula with S0 = 26, t = 0.25, σ = 0.3, δ = 0 and r = ln(1.055). Table 1: Spot Price Premium of a put option

22

24

26

28

30

0.1983495

0.6038701

1.377368

2.546227

4.045966

75/83

c





Example 13 Use the put premiums from Table 1, i = 5.5% and T = 0.25. An investor buys a put option for 500 shares. Find the profit function for the buyer of a put option with the following strike prices: $24, $26, $28.

76/83

c





Example 13 Use the put premiums from Table 1, i = 5.5% and T = 0.25. An investor buys a put option for 500 shares. Find the profit function for the buyer of a put option with the following strike prices: $24, $26, $28. Solution: Profit = 500 max(0, K − ST ) − 500Put(K , T )(1 + i)T . If K = 24, the profit is 500 max(0, 24 − ST ) − (500)(0.6038701)(1.055)0.25 =500 max(0, 24 − ST ) − 306.0036775. If K = 26, the profit is 500 max(0, 26 − ST ) − (500)(1.377368)(1.055)0.25 =500 max(0, 26 − ST ) − 697.9641. If K = 28, the profit is 500 max(0, 28 − ST ) − (500)(2.546227)(1.055)0.25 =500 max(0, 28 − ST ) − 1290.268926. c



77/83



When ST is low, the higher the strike price is, the higher the profit is. When ST is high, the higher the strike price is, the higher the loss is.

Figure 4: Profit for three puts. 78/83

c




I I I


A purchased put option is in–the–money, if ST < K . A purchased put option is out–the–money, if ST > K . A purchased put option is at–the–money, if ST = K .

79/83

c





If K is very small, the put option will almost certainly not be executed. Hence, if K is very small, Put(K , T ) = 0, i.e. lim Put(K , T ) = 0. If K is very large, the put option will almost K →0+

certainly be executed. If a put is executed its profit is K − ST . Hence, lim Put(K , T ) = ∞. As a function on K , Put(K , T ) is K →∞

an increasing function with lim Put(K , T ) = 0 and K →0+

lim Put(K , T ) = ∞.

K →∞

Figure 5 shows the graph of Put(K , T ) as a function of K . Put(K , T ) was found using the Black–Scholes formula with T = 1, S0 = 100, T = 1, σ = 0.25, r = ln(1.06) and δ = 0.0.

80/83

c





Figure 5: Graph of Put(K , T ) as a function of K . 81/83

c





We have the following strategies to speculate on the change of an asset

price will decrease price will increase

volatility will decrease

no volatility info

volatility will increase

sell a call

sell asset

buy a put

sell a put

buy asset

buy a call

82/83

c





If the price of asset will decrease, we can make a profit by either selling a call, or selling asset, or buying a put. If the volatility will decrease, the chances than option is executed decrease. Hence, if the price of asset and volatility will decrease, then the preferred strategy is to sell a call. By a similar argument, we have that: If the price of asset will decrease and the volatility will increase, then the preferred strategy is to buy a put. If the price of asset will increase and the volatility will decrease, then the preferred strategy is to sell a put. If the price of asset and volatility will increase, then the preferred strategy is to buy a call.

83/83

c




Manual for SOA Exam FM/CAS Exam 2. Chapter 7. Derivatives markets. Section 7.6. Put–call parity. c



1/51

c




Section 7.6. Put–call parity.

Put–call parity Recall that the actions and payoffs corresponding to a call/put are:

long call short call long put short put

long call short call long put short put

If ST < K no action no action sell the stock buy the stock If ST < K 0 0 K − ST −(K − ST )

If K < ST buy the stock sell the stock no action no action If K < ST ST − K −(ST − K ) 0 0 2/51

c





I

If we have a K –strike long call and a K –strike short put, we are able to buy the asset at time T for K . Hence, having both a K –strike long call and a K –strike short put is equivalent to have a K –strike long forward contract with price K .

I

Entering into both a K –strike long call and a K –strike short put is called a synthetic long forward.

I

Reciprocally, if we have a K –strike short call and a K –strike long put, we are able to sell the asset at time T for K . Having both a K –strike short call and a K –strike long put is equivalent to have a short forward contract with price K .

I

Entering into both a K –strike short call and a K –strike long put is called a synthetic short forward. 3/51

c





The no arbitrage cost at time T of buying an asset using a long forward contract is F0,T . The cost at time T for buying an asset using a K –strike long call and a K –strike short put is (Call(K , T ) − Put(K , T ))e rT + K . If there exists no arbitrage, then:

Theorem 1 (Put–call parity formula) (Call(K , T ) − Put(K , T ))e rT + K = F0,T . If we use effective interest, the put–call parity formula becomes: (Call(K , T ) − Put(K , T ))(1 + i)T + K = F0,T .

4/51

c





Often, F0,T = S0 (1 + i)T . This forward price applies to assets which have neither cost nor benefit associated with owning them. In the absence of arbitrage, we have the following relation between call and put prices:

Theorem 2 (Put–call parity formula) For a stock which does not pay any dividends, (Call(K , T ) − Put(K , T ))e rT + K = S0 e rT .

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Proof. Consider the portfolio consisting of buying one share of stock and a K –strike put for one share; selling a K –strike call for one share; and borrowing S0 − Call(K , T ) + Put(K , T ). At time T , we have the following possibilities: 1. If ST < K , then the put is exercised and the call is not. We finish without stock and with a payoff for the put of K . 2. If ST > K , then the call is exercised and the put is not. We finish without stock and with a payoff for the call of K . In any case, the payoff of this portfolio is K . Hence, K should be equal to the return in an investment of S0 + Put(K , T ) − Call(K , T ) in a zero–coupon bond, i.e. K = (S0 + Put(K , T ) − Call(K , T ))e rT .

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Example 1 The current value of XYZ stock is 75.38 per share. XYZ stock does not pay any dividends. The premium of a nine–month 80–strike call is 5.737192 per share. The premium of a nine–month 80–strike put is 7.482695 per share. Find the annual effective rate of interest.

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c





Example 1 The current value of XYZ stock is 75.38 per share. XYZ stock does not pay any dividends. The premium of a nine–month 80–strike call is 5.737192 per share. The premium of a nine–month 80–strike put is 7.482695 per share. Find the annual effective rate of interest. Solution: The put–call parity formula states that (Call(K , T ) − Put(K , T ))(1 + i)T + K = S0 (1 + i)T . So, (5.737192 − 7.482695)(1 + i)3/4 + 80 = 75.38(1 + i)T . 80 = (75.38 − (5.737192 − 7.482695))(1 + i)3/4 = (77.125503)(1 + i)3/4 , and i = 5%. 8/51

c





Example 2 The current value of XYZ stock is 85 per share. XYZ stock does not pay any dividends. The premium of a six–month K –strike call is 3.329264 per share and the premium of a one year K –strike put is 10.384565 per share. The annual effective rate of interest is 6.5%. Find K .

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Example 2 The current value of XYZ stock is 85 per share. XYZ stock does not pay any dividends. The premium of a six–month K –strike call is 3.329264 per share and the premium of a one year K –strike put is 10.384565 per share. The annual effective rate of interest is 6.5%. Find K . Solution: The put–call parity formula states that (Call(K , T ) − Put(K , T ))(1 + i)T + K = S0 (1 + i)T . So, (3.329264 − 10.384565)(1.065)0.5 + K = 85(1.065)0.5 and K = (85 − 3.329264 + 10.384565)(1.065)0.5 = 95.

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Example 3 XYZ stock does not pay any dividends. The price of a one year forward for one share of XYZ stock is 47.475. The premium of a one year 55–strike put option of XYZ stock is 9.204838 per share. The annual effective rate of interest is 5.5%. Calculate the price of a one year 55–strike call option for one share of XYZ stock.

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Example 3 XYZ stock does not pay any dividends. The price of a one year forward for one share of XYZ stock is 47.475. The premium of a one year 55–strike put option of XYZ stock is 9.204838 per share. The annual effective rate of interest is 5.5%. Calculate the price of a one year 55–strike call option for one share of XYZ stock. Solution: The put–call parity formula states that (Call(K , T ) − Put(K , T ))(1 + i)T + K = F0,T . So, (Call(55, 1) − 9.204838)(1.055) + 55 = 47.475 and Call(55, 1) = 9.204838 + (47.475 − 55)(1.055)−1 = 2.072136578.

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If prices of put options and call options do not satisfy the put–call parity, it is possible to do arbitrage. I

If (S0 − Call(K , T ) + Put(K , T ))e rT > K , we can make a profit by buying a call option, selling a put option and shorting stock. The profit of this strategy is = − K + (S0 − Call(K , T ) + Put(K , T ))e rT .

I

If (S0 − Call(K , T ) + Put(K , T ))e rT < K , we can do arbitrage by selling a call option, buying a put option and buying stock. At expiration time, we get rid of the stock by satisfying the options and make K − (S0 − Call(K , T ) + Put(K , T ))e rT . 13/51

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Example 4 XYZ stock trades at $54 per share. XYZ stock does not pay any dividends. The cost of an European call option with strike price $50 and expiration date in three months is $8 per share. The risk free annual interest rate continuously compounded is 4%.

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Example 4 XYZ stock trades at $54 per share. XYZ stock does not pay any dividends. The cost of an European call option with strike price $50 and expiration date in three months is $8 per share. The risk free annual interest rate continuously compounded is 4%. (i) Find the no–arbitrage price of a European put option with the same strike price and expiration time.

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Example 4 XYZ stock trades at $54 per share. XYZ stock does not pay any dividends. The cost of an European call option with strike price $50 and expiration date in three months is $8 per share. The risk free annual interest rate continuously compounded is 4%. (i) Find the no–arbitrage price of a European put option with the same strike price and expiration time. Solution: (i) With continuous interest, the put–call parity formula is (Call(K , T ) − Put(K , T ))e rT + K = S0 e rT . Hence, (8 − Put(50, 0.25))e 0.04(0.25) + 50 = 54e 0.04(0.25) and Put(50, 0.25) = 8 − 54 + 50e −0.04(0.25) = 3.502491687.

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Example 4 XYZ stock trades at $54 per share. XYZ stock does not pay any dividends. The cost of an European call option with strike price $50 and expiration date in three months is $8 per share. The risk free annual interest rate continuously compounded is 4%. (ii) Suppose that the price of an European put option with the same strike price and expiration time is $3, find an arbitrage strategy and its profit per share.

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Example 4 XYZ stock trades at $54 per share. XYZ stock does not pay any dividends. The cost of an European call option with strike price $50 and expiration date in three months is $8 per share. The risk free annual interest rate continuously compounded is 4%. (ii) Suppose that the price of an European put option with the same strike price and expiration time is $3, find an arbitrage strategy and its profit per share. Solution: (ii) A put option for $3 per share is undervalued. An arbitrage portfolio consists in selling a call option, buying a put option and stock and borrowing $−8 + 3 + 54 =$49, with all derivatives for one share of stock. At redemption time, we sell the stock and use it to execute the option which will be executed. We also repaid the loan. The profit is 50 − (49)e (0.04)(0.25) = 50 − 49.49245819 = 0.50754181.

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Example 5 Suppose that the current price of XYZ stock is 31. XYZ stock does not give any dividends. The risk free annual effective interest rate is 10%. The price of a three–month 30–strike European call option is $3. The price of a three–month 30–strike European put option is $2.25. Find an arbitrage opportunity and its profit per share.

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Example 5 Suppose that the current price of XYZ stock is 31. XYZ stock does not give any dividends. The risk free annual effective interest rate is 10%. The price of a three–month 30–strike European call option is $3. The price of a three–month 30–strike European put option is $2.25. Find an arbitrage opportunity and its profit per share. Solution: We have that (S0 + Put(K , T ) − Call(K , T ))(1 + i)T =(31 + 2.25 − 3)(1.1)0.25 = 30.97943909 > 30. We conclude that the put is overpriced relatively to the call. We can sell a put, buy a call and short stock. The profit per share is (2.25 − 3 + 31)(1.1)0.25 − 30 = 0.9794390948. Notice that at expiration time one of the options is executed and we get back the stock which we sold. c



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Synthetic forward. Definition 1 A synthetic long forward is the combination of buying a call and selling a put, both with the same strike price, amount of the asset and expiration date. The payments to get a synthetic long forward are I (Call(K , T ) − Put(K , T )) paid at time zero. I K paid at time T . The future value of these payments at time T is K + (Call(K , T ) − Put(K , T ))(1 + i)T . The payment of long forward is F0,T paid at time T . In the absence of arbitrage (put–call parity) F0,T = K + (Call(K , T ) − Put(K , T ))(1 + i)T . 21/51

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The premium of a synthetic long forward, i.e. the cost of entering this position, is (Call(K , T ) − Put(K , T )). I

If F0,T = K , the premium of a synthetic long forward is zero. You will buying the asset at the estimated future value of the asset.

I

If F0,T > K , the premium of a synthetic long forward is positive. You will buying the asset lower than the estimated future value of the asset.

I

If F0,T < K , the premium of a synthetic long forward is negative. You will buying the asset higher than the estimated future value of the asset.

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Constructive sale. An investor owns stock. He would like to sell his stock. But, he does not want to report capital gains to the IRS this year. So, instead of selling this stock, he holds the stock, buys a K –strike put, sells a K –strike call, and borrows K (1 + i)−T . The payoff which he gets at time T is ST + max(K − ST , 0) − max(ST − K , 0) − K = max(ST , K ) − max(ST , K ) = 0. At time zero, the investor gets Call(K , T ) − Put(K , T ) + K (1 + i)−T = F0,T (1 + i)−T . At expiration time, the investor can use the stock to meet the option which will be executed. Practically, the investor sold his stock at time zero for F0,T (1 + i)−T . According with current USA tax laws, this is considered a constructive sale. He will have to declare capital gains when the options are bought. c



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Floor. Suppose that you own some asset. If the asset losses value in the future, you lose money. A way to insure this long position is to buy a put position. The purchase of a put option is called a floor. A floor guarantees a minimum sale price of the value of an asset. I

I

The profit of buying an asset is ST − S0 (1 + i)T , which is the same as the profit of a long forward. The minimum profit of buying an asset is −S0 (1 + i)T . The profit for buying an asset and a put option is ST − S0 (1 + i)T + max(K − ST , 0) − Put(K , T )(1 + i)T = max(ST , K ) − (S0 + Put(K , T ))(1 + i)T . The minimum profit for buying an asset and a put option is K − (S0 + Put(K , T ))(1 + i)T .

We know that −S0 (1 + i)T < K − (S0 + Put(K , T ))(1 + i)T . c



24/51



Here is the graph of the profit for buying an asset and a put option:

Figure 1: Profit for a long position and a long put.

Notice that this is the graph of the profit of a purchased call. c



25/51



Here is a joint graph for the profits of the strategies: (i) buying an asset, (ii) buying an asset and a put.

Figure 2: Profit for long forward and buying an asset and a put. 26/51

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Table 1 was found using the Black–Scholes formula with S0 = 75, t = 1, σ = 0.20, δ = 0, r = log(1.05). Table 1: Prices of some calls and some puts.

K Call(K , T ) Put(K , T )

65 14.31722 1.221977

70 10.75552 2.422184

75 7.78971 4.218281

80 5.444947 6.635423

85 3.680736 9.633117

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Example 6 The current price of one share of XYZ stock is $75. Use the Table 1 for prices of puts and calls. Steve buys 500 shares of XYZ stock. Nicole buys 500 shares of XYZ stock and a put option on 500 shares of XYZ stock with a strike price 65 and an expiration date one year from now. The premium of a 65–strike put option is $1.221977. The risk free annual effective rate of interest is 5%.

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Example 6 The current price of one share of XYZ stock is $75. Use the Table 1 for prices of puts and calls. Steve buys 500 shares of XYZ stock. Nicole buys 500 shares of XYZ stock and a put option on 500 shares of XYZ stock with a strike price 65 and an expiration date one year from now. The premium of a 65–strike put option is $1.221977. The risk free annual effective rate of interest is 5%. (i) Calculate the profits for Steve and Nicole.

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Example 6 The current price of one share of XYZ stock is $75. Use the Table 1 for prices of puts and calls. Steve buys 500 shares of XYZ stock. Nicole buys 500 shares of XYZ stock and a put option on 500 shares of XYZ stock with a strike price 65 and an expiration date one year from now. The premium of a 65–strike put option is $1.221977. The risk free annual effective rate of interest is 5%. (i) Calculate the profits for Steve and Nicole. Solution: (i) Steve’s profit is (500)(ST − 75). Nicole’s profit is (500)(ST − 75 + max(65 − ST , 0) − 1.221977(1.05)) =(500)(max(65, ST ) − 76.28307585).

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Example 6 The current price of one share of XYZ stock is $75. Use the Table 1 for prices of puts and calls. Steve buys 500 shares of XYZ stock. Nicole buys 500 shares of XYZ stock and a put option on 500 shares of XYZ stock with a strike price 65 and an expiration date one year from now. The premium of a 65–strike put option is $1.221977. The risk free annual effective rate of interest is 5%. (ii) Find a table with Steve’s and Nicole’s profits for the following spot prices at expiration: 60, 65, 70, 75, 80 and 85.

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Example 6 The current price of one share of XYZ stock is $75. Use the Table 1 for prices of puts and calls. Steve buys 500 shares of XYZ stock. Nicole buys 500 shares of XYZ stock and a put option on 500 shares of XYZ stock with a strike price 65 and an expiration date one year from now. The premium of a 65–strike put option is $1.221977. The risk free annual effective rate of interest is 5%. (ii) Find a table with Steve’s and Nicole’s profits for the following spot prices at expiration: 60, 65, 70, 75, 80 and 85. Solution: (ii) Spot Price Steve’s prof Nicole’s prof

60 −7500 −5641.54

65 −5000 −5641.54

70 −2500 −3141.54

75 0 −641.54

80 2500 1858.46

85 5000 4358.46

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Example 6 The current price of one share of XYZ stock is $75. Use the Table 1 for prices of puts and calls. Steve buys 500 shares of XYZ stock. Nicole buys 500 shares of XYZ stock and a put option on 500 shares of XYZ stock with a strike price 65 and an expiration date one year from now. The premium of a 65–strike put option is $1.221977. The risk free annual effective rate of interest is 5%. (iii) Calculate the minimum and maximum profits for Steve and Nicole.

