Nov 2012 Mlc

Solutions to the November 2012 Course MLC Examination by Krzysztof Ostaszewski, http://www.krzysio.net Ostaszewski, http://www.krzysio.net, [email protected] Copyright © 2012 by Krzysztof Ostaszewski All rights reserved. No reproduction in any form is permitted without explicit permission of the copyright owner. Dr. Ostaszewski’s manual for Course MLC is available from Actex Publications (http://www.actexmadriver.com http://www.actexmadriver.com)) and the Actuarial Bookstore (http://www.actuarialbookstore.com http://www.actuarialbookstore.com)) Exam MLC seminar at Illinois State University: http://math.illinoisstate.edu/actuary/exams/prep_courses.shtml

November 2012 Course MLC Examination, Problem No. 1

For two lives, (80) and (90), with independent indepe ndent future lifetimes, you are given: k

p80

+

k

p90

+

k

0 0.9 0.6 1 0.8 0.5 2 0.7 0.4 Calculate the probability that the last survivor will die in the third year. A. 0.20

B. 0.21

C. 0.22

D. 0.23

E. 0.24

Solution. The probability sought is 2

p

80:90

− 3 p80:90

= Using Poohsticks

( 2 p80 + 2 p90 − 2 p80:90 ) − ( 3 p80 + 3 p90 − 3 p80:90 ) =

( 2 p80 + 2 p90 − 2 p80 ⋅ 2 p90 ) − ( 3 p80 + 3 p90 − 3 p80 ⋅ 3 p90 ) = = ( p80 p81 + p90 p91 − p80 p81 ⋅ p90 p91 ) − ( p80 p81 p82 + p90 p91 p92 − p80 p81 p82 ⋅ p90 p91 p92 ) = = ( 0.9 ⋅ 0.8 + 0.6 ⋅ 0.5 − 0.9 ⋅ 0.8 ⋅ 0.6 ⋅ 0.5 ) − − ( 0.9 ⋅ 0.8 ⋅ 0.7 + 0.6 ⋅ 0.5 ⋅ 0.4 − 0.9 ⋅ 0.8 ⋅ 0.7 ⋅ 0.6 ⋅ 0.5 ⋅ 0.4 ) = = ( 0.72 + 0.3 − 0.216 ) − ( 0.504 + 0.12 − 0.06048 ) = 0.24048. =

Answer E.


You are given: (i) An excerpt from a select and ultimate u ltimate life table with a select period of 3 years: x x + 3 l x l x 1 l x 2 l x 3 [ ] [ ] [ ] +

+

+

60 80,000 79,000 77,000 74,000 61 78,000 76,000 73,000 70,000 62 75,000 72,000 69,000 67,000 63 71,000 68,000 66,000 65,000 (ii) Deaths follow a constant force of mortality over each year of age.

63 64 65 66

Copyright © 2012 by Krzysztof Ostaszewski. All rights reserved. No reproduction in any form is permitted without explicit permission of the copyright owner.

Calculate

1000 2

A. 104

3

q 60 + 0.75 . [ ]

B. 117

C. 122

D. 135

E. 142

Solution. Under constant force over each year of age, for 0 ≤ k ≤ 1, and x an integer, l x

k

+

k

= k

l x

p x

=

e

− k µ µ x

(e

=

− µ x

⎛ l ⎞ ) = ⎝ ⎜ xl ⎟ ⎠ , x k

+1

so that l x+ k

=

1− k

(l ) (l ) x

x +1

k

.

We have 1000 2

3

q[ 60]

+ 0.75

= 1000 ⋅

l[ 60]

+ 2.75

−

l[ 60]

+ 5.75

l[60]

= 1000 ⋅

+ 0.75

= 1000 ⋅

(l

[ 60]+2

)

0.25

0.75

⋅

0.25

−

l65

0.25

(l ) ( l ⋅

[ 60]

= 1000 ⋅

l63

77000

0.25 ⋅

[ 60]+1

74000

)

+

−

+

=

+

0.75

l66

⋅

=

0.75

0.75

80000

l([ 60] 2 ) 0.75 l65.75 l[ 60] 0.75

−

0.25 ⋅

67000

79000

0.25 ⋅

65000

0.75 ≈

0.75

116.70.

Answer B.


You are given: 1

(i) (ii)

⎛ t ⎞ S ( t ) ⎜ 1 − ⎝ ω ⎟ ⎠

4

0

=

1

µ 65

=

, for 0 ≤ t ≤ ω .

.

180

Calculate

e 106

A. 2.2

, the curtate expectation of life at age 106.

B. 2.5

C. 2.7

D. 3.0

E. 3.2

Solution. We have

⎛ ⎞ d d t ⎞ ⎛ − ( ln S (t )) − ⎜ ln ⎜ 1 − ⎟ ⎟ ω ⎠ ⎟ dt dt ⎜ ⎝ ⎝ ⎠ 1 d ⎛ ⎛ 1 1 1 t ⎞ ⎞ − ln ⎜ 1 − − ⋅ t ⋅ ⎛ − ⎞ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ 4 dt ⎝ ⎝ 4 ω ⎠ ⎠ ω ⎠ 1− 1 4

µ t

=

=

0

=

=

=

1 =

4

⋅

1 ω

− t

.

ω

Therefore, Copyright © 2012 by Krzysztof Ostaszewski. All rights reserved. No reproduction in any form is permitted without explicit permission of the copyright owner.

1 180

µ 65

=

1 =

4

1

⋅

ω

− 65

,

so that 1 45

and

ω

=

1 =

ω

110.

− 65

,

Based on this 1

t

p106

=

⎛ 1 − S (106 + t ) ⎜⎝ 0

=

S 0 (106 )

106 + t 110

⎞ ⎟ ⎠

4

1

⎛ 1 − 106 ⎞ ⎜⎝ 110 ⎟ ⎠

4

1

⎛ 4 − t ⎞ =⎜ ⎝ 4 ⎟ ⎠

4

.

Therefore, 1

⎛ 4 − 1⎞ + ... = ⎜ ⎝ 4 ⎟ ⎠

4

e106 = 1 p106 +

2

p106 +

p106 + 3

4

p106

 =0

1

⎛ 4 − 2 ⎞ +⎜ ⎝ 4 ⎟ ⎠

4

1

⎛ 4 − 3 ⎞ ≈ 2.4786. +⎜ ⎝ 4 ⎟ ⎠ 4

Answer B.


For a special fully discrete whole life insurance on (40), you are given: (i) The death benefit is 50,000 in the first 20 years and 100,000 thereafter. (ii) Level benefit premiums of 1116 are payable for 20 years. (iii) Mortality follows the Illustrative Life Table. (iv) i = 0.06. Calculate 10V , the benefit reserve at the end of year 10 for this insurance. A. 13,340

B. 13,370

C. 13,400

D. 13,430

E. 13,460

Solution. Prospectively, we calculate the reserve as 10V =

( 50,000 A

1

50:10

+ 100,000 10

)

−

50:10 1116 a

50,000 ( A50

=

50,000 ( 0.24905 + 0.51081 0.36913) 1116 (13.2668

≈

⋅

⋅

A60

−

50 1116 ( a

=

=

+ 10 E 50

)

A50

⋅

−

10 E 50 −

⋅

60 a

)=

⋅

−

0.5108 11.1451) = ⋅

13,428.

Answer D.


A special fully discrete 2-year endowment insurance with a maturity value of 2000 is issued to ( x). The death benefit is 2000 plus the benefit reserve at the end of the year of death. For year 2, the benefit reserve is the benefit reserve just before the maturity benefit is paid. You are given: (i) i = 0.10. Copyright © 2012 by Krzysztof Ostaszewski. All rights reserved. No reproduction in any form is permitted without explicit permission of the copyright owner.

(ii) q x 0.150 and q x 1 = 0.165. Calculate the level annual benefit premium. =

+

A. 1070

B. 1110

C. 1150

D. 1190

E. 1230

Solution. The initial reserve at time 0 is 0V 0. The terminal reserve at time 2 is 2V 2000, as the policy endows at that time. Let us write P for the level annual benefit premium sought. We use the recursive reserve formula twice to find it: =

=

( V + P )(1 + i ) = q x ( 2000 + V ) + (1 − q x ) V , ( V + P )(1 + i ) = q x ( 2000 + V ) + (1 − q x ) V . 0

1

1

+1

2

1

+1

2

Substituting known values, we obtain

( 0 + P ) ⋅1.1 = 0.150 ⋅ ( 2000 + 1V ) + 0.850 ⋅ 1V , ( V + P ) ⋅1.1 = 0.165 ⋅ ( 2000 + 2000 ) + 0.835 ⋅ 2000. 1

From the first equation,

V

1

=

1.1P − 300, and after we put this in the second equation

(1.1P − 300 + P ) ⋅1.1 = 0.165 ⋅ 4000 + 0.835 ⋅ 2000, or P

2660 =

2.31

≈

1151.52.

Answer C.