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Example 6 The current price of one share of XYZ stock is $75. Use the Table 1 for prices of puts and calls. Steve buys 500 shares of XYZ stock. Nicole buys 500 shares of XYZ stock and a put option on 500 shares of XYZ stock with a strike price 65 and an expiration date one year from now. The premium of a 65–strike put option is $1.221977. The risk free annual effective rate of interest is 5%. (iii) Calculate the minimum and maximum profits for Steve and Nicole. Solution: (iii) The minimum and maximum of Steve’s profit are (500)(−75) = −37500 and ∞, respectively. The minimum Nicole’s profit is (500)(65 − 76.28307585) = −5641.537925. The maximum Nicole’s profit is ∞.

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Example 6 The current price of one share of XYZ stock is $75. Use the Table 1 for prices of puts and calls. Steve buys 500 shares of XYZ stock. Nicole buys 500 shares of XYZ stock and a put option on 500 shares of XYZ stock with a strike price 65 and an expiration date one year from now. The premium of a 65–strike put option is $1.221977. The risk free annual effective rate of interest is 5%. (iv) Find spot prices at expiration at which each of Steve makes a profit. Answer the previous question for Nicole.

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Example 6 The current price of one share of XYZ stock is $75. Use the Table 1 for prices of puts and calls. Steve buys 500 shares of XYZ stock. Nicole buys 500 shares of XYZ stock and a put option on 500 shares of XYZ stock with a strike price 65 and an expiration date one year from now. The premium of a 65–strike put option is $1.221977. The risk free annual effective rate of interest is 5%. (iv) Find spot prices at expiration at which each of Steve makes a profit. Answer the previous question for Nicole. Solution: (iv) Steve’s profit is positive if (500)(ST − 75) > 0, i.e. if ST > 75. Nicole’s profit is positive if (500)(max(65, ST ) − 76.28307585) > 0, ie. if ST > 76.28307585.

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Next, we proof that the profit of buying an asset and a put is the same as the profit of buying a call option. The payoff for buying a call option and a zero–coupon bond which pays the strike price at expiration date is max(0, ST − K ) + K = max(ST , K ) The payoff for buying stock and a put option is ST + max(K − ST , 0) = max(ST , K ). Since the two strategies have the same payoff in the absence of arbitrage, they have the same profit, i.e. max(ST , K ) − K − (Call(K , T ))(1 + i)T = max(ST , K ) − (S0 + Put(K , T ))(1 + i)T . This equation is equivalent to the put–call parity: (S0 + Put(K , T ))(1 + i)T = K + Call(K , T )(1 + i)T . 37/51

c





Example 7 Michael buys 500 shares of XYZ stock and a 45–strike four–year put for 500 shares of XYZ stock. Rita buys a 45–strike four–year call for 500 shares of XYZ stock and invests P into a zero–coupon bond. The annual rate of interest continuously compounded is 4.5%. Find P so that Michael and Rita have the same payoff at expiration.

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Example 7 Michael buys 500 shares of XYZ stock and a 45–strike four–year put for 500 shares of XYZ stock. Rita buys a 45–strike four–year call for 500 shares of XYZ stock and invests P into a zero–coupon bond. The annual rate of interest continuously compounded is 4.5%. Find P so that Michael and Rita have the same payoff at expiration. Solution: Michael’s payoff at expiration is 500(S4 + max(45 − S4 , 0)) = 500 max(45, S4 ). Rita’s payoff at expiration is 500 max(S4 − 45, 0) + Pe (0.045)(4) =500 max(45, S4 ) − (500)(45) + Pe (0.045)(4) . Hence, P = (500)(45)e −(0.045)(4) = 18793.57976. 39/51

c





Cap. If you have an obligation to buy stock in the future, you have a short position on the stock. You will experience a loss, when the price of the stock price rises. You can insure a short position by purchasing a call option. Buying a call option when you are in a short position is called a cap. The payoff of having a short position and buying a call option is −ST + max(ST − K , 0) = max(−ST , −K ) = − min(ST , K ). The payoff of having a purchased put combined with borrowing the strike price at closing is max(K − ST , 0) − K = max(−ST , −K ) = − min(ST , K ), which is the same as before. 40/51

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Suppose that you have a short position on a stock. Consider the following two strategies: 1. Buy a call option. The payoff at expiration is −ST + max(ST − K , 0) = max(−ST , −K ). The profit at expiration is max(−ST , −K ) − Call(K , T )(1 + i)T . 2. Buy stock (to cover the short) and a put option. The payoff at expiration is max(K − ST , 0) = K + max(−ST , −K ). The profit at expiration is K + max(−ST , −K ) − (S0 + Put(K , T ))(1 + i)T . By the put–call parity formula, both strategies have the same profit. c



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Example 8 Heather has a short position in 1200 shares of stock XYZ. She is supposed to return this stock one year from now. To insure her position, she buys a $65–strike call. The premium of a $65–strike call is $14.31722. The current price of one share of XYZ stock is $75. The risk free annual effective rate of interest is 5%.

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Example 8 Heather has a short position in 1200 shares of stock XYZ. She is supposed to return this stock one year from now. To insure her position, she buys a $65–strike call. The premium of a $65–strike call is $14.31722. The current price of one share of XYZ stock is $75. The risk free annual effective rate of interest is 5%. (i) Make a table with Heather’s profit when the spot price at expiration is $40, $50, $60, $70, $80, $90.

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Example 8 Heather has a short position in 1200 shares of stock XYZ. She is supposed to return this stock one year from now. To insure her position, she buys a $65–strike call. The premium of a $65–strike call is $14.31722. The current price of one share of XYZ stock is $75. The risk free annual effective rate of interest is 5%. (i) Make a table with Heather’s profit when the spot price at expiration is $40, $50, $60, $70, $80, $90. Solution: (i) Heather’s profit is (1200) max(−ST , −K ) − Call(K , T )(1 + i)T =1200(max(−ST , −65) − 14.31722(1.05)1 ) =1200(max(−ST , −65) − 15.033081200). Spot Price H’s profit

40 −66039.70

50 −78039.70

60 −90039.70

70 −96039.70

80 −96039.70

90 −96039.70

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Example 8 Heather has a short position in 1200 shares of stock XYZ. She is supposed to return this stock one year from now. To insure her position, she buys a $65–strike call. The premium of a $65–strike call is $14.31722. The current price of one share of XYZ stock is $75. The risk free annual effective rate of interest is 5%. (ii) Assuming that she does not buy the call, make a table with her profit when the spot price at expiration is $40, $50, $60, $70, $80, $90.

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Example 8 Heather has a short position in 1200 shares of stock XYZ. She is supposed to return this stock one year from now. To insure her position, she buys a $65–strike call. The premium of a $65–strike call is $14.31722. The current price of one share of XYZ stock is $75. The risk free annual effective rate of interest is 5%. (ii) Assuming that she does not buy the call, make a table with her profit when the spot price at expiration is $40, $50, $60, $70, $80, $90. Solution: (ii) Heather’s profit is −(1200)ST . Spot Price H’s profit

40 −48000

50 −60000

60 −72000

70 −84000

80 −96000

90 −108000

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Example 8 Heather has a short position in 1200 shares of stock XYZ. She is supposed to return this stock one year from now. To insure her position, she buys a $65–strike call. The premium of a $65–strike call is $14.31722. The current price of one share of XYZ stock is $75. The risk free annual effective rate of interest is 5%. (iii) Draw the graphs of the profit versus the spot price at expiration for the strategies in (i) and in (ii).

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Example 8 Heather has a short position in 1200 shares of stock XYZ. She is supposed to return this stock one year from now. To insure her position, she buys a $65–strike call. The premium of a $65–strike call is $14.31722. The current price of one share of XYZ stock is $75. The risk free annual effective rate of interest is 5%. (iii) Draw the graphs of the profit versus the spot price at expiration for the strategies in (i) and in (ii). Solution: (iii) The graph of profits is on Figure 3.

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Notice that the possible loss for the short position can be arbitrarily large. By buying the call Heather has limited her possible losses.

Figure 3: Profit for a long position and a long put. 49/51

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Selling calls and puts.

Selling a option when there is a corresponding long position in the underlying asset is called covered writing or option overwriting. Naked writing occurs when the writer of an option does not have a position in the asset. I

A covered call is achieved by writing a call against a long position on the stock. An investor holding a long position in an asset may write a call to generate some income from the asset.

I

A covered put is achieved by writing a put against a short position on the stock.

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An arbitrageur buys and sells call and puts. A way to limit risks in the sales of options is to cover these positions. He also can match opposite options. For example, I

A purchase of K –strike call option, a sale of a K –strike put option and a short forward cancel each other.

I

A sale of K –strike call option, a purchase of a K –strike put option and a long forward cancel each other.

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c




Manual for SOA Exam FM/CAS Exam 2. Chapter 7. Derivatives markets. 7.7. Equity linked certificates of deposit. c



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Section 7.7. Equity linked certificates of deposit.

Equity linked certificates of deposit

To estimate the evolution of the stock market, stock indexes are used. A stock market index is a listing of stocks and a way to obtain the composite value of its components. The three most used stock indexes are the (Dow Jones) Dow Jones Industrial Average, (S & P 500) Standard & Poor’s 500 index, and the NASDAQ Composite Index. Usually the composite value of an index is a sort of average of the stocks in the index. However, there are different ways to find this average. Every stock market index has its rules to find its composite value.

2/21

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The S & P 500 index consisting of 500 stocks of large corporations selected by Standard & Poor. Standard & Poor is a financial company which specializes in providing independent credit rating and index evaluation. The stocks on the S & P 500 trade in stock markets, like the (NYSE) NYSE New York Stock Exchange and (National Association of Securities Dealers Automated Quotations system) NASDAQ. The New York Stock Exchange is the largest equities marketplace in the world. NASDAQ is an electronic stock exchange.

3/21

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The Dow Jones Industrial Average is obtained ”averaging” the value of 30 stocks selected by the Dow Jones & Company. These 30 stocks are selected from largest and the most widely held public companies in the USA across a range of industries except for transport and utilities. Dow Jones & Company publishes the The Wall Street Journal. The Wall Street Journal editors have a lot input on the selection of the stocks in the Dow Jones Industrial Average. The NASDAQ Composite Index consists of all securities listed on NASDAQ. It contains mainly stocks of technology and growth companies. Roughly, the difference between the three indexes is on the type of stocks which they represent. The S & P 500 focuses on all large–cap stocks in the market. The Dow Jones Industrial Average focuses on a very selected group of large companies. NASDAQ focuses on technology and fast growing companies. 4/21

c





An (ELCD) equity linked CD (or equity linked note, or equity indexed CD, or market index linked CD) is an FDIC–insured certificate of deposit that ties the rate of return to the performance of a stock index such as the S & P 500 and guarantee a certain payment. Chase Manhattan Bank first introduced ELCD’s in 1987. But, now many financial institutions offer ELCD’s. Usually, the guarantee payment is the original principal. The investor at expiration also gets a payment depending on the performance of the stock index. Usually, there exists a participation rate r , 0 < r ≤ 1, such investor gets the that the ST guarantee payment plus rP max S0 − 1, 0 , where P is the principal invested, ST is the index price at expiration, S0 is the spot price. Hence, usually, the payoff of an ELCD is ST P 1 + r max − 1, 0 . S0 5/21

c





An investor is attracted to ELCD’s because it has the potential for market appreciation and diversification without risking capital. Diversification is attained by using market indices, which combine several stocks. Usually the return of the capital is FDIC insured. One disadvantage is the possible loss of interest on the invested principal. Notice that the smallest payoff which the investor may get is his invested principal, i.e. he does not get any interest. The instrument is appropriate for conservative equity investors or fixed income investors who desire equity exposure with controlled risk.

6/21

c





Example 1 Aaron deposits $15000 in an ELCD, which provides 100% principal protection and pays 80% of the appreciation of the S & P 500 three years from now. The index closes at 1500 on the day the ELCD is issued.

7/21

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Example 1 Aaron deposits $15000 in an ELCD, which provides 100% principal protection and pays 80% of the appreciation of the S & P 500 three years from now. The index closes at 1500 on the day the ELCD is issued. (i) Find Aaron’s payoff as a function of S3 .

8/21

c





Example 1 Aaron deposits $15000 in an ELCD, which provides 100% principal protection and pays 80% of the appreciation of the S & P 500 three years from now. The index closes at 1500 on the day the ELCD is issued. (i) Find Aaron’s payoff as a function of S3 . Solution: (i) Aaron’s payoff is ST P 1 + r max − 1, 0 S0 ST − 1, 0 . =(15000) 1 + (0.80) max 1500

9/21

c





Example 1 Aaron deposits $15000 in an ELCD, which provides 100% principal protection and pays 80% of the appreciation of the S & P 500 three years from now. The index closes at 1500 on the day the ELCD is issued. (ii) Find the Aaron’s payoff in the forward contract if S3 is $1300, $1400, $1500, $1600, $1700, $1800.

10/21

c





Example 1 Aaron deposits $15000 in an ELCD, which provides 100% principal protection and pays 80% of the appreciation of the S & P 500 three years from now. The index closes at 1500 on the day the ELCD is issued. (ii) Find the Aaron’s payoff in the forward contract if S3 is $1300, $1400, $1500, $1600, $1700, $1800. Solution: (ii) Using that Aaron’s payoff is ST (15000) 1 + (0.80) max − 1, 0 , 1500 Payoff S3

15000 1300

15000 1400

15000 1500

15800 1600

16600 1700

17400 1800

11/21

c





Example 1 Aaron deposits $15000 in an ELCD, which provides 100% principal protection and pays 80% of the appreciation of the S & P 500 three years from now. The index closes at 1500 on the day the ELCD is issued. (iii) Graph Aaron’s payoff as a function of S3 .

12/21

c





Example 1 Aaron deposits $15000 in an ELCD, which provides 100% principal protection and pays 80% of the appreciation of the S & P 500 three years from now. The index closes at 1500 on the day the ELCD is issued. (iii) Graph Aaron’s payoff as a function of S3 . Solution: (iii) Aaron’s payoff is Figure 1.

13/21

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Figure 1: Example 1. ELCD payoff. 14/21

c





The payoff of an ELCD is a combination of the payoff of a long call option and a long bond position. It is possible to create a synthetic ELCD by buying a long call option and a zero–coupon bond.

15/21

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The return on a ELCD is ST Pr P 1 + r max − 1, 0 =P+ max(ST − S0 , 0), S0 S0 Suppose that an investor buys a zero–coupon bond with face value P and a S0 –strike call option for Pr S0 shares. His payoff at expiration date is P+

Pr max(ST − S0 , 0). S0

The cost of making this investment is P(1 + i)−T +

Pr Call(S0 , T ). S0

If there exist no arbitrage, P = P(1 + i)−T + Hence, r =

Pr Call(S0 , T ). S0

(1−(1+i)−T )S0 Call(S0 ,T ) .

c


16/21




Example 2 The risk–free effective rate of interest is 7%. The current price of the S & P 500 is 1500. The price of an European call with strike 1500 and expiration date in 3 years is 400. Find the participation rate of a three–year ELCD which provides 100% principal protection.

17/21

c





Example 2 The risk–free effective rate of interest is 7%. The current price of the S & P 500 is 1500. The price of an European call with strike 1500 and expiration date in 3 years is 400. Find the participation rate of a three–year ELCD which provides 100% principal protection. Solution: We have that (1 − (1 + i)−T )S0 Call(S0 , T ) (1 − (1.07)−3 )1500 = = 0.6888829617 = 68.88829617%. 400 r=

18/21

c





Suppose that an ELCD return all the principal invested at expiration, offers a guaranteed rate of return g , where g < i, and a participation rate r on the S & P 500 stock index. The payoff of this ELCD is Pr ST − S0 T , 0 = P(1+g )T + max(ST −S0 , 0). P(1+g ) +Pr max S0 S0 We can get this payoff by buying a zero–coupon bond with face value P(1 + g )T and a S0 –strike call option for Pr S0 shares. The cost of this portfolio is Pr Call(S0 , T ). P(1 + g )T (1 + i)−T + S0 If there exist no arbitrage, Pr Call(S0 , T ) P = P(1 + g )T (1 + i)−T + S0 and (1 − (1 + g )T (1 + i)−T )S0 r= . Call(S0 , T ) c



19/21



Example 3 The risk–free effective rate of interest is 5%. The current price of the S & P 500 is 1500. The price of an European call with strike 1500 and expiration date in six month is 125. Find the participation rate of a ELCD which provides 100% principal protection and a guaranteed annual interest rate of 1.5%.

20/21

c





Example 3 The risk–free effective rate of interest is 5%. The current price of the S & P 500 is 1500. The price of an European call with strike 1500 and expiration date in six month is 125. Find the participation rate of a ELCD which provides 100% principal protection and a guaranteed annual interest rate of 1.5%. Solution: (1 − (1 + g )T (1 + i)−T )S0 Call(S0 , T ) (1 − (1.015)0.5 (1.05)−0.5 )1500 = = 0.201695 = 20.1695%. 125 r=

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Manual for SOA Exam FM/CAS Exam 2. Chapter 7. Derivatives markets. Section 7.8. Spreads. c



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Section 7.8. Spreads.

Spreads

An option spread (or a vertical spread) is a combination of only calls or only puts, in which some options are bought and some others are sold. By buying/selling several call/puts we can create portfolios useful for many different objectives. A ratio spread is a combination of buying m calls at one strike price and selling n calls at a different strike price.

2/97

c





Speculating on the increase of an asset price. Bull spread.

Definition 1 A bull spread consists on buying a K1 –strike call and selling a K2 –strike call, both with the same expiration date T and nominal amount, where 0 < K1 < K2 . A way to speculate on the increase of an asset price is buying the asset. This position needs a lot of investment. Another way to speculate on the increase of an asset price is to buy a call option. A bull spread allows to speculate on increase of an asset price by making a limited investment.

3/97

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The payoff for buying a K1 –strike call is max(ST − K1 , 0). The payoff for selling a K2 –strike call is − max(ST − K2 , 0). − max(ST − K2 , 0)

max(ST − K1 , 0) 6

6

-

K1

K2

ST

-

K1

K@ 2

ST

@ @ @ @

Figure 1: Payoff of the calls in a bull spread 4/97

c





The bull spread payoff is max(ST − K1 , 0) − max(ST − K2 , 0) = max(ST − K1 , 0) + K2 − K1 − max(ST − K1 , K2 − K1 ) = max(ST − K1 , 0) + K2 − K1 − max(ST − K1 , 0, K2 − K1 ) = max(ST − K1 , 0) + K2 − K1 − max(max(ST − K1 , 0), K2 − K1 ) = min(max(ST   0 = ST − K1   K2 − K1

− K1 , 0), K2 − K1 ) if ST < K1 , if K1 ≤ ST < K2 , if K2 ≤ ST .

5/97

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max(ST − K1 , 0) − max(ST − K2 , 0) 6

-

K1

K2

ST

Figure 2: Payoff of a bull spread

6/97

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The profit of a bull spread is min(max(ST − K1 , 0), K2 − K1 ) − (Call(K1 , T ) − Call(K2 , T ))(1 + i)T  T  −(Call(K1 , T ) − Call(K2 , T ))(1 + i) = ST − K1 − (Call(K1 , T ) − Call(K2 , T ))(1 + i)T   K2 − K1 − (Call(K1 , T ) − Call(K2 , T ))(1 + i)T

if ST < K1 , if K1 ≤ ST < K2 , if K2 ≤ ST .

Figure 3 shows a graph of the profit of a bull spread. Notice that the profit is positive for values of ST large enough.

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Figure 3: Profit for a bull spread. 8/97

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In this section, we will use the values of calls/puts in Table 1. Table 1: Prices of some calls and some puts.

K Call(K , T ) Put(K , T )

65 14.31722 1.221977

70 10.75552 2.422184

75 7.78971 4.218281

80 5.444947 6.635423

85 3.680736 9.633117

9/97

c





Example 1 (Use Table 1) Ronald buys a $70–strike call and sells a $85–strike call for 100 shares of XYZ stock. Both have expiration date one year from now. The current price of one share of XYZ stock is $75. The risk free annual effective rate of interest is 5%. The premium per share of $70–strike and $85–strike calls are 10.75552 and 3.680736 respectively.

10/97

c





Example 1 (Use Table 1) Ronald buys a $70–strike call and sells a $85–strike call for 100 shares of XYZ stock. Both have expiration date one year from now. The current price of one share of XYZ stock is $75. The risk free annual effective rate of interest is 5%. The premium per share of $70–strike and $85–strike calls are 10.75552 and 3.680736 respectively. (i) Find the profit at expiration as a function of the strike price.

11/97

c





Example 1 (Use Table 1) Ronald buys a $70–strike call and sells a $85–strike call for 100 shares of XYZ stock. Both have expiration date one year from now. The current price of one share of XYZ stock is $75. The risk free annual effective rate of interest is 5%. The premium per share of $70–strike and $85–strike calls are 10.75552 and 3.680736 respectively. (i) Find the profit at expiration as a function of the strike price. Solution: (i) Ronald’s profit is (100) (min(max(ST − 70, 0), 85 − 70) − (10.75552 − 3.680736)(1.05)) =(100) (min(max(ST − 70, 0), 85 − 70) − 7.428523)   if ST < 70, −(100)(7.428523) = (100)(ST − 70 − 7.428523) if 70 ≤ ST < 85,   (100)(85 − 70 − 7.428523) if 85 ≤ ST . 12/97

c





Example 1 (Use Table 1) Ronald buys a $70–strike call and sells a $85–strike call for 100 shares of XYZ stock. Both have expiration date one year from now. The current price of one share of XYZ stock is $75. The risk free annual effective rate of interest is 5%. The premium per share of $70–strike and $85–strike calls are 10.75552 and 3.680736 respectively. (ii) Make a table with Ronald’s profit when the spot price at expiration is $65, $70, $75, $80, $85, $90.

13/97

c





Example 1 (Use Table 1) Ronald buys a $70–strike call and sells a $85–strike call for 100 shares of XYZ stock. Both have expiration date one year from now. The current price of one share of XYZ stock is $75. The risk free annual effective rate of interest is 5%. The premium per share of $70–strike and $85–strike calls are 10.75552 and 3.680736 respectively. (ii) Make a table with Ronald’s profit when the spot price at expiration is $65, $70, $75, $80, $85, $90. Solution: (ii) Spot Price Ronald’s profit

65 −742.8523

Spot Price Ronald’s profit

80 257.1477

70 −742.8523 85 757.1477

75 −242.8523 90 757.1477 14/97

c





Example 1 (Use Table 1) Ronald buys a $70–strike call and sells a $85–strike call for 100 shares of XYZ stock. Both have expiration date one year from now. The current price of one share of XYZ stock is $75. The risk free annual effective rate of interest is 5%. The premium per share of $70–strike and $85–strike calls are 10.75552 and 3.680736 respectively. (iii) Draw the graph of Ronald’s profit.

15/97

c





Example 1 (Use Table 1) Ronald buys a $70–strike call and sells a $85–strike call for 100 shares of XYZ stock. Both have expiration date one year from now. The current price of one share of XYZ stock is $75. The risk free annual effective rate of interest is 5%. The premium per share of $70–strike and $85–strike calls are 10.75552 and 3.680736 respectively. (iii) Draw the graph of Ronald’s profit. Solution: (ii) The graph of the profit is Figure 3.

16/97

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Let 0 < K1 < K2 . We have that [long K1 − strike call ] + [short K1 − strike put ] ≡ [buying asset for K1 at time T ], [long K2 − strike call ] + [short K2 − strike put ] ≡ [buying asset for K2 at time T ]. Since the two investment strategies differ by a constant, they have the same profit. Hence, so do the following strategies, [long K1 − strike call ] + [short K2 − strike call ], [long K1 − strike put ] + [short K2 − strike put ]. In other words, buying a K1 –strike call and selling a K2 –strike call has the same profit as buying a K1 –strike put and selling a K2 –strike put. We can form a bull spread either buying a K1 –strike call and selling a K2 –strike call, or buying a K1 –strike put and selling a K2 –strike put. c



17/97



Example 2 The current price of XYZ stock is $75 per share. The effective annual interest rate is 5%. Elizabeth, Daniel and Catherine believe that the price of XYZ stock is going to appreciate significantly in the next year. Each person has $10000 to invest. The premium of a one–year 85–strike call option is 3.680736 per share. The premium of a one–year 75–strike call option is 7.78971 per share. Elizabeth buys a one–year zero–coupon bond for $10000. She also enters into a one–year forward contract on XYZ stock worth equal to the her bond payoff at redemption. Daniel buys a one–year 85–strike call option which costs $10000. Catherine buys a one–year 75–strike call option and sells a one–year 85–strike call option. The nominal amounts on both calls are the same. The difference between the cost of the 85–strike call option and the 75–strike call option is 10000. Suppose that the stock price at redemption is 90 per share. Calculate the profits and the yield rates for Elizabeth, Daniel and Catherine. Which one makes a bigger profit? c



18/97



Solution: The price of a long forward is 75(1.05) = 78.75. So, Elizabeth long forward is for (10000)(1.05) = 133.333333 shares. 78.75 Elizabeth’s profit is 133.333333(90 − 78.75) = 1500. Her yield rate 1500 = 15%. is 10000 10000 The nominal amount in Daniel’s call is 3.680736 = 2716.847935 shares. Daniel’s profit is 2716.847935(90 − 85) = 13584.24. Daniel’s yield of return is 13584.24 10000 = 135.8424%. The nominal amount in Catherine’s calls is 10000 7.78971−3.680736 = 2433.697561 shares. Catherine’s profit is 2433.697561(85 − 75) = 24336.97561. Catherine’s yield of return is 24336.97561 = 243.3697561%. 10000 Catherine’s profit is biggest.

19/97

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Speculating on the decrease of an asset price. Bear spread. A bear spread is precisely the opposite of a bull spread. Suppose that you want to speculate on the price of an asset decreasing. Let 0 < K1 < K2 . Consider selling a K1 –strike call and buying a K2 –strike call, both with the same expiration date T . The profit is − min(max(ST −K1 , 0), K2 −K1 )+(Call(K1 , T )−Call(K2 , T ))(1+i)T or  T  (Call(K1 , T ) − Call(K2 , T ))(1 + i) K1 − ST + (Call(K1 , T ) − Call(K2 , T ))(1 + i)T   K1 − K2 + (Call(K1 , T ) − Call(K2 , T ))(1 + i)T


A graph of this profit is in Figure 4. Notice that the profit is positive for values of ST small enough. 20/97

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Figure 4: Profit for a bear spread. 21/97

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Example 3 (Use Table 1) Rebecca sells a $65–strike call and buys a $80–strike call for 1000 shares of XYZ stock. Both have expiration date one year from now. The current price of one share of XYZ stock is $75. The risk free annual effective rate of interest is 5%. The premium per share of $65–strike and $80–strike calls are 14.31722 and 5.444947 respectively.

22/97

c





Example 3 (Use Table 1) Rebecca sells a $65–strike call and buys a $80–strike call for 1000 shares of XYZ stock. Both have expiration date one year from now. The current price of one share of XYZ stock is $75. The risk free annual effective rate of interest is 5%. The premium per share of $65–strike and $80–strike calls are 14.31722 and 5.444947 respectively. (i) Find Rebecca’s profit as a function of the spot price at expiration.

23/97

c





Example 3 (Use Table 1) Rebecca sells a $65–strike call and buys a $80–strike call for 1000 shares of XYZ stock. Both have expiration date one year from now. The current price of one share of XYZ stock is $75. The risk free annual effective rate of interest is 5%. The premium per share of $65–strike and $80–strike calls are 14.31722 and 5.444947 respectively. (i) Find Rebecca’s profit as a function of the spot price at expiration. Solution: (i) Rebecca’s profit is − (100) max(ST − 65, 0) + (100) max(ST − 80, 0) + (100)(14.31722 − 5.444947)(1.05) = − (100) min(max(ST − 65, 0), 80 − 65) + (100)(9.31588665)   if ST < 65 (100)(9.31588665) = (100)(65 − ST + 9.31588665) if 65 ≤ ST < 80   (100)(65 − 80 + 9.31588665) if 80 ≤ ST . 24/97

c





Example 3 (Use Table 1) Rebecca sells a $65–strike call and buys a $80–strike call for 1000 shares of XYZ stock. Both have expiration date one year from now. The current price of one share of XYZ stock is $75. The risk free annual effective rate of interest is 5%. The premium per share of $65–strike and $80–strike calls are 14.31722 and 5.444947 respectively. (ii) Make a table with Rebecca’s profit when the spot price at expiration is $60, $65, $70, $75, $80, $85, $90.

25/97

c





Example 3 (Use Table 1) Rebecca sells a $65–strike call and buys a $80–strike call for 1000 shares of XYZ stock. Both have expiration date one year from now. The current price of one share of XYZ stock is $75. The risk free annual effective rate of interest is 5%. The premium per share of $65–strike and $80–strike calls are 14.31722 and 5.444947 respectively. (ii) Make a table with Rebecca’s profit when the spot price at expiration is $60, $65, $70, $75, $80, $85, $90. Solution: (ii) Spot Price Rebecca’s profit Spot Price Rebecca’s profit

60 931.59

75 −68.41

65 931.59 80 −568.41

70 431.59 85 −568.41

75 −68.41 90 −568.41 26/97

c





Example 3 (Use Table 1) Rebecca sells a $65–strike call and buys a $80–strike call for 1000 shares of XYZ stock. Both have expiration date one year from now. The current price of one share of XYZ stock is $75. The risk free annual effective rate of interest is 5%. The premium per share of $65–strike and $80–strike calls are 14.31722 and 5.444947 respectively. (iii) Draw the graph of Rebecca’s profit.

27/97

c





Example 3 (Use Table 1) Rebecca sells a $65–strike call and buys a $80–strike call for 1000 shares of XYZ stock. Both have expiration date one year from now. The current price of one share of XYZ stock is $75. The risk free annual effective rate of interest is 5%. The premium per share of $65–strike and $80–strike calls are 14.31722 and 5.444947 respectively. (iii) Draw the graph of Rebecca’s profit. Solution: (iii) Figure 4 shows the graph of the profit.

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Collar A collar is the purchase of a put option at a strike price and the sale of a call option at a higher strike price. Let K1 be the strike price of the put option. Let K2 be the strike price of the call option. Assume that K1 < K2 . The profit of this strategy is max(K1 − ST , 0) − max(ST − K2 , 0) − (Put(K1 , T ) − Call(K2 , T ))(1 + i)T  T  K1 − ST − (Put(K1 , T ) − Call(K2 , T ))(1 + i) = −(Put(K1 , T ) − Call(K2 , T ))(1 + i)T   K2 − ST − (Put(K1 , T ) − Call(K2 , T ))(1 + i)T


A collar can be use to speculate on the decrease of price of an asset. The collar width is the difference between the call strike and the put strike. c



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Example 4 (Use Table 1) Toto buys a $65–strike put option and sells a $80–strike call for 100 shares of XYZ stock. Both have expiration date one year from now. The current price of one share of XYZ stock is $75. The risk free annual effective rate of interest is 5%. The premium of a $65–strike put option is 1.221977 per share. The premium of a $80–strike call option is 5.444947 per share.

30/97

c





Example 4 (Use Table 1) Toto buys a $65–strike put option and sells a $80–strike call for 100 shares of XYZ stock. Both have expiration date one year from now. The current price of one share of XYZ stock is $75. The risk free annual effective rate of interest is 5%. The premium of a $65–strike put option is 1.221977 per share. The premium of a $80–strike call option is 5.444947 per share. (i) Find Toto’s profit as a function of the spot price at expiration.

31/97

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Example 4 (Use Table 1) Toto buys a $65–strike put option and sells a $80–strike call for 100 shares of XYZ stock. Both have expiration date one year from now. The current price of one share of XYZ stock is $75. The risk free annual effective rate of interest is 5%. The premium of a $65–strike put option is 1.221977 per share. The premium of a $80–strike call option is 5.444947 per share. (i) Find Toto’s profit as a function of the spot price at expiration. Solution: (i) Toto’s profit is (100) (max(65 − ST , 0) − max(ST − 80, 0) −(1.221977 − 5.444947)(1.05)) =(100) (max(65 − ST , 0) − max(ST − 80, 0) + 4.4341185)   (100)(65 − ST + 4.4341185) if ST < 65, = 100(4.4341185) if 65 ≤ ST < 80,   100(80 − ST + 4.4341185) if 80 ≤ ST . 32/97

c





Example 4 (Use Table 1) Toto buys a $65–strike put option and sells a $80–strike call for 100 shares of XYZ stock. Both have expiration date one year from now. The current price of one share of XYZ stock is $75. The risk free annual effective rate of interest is 5%. The premium of a $65–strike put option is 1.221977 per share. The premium of a $80–strike call option is 5.444947 per share. (ii) Make a table with Toto’s profit when the spot price at expiration is $55, $60, $65, $70, $75, $80, $85.

33/97

c





Example 4 (Use Table 1) Toto buys a $65–strike put option and sells a $80–strike call for 100 shares of XYZ stock. Both have expiration date one year from now. The current price of one share of XYZ stock is $75. The risk free annual effective rate of interest is 5%. The premium of a $65–strike put option is 1.221977 per share. The premium of a $80–strike call option is 5.444947 per share. (ii) Make a table with Toto’s profit when the spot price at expiration is $55, $60, $65, $70, $75, $80, $85. Solution: (ii) A table with Toto’s profit is Spot Price Toto’s profit

55 1443.41

Spot Price Toto’s profit

75 443.41

60 943.41 80 443.41

65 443.41 85 −56.59

70 443.41 90 −556.59 34/97

c





Example 4 (Use Table 1) Toto buys a $65–strike put option and sells a $80–strike call for 100 shares of XYZ stock. Both have expiration date one year from now. The current price of one share of XYZ stock is $75. The risk free annual effective rate of interest is 5%. The premium of a $65–strike put option is 1.221977 per share. The premium of a $80–strike call option is 5.444947 per share. (iii) Draw the graph of Toto’s profit.

35/97

c





Example 4 (Use Table 1) Toto buys a $65–strike put option and sells a $80–strike call for 100 shares of XYZ stock. Both have expiration date one year from now. The current price of one share of XYZ stock is $75. The risk free annual effective rate of interest is 5%. The premium of a $65–strike put option is 1.221977 per share. The premium of a $80–strike call option is 5.444947 per share. (iii) Draw the graph of Toto’s profit. Solution: (iii) The graph of the profit is on Figure 5.

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Figure 5: Profit for a collar. 37/97

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Written collar

A written collar is a reverse collar.