For a special fully continuous 10-year increasing term insurance, you are given: (i) The death benefit is payable at the moment of death and increases linearly from 10,000 to 110,000. (ii) µ 0.01. (iii) δ 0.05. (iv) The annual premium rate is 450. (v) Premium-related expenses equal 2% of premium, incurred continuously. (vi) Claims-related expenses equal 200 at the moment of death. (vii) t V denotes the gross premium reserve at time t for this insurance. (viii) You estimate 9.6V using Euler’s method with step size 0.2 and the derivative of t V at time 9.6. (ix) Your estimate of 9.8V is 126.68. Calculate the estimate of 9.6V . =

=

A. 230

B. 250

C. 270

D. 280

E. 290

Solution. Using Euler’s method with step size 0.2 and the derivative of t V at time 9.6, we obtain


V

≈

9.6

V

9.8

−

0.2

⋅

d t V dt

. t 9.6 =

Recall the Thiele’s differential equation when expenses are included: d dt

(

)

− et − S t + E t − t V µ x +t . V = δ t t V + P t

t

In this case, t = 9.6, δ 9.6

=

δ

=

0.05,

premium with the annual rate of 450, S t

is constant with d t V dt

=

S 9.6

=

106,000,

V

is denoted by 9.6V and it is unknown, P t is a gross

e

are premium related expenses at 2% of premium,

t

t

E 9.6

=

200, so that

0.05 ⋅ 9.6V + 450 − 0.02 ⋅ 450 − (106,000 + 200 −

V

9.6

) ⋅ 0.01 = −621 + 0.06 ⋅

V .

9.6

t = 9.6

Substituting into the Euler’s formula, we obtain V

9.6

≈

V

9.8

−

0.2

⋅

d t V dt

= 126.68 −

0.2

⋅

(

612 + 0.06

−

⋅

V

9.6

) = 250.88

−

0.012

⋅

V .

9.6

t = 9.6

This results in V 9.6

250.88 ≈

1.012

≈

247.91.

Answer B.


You are calculating asset shares for a universal life insurance policy with a death benefit of 1000 on (x) with death benefits payable at the end of the year of death. You are given: (i) The account value at the end of year 4 is 30. (ii) The asset share at the end of year 4 is 20. (iii) At the beginning of year 5: a. A premium of 20 is paid. b. Annual cost of insurance charges of 2 and annual expense charges of 7 are deducted from the account value. (iv) At the beginning of year 5, the insurer incurs expenses of 2. (v) All withdrawals occur at the end of the policy year; the withdrawal benefit is the account value less a surrender charge of 20. (vi) q x( )4 = 0.001 and q x( )4 = 0.050 are the probabilities of death and withdrawal, respectively. (vii) The annual interest rate for asset shares is 0.08; the annual interest rate credited to the universal life insurance policy is 0.06. Calculate the asset share at the end of year 5. A. 39

d

w

+

+

B. 40

C. 41

D. 42

E. 43

Solution. The account value (a.k.a., reserve) at the end of year 5 is Copyright © 2012 by Krzysztof Ostaszewski. All rights reserved. No reproduction in any form is permitted without explicit permission of the copyright owner.

⎛ ⎞ ⎜ ⎟ + − 30 20 9 ⎜ ⎟ ⋅ 1.06 ≈ 43.46. Cost of insurance of 2 ⎟ Accumulated at value at Premium ⎜ Account plus annual expense the end of year 4 the interest rate ⎝ ⎠ credited charge of 7 (i.e., 6%) 







The surrender value at the end of year 5 is therefore this account value of 43.46 reduced by surrender charge of 20, i.e., 43.46 – 20 = 23.46. The asset share at the end of year 5 is Expected payout Accumulated ⎛ ⎛ Premium Expenses at withdrawal Expected death benefit payout incurred ⎞ ⎞ at asset share rate ⎜ 20 + ⎜ 20 − 2 ⎟ ⎟ ⋅ 1.08 − 0.001 ⋅1000 − 0.05 ⋅ 23.46 ⎜ ⎜ ⎟ ⎟ ⎟ ⎜⎝ ⎜⎝ ⎠ ⎟ ⎠ ≈ 40.96. 1 − 0.001 − 0.05 AS 4





  





  

   

Probability of policy continuing

Answer C.


For a universal life insurance policy with a death benefit of 50,000 plus the account value, you are given: (i) Policy Year Monthly Percent of Monthly Monthly Surrender Premium Premium Cost of Expense Charge Charge Insurance Charge Rate per 1000 1 300 2 10 500 W % 2 300 15% 3 10 125 (12 )

0.054. (ii) The credited interest rate is i (iii) The cash surrender value at the end of month 11 is 1200.00. (iv) The cash surrender value at the end of month 13 is 1802.94. Calculate W %, the percent of premium charge in policy year 1.

A. 25%

B. 30%

=

C. 35%

D. 40%

E. 45%

Solution. Let us write CSV k for the cash surrender value at time k, with time counted in months, and SC k for the surrender charge at time k. Let be COI k be the cost of insurance for month k. Since j

i =

i

(12 )

(12)

12

=

0.054, the effective monthly interest rate is

0.054 =

12

=

0.0045

=

0.45%.

Note that the cash surrender value equals the account value (i.e., reserve) minus the surrender charge. We are given that CSV 13 1802.94. Based on this, =

1802.94

=

CSV 13

=

AV 13

− SC 13

=

AV 13

− 125,

so that Copyright © 2012 by Krzysztof Ostaszewski. All rights reserved. No reproduction in any form is permitted without explicit permission of the copyright owner.

= 1802.94 + 125 = 1927.94.

AV 13

Similarly, AV 11

CSV 11

=

+

SC 11

= 1200 +

500

= 1700.

Now we use the recursive formula for the account value (reserve) for universal life with increasing total benefit and fixed ADB

( AV

t −1

+

) (1 i )

Gt (1 − r t ) − et

+

q x t ⋅ ( B + AV t 1 ) + p x t ⋅ AV t = q x t ⋅ B + AV t .

=

t

+

+

+

+

The cost of insurance in this formula is COI t = q x t ⋅ B so that ,

+

(

AV t = AV t −1

) − e ) (1 i ) − COI .

(

+

Gt 1 − r t

+

G13

+

G12

+

t

t

t

In this case AV 13

=

AV 12

=

( AV ( AV

12

11

(1 − r ) − e )(1 (1 − r ) − e ) (1 13

13

12

12

) − COI , ) − COI .

+

i13

+

i12

13

12

Substituting known values (note that the cost of insurance figures are given per thousand), we obtain

(

) − 10) ⋅1.0045 − 50 ⋅ 3,

(

1927.94 = AV 12 + 300 1 − 0.15 AV 12

=

(1700

+

) − 10) ⋅1.0045 − 50 ⋅ 2.

(

300 1 − W

From the first equation AV 12

1927.94 + 50 3 ⋅

=

1.0045

+ 10 −

(

)

300 1 0.15 −

≈

1823.63116,

and substituting this into the second equation, we obtain 1823.63116 W

≈

1

−

+ 50 ⋅ 2 −

1.0045 300

1700 + 10 ≈

0.25.

Answer A.


You are given the following about a universal life insurance policy on (60): (i) The death benefit equals the account value plus 200,000. (ii) Age x Annual Premium Annual Cost of Annual Expense Insurance Rate per Charges 1000 60 5000 5.40 100 61 5000 6.00 100 (iii) Interest is credited at 6%. (iv) Surrender value equals 93% of account value during the first two years. Surrenders occur at the ends of policy years. (v) Surrenders are 6% per year of those who survive. (vi) Mortality rates are 1000 q60 3.40 and 1000 q61 3.80. (vii) i = 7%. =

=


Calculate the present value at issue of the insurer’s expected surrender benefits paid in the second year. A. 380

B. 390

C. 400

D. 410

E. 420

Solution. The account value at time 0, just before the first premium is paid is zero. Then, since death benefit equals fixed ADB plus account value,

(

AV 1 = AV 0 + G1 (1 − r 1 ) − e1

) (1 i ) − q B (0 +

1

=

60

+ 5000

⋅1 − 100) ⋅1.06 − 200 ⋅ 5.40 = 4114.

Following into the second year, AV 2

=

( AV

=

( 4114

1

+

G2 (1 r 2 ) e2 −

−

)(1 i ) +

+ 5000 ⋅ 1 − 100

2

) 1.06 ⋅

−

−

q61 B =

200 6.00 ⋅

≈

8354.84.

If a policy is surrendered in year 2 (and for a policy in existence at the end of year 1 this happens with probability 0.06), then it was not surrendered in year 1 (and for a policy issued at time 0 this happens with probability 1 – 0.06 = 0.94), so that the present value at time 0 of the expected surrender benefits in year 2 is 8354.84     Account value available for surrenders

≈

0.93

⋅

0.06

⋅



Adjusted for surrender charge



Fraction that surrenders in year 2

⋅

(1

−

0.06 ) (1 0.0034 ) (1 0.0038 ) 1.07

   

⋅

−

⋅

 

Fraction that did not surrender in year 1

1

−

−

⋅

 

q60

1

−

q61

2

−



≈

Present value factor

380.01.

Answer A.

November 2012 Course MLC Examination, Problem No. 10 For a special 3-year term life insurance policy on ( x) and ( y) with dependent future

lifetimes, you are given: (i) A death benefit of 100,000 is paid at the end of the year of death if both ( x) and ( y) die within the same year. No death benefits are payable otherwise. (ii) p x k = 0.84366, k = 0, 1, 2. (iii) p y k = 0.86936, k = 0, 1, 2. +

+

(iv) p x

+

k : y+ k =

0.77105, k = 0, 1, 2.