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Collars are used to insure a long position on a stock. This position is called a collared stock. A collared stock involves buying the index, buy a K1 –strike put option and selling a K2 –strike call option, where K1 < K2 . The payoff per share of this strategy is ST + max(K1 − ST , 0) − max(ST − K2 , 0) = max(K1 , ST ) − max(ST , K2 ) + K2 = max(K1 , ST ) − max(ST , K1 , K2 ) + K2 = min(max(K1 , ST ), K2 ) The profit per share of this portfolio is min(max(K1 , ST ), K2 ) − (S0 + Put(K1 , T ) − Call(K2 , T ))(1 + i)T  T  if ST < K1 , K1 − (S0 + Put(K1 , T ) − Call(K2 , T ))(1 + i) = ST − (S0 + Put(K1 , T ) − Call(K2 , T ))(1 + i)T if K1 ≤ ST < K2 ,   K2 − (S0 + Put(K1 , T ) − Call(K2 , T ))(1 + i)T if K2 ≤ ST . 39/97

c





Example 5 (Use Table 1) Maggie buys 100 shares of XYZ stock, a $65–strike put option and sells a $80–strike call. Both options are for 100 shares of XYZ stock and have expiration date one year from now. The current price of one share of XYZ stock is $75. The risk free annual effective rate of interest is 5%. The premium per share of a $65–strike put option is 1.221977. The premium per share of a $80–strike call is 5.444947.

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Example 5 (Use Table 1) Maggie buys 100 shares of XYZ stock, a $65–strike put option and sells a $80–strike call. Both options are for 100 shares of XYZ stock and have expiration date one year from now. The current price of one share of XYZ stock is $75. The risk free annual effective rate of interest is 5%. The premium per share of a $65–strike put option is 1.221977. The premium per share of a $80–strike call is 5.444947. (i) Find Maggie’s profit as a function of the spot price at expiration.

41/97

c





Example 5 (Use Table 1) Maggie buys 100 shares of XYZ stock, a $65–strike put option and sells a $80–strike call. Both options are for 100 shares of XYZ stock and have expiration date one year from now. The current price of one share of XYZ stock is $75. The risk free annual effective rate of interest is 5%. The premium per share of a $65–strike put option is 1.221977. The premium per share of a $80–strike call is 5.444947. (i) Find Maggie’s profit as a function of the spot price at expiration. Solution: (i) Maggie’s profit is (100) min(max(65, ST ), 80) − (75 + 1.221977 − 5.444947)(1.05)1

=(100) (min(max(65, ST ), 80) − 74.315882)   100(65 − 74.315882) if ST < 65, = 100(ST − 74.315882) if 65 ≤ ST < 80,   100(80 − 74.315882) if 80 ≤ ST . 42/97

c





Example 5 (Use Table 1) Maggie buys 100 shares of XYZ stock, a $65–strike put option and sells a $80–strike call. Both options are for 100 shares of XYZ stock and have expiration date one year from now. The current price of one share of XYZ stock is $75. The risk free annual effective rate of interest is 5%. The premium per share of a $65–strike put option is 1.221977. The premium per share of a $80–strike call is 5.444947. (ii) Make a table with Maggie’s profit when the spot price at expiration is $60, $65, $70, $75, $80, $85.

43/97

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Example 5 (Use Table 1) Maggie buys 100 shares of XYZ stock, a $65–strike put option and sells a $80–strike call. Both options are for 100 shares of XYZ stock and have expiration date one year from now. The current price of one share of XYZ stock is $75. The risk free annual effective rate of interest is 5%. The premium per share of a $65–strike put option is 1.221977. The premium per share of a $80–strike call is 5.444947. (ii) Make a table with Maggie’s profit when the spot price at expiration is $60, $65, $70, $75, $80, $85. Solution: (ii) A table with Maggie’s profit for the considered spot prices is Spot Price Maggie’s profit

60 −931.59

Spot Price Maggie’s profit

75 68.41

c


65 −931.59 80 568.41

70 −431.59 85 568.41


44/97



Example 5 (Use Table 1) Maggie buys 100 shares of XYZ stock, a $65–strike put option and sells a $80–strike call. Both options are for 100 shares of XYZ stock and have expiration date one year from now. The current price of one share of XYZ stock is $75. The risk free annual effective rate of interest is 5%. The premium per share of a $65–strike put option is 1.221977. The premium per share of a $80–strike call is 5.444947. (iii) Draw the graph of Maggie’s profit.

45/97

c





Example 5 (Use Table 1) Maggie buys 100 shares of XYZ stock, a $65–strike put option and sells a $80–strike call. Both options are for 100 shares of XYZ stock and have expiration date one year from now. The current price of one share of XYZ stock is $75. The risk free annual effective rate of interest is 5%. The premium per share of a $65–strike put option is 1.221977. The premium per share of a $80–strike call is 5.444947. (iii) Draw the graph of Maggie’s profit. Solution: (iii) The graph of the profit is on Figure 6.

46/97

c





Figure 6: Profit for a collar plus owning the stock. 47/97

c





Suppose that you worked five years for Microsoft and have $100000 in stock of this company. You would like to insure this position buying a collar with expiration T years from now. You can choose K1 and K2 so that the combined premium is zero. In this case, no matter what the price of the stock T years form now, you will have $100000 or more. The cost of buying this insurance is zero. Notice that you have not got anything from free, you have loss interest in the stock. You also can make a collar with premium zero, by taking K1 = K2 = F0,T . In this case you will receive F0,T at time T , i.e. you are entering into a synthetic forward. Suppose that a zero–cost collar consists of buying a K1 –strike put option and selling a K2 –strike call option, where K1 < K2 . The profit of a zero–cost collar is   K1 − ST if ST < K1 , max(K1 −ST , 0)−max(ST −K2 , 0) = 0 if K1 ≤ ST < K2 ,   K2 − ST if K2 ≤ ST . 48/97

c





Speculating on volatility. Straddle A straddle consists of buying a K –strike call and a K –strike put with the same time to expiration. The payoff of this strategy is max(K − ST , 0) + max(ST − K , 0) = max(K − ST , 0) − min(K − ST , 0) =|ST − K |. Its profit is |ST − K | − (Put(K , T ) + Call(K , T ))(1 + i)T . A straddle is used to bet that the volatility of the market is higher than the market’s assessment of volatility. Notice that the prices of the put and call use the market’s assessment of volatility. 49/97

c





Example 6 (Use Table 1) Pam buys a $80–strike call option and a $80–strike put for 100 shares of XYZ stock. Both have expiration date one year from now. The current price of one share of XYZ stock is $75. The risk free annual effective rate of interest is 5%. The premium per share of a $80–strike call option is 5.444947. The premium per share of a $80–strike put option is 6.635423.

50/97

c





Example 6 (Use Table 1) Pam buys a $80–strike call option and a $80–strike put for 100 shares of XYZ stock. Both have expiration date one year from now. The current price of one share of XYZ stock is $75. The risk free annual effective rate of interest is 5%. The premium per share of a $80–strike call option is 5.444947. The premium per share of a $80–strike put option is 6.635423. (i) Calculate Pam’s profit as a function of the spot price at expiration.

51/97

c





Example 6 (Use Table 1) Pam buys a $80–strike call option and a $80–strike put for 100 shares of XYZ stock. Both have expiration date one year from now. The current price of one share of XYZ stock is $75. The risk free annual effective rate of interest is 5%. The premium per share of a $80–strike call option is 5.444947. The premium per share of a $80–strike put option is 6.635423. (i) Calculate Pam’s profit as a function of the spot price at expiration. Solution: (i) The future value per share of the cost of entering the option contracts is (Call(80, T ) + Put(80, T ))(1 + i)T =(5.444947 + 6.635423)(1.05) = 12.6843885. Pam’s profit is (100)(|ST − 80| − 12.6843885). c



52/97



Example 6 (Use Table 1) Pam buys a $80–strike call option and a $80–strike put for 100 shares of XYZ stock. Both have expiration date one year from now. The current price of one share of XYZ stock is $75. The risk free annual effective rate of interest is 5%. The premium per share of a $80–strike call option is 5.444947. The premium per share of a $80–strike put option is 6.635423. (ii) Make a table with Pam’s profit when the spot price at expiration is $65, $70, $75, $80, $85, $90, $95.

53/97

c





Example 6 (Use Table 1) Pam buys a $80–strike call option and a $80–strike put for 100 shares of XYZ stock. Both have expiration date one year from now. The current price of one share of XYZ stock is $75. The risk free annual effective rate of interest is 5%. The premium per share of a $80–strike call option is 5.444947. The premium per share of a $80–strike put option is 6.635423. (ii) Make a table with Pam’s profit when the spot price at expiration is $65, $70, $75, $80, $85, $90, $95. Solution: (ii) Pam’s profit table is Spot Price Pam’s profit

65 231.56

70 −268.44

Spot Price Pam’s profit

80 −1268.44

75 −768.44

85 −768.44

80 −1268.44

90 −268.44

95 231.56 54/97

c





Example 6 (Use Table 1) Pam buys a $80–strike call option and a $80–strike put for 100 shares of XYZ stock. Both have expiration date one year from now. The current price of one share of XYZ stock is $75. The risk free annual effective rate of interest is 5%. The premium per share of a $80–strike call option is 5.444947. The premium per share of a $80–strike put option is 6.635423. (iii) Draw the graph of Pam’s profit.

55/97

c





Example 6 (Use Table 1) Pam buys a $80–strike call option and a $80–strike put for 100 shares of XYZ stock. Both have expiration date one year from now. The current price of one share of XYZ stock is $75. The risk free annual effective rate of interest is 5%. The premium per share of a $80–strike call option is 5.444947. The premium per share of a $80–strike put option is 6.635423. (iii) Draw the graph of Pam’s profit. Solution: (iii) The graph of the profit is on Figure 7.

56/97

c





Example 6 (Use Table 1) Pam buys a $80–strike call option and a $80–strike put for 100 shares of XYZ stock. Both have expiration date one year from now. The current price of one share of XYZ stock is $75. The risk free annual effective rate of interest is 5%. The premium per share of a $80–strike call option is 5.444947. The premium per share of a $80–strike put option is 6.635423. (iv) Find the values of the spot price at expiration at which Pam makes a profit.

57/97

c





Example 6 (Use Table 1) Pam buys a $80–strike call option and a $80–strike put for 100 shares of XYZ stock. Both have expiration date one year from now. The current price of one share of XYZ stock is $75. The risk free annual effective rate of interest is 5%. The premium per share of a $80–strike call option is 5.444947. The premium per share of a $80–strike put option is 6.635423. (iv) Find the values of the spot price at expiration at which Pam makes a profit. Solution: (iv) Pam makes a profit if (100)(|ST − 80| − 12.6843885) > 0, i.e. if |ST − 80| > 12.6843885. This can happen if either ST − 80 < −12.6843885, or ST − 80 > 12.6843885. We have that ST − 80 < −12.6843885 is equivalent to ST < 80 − 12.6843885 = 67.3156115. ST − 80 > 12.6843885 is equivalent to ST > 92.6843885. Hence, Pam makes a profit if either ST < 67.3156115 or ST > 92.6843885. 58/97

c





Figure 7: Profit for a straddle. 59/97

c





Strangle

A strangle consists of buying a K1 –strike put and a K2 –strike call with the same expiration date, where K1 < K2 . The profit of this portfolio is max(K1 − ST , 0) + max(ST − K2 , 0) − ((Put(K1 , T ) + Call(K2 , T ))(1 + i)T  T  if ST < K1 , K1 − ST − ((Put(K1 , T ) + Call(K2 , T ))(1 + i) T = −((Put(K1 , T ) + Call(K2 , T ))(1 + i) if K1 ≤ ST < K2   T ST − K2 − ((Put(K1 , T ) + Call(K2 , T ))(1 + i) . if K2 ≤ ST .

60/97

c





Example 7 (Use Table 1) Beth buys a $75–strike put option and a $85–strike call for 100 shares of XYZ stock. Both have expiration date one year from now. The current price of one share of XYZ stock is $75. The risk free annual effective rate of interest is 5%. The premium per share of a $75–strike put option is 4.218281. The premium per share of a $85–strike call option is 3.680736.

61/97

c





Example 7 (Use Table 1) Beth buys a $75–strike put option and a $85–strike call for 100 shares of XYZ stock. Both have expiration date one year from now. The current price of one share of XYZ stock is $75. The risk free annual effective rate of interest is 5%. The premium per share of a $75–strike put option is 4.218281. The premium per share of a $85–strike call option is 3.680736. (i) Calculate Beth’s profit as a function of ST .

62/97

c





Example 7 (Use Table 1) Beth buys a $75–strike put option and a $85–strike call for 100 shares of XYZ stock. Both have expiration date one year from now. The current price of one share of XYZ stock is $75. The risk free annual effective rate of interest is 5%. The premium per share of a $75–strike put option is 4.218281. The premium per share of a $85–strike call option is 3.680736. (i) Calculate Beth’s profit as a function of ST . Solution: (i) Beth’s profit is 100(max(75 − ST , 0) + max(ST − 85, 0) − (4.218281 + 3.680736)(1.05)) =100(max(75 − ST , 0) + max(ST   100(75 − ST − 8.29396785) = 100(−8.29396785)   100(ST − 85 − 8.29396785)

− 85, 0) − 8.29396785) if ST < 75, if 75 ≤ ST < 85, if 85 ≤ ST . 63/97

c





Example 7 (Use Table 1) Beth buys a $75–strike put option and a $85–strike call for 100 shares of XYZ stock. Both have expiration date one year from now. The current price of one share of XYZ stock is $75. The risk free annual effective rate of interest is 5%. The premium per share of a $75–strike put option is 4.218281. The premium per share of a $85–strike call option is 3.680736. (ii) Make a table with Beth’s profit when the spot price at expiration is $50, $60, $70, $80, $90, $100, $110.

64/97

c





Example 7 (Use Table 1) Beth buys a $75–strike put option and a $85–strike call for 100 shares of XYZ stock. Both have expiration date one year from now. The current price of one share of XYZ stock is $75. The risk free annual effective rate of interest is 5%. The premium per share of a $75–strike put option is 4.218281. The premium per share of a $85–strike call option is 3.680736. (ii) Make a table with Beth’s profit when the spot price at expiration is $50, $60, $70, $80, $90, $100, $110. Solution: (ii) Spot Price Beth’s profit

50 1670.60

Spot Price Beth’s profit

60 670.60

90 −329.40

70 −329.40

100 670.60

80 −829.40

110 1670.60 65/97

c





Example 7 (Use Table 1) Beth buys a $75–strike put option and a $85–strike call for 100 shares of XYZ stock. Both have expiration date one year from now. The current price of one share of XYZ stock is $75. The risk free annual effective rate of interest is 5%. The premium per share of a $75–strike put option is 4.218281. The premium per share of a $85–strike call option is 3.680736. (iii) Draw the graph of Beth’s profit.

66/97

c





Example 7 (Use Table 1) Beth buys a $75–strike put option and a $85–strike call for 100 shares of XYZ stock. Both have expiration date one year from now. The current price of one share of XYZ stock is $75. The risk free annual effective rate of interest is 5%. The premium per share of a $75–strike put option is 4.218281. The premium per share of a $85–strike call option is 3.680736. (iii) Draw the graph of Beth’s profit. Solution: (iii) The graph of the profit is Figure 8.

67/97

c





Figure 8: Profit for a strangle. 68/97

c





The straddle and the strangle bet in volatility of the market in a similar way. Suppose that a straddle and a strangle are centered around the same strike price. The maximum loss of the strangle is smaller than the maximum loss of the straddle. However, the strangle needs more volatility to attain a profit. The possible profit of the strangle is smaller than that of the straddle. See Figure 9.

69/97

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Figure 9: Profit for a straddle and a strangle. 70/97

c





Written strangle

A written strangle consists of selling a K1 –strike call and a K2 –strike put with the same time to expiration, where 0 < K1 < K2 . A written strangle is a bet on low volatility.

71/97

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Speculating on low volatility: Butterfly spread Given 0 < K1 < K2 < K3 , a butterfly spread consists of: (i) selling a K2 –strike call and a K2 –strike put, all options for the notional amount. (ii) buying a K1 –strike call and a K3 –strike put, all options for the notional amount. The profit per share of this strategy is max(ST − K1 , 0) + max(K3 − ST , 0) − max(ST − K2 , 0) − max(K2 − ST , 0) − FVP, where FVP = (Call(K1 , T ) − Call(K2 , T ) −Put(K2 , T ) + Put(K3 , T )) (1 + i)T . 72/97

c





The profit per share of a butterfly spread is max(ST − K1 , 0) + max(K3 − ST , 0) − max(ST − K2 , 0) − max(K2 − ST , 0) − FVP   K3 − K2 − FVP if ST < K1 ,    S − K − K + K − FVP if K1 ≤ ST < K2 , 1 2 3 T =  −ST − K1 + K2 + K3 − FVP if K2 ≤ ST < K3 ,    K − K − FVP if K3 ≤ ST , 2 1 The graph of this profit is Figure 10

73/97

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Figure 10: Profit for a butterfly. 74/97

c





Example 8 (Use Table 1) Steve buys a 65–strike call and a 85–strike put and sells a 75–strike call and a 75–strike put. All are for 100 shares and have expiration date one year from now. The current price of one share of XYZ stock is $75. The risk free annual effective rate of interest is 5%.

75/97

c





Example 8 (Use Table 1) Steve buys a 65–strike call and a 85–strike put and sells a 75–strike call and a 75–strike put. All are for 100 shares and have expiration date one year from now. The current price of one share of XYZ stock is $75. The risk free annual effective rate of interest is 5%. (i) Find Steve’s profit as a function of the spot price at expiration.