(v) Maturity (in years) Annual Effective Spot Rate 1 3% 2 8% 3 10% Calculate the expected present value of the death benefit. A. 9,500

B. 10,100

C. 12,100

D. 12,500

E. 14,100

Solution. The death benefit of $100,000 is paid at the end of year 1 if both ( x) and ( y) die during the first year, or it is paid at the end of year 2 if both ( x) and ( y) are alive at time 1 and Copyright © 2012 by Krzysztof Ostaszewski. All rights reserved. No reproduction in any form is permitted without explicit permission of the copyright owner.

both die during the second year, or it is paid at the end of year 3 if both ( x) and ( y) are alive at time 2 and both die during the third year. Let us note the following, true for k = 0, 1, or 2: q x

+

k : y+ k

)

− p x

=1

− ( 0.84366 + 0.86936 − 0.77105 ) = 0.05803.

+

1 = k : y+ k Poohsticks

− ( p x

=1

+

k +

p y

+

k

− p x

+

k : y+ k

=

Also, 2

⋅p = 0.77105 ⋅ 0.77105 = 0.77105 . The expected present value we are looking for is therefore 2

p x: y = p x: y

x +1: y +1

⋅1.03−1 100,000   

⋅

+

q

x: y



Present value of death benefit Probability that payment made at time 1 both ( x ) and ( y ) die during first year

+

−3 ⋅ 100,000 1.10   

2

−2 ⋅1.08 100,000  

p x: y q

x +1: y +1

+

   

Present value of death benefit Probability that payment made at time 2 both ( x ) and ( y ) die during second year

p x: y q

x + 2: y + 2

=

  

Present value of death benefit Probability that payment made at time 2 both ( x ) and ( y ) die during second year

⎛ 0.05803 0.77105 ⋅ 0.05803 0.77105 2 ⋅ 0.05803 ⎞ ≈ 12,062.09. = 100,000 ⋅ + ⎜⎝ 1.03 + ⎟ 1.08 2 1.10 3 ⎠ Answer C.


For a whole life insurance of 1000 on (70), you are given: (i) Death benefits are payable at the end of the year of death. (ii) Mortality follows the Illustrative Life Table. (iii) Maturity (in years) Annual Effective Spot Rate 1 1.6% 2 2.6% (iv) For the year starting at time k − 1 and ending at time k , k = 3, 4, 5, ... , the one-year forward rate is 6%. Calculate the expected present value of the death benefits. A. 520

B. 530

C. 550

D. 570

E. 600

Solution. It seems like it will be a lot of calculations to go from age 70 to the limiting age if we do every year calculation separately. But the Illustrative Life Table of course gives us actuarial present values of whole life insurance death benefits, except that it does so for the valuation interest rate of 6%. Here, the spot rates for the first two years are well below 6%, but beyond that all one-year forward rates are 6% (so that all rates for all periods starting at time 2 or later are also 6%). Th us starting from age 72, we can determine actuarial present values from the Illustrative Life Table. We have


1000 A70

= 1000

⎛ q + p ⋅ q ⎜⎝ 1.016 1.026 70

71

70

2

+

p70 ⋅ p71 1.026

2

, ⋅ A ⎞ ⎟ ⎠ 72

where all values on the right-hand side can now be looked up in the Illustrative Life Table. From the table: q70 0.03318, q71 0.03626, A72 0.54560, and therefore p70 0.96682, p71 0.96374, so that =

=

=

=

=

1000 A70

= 1000

⎛ 0.03318 + 0.96682 ⋅ 0.03626 + 0.96682 ⋅ 0.96374 ⋅ 0.54560 ⎞ ≈ ⎜⎝ 1.016 ⎟ ⎠ 1.026 1.026 2

2

≈ 548.89. Answer C.


A party of scientists arrives at a remote island. Unknown to them, a hungry tyrannosaur lives on the island. You model the future lifetimes of the scientists as a three-state model, where: State 0: no scientists have been eaten. State 1: exactly one scientist has been eaten. State 2: at least two scientists have been eaten. You are given: (i) Until a scientist is eaten, they suspect nothing, so µ 01 = 0.01 + 0.02 ⋅ 2 , t > 0. (ii) Until a scientist is eaten, they suspect nothing, so the tyrannosaur may come across two together and eat both, with µ 02 0.5 µ 01 , t > 0. t

t

=

t

t

(iii) After the first death, scientists become much more careful, so µ 12 Calculate the probability that no scientists are eaten in the first year. t

A. 0.928

B. 0.943

C. 0.951

D. 0.956

=

0.01, t > 0.

E. 0.962

Solution. Note that because it is impossible to return to state 0, the probability we are looking for, 00

1

p0

00

is the same as

1

p0 .

We calculate

1

− 00

p0

1

=

00

p0

=

1

e

1

∫ (µ t

01

02 + µ t

) dt

0

− =

t =1

=

e

0.03 t ⎞ −⎛ ⎜⎝ 0.015 t + ln2 ⋅2 ⎟ ⎠ t 0 =

e

1

∫ ( µ t

01

01 + 0.5 µ t

0

e

− =

⎞ −⎛ ⎜⎝ 0.015+ ln2 − 0− ln 2 ⎟ ⎠ 0.06

=

) dt e

0.03

=

e

∫

(

⋅

1.5 0.01+ 0.02 2

t

0

) dt =

−0.015−

0.03 ln2

≈ 0.94338496.

We could also note that 01

µ t

+

02

µ t

=

( 0.01

+

0.02 ⋅ 2

t

)

+

(

0.5 0.01 + 0.02 ⋅ 2

t

)

=

t

0.015 + 0.03 ⋅ 2 ,

and this is the Makeham’s Law with A = 0.015, B = 0.03, c = 2. Recall that under x

Makeham’s Law,

n

p x

=

n

s g

c

( c −1)

0.03

B

n

, where g

=

e

−

ln c

and

s

=

e

− A

. Here,

−

g

=

e

ln2

,


s

=

e

−0.015

, and

is the same as

p x

n

1

e

=

⎛ − ⎞ ⋅ ⎜ e ⎟ ⎝ ⎠

−0.015 n

00

00

=

1

n

, so that the quantity sought, i.e.,

00

1

p0

=

00

1

p0

,

under this Makeham model, i.e.,

p0

(

p0

e

=

0.015 1

−

⋅

⋅

)

1 2 1 −

0.03

p0

⋅( 2 −1)

ln 2

−

1

x

2

0.03

ln2

e

≈

0.94338496.

Answer B.


You are given: (i) The following excerpt from a triple decrement table: x

(τ )

(1)

( 2)

d x

l x

( 3)

d x

d x

50 100,000 490 8,045 1,100 51 90,365 -8,200 -52 80,000 ---(ii) All decrements are uniformly distributed over each year of a ge in the triple decrement table. (iii)

q x′

( 3)

is the same for all ages.

Calculate 10,000 q51( ) . 1

′

A. 130

B. 133

C. 136

D. 138

E. 141

Solution. Recall the key formula for UDD over each year of age in the multiple decrement table: ( j )

′

p x

( j )

=

(p ) (τ )

x

q x

(τ )

q x

.

We have (τ )

p50

( 3)

q50

(τ ) q50

(τ )

l51

=

(τ )

l50

90,365 =

100,000

( 3)

d 50

=

=

1,100 =

(τ )

l50

0.90365,

=

100,000

(τ )

1 − p50

=

=

0.011,

1 − 0.90365

=

0.09635.

We calculate therefore ( 3)

( 3)

q51 ′

=

( 3)

q50 ′

This means that

=

( 3)

1 − p50 ′

=

(

1 − p50

( 3) ≈ 0.9885 ′ p51

( 3)

( p( ) ) τ

=

51

0.011

q50

=

1 − 0.90365

0.09635

≈ 0.0115.

( 3)

q51

(τ )

q51

)

(τ )

and

( 3)

0.9855 ≈ p51 ′

(τ )

q50

⎛ 80,000 ⎞ − ⎜⎝ 90,365 ⎟ ⎠ 1

=

q51

80,000 90,365

.

From this we calculate Copyright © 2012 by Krzysztof Ostaszewski. All rights reserved. No reproduction in any form is permitted without explicit permission of the copyright owner.

⎛ 1 − 80,000 ⎞ ⋅ ln 0.9855 ⎜⎝ 90,365 ⎟ ⎠ ( ) q ≈ ≈ 0.010889815. 80,000 ⎞ ⎛ ln ⎜ ⎝ 90,365 ⎟ ⎠ 3 51

This means that ( 3)

d 51

(1)

d 51

(1)

q51

=

( 3)

(τ )

q51 l51 ⋅

(τ )

(2)

=

d 51

≈

( 90,365

−

d 51

(1)

=

0.010889815 90,365

≈

⋅

−

−

( 3)

d 51

80,000 )

1,181

d 51

≈

(τ )

l51

(l ( ) τ

=

90,365

≈

50

−

−

(τ )

l51

8200

)

−

−

≈

(2)

d 51

984

984.0581,

−

=

( 3)

d 51

≈

1,181,

0.01306922,

and we also have 80,000

(τ ) p51

=

(τ ) q51

=

90,365 1

−

(τ ) p51

≈

=

0.88529851,

1

80,000 −

90,365

≈

0.11470149.

We therefore conclude that (1)

(1)

p51 ′

=

(p ) (τ )

51

0.01306922

q51

(τ )

q51

≈ 0.885298510.11470149 ≈ 0.98621441.

The quantity sought is (1)

10,000 q51 ′

=

(

(1)

10,000 1 − p51 ′

)

≈ 10,000

(1 − 0.98621441) ≈ 138.

Answer D.


For a special whole life insurance policy issued on (40), you are given: (i) Death benefits are payable at the end of the year of death. (ii) The amount of benefit is 2 if death occurs within the first 20 years and is 1 thereafter. (iii) Z is the present value random variable for the payments under this insurance. (iv) i = 0.03. (v) x

40 60 (vi) E ( Z 2 )

=

A x

20

0.36987 0.62567

0.51276 0.17878

E x

0.24954.