76/97

c





Example 8 (Use Table 1) Steve buys a 65–strike call and a 85–strike put and sells a 75–strike call and a 75–strike put. All are for 100 shares and have expiration date one year from now. The current price of one share of XYZ stock is $75. The risk free annual effective rate of interest is 5%. (i) Find Steve’s profit as a function of the spot price at expiration. Solution: (i) We have that (Call(K1 , T ) − Call(K2 , T ) − Put(K2 , T ) + Put(K3 , T )) (1 + i)T =(14.31722 − 9.633117 − 7.78971 + 4.218281)(1.05) = 12.539459. Steve’s profit is (100) max(ST − 65, 0) + (100) max(85 − ST , 0) − (100) max(ST − 75, 0) − (100) max(75 − ST , 0) − (100)(12.539459) 77/97

c





Example 8 (Use Table 1) Steve buys a 65–strike call and a 85–strike put and sells a 75–strike call and a 75–strike put. All are for 100 shares and have expiration date one year from now. The current price of one share of XYZ stock is $75. The risk free annual effective rate of interest is 5%. (i) Find Steve’s profit as a function of the spot price at expiration. Solution: (i) (continuation) (100) max(ST − 65, 0) + (100) max(85 − ST , 0) − (100) max(ST − 75, 0) − (100) max(75 − ST , 0) − (100)(12.539459)   100(10 − 12.539459) if ST < 65,    100(S − 55 − 12.539459) if 65 ≤ ST < 75, T =  100(−ST + 95 − 12.539459) if 75 ≤ ST < 85,    100(10 − 12.539459) if 85 ≤ ST . 78/97

c





Example 8 (Use Table 1) Steve buys a 65–strike call and a 85–strike put and sells a 75–strike call and a 75–strike put. All are for 100 shares and have expiration date one year from now. The current price of one share of XYZ stock is $75. The risk free annual effective rate of interest is 5%. (ii) Make a table with Steve’s profit when the spot price at expiration is $60, $65, $70, $75, $80, $85, $90.

79/97

c





Example 8 (Use Table 1) Steve buys a 65–strike call and a 85–strike put and sells a 75–strike call and a 75–strike put. All are for 100 shares and have expiration date one year from now. The current price of one share of XYZ stock is $75. The risk free annual effective rate of interest is 5%. (ii) Make a table with Steve’s profit when the spot price at expiration is $60, $65, $70, $75, $80, $85, $90. Solution: (ii) table with Steve’s profit is Spot Price Steve’s profit

60 −253.95

Spot Price Steve’s profit

75 746.05

65 −253.95 80 246.05

70 246.05

85 −253.95

75 746.05 90 −253.95

80/97

c





Example 8 (Use Table 1) Steve buys a 65–strike call and a 85–strike put and sells a 75–strike call and a 75–strike put. All are for 100 shares and have expiration date one year from now. The current price of one share of XYZ stock is $75. The risk free annual effective rate of interest is 5%. (iii) Draw the graph of Steve’s profit.

81/97

c





Example 8 (Use Table 1) Steve buys a 65–strike call and a 85–strike put and sells a 75–strike call and a 75–strike put. All are for 100 shares and have expiration date one year from now. The current price of one share of XYZ stock is $75. The risk free annual effective rate of interest is 5%. (iii) Draw the graph of Steve’s profit. Solution: (iii) The graph of the profit is Figure 10.

82/97

c





Asymmetric Butterfly spread. Given 0 < K1 < K2 < K3 and 0 < λ < 1, a butterfly spread consists of: buying λ K1 –strike calls, buying (1 − λ) K3 –strike calls; and selling one K2 –strike call. The profit of this strategy is (λ) max(ST − K1 , 0) + (1 − λ) max(ST − K3 , 0) − max(ST − K2 , 0)   0 − FVPrem if ST < K1 ,    λ(S − K ) − FVPrem if K1 ≤ ST < K2 , 1 T = −λK1 + K2 − (1 − λ)ST − FVPrem if K2 ≤ ST < K3 ,    −λK + K − (1 − λ)K − FVPrem if K ≤ S , 1 2 3 3 T where FVPrem = (λCall(K1 , T ) − Call(K2 , T ) + (1 − λ)Call(K3 , T )) (1+i)T . 83/97

c





The profit of an symmetric butterfly spread is (λ) max(ST − K1 , 0) + (1 − λ) max(ST − K3 , 0) − max(ST − K2 , 0)   if ST < K1 , 0 − FVPrem   λ(S − K ) − FVPrem if K1 ≤ ST < K2 , 1 T =  −λK1 + K2 − (1 − λ)ST − FVPrem if K2 ≤ ST < K3 ,    −λK + K − (1 − λ)K − FVPrem if K ≤ S , 1 2 3 3 T For an asymmetric butterfly spread, the profit functions increases in one interval and decreases in another interval. The rate of increase is not necessarily equal to the rate of decrease.

84/97

c





Example 9 (Use Table 1) Karen buys seven 65–strike calls, buys three 85–strike calls and sells ten 75–strike calls. Each option involves 100 shares of XYZ stock. Both have expiration date one year from now. The risk free annual effective rate of interest is 5%.

85/97

c





Example 9 (Use Table 1) Karen buys seven 65–strike calls, buys three 85–strike calls and sells ten 75–strike calls. Each option involves 100 shares of XYZ stock. Both have expiration date one year from now. The risk free annual effective rate of interest is 5%. (i) Find Karen’s profit as a function of the spot price at expiration.

86/97

c





Example 9 (Use Table 1) Karen buys seven 65–strike calls, buys three 85–strike calls and sells ten 75–strike calls. Each option involves 100 shares of XYZ stock. Both have expiration date one year from now. The risk free annual effective rate of interest is 5%. (i) Find Karen’s profit as a function of the spot price at expiration. Solution: (i) The future value per share of the cost of entering the option contracts is (7)(14.31722)(1.05) + (3)(7.78971)(1.05) − (3.680736)(1.05) =35.0339184. Karen’s profit is (100) ((7) max(ST − 65, 0) + (3) max(ST − 85, 0) −(10) max(ST − 75, 0) − (35.0339184)) 87/97

c





Example 9 (Use Table 1) Karen buys seven 65–strike calls, buys three 85–strike calls and sells ten 75–strike calls. Each option involves 100 shares of XYZ stock. Both have expiration date one year from now. The risk free annual effective rate of interest is 5%. (i) Find Karen’s profit as a function of the spot price at expiration. Solution: (i) (continuation) (100) ((7) max(ST − 65, 0) + (3) max(ST − 85, 0) −(10) max(ST − 75, 0) − (35.0339184))   if 100(−35.0339184)     100((7)(ST − 65) − 35.0339184) if    100((7)(S − 65) − (10)(S − 75) T T =  −35.0339184) if      100((7)(ST − 65) − (10)(ST − 75)    +(3)(S − 85) − 35.0339184) if T c


ST < 65, 65 ≤ ST < 75, 75 ≤ ST < 85, 85 ≤ ST ,


88/97



Example 9 (Use Table 1) Karen buys seven 65–strike calls, buys three 85–strike calls and sells ten 75–strike calls. Each option involves 100 shares of XYZ stock. Both have expiration date one year from now. The risk free annual effective rate of interest is 5%. (ii) Make a table with Karen’s profit when the spot price at expiration is $60, $65, $70, $75, $80, $85, $90.

89/97

c





Example 9 (Use Table 1) Karen buys seven 65–strike calls, buys three 85–strike calls and sells ten 75–strike calls. Each option involves 100 shares of XYZ stock. Both have expiration date one year from now. The risk free annual effective rate of interest is 5%. (ii) Make a table with Karen’s profit when the spot price at expiration is $60, $65, $70, $75, $80, $85, $90. Solution: (ii) A table with Karen’s profit is Spot Price Karen’s profit Spot Price Karen’s profit

60 −3503.39 75 3496.61

65 −3503.39 80 1996.61

70 −3.39 85 496.61

75 3496.61 90 496.61

90/97

c





Example 9 (Use Table 1) Karen buys seven 65–strike calls, buys three 85–strike calls and sells ten 75–strike calls. Each option involves 100 shares of XYZ stock. Both have expiration date one year from now. The risk free annual effective rate of interest is 5%. (iii) Draw the graph of Karen’s profit.

91/97

c





Example 9 (Use Table 1) Karen buys seven 65–strike calls, buys three 85–strike calls and sells ten 75–strike calls. Each option involves 100 shares of XYZ stock. Both have expiration date one year from now. The risk free annual effective rate of interest is 5%. (iii) Draw the graph of Karen’s profit. Solution: (iii) The graph of the profit is on Figure 11.

92/97

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Figure 11: Profit for an asymmetric butterfly. 93/97

c





Box spreads. A box spread is a combination of options which create a synthetic long forward at one price and a synthetic short forward at a different price. Let K1 be the price of the synthetic long forward. Let K2 be the price of the synthetic short forward. With a box spread, you are able to buy an asset for K1 at time T and sell it for K2 at time T not matter the spot price at expiration. At time T , a payment of K2 − K1 per share is obtained. A box spread can be obtained from: (i) buy a K1 –strike call and sell a K1 –strike put. (ii) sell a K2 –strike call and buy a K2 –strike put. If put–call parity holds, the premium per share to enter these option contract is Call(K1 , T ) − Put(K1 , T ) − (Call(K2 , T ) − Put(K2 , T )) =(K2 − K1 )(1 + i)−T . 94/97

c





I

If K1 < K2 , a box spread is a way to lend money. An investment of (K2 − K1 )(1 + i)−T per share is made at time zero and a return of K2 − K1 per share is obtained at time T .

I

If K1 > K2 , a box spread is a way to borrow money. A return of (K1 − K2 )(1 + i)−T per share is received at time zero and a loan payment of K1 − K2 per share is made at time T .

95/97

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Example 10 Mario needs $50,000 to open a pizzeria. He can borrow at the annual effective rate of interest of 8.5%. Mario also can buy/sell three–year options on XYZ stock with the following premiums per share: Call(K , T ) Put(K , T ) K

14.42 7.37 70

7.78 17.29 90

Mario buys a 90–strike call and a 70–strike put, and sells a 90–strike put and a 70–strike call. All the options are for the same nominal amount. Mario receives a total of $50000 from these sales. Find Mario’s cost at expiration time to settle these options. Find the annual rate of return that Mario gets on this ”loan”. 96/97

c





Solution: The price per share of Mario’s portfolio is 7.78 + 7.37 − 17.29 − 14.42 = −16.56. So, the nominal amount of each option is 50000 16.56 = 3019.3237. Initially, Mario gets $50000 for entering these option contracts. In three years, Mario buys at $90 per share and sells at $70 per share. Hence, Mario pays (3019.3237)(90 − 70) = 60386.474 for settling the option contracts. Let i be the annual effective rate of interest on this loan. We have that 50000(1 + i)3 = 60386.474 and i = 6.493529844%.

97/97

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Manual for SOA Exam FM/CAS Exam 2. Chapter 7. Derivatives markets. Section 7.9. Risk Management. c



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Section 7.9. Risk Management.

Risk Management

The main reasons why firms enter derivatives are: to hedge, to speculate, and to reduce transactions costs. Managers use derivatives taking in account their view of the market. So, a speculative component is added to each decision. Firms convert inputs, such a labor and raw materials, into goods and services. A firm makes money if its income exceeds its costs. A change in the price of raw materials could make the firm unprofitable. A firm can use derivatives to alter its risk and protect its profitability. To do this is to do risk management.

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Firms do risk management to attain financial stability. There are many reasons to avoid large losses. After a large loss, interest rates on loans will be obtained at a higher rate. Large losses can cause bankruptcy and distress costs. After a large loss, a company can face low cashflows and difficulty making fixed obligations such as wages and payments to banks and suppliers. This makes more costly to find employees, debtors and suppliers. Suppose that the profit can be modeled by a random variable X . Let f (X ) be the the profit after taking in account the effects of this profit. Because of the reasons before, f can be modeled using a concave function.

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Suppose that the financial total   1.5x f (x) = 1.1x   x

impact of the profit is

Example 1

Since

  1.5 f 0 (x) = 1.1   1

if x ≤ −2, if − 2 < x ≤ 0 if x > 0. if x ≤ −2, if − 2 < x ≤ 0 if x > 0.

is a nonincreasing function, f is a concave function.

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Example 2 Suppose that the a company profit before taxes can be modeled by a random variable X . The company pays 35% of its profit in taxes, if the profit is positive. It does not pay any taxes if its profit is negative. The company’s profit after taxes is f (X ), where ( (0.65)x if x > 0, f (x) = x if x ≤ 0. f (x) is a concave function.

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If the company is able to hedge and attain a constant profit of E [X ], then after taxes its profit is f (E [X ]). If the company does not hedge its profit, its expected profit after taxes is E [f (X )]. Since f is a concave function, by the Jensen’s inequality, E [f (X )] ≤ f (E [X ]).

Theorem 1 (Jensen’s inequality) Let X be a random variable. Let f : R → R be a function. Then, (i) If f is convex, then f (E [X ]) ≤ E [f (X )]. (ii) If f is concave, then E [f (X )] ≤ f (E [X ]). By using hedging, the random profit X is changed into a random profit with less variation (taking big losses with less probability). The expectation of this new random profit is larger than the expectation of the original profit. 6/44

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Example 3 Hank is a wheat farmer. He will produce 50000 bushels of wheat at the end of one year. The cost of producing this wheat is $5.7 per bushel. The price of wheat per bushel in one year will be: S1 Probability

5.5 0.5

6.5 0.5

A speculator offers short forward contracts for a price equal to 1% less than the expected value of the price of the wheat. It also offers a 6–strike put option with a price equal to 1% more than the present value of the expected payoff of this put. The annual effective rate of interest is 5%. Hank pays 30% of its profits on taxes and has no tax benefits for losses.

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(i) Calculate the prices of the forward contract and the put option.

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(i) Calculate the prices of the forward contract and the put option. Solution: (i) The expected price of wheat is E [S1 ] = (0.5)(5.5) + (0.5)(6.5) = 6.00/bush. The price of a forward contract is 6(0.99) = 5.94/bush. The expected payoff of the put is E [max(6 − S1 , 0)] = (0.5) max(6 − 5.5, 0) + (0.5) max(6 − 6.5, 0) =(0.5)(6 − 5.5) + (0.5)(0) = 0.25. The price of a put contract is (0.25)(1.01)(1.05)−1 = 0.2404761905/bush.

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(ii) Calculate Hank’s expected profit before taxes if (a) he does not buy any derivative, (b) he enters a short forward contract, (c) he buys a put option. Which strategy has the biggest expected profit before taxes?

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(ii) Calculate Hank’s expected profit before taxes if (a) he does not buy any derivative, (b) he enters a short forward contract, (c) he buys a put option. Which strategy has the biggest expected profit before taxes? Solution: (ii) For an uninsured position, Hank’s profit before taxes is 50000(S1 − 5.7). Hank’s expected profit before taxes is 50000(E [S1 ] − 5.7) = 50000(6 − 5.7) = 15000. Entering the short forward contract, Hank’s profit before taxes is 50000(S1 − 5.7 + 5.94 − S1 ) = (50000)(5.94 − 5.7) = 12000. Entering the short forward contract, Hank’s expected profit before taxes is 12000.

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Buying the put option, Hank’s profit before taxes is 50000 (S1 − 5.7 + max(6 − S1 , 0) − (0.2404761905)(1.05)) =50000 (max(6, S1 ) − 5.7 − (0.2404761905)(1.05)) =50000(max(6, S1 ) − 5.9525). Buying the put option, Hank’s expected profit before taxes is E [50000 (max(6, S1 ) − 5.9525)] =50000 (E [max(6, S1 )] − 5.9525) =(50000)((0.5)(6) + (0.5)(6.5) − 5.9525)) = 14875. The biggest expected profit before taxes is attained for an uninsured position.

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(iii) Calculate Hank’s expected profit after taxes for each of three strategies in (ii). Which strategy has the biggest expected profit after taxes?

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(iii) Calculate Hank’s expected profit after taxes for each of three strategies in (ii). Which strategy has the biggest expected profit after taxes? Solution: (iii) For an uninsured position, Hank’s profit before taxes is ( 50000(5.5 − 5.7) if S1 = 5.5, 50000(S1 − 5.7) = 50000(6.5 − 5.7) if S1 = 6.5, Hank’s profit after taxes is ( 50000(5.5 − 5.7) 50000(S1 − 5.7) = 50000(6.5 − 5.7)(0.7)

if S1 = 5.5, if S1 = 6.5,

Hank’s expected profit after taxes is (0.5)(50000)(5.5 − 5.7) + (0.5)(50000)(6.5 − 5.7)(0.7) = 9000. Entering the forward contract, Hank’s expected profit after taxes is 12000(0.7) = 8400. c



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Buying the put option, Hank’s profit before taxes is =50000 (max(6, S1 ) − 5.9525) ( 50000(6 − 5.9525) if S1 = 5.5, == . 50000(6.5 − 5.9525) if S1 = 6.5, Hank’s profit after taxes is 50000 (max(6, S1 ) − 5.9525) ( 50000(6 − 5.9525)(0.7) = 50000(6.5 − 5.9525)(0.7)

if S1 = 5.5, if S1 = 6.5,

Hank’s expected profit after taxes is

(0.5)(50000)(6−5.9525)(0.7)+(0.5)(50000)(6.5−5.9525)(0.7) = 10412.5. The biggest expected profit after taxes is attained when buying the put. c



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There are several reasons not to hedge: I

paying transaction costs.

I

need expertise to asses costs and benefits of a given strategy.

I

need expertise to do accounting and taxes in derivative transactions.

In the real world, small companies are discouraged to do derivatives because the reasons above. However, large companies have financial, accounting and legal departments which allow them to take advantage of the opportunities on market derivatives. The financial department of a large company can asses derivatives as well or better than the market does. Their legal and accounting departments allow them to take advantage of the current tax laws.

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A producer perspective on risk management. (URMC) Utah Red Mountain Company is a copper–mining company. It plans to mine and sell 1,000,000 pounds of copper over the next year. Suppose that it sells all the next year’s production, precisely one year from today. Suppose that the firm incurs in two types of costs: fixed costs and variable costs. Fixed costs are assumed whether its mine is open or closed. Variable costs are assumed only its mine is open. Suppose that the total fixed costs add to $0.75/lb. and the total variable costs add to $2.25/lb. Let S1 be the price of copper per pound in one year. I

The net income per pound if the mine is open is S1 − 0.75 − 2.25 = S1 − 3.00.

I

The net income per pound if the mine is closed is −0.75/lb. 17/44

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Utah Red Mountain Company would be better to close its mine if S1 − 3 ≤ −0.75, which is equivalent to S1 ≤ 2.25. URMC would be better keep its mine open if S1 ≥ 2.25. But, if 2.25 ≤ S1 ≤ 3, URMC assumes the loss 3 − S1 . If S1 ≥ 3 and URMC keeps its mine open, its (positive) net income is S1 − 3. The following table shows the net income of URMC (see also Figure 1).

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Figure 1: Profit according whether the mine is closed, insured open and open insured with a forward. 19/44

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Suppose that URMC can enter a short forward contract agreeing to sell its copper one year from now. Suppose that F0,1 = 3.5/lb. If Utah Red Mountain Company enters this contract, its profit is $0.5/lb. In this way Utah Red Mountain Company reduces risk. Figure 1 shows the graph of the profit under the three considered alternatives. For an uninsured position, the possible losses can be very high. However, by entering the forward, the company has a fixed benefit.

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However, URMC would like to benefit if the price of the copper goes higher than $3.5/lb. It could use options. Suppose that the continuous rate of interest is 0.05 and the variability is σ = 0.25. Using the Black–Scholes formula, we have the following table of premiums of options: Table 1: K Call(K , T ) Put(K , T )

3.3 0.42567 0.23542

3.4 0.3762 0.28107

3.5 0.33119 0.33119

3.6 0.29048 0.3856

3.7 0.25386 0.44411

3.8 0.22111 0.50648

3.9 0.19196 0.57245

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Utah Red Mountain Company could buy a 3.7–strike put. URMC’s profit per pound is S1 − 3 + max(3.7 − S1 , 0) − (0.44411)e 0.05 = max(3.7, S1 ) − 3.466880007 ( 0.2331199934 if S1 < 3.7, = S1 − 3.466880007 if 3.7 ≤ S1 .

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Under this strategy, URMC does not benefit much if the price of copper is high. Another strategy is to buy a 3.4–strike put. The profit per pound is S1 − 3 + max(3.4 − S1 , 0) − (0.28107)e 0.05 = max(3.4, S1 ) − 3.295480767 ( 0.104519233 if S1 < 3.4, = S1 − 3.295480767 if 3.4 ≤ S1 . Under this strategy, the company makes $0.17139924/lb more than before if the price of copper is over $3.7/oz. However, its guaranteed profit is 0.104519233, which is smaller than the guaranteed profit under a 3.7–strike put.

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See Figure 2 for the profit under a short forward, a 3.7–strike put and 3.4–strike put.

Figure 2: Profit for forward, 3.4–strike put and 3.7–strike put. 24/44

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In other words, buying a put is like buying insurance against small spot prices. As bigger the strike price as bigger the price of the insurance. As bigger the strike price as bigger the obtained payment when the insurance is needed. A producer company needs to buy a put with a strike large enough strike to cover low spot prices. If the strike put is too large, the company will be wasting money in insurance which it does not need.

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Example 4 Utah Red Mountain Company has the following profits: (i) If it does not insure, S1 − 3.00. (ii) If it enters a short forward, 0.5. (iii) If it buys a 3.4–strike put, max(3.4, S1 ) − 3.295480767. Suppose that an actuary consulting for Utah Red Mountain Company estimates that the price of cooper in one year will be either 2.6, or 3.6, or 4.6, with respective probabilities 0.36, 0.33 and 0.31. (i) Compute the URMC’s expected profit before taxes for each of the above strategies. Find the strategy with the biggest expected profit before taxes. (ii) The URMC pays a 40% tax rate, and has no tax benefits for losses. Compute the URMC’s expected profit after taxes for each of the above strategies. Find the strategy with the biggest expected profit after taxes. 26/44

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Solution: (i) uninsured profit profit under a short forward profit under a 3.4–strike put ST Probability

−0.4 0.5 0.104519233 2.6 0.36

0.6 0.5 0.304519233 3.6 0.33

1.6 0.5 1.304519233 4.6 0.31

The expected profit under an uninsured position is (−0.40)(0.36) + (0.60)(0.33) + (1.60)(0.31) = 0.55. The expected profit under a short forward is 0.5. The expected profit under a 3.7–strike put is (0.104519233)(0.36) + (0.304519233)(0.33)(1.304519233)(0.31) =0.542519233. The strategy with the biggest expected profit before taxes is the uninsured position. c



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Solution: (ii) After taxes, the profits are given by the following table: uninsured profit profit with a forward profit with a 3.4–strike put Probability

−0.40 0.3 0.0627115398 0.36

0.36 0.3 0.1827115398 0.33

0.96 0.3 0.7827115398 0.31

The expected profit after taxes for an uninsured position is (−0.40)(0.36) + (0.36)(0.33) + (0.96)(0.31) = 0.2724. The expected profit after taxes under a short forward is (0.6)(0.5) = 0.3. The expected profit after taxes under a a 3.4–strike put is (0.6)(0.542519233) = 0.3255115398.

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Another strategy for URMC is to buy a 3.6–strike put and sell a 3.9–strike call option. Buying these two options, URMS will sell cooper at min(max(3.6, S1 ), 3.9). This strategy is called a collar. URMC’s profit per pound is S1 − 3 + max(3.6 − S1 , 0) − max(S1 − 3.9, 0) − (0.3856 − 0.19196)e 0.05 = − 3 + 3.9 + max(3.6, S1 ) − max(S1 , 3.9) − (0.3856 − 0.19196)e 0.05

= − 3 + 3.9 + max(3.6, S1 ) − max(S1 , 3.9, 3.6) − (0.3856 − 0.19196)e 0.05 = min(max(3.6, S1 ), 3.9) − 3 − (0.3856 − 0.19196)e 0.05 = min(max(3.6, S1 ), 3.9) − 3.203568135   if S1 < 3.6, 0.396431865 = S1 − 3.203568135 if 3.6 ≤ S1 < 3.9,   0.696431865 if 3.9 < ST . 29/44

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Under this strategy the profit of the company is very close to that of a forward contract. But, instead of winning a constant of $0.5/lb. in the forward contract, the company’s profit varies with the future price of copper, although not much.

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Another type of strategies are the ones called paylater strategies. By a buying a put, URMC hedges against low prices. But, if the prices are high, its profit is not as high as when it does not hedge. A paylater strategy allows to have the usual profit for high enough spot prices. It is like the price of the insurance does not need to be paid. Consider the strategy of buying two 3.6–strike puts and buying a K –strike call, such that the cost of the portfolio is zero. The cost of two 3.6–strike puts is (2)(0.3856) = 0.7712. By numerical methods, we have that the cost of a 4.1763–strike put is 0.7712. Hence, K = 4.1763. The profit of this strategy is S1 − 3 + 2 max(3.6 − S1 , 0) − max(4.1763 − S1 , 0) = − 3 + 2 max(3.6, S1 ) − max(4.1763, S1 )   if S1 < 3.6, 0.0237 = 2S1 − 7.1763 if 3.6 ≤ S1 < 4.1763,   S1 − 3 if 4.1763 < S1 , 31/44

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Notice that under the previous strategy Company URMC always has a positive profit and if the spot price is bigger than 4.1763, its profit is the same as it would not hedge.

Figure 4: Profit for a paylater. 32/44

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The buyer’s perspective on risk management.

Toughminum makes fridges. Suppose that: (i) The fixed cost per fridge is $100. (ii) Toughminum sells fridges for $350. (iii) To manufacture a fridge Toughminum need 5 pounds of aluminum. Let ST be the price of a pound of aluminum at time T . The profit one year from now is 350 − (5)(S1 ) − 100 = 250 − 5S1 .

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Figure 5 shows the graph of this profit (”uninsured” line).

Figure 5: Profit uninsured, long forward and call. 34/44

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Toughminum could faces severe loses if the price of the aluminum goes very high. To hedge risk, it could enter a long forward. If aluminum is selling at $40 a pound in the forward market, the profit of entering a long forward is 350 − (5)(40) − 100 = 50. Figure 5 shows the graph of this profit (”forward” line).

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Another alternative is to buy a call. Suppose that Toughminum buys a 35–strike call with an expiration date one year from now and nominal amount 5 lbs. Assume that σ = 0.35 and r = 0.06. Then, Call(35, 1) = 7.609104. The profit is 250 − 5S1 + 5 max(S1 − 35, 0) − 5Call(35, 1)e r =250 + 5 max(−35, −S1 ) − 5(7.609104)e 0.06 =209.6018764 − 5 min(35, S1 ) ( 209.6018764 − 5S1 if S1 < 35, = 34.6018764 if 35 ≤ S1 . Figure 5 shows the graph of this profit (”call” line).

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Example 5 Toughminum has the following profits: (i) If it does not insure, 250 − 5S1 . (ii) If it enters a short forward 50. (iii) If it buys a 35–strike call, 209.6018 − 5 min(35, S1 ). Suppose that an actuary consulting for Toughminum estimates that the price of aluminum in one year will be either 30, 35, or 40, or 55, with respective probabilities 0.25, 0.25, 0.25, and 0.25. (i) Compute the Toughminum’s expected profit before taxes for each of the above strategies. Find the strategy with the biggest expected profit before taxes. (ii) The Toughminum pays a 35% tax rate, and has no tax benefits for losses. Compute the Toughminum’s expected profit after taxes for each of the above strategies. Find the strategy with the biggest expected profit after taxes. 37/44

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Solution: (i) uninsured profit profit under a long forward profit under a 35–strike call ST Probability

100 50 59.6019 30 0.25

75 50 34.6019 35 0.25

50 50 34.6019 40 0.25

−25 50 34.6019 55 0.25

The expected profit under an uninsured position is (100)(0.25) + (75)(0.25) + (50)(0.25) + (−25)(0.25) = 50. The expected profit under a long forward is 50. The expected profit under a 35–strike call is

(59.6019)(0.25) + (34.6019)(0.25) + (34.6019)(0.25) + (34.6019)(0.25) = The strategies with the biggest expected payoff are the uninsured position and the long forward position. c



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Solution: (ii) After taxes, the profits are given by the following table: uninsured profit profit with a long forward profit under a 35–strike call ST Probability

65.00 32.5 38.7412 30 0.25

48.75 32.5 22.4912 35 0.25

32.50 32.5 22.4912 40 0.25

−25.00 32.5 22.4912 55 0.25

The expected profit for an uninsured position is (65)(0.25) + (48.75)(0.25) + (32.5)(0.25) + (−25)(0.25) = 30.3125. The expected profit under a long forward is 32.5. The expected profit under a 35–strike call is

(38.7412)(0.25) + (22.4912)(0.25) + (22.4912)(0.25) + (22.4912)(0.25) = The strategy with the biggest expected payoff is the long forward position. c



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Toughminum could sell a 30–strike put option and buy a 45–strike call option. Both with nominal amount 5 lbs. This position is called a 30–45 reverse collar. Under this position, Toughminum will buy aluminum at min(max(30, S1 ), 45) per lb. The cost of a 30–strike put is $1.308289/lb. The cost of a 45–strike call is $3.514166/lb. The profit per ounce is 350 − 100 − 5S1 − 5 max(30 − S1 , 0) + 5 max(S1 − 45, 0) + 5Put(30, 1)e r − 5Call(45, 1)e r =250 − 5 max(30, S1 ) − 5(45) + 5 max(S1 , 45) + 5(1.308289)e 0.06 − 5(3.514166)e 0.06 =238.2886 − 5 max(30, S1 ) − 5(45) + 5 max(S1 , 45, 30) =238.2886 − 5 min(max(30, S1 ), 45) or

  88.2886 profit = 238.2886 − 5S1   13.2886 c


if S1 < 30, if 30 ≤ S1 < 45, if 45 ≤ S1 .


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The graph of this profit in Figure 6.

Figure 6: Profit for a 30–45 reverse collar. 41/44

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Importer/exporter’s perspective.

Suppose that Company ABCD imports electronics to Canada. It gets paid in Canadian dollars. Whenever this company gets a payment, it needs to exchange Canadian dollars into US dollars. Suppose that in two months, ABCD expects to get 100,000 Canadian dollars. If the price of a Canadian dollar at time T is ST , the amount of US dollars the company ABCD will get is 100000 S2/12 . To hedge against possible changes in exchange rates, company ABCD can sell a forward on 100000 Canadian dollars at the current price. It also can buy a put on Canadian dollars.

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Example 6 Suppose that Company ABCD buys a put on (Canadian $’s) CAD100000 with a strike price of (U.S.A. dollar) USD0.85 per CAD for USD0.01 per CAD. Two months later, ABCD receives CAD100000. At this moment the exchange rate is USD0.845 per CAD. (i) How many US dollars does company ABCD gets in this transaction? (ii) How many US dollars would company ABCD have gotten in the exchange if it would not have signed the put?

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Example 6 Suppose that Company ABCD buys a put on (Canadian $’s) CAD100000 with a strike price of (U.S.A. dollar) USD0.85 per CAD for USD0.01 per CAD. Two months later, ABCD receives CAD100000. At this moment the exchange rate is USD0.845 per CAD. (i) How many US dollars does company ABCD gets in this transaction? (ii) How many US dollars would company ABCD have gotten in the exchange if it would not have signed the put? Solution: (i) Since the strike price is bigger than the current spot price, the company exercises the put. It gets (100000)(0.85 − 0.01) = 84000 US dollars. (i) Company ABCD would have got 100000(0.845) = 84500 US dollars 44/44

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Manual for SOA Exam FM/CAS Exam 2. Chapter 7. Derivatives markets. Section 7.10. Swaps. c



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Section 7.10. Swaps.

Swaps Definition 1 A swap is a contract between two counterparts to exchange two similar financial quantities which behave differently. I

The two things exchanged are called the legs of the swap.

I

A common type of swap involves a commodity. Another common type of swap is an interest rate swap of a fixed interest rate in return for receiving an adjustable rate.

I

Usually, one leg involves quantities that are known in advance, known as the fixed leg. The other involves quantities that are (uncertain) not known in advance, known as the floating leg.

I

Usually, a swap entails the exchange of payments over time. 2/78

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LIBOR.

The (London Interbank office rate) LIBOR is the most widely used reference rate for short term interest rates world–wide. The LIBOR is published daily the (British Bankers Association) BBA. It is based on rates that large international banks in London offer each other for inter–bank deposits. Rates are quoted for 1–month, 3–month, 6–month and 12–month deposits.

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The following table shows the LIBOR interest rates for a loan in dollars during a week on June, 2007: Date 6–18–2007 6–19–2007 6–20–2007 6–21–2007 6–22–2007

1–month 5.3200% 5.3200% 5.3200% 5.3200% 5.3200%

3–month 5.3600% 5.3600% 5.3600% 5.3600% 5.3600%

6–month 5.4000% 5.3981% 5.3931% 5.3934% 5.3900%

9–month 5.4400% 5.4369% 5.4194% 5.4273% 5.4200%

12–month 5.4741% 5.4650% 5.4387% 5.4531% 5.4494%

The previous rates are from http://www.bba.org.uk/bba.

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A LIBOR loan is an adjustable loan on which the interest rate is tied to a specified Libor. The interest rate is the most recent value of the LIBOR plus a margin, subject to any adjustment cap. LIBOR is used in determining the price of interest rate futures, swaps and Eurodollars. The most important financial derivatives related to LIBOR are Eurodollar futures. Traded at the Chicago Mercantile Exchange (CME), Eurodollars are US dollars deposited at banks outside the United States, primarily in Europe. The interest rate paid on Eurodollars is largely determined by LIBOR. Eurodollar futures provide a way of betting on or hedging against future interest rate changes.

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Structure of interest rates.

Let P(0, t) be the price of a $1–face value zero coupon bond 1 maturing on date t. Notice that P(0,t) is the interest factor from time zero to time t, i.e. $1 invested at time 0 accumulates to 1 P(0,t) at time t. P(0, t) is the discount factor from time zero to time t. The implied interest factor from time tj−1 to time tj is P(0,tj−1 ) P(0,tj ) . The implied forward rate from time tj−1 to time tj is P(0,tj−1 ) P(0,tj ) 1 P(0,n) .

r0 (tj−1 , tj ) = (1 + sn

)n

=

− 1. Let sn be the n–year spot rate. Then,

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A quantity which appear later is the coupon rate R for a n–year bond with annual coupons and face value, redemption value and price all equal to one. The price of this bond is 1=

n X

RP(0, j) + P(0, n).

j=1

Hence, 1 − P(0, n) R = Pn . j=1 P(0, j)

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Example 1 The following table lists prices of zero–coupon $1–face value bonds with their respective maturities: Number of years to maturity 1 2 3 4

Price $0.956938 $0.907029 $0.863838 $0.807217

(i) Calculate the 1–year, 2–year, 3–year, and 4–year spot rates of interest. (ii) Calculate the 1–year, 2–year, and 3–year forward rates of interest. (iii) Calculate the coupon rate R for a j–year bond with annual coupons whose face value, redemption value and price are all one. 8/78

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Solution: (i) Note that the price of j–th bond is 1 P(0, j) = (1 + sj )−j . Hence, sj = P(0,j) 1/j − 1. In particular, 1 1 −1= P(0, 1) 0.956938 1 1 s2 = −1= P(0, 2)1/2 0.9070291/2 1 1 −1= s3 = 1/3 P(0, 3) 0.8638381/3 1 1 −1= s4 = 1/4 P(0, 4) 0.8072171/4 s1 =

− 1 = 4.499967135% − 1 = 5.000027694% − 1 = 4.999983734% − 1 = 5.499991613%

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Solution: (ii) To get the j − 1 year forward rate fj , we do fj =

(1+sj )j (1+sj−1 )j−1

−1= f2 = f3 = f4 =

P(0,j−1) P(0,j)

0.956938 0.907029 0.907029 0.863838 0.863838 0.807217

− 1. We get that:

− 1 = 5.502470153%, − 1 = 4.999895814%, − 1 = 7.014346824%.