Calculate the standard deviation of Z. A. 0.27

B. 0.32

C. 0.37

D. 0.42

E. 0.47

Solution. The present value of death benefit random variable is Copyright © 2012 by Krzysztof Ostaszewski. All rights reserved. No reproduction in any form is permitted without explicit permission of the copyright owner.

⎧⎪ 2 v 1 , Z = ⎨ 1 ⎪⎩ v , K +

K <

20,

K ≥

20.

K +

Therefore,

( )

E Z

=

2 A40

−

20

E 40 A60

2 0.36987

=

⋅

⋅

−

0.51276 0.62567 ⋅

≈

0.41892.

We conclude that the variance of Z is Var ( Z )

=

( ) ( E ( Z )) 2

E Z

2

−

≈

0.24954

0.41892

−

2

0.07405,

≈

and the standard deviation of Z is σ

Z

( )

Var Z

=

0.07405

≈

≈

0.27212.

Answer A.

November 2012 Course MLC Examination, Problem No. 15 For a special 2-year term insurance policy on ( x), you are given:

(i) Death benefits are payable at the end of the half-year of death. (ii) The amount of the death benefit is 300,000 for the first half-year and increases by 30,000 per half-year thereafter. (iii) q x 0.16 and q x 1 = 0.23. =

(2)

+

0.18. (iv) i (v) Deaths are assumed to follow a constant force of mortality between integral ages. (vi) Z is the present value random variable for this insurance. Calculate Pr( Z > 277,000). =

A. 0.08

B. 0.11

C. 0.14

D. 0.18

E. 0.21

Solution. Note that the effective half a year interest rate is 9%. Also, under the constant force of mortality between integral ages assumption, if we write µ for the constant force between ages x and x + 1, and µ 1 for the constant force between ages x +1 and x + 2, then x

x +

1

0.5

q x

0.5

q x +0.5 = 1

0.5

q x +1 = 1

0.5

q x +1.5 = 1

=

−

p x

0.5 −

−

1

−

e

p x +1 = 1

0.5

0.5 µ x

−

p x + 0.5 = 1

0.5

0.5

−

=

−

p x +1.5 = 1

1

−

0.5

p x

0.5

= 1−

0.5 µ x+1

= 1−

p x

−

e

=

−

e

−

0.5 µ x +1

−

=

1

−

1 0.16 −

1 0.16 −

0.5

p x +1 = 1

= 1−

≈

−

0.083485,

0.083485,

1 0.23 −

0.5

p x +1 = 1

≈

−

1

−

≈

0.23

0.122504, ≈

0.122504.

Based on the above, probabilities of the insured dying in each of the four half-years are: 0 0.5

q x

=

q 0.5 0.5 x

0.5

q x

= 0.5

p x

1 0.5

q x = 1 p x

⋅

1 0.5

q x = 1 p x

⋅

0.083485,

≈

⋅

0.5 q x + 0.5

0.5

q x +1

0.5

p x +1 ⋅

≈

(1

0.5

≈

0.916515 0.083485

−

0.16 ) 0.122504

⋅

q x +1.5

⋅

≈

(1

−

≈

≈

0.076515,

0.102903,

0.16 ) 0.877496 0.122504 ⋅

⋅

≈

0.090297.

The random present value of the death benefit is Copyright © 2012 by Krzysztof Ostaszewski. All rights reserved. No reproduction in any form is permitted without explicit permission of the copyright owner.

⎧ ⎪ ⎪ ⎪ ⎪⎪ Z ⎨ ⎪ ⎪ ⎪ ⎪ ⎪⎩ =

300,000

≈ 275, 229.36

1.9 330,000

with probability

≈ 277, 754.40 with probability

1.9 2 360,000 1.9 3 390,000 1.9 4

q ≈ 0.083485,

0.5 x

0.5 0.5

≈ 277, 986.05

with probability

≈ 276, 285.83

with probability

q x ≈ 0.076515,

q ≈ 0.102903,

1 0.5 x

q ≈ 0.090297.

1 0.5 x

Also, Z = 0 otherwise. We conclude that Pr ( Z > 277,000 ) =

0.5 0.5

q x

+

1 0.5

q x

≈

0.076515 + 0.102903

≈

0.179418.

Answer D.


You are evaluating the financial strength of companies based on the following multiple state model: State 0 Solvent

State 1 Bankrupt

State 2 Liquidated

For each company, you assume the following constant transition intensities: (i) µ 01 0.02. =

(ii) (iii)

10

µ

=

0.06.

12

µ

=

0.10.