10/78

c




Solution: (iii) We have that Rj = R1 R2 R3 R4

= = = =


1−P(0,j) Pj . k=1 P(0,k)

Hence, we get that:

1−0.956938 0.956938 − 1 = 4.499978055%, 1−0.907029 0.956938+0.907029 − 1 = 4.987802896%, 1−0.863838 0.956938+0.907029+0.863838 − 1 = 4.991632466%, 1−0.807217 0.956938+0.907029+0.863838+0.807217 − 1 = 5.453516272%.

11/78

c





Example 2 Suppose the current LIBOR discount factors P(0, tj ) are given by the table below. LIBOR discout rates P(0, tj ) time in months

0.986923

0.973921

0.961067

0.948242

3

6

9

12

Calculate the annual nominal interest rate compounded quarterly for a loan for the following maturity dates: 3, 6, 9 and 12 months.

12/78

c





(4)

Solution: Let sj be the annual nominal interest rate compounded quarterly for a loan maturing in 3j months, (2) 1 . So, j = 1, 2, 3, 4. Then, (1 + sj /4)j = P(0,j/4)

(4)

s2

(4)

s3

(4)

s4

1 =4 − 1 = 5.300109532%, 0.986923 ! 1/2 1 =4 − 1 = 5.320086032%, 0.973921 ! 1/3 1 =4 − 1 = 5.330019497%, 0.961067 ! 1/4 1 =4 − 1 = 5.350015985%. 0.948242 (4) s1

13/78

c





We denote by rt0 (t1 , t2 ) to the nonannualized interest rate for the period from t1 to t2 using the interest rates at t0 , i.e. 1 + rt0 (t1 , t2 ) is the interest factor for the period from t1 to t2 using the interest rates at t0 . We denote by Pt0 (t0 , t1 ) to the price at time t0 of a zero–coupon bond with face value $1 and redemption time t1 . So, 1 + rt0 (t1 , t2 ) =

Pt0 (t0 , t1 ) . Pt0 (t0 , t2 )

Notice that we abbreviate P0 (0, t) = P(0, t).

14/78

c





Forward rate agreement.

Suppose that a borrower plans to take a loan of $L at time t1 , where t1 > 0. He will repaid the loan at time t2 , where t2 > t1 . The amount of the loan payment depends on the term structure of interest rates at time t1 . Let rt1 (t1 , t2 ) be interest rate from t1 to t2 with respect to the structure of interest rates at time t1 , i.e. a zero–coupon bond with face value F and redemption time t2 costs F 1+rt1 (t1 ,t2 ) at time t1 . To pay the loan, the borrower needs to pay $L(1 + rt1 (t1 , t2 )) at time t2 .

15/78

c





Since rt1 (t1 , t2 ) is unknown at time zero, the borrower does not know how much it will have to pay for the loan. In order to hedge against increasing interest rates, the borrower can enter into a (FRA) forward rate agreement. A FRA is a financial contract to exchange interest payments for a notional principal on settlement date for a specified period from start date to maturity date. Usually one of the interest payments is relative to a benchmark such as the LIBOR. This is a floating interest rate, which was described in Subsection 2. The other interest payment is with respect to a fixed rate of interest. An FRA contract is settled in cash. The settlement can be made either at the beginning or at the end of the considered period, i.e. either at the borrowing time or at the time of repayment of the loan.

16/78

c





The two payments involved in an FRA are called legs. Both payments are made at time t2 . Usually FRA’s are floating–against–fixed. One leg consists of an interest payment with respect to a floating rate. The interest payment of the floating rate leg is Lrt1 (t1 , t2 ). The side making the floating–rate leg payment is called either the floating–rate leg party, or the floating–rate side, or the floating–rate payer. The interest payment of the fixed rate leg is LrFRA , where rFRA is an interest rate specified in the contract. The side making the fixed–rate leg payment is called either the fixed–rate leg party, or the fixed–rate side, or the fixed–rate payer.

17/78

c





If the FRA is settled at the time of the repayment of the loan, we say that the FRA is settled in arrears. Suppose that the FRA is settled at time t2 (in arrears). The FRA is settled in two different ways: 1. If L(rFRA − rt1 (t1 , t2 )) > 0, the fixed–rate side makes a payment of L(rFRA − rt1 (t1 , t2 )) to the floating–rate side. 2. If L(rFRA − rt1 (t1 , t2 )) < 0, the floating–rate side makes a payment of L(rt1 (t1 , t2 ) − rFRA ) to the fixed–rate side.

18/78

c





Usually an FRA is mentioned as an exchange of (interest payments) legs. By interchanging their legs, it is meant that: 1. The floating–rate leg party makes a payment of Lrt1 (t1 , t2 ) to its counterpart. 2. The fixed–rate leg party makes a payment of LrFRA to its counterpart. The combination of these two payments is: the fixed–rate leg party makes a payment of L(rFRA − rt1 (t1 , t2 )) to the floating–rate leg party. This means that if L(rFRA − rt1 (t1 , t2 )) is a negative number, the floating–rate side makes a payment of L(rt1 (t1 , t2 ) − rFRA ) to the fixed–rate side.

19/78

c





Example 3 Company A pays $75,000 in interest payments at the end of one year. Company B pays the then–current LIBOR plus 50 basis points on a $1,000,000 loan at the end of the year. Suppose that the two companies enter into an interest payment swap. Suppose that in one year the current LIBOR rate is 6.45%. Find which company is making a payment at the end of year and its amount.

20/78

c





Example 3 Company A pays $75,000 in interest payments at the end of one year. Company B pays the then–current LIBOR plus 50 basis points on a $1,000,000 loan at the end of the year. Suppose that the two companies enter into an interest payment swap. Suppose that in one year the current LIBOR rate is 6.45%. Find which company is making a payment at the end of year and its amount. Solution: Company B’s interest payment is (1000000)(0.0645 + 0.0050) = 69500. To settle the forward interest agreement, company A must make a payment of 75000 − 69500 = 5500 to Company B.

21/78

c





The fixed–rate side payment is L(rFRA − rt1 (t1 , t2 )). Assuming that the fixed–rate side borrows L at time t1 his total interest payment at time t2 is Lrt1 (t1 , t2 ) + L(rFRA − rt1 (t1 , t2 )) = LrFRA . A borrower can enter into an FRA as a fixed–rate side to hedge against increasing interest rates. If the FRA is settled at time t1 (at borrowing time), to settle the L(r 1 (t1 ,t2 )−rFRA ) to FRA the floating–rate side makes a payment of t1+r t1 (t1 ,t2 ) the fixed–rate side. This number could be negative. If rFRA > rt1 (t1 , t2 ), the fixed–rate side makes a (positive) payment L(r −rt1 (t1 ,t2 )) of FRA to the floating–rate side. 1+rt (t1 ,t2 ) 1

22/78

c





Since the interest factor from time t1 to time t2 is 1 + rt1 (t1 , t2 ), the previous payoffs are equivalent to the ones for an FRA paid in arrears. In this case, the fixed–rate side can apply the FRA payment to the principal he borrows. He takes a loan of L+

L(1 + rFRA ) L(rFRA − rt1 (t1 , t2 )) = 1 + rt1 (t1 , t2 ) 1 + rt1 (t1 , t2 )

at time t1 . The principal of the loan at time t2 is (1 + rt1 (t1 , t2 ))

L(1 + rFRA ) = L(1 + rFRA ). 1 + rt1 (t1 , t2 )

Again, it is like the fixed–rate side is able to borrow at the rate rFRA .

23/78

c





Usually the floating–rate side is a market maker. The FRA agreement transfers interest rate risk from the fixed–rate side to the floating–rate side. In order to hedge this interest rate risk, the market maker could create a synthetic reverse FRA.

24/78

c





Suppose that the FRA is settled in arrears. The scalper buys a zero–coupon bond maturing at t1 with face value L and short sells 1) a zero–coupon bond with face value LP(0,t P(0,t2 ) maturing at t2 . The scalper cashflow at time zero is LP(0, t1 ) −

LP(0, t1 ) P(0, t2 ) = 0. P(0, t2 )

At time t1 , the scalper gets L from the first bond, which invests at the current interest rate. He gets L(1 + rt1 (t1 , t2 )) at time t2 from this investment. His total cashflow at time t2 is LP(0, t1 ) P(0, t2 ) =L(1 + rt1 (t1 , t2 )) + L(rFRA − rt1 (t1 , t2 )) − L(1 + r0 (t1 , t2 )) L(1 + rt1 (t1 , t2 )) + L(rFRA − rt1 (t1 , t2 )) −

=L(rFRA − r0 (t1 , t2 )). Hence, the no arbitrage rate of an FRA is r0 (t1 , t2 ), which is the current nonannualized interest rate from t1 to t2 . c



25/78



Example 4 Suppose that the current spot rates are given in the following table (as annual nominal rates convertible semiannually) spot rate maturity (in months)

6% 6

7.5% 12

Timothy and David enter into separate forward rate agreements as fixed–rate sides for the period of time between 6 months and 12 months. Both FRA’s are for a notional amount $10000. Timothy’s FRA is settled in 12 months. David’s FRA is settled in 6 months. In six months, the annual nominal interest rate compounded semiannually for a six month loan is 7%. (i) Find the no arbitrage six month rate for an FRA for the period of time between 6 months and 12 months. (ii) Calculate Timothy’s payoff from his FRA. (iii) Calculate David’s payoff from his FRA. 26/78

c





Solution: (i) The no arbitrage six month rate for a FRA for the period of time between 6 months and 12 months is rFRA =

0.075 2 2 + 0.06 2

1+ 1

− 1 = 4.5054612%.

(ii) Timothy’s payoff is (10000)(0.035 − 4.5054612) = 447.04612 (iii) David’s payoff is (10000)

(0.035 − 4.5054612) = 431.9286184. 1 + 0.07 2 )

27/78

c





Interest rate swaps.

An interest rate swap is a contract in which one party exchanges a stream of interest payments for another party’s stream. Interest rate swaps are normally ”fixed–against–floating”. Interest rate swaps are valued using a notional amount. This nominal amount can change with time. We only consider constant nominal amounts. The fixed stream of payments are computed with respect to a rate determined by the contract. The floating stream of payments are determined using a benchmark, such as the LIBOR.

28/78

c





Suppose that a firm is interested in borrowing a large amount of money for a long time. One way to borrow is to issue bonds. Unless its credit rating is good enough, the firm may have trouble finding buyers. Lenders are unwilling to absorb long term loans from a firm with a so and so credit rating. So, the firm may have to borrow short term. Even if a company does not need to borrow short term, usually short term interest rates are lower than long term interest rates. As longer the maturity as more likely the default. Anyhow, suppose that a firm is interested in borrowing short term, but needs the cash long term. The firm takes a short term loan. At maturity, the firm pays this loan and takes another short term loan. This process will be repeated as many times as needed. Current short term rates are known. But, the short term interest rates which the firm may need to take in the future are uncertain. The firm has an interest rate risk. If short term rates increase, the company may get busted. To hedge this risk, the firm may enter into an interest rate swap, which we describe next. 29/78

c





Suppose that a borrower takes a loan of L paying a floating interest rate according a benchmark such as the LIBOR. Suppose that the interest is paid at times t1 < t2 < · · · < tn . The principal owed after each payment is L. This means that at time tj , the borrower pays Lrtj−1 (tj−1 , tj ) in interest, where 1 + rtj−1 (tj−1 , tj ) is the interest factor from time tj−1 to time tj , calculated using the LIBOR at time tj−1 . This rate is the rate which the borrower would pay, if he borrows at time tj−1 , pays this loan at time tj and takes a new loan for L at time tj . The borrower is paying a stream of floating interest rate payments. The borrower can hedge by taking several FRA’s. If the borrower would like to have a single contract, he enters into a interest rate swap.

30/78

c





The borrower would like to enter into an interest rate swap so that the current interest payments plus the payments to the swap will add to a fixed payment. This situation is similar to that of having several FRA’s. The borrower can enter an interest rate swap with notional amount L. The borrower would like to have a fixed–rate leg in its contract: Payment Time

LR t1

LR t2

··· ···

LR tn

where R is the swap interest rate in the contract. The borrower would like that its counterpart has a floating–rate leg: Payment Time

Lr0 (0, t1 ) t1

Lrt1 (t1 , t2 ) t2

··· ···

Lrtn−1 (tn−1 , tn ) tn

31/78

c





An interest rate swap consists of an interchange of interest payments. The total outcome of this interchange is that at every time tj the floating–leg side makes a payment of L(rtj−1 (tj−1 , tj ) − R) to the fixed–leg side. Again L(rtj−1 (tj−1 , tj ) − R) could be negative. If L(rtj−1 (tj−1 , tj ) − R) < 0, the fixed–leg side makes a payment at each time tj of L(R − rtj−1 (tj−1 , tj )) to the floating–leg side.

32/78

c





By exchanging legs, the borrower makes a payment at each time tj when L(R − rtj−1 (tj−1 , tj )) to his counterpart. The borrower is also making interest payments of Lrtj−1 (tj−1 , tj ). The total borrower’s interest payments add to LR. By entering a swap, a borrower is hedging against increasing interest rates. The total borrower’s cashflow is that of a company issuing bonds. Often the borrower has poor credit rating and it is unable to issue bonds. In some sense, some borrowers enter into an interest rate swap so that its counterpart issues a ”bond” to them. Sometimes the borrower uses a interest rate swap to avoid to issue a fixed rate long term loan. Takers of this loan could require a higher interest to borrow.

33/78

c





Usually, the borrower’s counterpart is a market–maker, which must hedge its interest rate risk. The (market–maker) fixed–rate payer gets a payment of L(rtj−1 (tj−1 , tj ) − R) at each time tj . The fixed–rate payer profit by entering the swap is n X

P(0, tj )L(rtj−1 (tj−1 , tj ) − R).

j=1

34/78

c





By using bonds, a market–maker can create a synthetic cashflow of payments equal to the swap payments. The cost of these bonds is its present value according with the current term structure of interest rates. Hence, if there is no arbitrage, a market–maker can arrange so that the cost of payments he receives is n X

LP(0, tj )(r0 (tj−1 , tj ) − R),

j=1

i.e. instead of using the uncertain rates rtj−1 (tj−1 , tj ), the scalper can use the current forward rates. Therefore, the no arbitrage swap rate is Pn j=1 P(0, tj )r0 (tj−1 , tj ) Pn . R= j=1 P(0, tj ) The swap rate R when there is no arbitrage is called the par swap rate. Notice that the par swap rate R is a weighted average of implied forward rates r0 (tj−1 , tj ). The weights depend on the present value of a payment made at time tj . c



35/78


Using that

P(0,tj−1 ) P(0,tj )

Pn R=

= 1 + r0 (tj−1 , tj ), we get that

j=1 P(0, tj ) Pn

P(0,tj−1 ) P(0,tj )

−1

j=1 P(0, tj ) Pn j=1 P(0, tj−1 ) − i=1 P(0, tj ) Pn j=1 P(0, tj )

Pn =


1 − P(0, tn ) = Pn . j=1 P(0, tj )

Notice that R is the coupon rate for a bond with price, face value and redemption all equal, using the current term structure of interest rates. It is like that the floating–rate party enters the swap to use the market maker credit rating to issue a bond.

36/78

c





Example 5 Suppose the LIBOR discount factors P(0, tj ) are given in the table below. Consider a 3–year swap with semiannual payments whose floating payments are found using the LIBOR rate compiled a semester before the payment is made. The notional amount of the swap is 10000. LIBOR discount rates P(0, tj ) time (months)

0.9748

0.9492

0.9227

0.8960

0.8687

0.8413

6

12

18

24

30

36

(i) Calculate the par swap rate. (ii) Calculate net payment made by the fixed–rate side in 18 months if the six–month LIBOR interest rate compiled in 12 months is 2.3%. 37/78

c





Solution: (i) The par swap rate is R=

1−P(0,tn ) P n j=1 P(0,tj )

1−0.8413 = 0.9748+0.9492+0.9227+0.8960+0.8687+0.8413 = 0.02910484714 = 2.910484714%.

(ii) The payment made by the fixed–rate side is L(R−rtj−1 (tj−1 , tj )) = 10000(0.02910484714−0.023) = 61.0484714.

38/78

c





Notice that the floating interest payment use the interest rates compiled one period before the payment. These interest rates are called realized interest rates. Since interest rates change daily, we may be interested in the market value of a swap contract. One of the parties in the swap contract may sell/buy his position in the contract. The market value of a swap contract for the fixed–rate payer is the present value of the no arbitrage estimation of the payments which he will receive. The market value of a swap contract for the fixed–rate payer immediately after the k–the payment is n X P(tk , tj )L(rtk (tj−1 , tj ) − R). j=k+1

If this value is positive, the fixed–rate payer has exposure to interest rates. Current interest rates are higher than when the swap was issued. The market value of a swap is the no arbitrage price to enter this contract. One of the counterparts in the contract may be interested in selling/buying his position on the contract. c



39/78



Example 6 Suppose current LIBOR discount factors P(0, tj ) are given by the table below. An interest rate swap has 6 payments left. The swap rate is 3.5% per period. The notional principal is two million dollars. The floating payments of this swap are the realized LIBOR interest rates. LIBOR discount rates P(0, tj ) time (months)

0.9748

0.9492

0.9227

0.8960

0.8687

0.8413

6

12

18

24

30

36

Calculate market value of this swap for the fixed–rate payer. 40/78

c





Solution: The market value of the swap per dollar is n X

P(0, tj )(rtk (tj−1 , tj ) − R)

j=k+1

1 0.9748 − 1 − 0.035 + 0.9492 − 1 − 0.035 0.9748 0.9492 0.9227 0.9492 − 1 − 0.035 + 0.8960 − 1 − 0.035 + 0.9227 0.9227 0.8960 0.8960 0.8687 + 0.8687 − 1 − 0.035 + 0.8413 − 1 − 0.035 0.8687 0.8413 =0.9748 (0.02585145671 − 0.035) + 0.9492 (0.02697008007 − 0.035) =0.9748

+ 0.9227 (0.02872006069 − 0.035) + 0.8960 (0.02979910714 − 0.035) + 0.8687 (0.03142626914 − 0.035) + 0.8413 (0.03256864377 − 0.035) = − 0.0321445. The market value of the swap is (2000000)(−0.0321445) = −64289. c



41/78



A deferred swap is a swap which begins in k periods. The swap par rate is computed as Pn P(0,tj−1 ) Pn P(0, t ) − 1 j j=k P(0,tj ) j=k P(0, tj )r0 (tj−1 , tj ) Pn Pn R= = j=k P(0, tj ) j=k P(0, tj ) Pn Pn P(0, tk−1 ) − P(0, tn ) j=k P(0, tj−1 ) − j=k P(0, tj ) Pn Pn = . = j=k P(0, tj ) j=k P(0, tj )

42/78

c





Example 7 Suppose the current annual nominal interest rates compounded quarterly from the LIBOR are given by the table below. Two counterparts enter into a fixed against floating swap using the LIBOR rate compiled a quarter before the payment is made. The notional principal is $50000. The times of the swap are in 12, 15 and 18 months. i (4) maturation time in months

4.5%

4.55%

4.55%

4.6%

4.6%

4.65%

3

6

9

12

15

18

(i) Calculate the par swap rate. (ii) Calculate the payment made by the fixed–rate party in 18 months if in 15 months the spot annual nominal interest rate compounded quarterly is 4.65%. 43/78

c





Solution: (i) The par swap rate is R=

P(0, tk−1 ) − P(0, tn ) Pn j=k P(0, tj )

(1 + 0.0455/4)−3 − (1 + 0.0465/4)−6 (1 + 0.046/4)−4 + (1 + 0.046/4)−5 + (1 + 0.0465/4)−6 =0.01187359454 = 1.187359454%. =

(ii) The payment made by the fixed–rate party is L(R−rtj−1 (tj−1 , tj )) = (50000)(0.01187359454−0.0465/4) = 12.429727.

44/78

c





1−P(0,tn ) P n j=1 P(0,tj )

The par swap rate R = is a weighted average of implied forward rates. If the current interest rate does depend on the maturing time, then P(0, t) = (1 + i)−t , for some constant i > 0. In this case, 1 − P(0, tn ) 1 − (1 + i)−tn R = Pn = Pn −tj . j=1 P(0, tj ) j=1 (1 + i) If the periods in the swap have the same length, then tj = jh, 1 ≤ j ≤ n, for some h > 0, and 1 − (1 + i)−nh = R = Pn −nj j=1 (1 + i) =

1 (1+i)−h 1−(1+i)−h

=

1

1 − (1 + i)−nh

(1+i)−h −(1+i)−(n+1)h 1−(1+i)−h − (1 + i)−h = (1 + i)h − 1. (1 + i)−h

(1 + i)h − 1 is the effective rate for a period of length h. Notice that the assumption P(0, t) = (1 + i)−t , for some constat i > 0, almost never happens. c



45/78



Example 8 Suppose the current annual nominal interest rates compounded quarterly from the LIBOR is 5.4% independently of the maturity of the loan. Two counterparts enter into a fixed against floating swap using realized LIBOR rates. The times of the swap are in 3, 6 and 12 months. Calculate the par swap rate.

46/78

c





Example 8 Suppose the current annual nominal interest rates compounded quarterly from the LIBOR is 5.4% independently of the maturity of the loan. Two counterparts enter into a fixed against floating swap using realized LIBOR rates. The times of the swap are in 3, 6 and 12 months. Calculate the par swap rate. Solution: The par swap rate is 1 − P(0, tn ) R = Pn j=1 P(0, tj ) 1 − (1 + 0.054/4)−4 (1 + 0.054/4)−1 + (1 + 0.054/4)−2 + (1 + 0.054/4)−4 =0.017959328 = 1.7959328%.

=

47/78

c





Notice that in the previous problem the swap periods do not have the same length. Even if annual interest rate is constant over time, the interest payments vary over time. If the annual interest rate remains constant over time, the floating–rate payments are L ((1 + i)t1 − 1) t1

L ((1 + i)t2 −t1 − 1) t2

··· ···

L ((1 + i)tn −tn−1 − 1) tn

Suppose that we take a loan of L at time zero. We make payments of LR at tj , for 1 ≤ j ≤ n. The par swap rate R is the constant periodic rate such that the final outstanding in this loan Pn is L. Notice that the present value of the payments is LR j=1 P(0, tj ). If the final principal is L =

L−LR

Pn

j=1 P(0,tj ) , P(0,tn )

then, R =

1−P(0,tn ) P . n j=1 P(0,tj )

48/78

c





Commodity swaps. A commodity swap is a swap where one of the legs is a commodity and the other one is cash. Hence, there are two counterparts in a swap: a party with a commodity leg and another party with a cash leg. Since the spot price of a commodity changes over time, the commodity leg is floating. Certain amount of a commodity is delivered at certain times. In some sense is like to combine several forward contracts. But, swaps are valuated considering the total deliveries. Hence, changes in interest rates change the value of a swap. Usually, the swap payment for this commodity is constant. The commodity leg party is called the short swap side. The cash leg party is called the long swap side.

49/78

c





Suppose that a party would like to sell a commodity at times 0 < t1 < t2 < · · · < tn . Suppose that the nominal amounts of the commodity are Q1 , Q2 , · · · , Qn , respectively. The commodity leg is Amount of delivered commodity Time

Q1 t1

Q2 t2

··· ···

Qn tn

Usually the cash leg is either Payments Time

C0 0

0 t1

0 t2

··· ···

0 tn

or Payments Time

C1 t1

C2 t2

··· ···

Cn tn

50/78

c





If two parties enter into a commodity swap, one will be designed the party with the commodity leg and the other the party with the cash leg. The party with the commodity leg will deliver commodity to its counterpart according with the table above. The party with the cash leg will pay cash payments to its counterpart. From a practical point of view, the party with the commodity leg is a seller. The party with the cash leg is a buyer. A commodity swap contract needs to specify the type and quality of the commodity, how to settle the contract, etc. A swap can be settled either by physical settlement or by cash settlement. If a swap is settled physically, the commodity leg side delivers the stipulated notional amount to the cash leg side, and the cash leg side pays to the commodity leg side. If a swap is cash settled, one of the parties will make a payment to the other party.

51/78

c





Suppose that the current forward price of this commodity with delivery in T years is F0,T . Let P(0, T ) be the price of a zero–coupon with face value $1 and expiration time T . Then, the present value of the commodity delivered is n X

P(0, tj )Qj F0,tj .

j=1

In a prepaid swap the buyer makes a unique payment at time zero. Pn If there exists no arbitrage, the price of a prepaid swap is j=1 P(0, tj )Qj F0,tj .

52/78

c





Usually the cash leg consists of series of payments made at the times when the commodity is delivered. Usually each swap payment per unit of commodity is a fixed amount. Hence, the cashflow of payments is Payment Time

Q1 R t1

Q2 R t2

··· ···

Qn R t2

where R is the swap price per unit ofP commodity. The present value of the cashflow of payments is nj=1 P(0, tj )Qj R. The no arbitrage price of a swap per unit of commodity is Pn j=1 P(0, tj )Qj F0,tj . R = Pn j=1 P(0, tj )Qj R is a weighted average of forward prices. 53/78

c





Suppose that at time of delivery, the buyer pays a level payment of R at each of the deliveryP times. Then, the present value of the cashflow of payments is nj=1 P(0, tj )R. The no arbitrage level payment is Pn j=1 P(0, tj )Qj F0,tj Pn . R= j=1 P(0, tj ) A commodity swap allows to lock the price of a sale. It can be used by a producer of a commodity to hedge by fixing the price that he will get in the future for this commodity. It also can be used by a manufacturer to hedge by fixing the price that he will pay in the future for a commodity. A commodity swap can be used instead of several futures/forwards. Since the price of a swap involves all the deliveries, a commodity swap involves loaning/lending somehow.

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Example 9 Suppose that an airline company must buy 10,000 barrels of oil every six months, for 2 years, starting 6 months from now. The company enters into a long oil swap contract to buy this oil. The cash leg swap consists of four level payments made at the delivery times. The following table shows the annual nominal rate convertible semiannually of zero–coupon bonds maturing in 6, 12, 18, 24 months and the forward price of oil with delivery at those times. F0,T annual nominal rate expiration in months

$50 4.5% 6

$55 5% 12

$55 5% 18

$60 5.5% 24

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(i) Find the no arbitrage price of a prepaid swap with the commodity leg which the airline needs.

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(i) Find the no arbitrage price of a prepaid swap with the commodity leg which the airline needs. Solution: (i) The present value of the cost of oil is n X

Qj F0,tj P(0, tj )

j=1

55 50 55 + (10000) 2 + (10000) 3 0.045 0.05 1+ 2 1+ 2 1 + 0.05 2 60 + (10000) 4 1 + 0.055 2

=(10000)

=488997.555 + 523497.9179 + 510729.676 + 538299.4402 = 2061524.58

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(ii) Suppose that the airline company pays the swap by a level payment of R at each of the delivery times. Calculate R.

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(ii) Suppose that the airline company pays the swap by a level payment of R at each of the delivery times. Calculate R. Solution: (ii) We have that Pn j=1 P(0, tj ) 1 1 1 = 1+ 10.045 + 2 + 3 + 4 (1+ 0.05 (1+ 0.05 (1+ 0.055 2 2 ) 2 ) 2 ) = 0.97799511 + 0.9518143962 + 0.9285994109 + 0.8971657337 = 3.755574651 and Pn R=

j=1 Qj F0,tj P(0, tj ) Pn j=1 P(0, tj )

=

2061524.589 = 548923.8747. 37555.74651

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(iii) Suppose that the airline company pays the swap by unique price per barrel. Calculate the price per barrel.

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(iii) Suppose that the airline company pays the swap by unique price per barrel. Calculate the price per barrel. = Solution: (iii) The price of the swap per barrel is 548923.8747 10000 54.89238747.

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Example 10 Suppose that an airline company must buy 10,000 barrels of oil in two months, 12,000 barrels of oil in four months and 15,000 barrels of oil in six months. The company enters into a long oil swap. The payment of the swap will be made at the delivery times. The following table shows the annual nominal rate convertible monthly of zero–coupon bonds maturing in 6, 12, 18, 24 months and the forward price of oil with delivery at those times. Barrels of oil F0,T annual nominal rate expiration in months

$10000 $55 4.5% 2

$12000 $56 4.55% 4

$15000 $58 4.65% 6

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(i) Suppose that the cash leg swap consists of a level payment of R at each of the delivery times. Calculate R.

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Solution: (i) We have that n X

Qj F0,tj P(0, tj )

j=1

55

=(10000) 1+

+ (12000) 0.045 2 12

56 1+

+ (15000) 0.0455 4 12

58 1+

0.0465 6 12

=545898.0877 + 661903.8839 + 850044.0252 = 2057845.997, Pn =

j=1 P(0, tj ) 1 1 2 + 4 (1+ 0.045 (1+ 0.0455 12 ) 12 )

1

+

6

(1+ 0.0465 12 ) = 0.9925419775 + 0.9849760176 + 0.9770620979 = 2.954580093

and Pn R=

j=1 Qj F0,tj P(0, tj ) Pn j=1 P(0, tj )

=

2057845.997 = 696493.5564. 2.954580093 64/78

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(ii) Suppose that the cash leg swap consists of a unique payment per barrel made at each of the delivery times. Calculate the no arbitrage swap price per barrel.

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(ii) Suppose that the cash leg swap consists of a unique payment per barrel made at each of the delivery times. Calculate the no arbitrage swap price per barrel. Solution: (ii) We have that Pn j=1 P(0, tj )Qj 12000 15000 10000 = 2 + 4 + 6 (1+ 0.045 (1+ 0.0455 (1+ 0.0465 12 ) 12 ) 12 ) = 9925.419775 + 11819.712212 + 14655.931469 = 36401.06346 and the swap price per barrel is Pn 2057845.997 j=1 P(0, tj )Qj F0,tj = = 56.53257903. R = Pn 36401.06346 j=1 P(0, tj )Qj

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If a swap is cashed settled, then the commodity is valued at the current spot price. If the current value of the commodity is bigger than the value of the cash payment, then the (cash leg side) long swap pays this difference to the (commodity leg side) short swap. Reciprocally, if the current value of the commodity is smaller than the value of the cash payment, the (commodity leg side) short swap side pays this difference to the (cash leg side) long swap.

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Suppose that the swap involves the sale of a commodity at times t1 < t2 < · · · < tn . with notional amounts of Q1 , Q2 , · · · , Qn , respectively. Suppose the swap payment is a fixed amount per unit of commodity. We saw that this amount is Pn j=1 P(0, tj )Qj F0,tj R = Pn . j=1 P(0, tj )Qj When a swap is cash settled, the value of the commodity is found using the current spot price Stj . At time tj the long swap pays Qj R − Qj Stj to its counterpart. Notice that Qj R − Qj Stj could be a negative real number. If Qj R − Qj Stj < 0, the short swap pays Qj Stj − Qj R to its counterpart.

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Changes in the forward contracts and interest rates alter the value of the swap. The market value of a swap is found using the present values of its legs using the current structure of interest rates. The market value of a long swap immediately after the settlement at time tk is n X P(tk , tj )Qj (Ftk ,tj − R). j=k+1

This is the price which an investor would pay to enter the swap as a long swap side. Immediately after the swap is undertaken the market value of the contract is n X

˜ tj )F˜0,t − P(0, tj )F0,t ), Qj (P(0, j j

j=1

˜ tj ) and F˜0,t are the new market values. where P(0, j 69/78

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Example 11 Suppose that an airline company must buy 1,000 barrels of oil every six months, for 3 years, starting 6 months from now. Instead of buying six separate long forward contracts, the company enters into a long swap contract. According with this swap the company will pay a level payment per barrel at delivery. The following table shows the annual nominal rate convertible semiannually of zero–coupon bonds maturing in 6, 12, 18, 24 months and the forward price of oil at those times. F0,T annual nominal rate expiration in months

$55 5.5% 6

$57 5.6% 12

$57 5.65% 18

$60 5.7% 24

$62 5.7% 30

$64 5.75% 36

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(i) Find the price per barrel of oil using the swap.

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(i) Find the price per barrel of oil using the swap. Solution: (i) We have that n X

P(0, tj )Qj F0,tj

j=1

55 57 57 + (1000) 2 + (1000) 3 0.055 0.056 1+ 2 1+ 2 1 + 0.0565 2 62 64 60 + (1000) + (1000) + (1000) 6 0.057 4 0.057 5 1+ 2 1+ 2 1 + 0.0575 2

=(1000)

=54254.00740 + 55436.90114 + 54651.28637 + 56698.44979 + 57765.24340 + 58747.41357 = 337553.3017,

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(i) Find the price per barrel of oil using the swap. Solution: (i) n X

P(0, tj )Qj

j=1

=

1000 1 1000 + 2 + 3 0.055 0.056 1+ 2 1+ 2 1 + 0.0565 2 1000 1000 1000 + + + 6 0.057 4 0.057 5 1+ 2 1+ 2 1 + 0.0575 2

=986.4364982 + 972.5772129 + 958.7944977 + 944.9741632 + 931.6974742 + 917.9283370 = 5712.408183. We have that Pn R=

j=1 P n

P(0, tj )Qj F0,tj

j=1 P(0, tj )Qj

=

c


337553.3017 = 59.09124329. 5712.408183 Manual for SOA Exam FM/CAS Exam 2.

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(ii) Suppose the swap is settled in cash. Assume that the spot rate for oil in 18 months is $57. Calculate the payment which the airline receives.

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(ii) Suppose the swap is settled in cash. Assume that the spot rate for oil in 18 months is $57. Calculate the payment which the airline receives. Solution: (ii) The airline gets a payment of Qj Stj − Qj R = (1000)(57) − (1000)(59.09124329) = −2091.24329.

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(iii) Suppose that immediately after the swap is signed up, the future prices of oil are given by the following table F˜0,T expiration in months

$55 6

$58 12

$59 18

$61 24

$62 30

$63 36

Calculate the value of the swap for the cash leg party.

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Solution: (iii) The market value of the swap for the long party is n X

P(0, tj )(Qj F˜t0 ,tj − Qj R)

j=1

=

1000 1 (55 − 59.09124329) + 2 (58 − 59.09124329) 0.055 1+ 2 1 + 0.056 2 1000 1000 + (59 − 59.09124329) + (61 − 59.09124329) 3 0.057 4 1 + 0.0565 1 + 2 2 1000 1000 + 5 (62 − 59.09124329) + (63 − 59.09124329) 0.0575 6 1 + 0.057 1 + 2 2

= − 4035.7517041 − 1061.3183576 − 87.4835644 + 1803.7257747 + 2710.0812797 + 3587.9585464 = 2917.211975.

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Suppose that immediately after the forward is signed, every future price increases by K . Then, the market value of the swap is n X

P(0, tj )(Qj (F0,tj + K ) − Qj R) =

j=1

n X

P(0, tj )Qj K .

j=1

The swap counterpart is a scalper which hedges the commodity risk resulting from the swap. The scalper has also interest rate exposure. The scalper needs to hedge changes in the price of the commodity and in interest rates.

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SOA exam FM CAS

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