Using Kolmogorov’s forward equations with step

h

1 =

2

, calculate the probability that a

company currently Bankrupt will be Solvent at the end of one year. A. 0.048

B. 0.051

C. 0.054

D. 0.057

E. 0.060

Solution. We have, using the Markov process notation: λ 01 ( t )

=

λ 01

λ 02 ( t )

=

λ 02

λ 0 ( t ) = λ 0

=

=

=

01

µ

02

µ

=

=

0.02, 0.00,

λ 01 + λ 02

=

0.02 + 0.00 = 0.02,


λ 10 ( t )

=

λ 10

=

µ

λ 12 ( t )

=

λ 12

=

µ

λ 1 ( t ) = λ 1 =

λ 20

λ 21 ( t )

=

λ 21

λ 2 ( t ) = λ 2

12

λ 10

=

λ 20 ( t )

10

=

=

0.06,

=

0.10,

λ 12

+ 20

µ

21

µ

λ 20

=

=

=

=

+

0.06 + 0.10 = 0.16,

=

0.00, 0.00,

λ 21

=

0+0

=

0.

This tells us that the generator matrix is:

⎡ − λ (t + t ) λ ( t + r ) λ ( t + r ) 0 01 02 ⎢ Qt r ⎢ λ 10 ( t + r ) − λ 1 ( t + t ) λ 12 ( t + r ) ⎢ ⎢⎣ λ 20 ( t + r ) λ 21 ( t + r ) − λ 2 ( t + t )

⎤ ⎥ ⎡ −0.02 0.02 ⎥ = ⎢ 0.06 −0.16 ⎥ ⎢ 0 0 ⎥⎦ ⎢⎣

+

0 0.1 0

⎤ ⎥. ⎥ ⎥⎦

For

r

Pt

=

⎡ ⎢ ⎢ ⎢ ⎢ ⎣

( t )

r

(t )

p00

r

( t )

r

p10

r

p20

p01

r

(t )

r

p11

r

p21

( t )

( t ) p02 ⎤ ( t )

r

p12

r

p22

(t )

( t )

⎥ ⎥, ⎥ ⎥ ⎦

the resulting Kolmogorov Forward Equation, in matrix form, is

d dr

r

Pt =

⎡ ⎢ d ⎢ dr ⎢ ⎢ ⎣

⎡ ⎢ = ⎢ ⎢ ⎢ ⎣

( t )

( t )

p00

r

p01

r

p02

r

() p10

r

() p11

r

() p12

t

( t )

r p20

() p00 t

r

( t )

r

t

( t )

r p21

() p01 t

r

(t )

( t )

r

p10

r

p11

r

() p20

r

() p21

t

t

( t )

r p22

⎡ ⎤ ⎢⎢ ⎥ ⎢ ⎥=⎢ ⎥ ⎢ ⎥ ⎢ ⎦ ⎢ ⎣

d dr d

() p22 t

r

⎡ −0.02 p(t ) + 0.06 p(t ) r 00 r 01 ⎢ ( t ) ( t ) = ⎢ −0.02 p r 10 + 0.06 r p11 ⎢ ( t ) ( t ) ⎢ −0.02 r p20 + 0.06 r p21 ⎣

( t )

dr d

dr d

p10

dr

r

p20

( t )

( t ) ⎤ p02 ⎥ ⎡ −0.02 ( t ) ⎥ ⎢ ⋅ 0.06 r p12 ⎢

t

d

r

d

(t )

dr d

dr d

r

p11

dr

r

p21

(t )

0

−0.16

0.1

0

0

0

t

() r p01

0.02

r

⎥ ⎥ ⎢⎣ ⎦

t

() r p00

() () p00 − 0.16 r p01

t

( t )

dr d

r

p12

dr

r

p22

( t )

(t ) ⎤ p01

t

0.02 r

(t )

( t )

p10 − 0.16 r p11

0.1 r

p11

0.02 r

() () p20 − 0.16 r p21

0.1 r

() p21

t

t

⎥ ⎥ ⎥= ⎥ ⎥ ⎥ ⎦

⎤ ⎥= ⎥ ⎥⎦

t

0.02 r

⎤

() r p02 ⎥

0.1 r

(t ) t

⎥ ⎥. ⎥ ⎥ ⎦

(0) We are looking for an approximate value of the probability 1 p10 . We can actually remove the superscript related to the starting point in time, because due to constant intensities of (0) transition, the starting time does not affect probabilities, so 1 p10 p , and similarly we 1 10 will drop the superscript for other probabilities. The equation =

d dr

( t )

r

p10

=

( t ) ( t ) + 0.06 r p11 −0.02 r p10


gives us this approximation 1

p10

−

0.5

p10

0.5

0.02 0.5 p10

≈ −

0.06 0.5 p11 ,

+

or 1

p10

≈

0.99 0.5 p10

+

0.03 0.5 p11 .

Additionally d

() p10 t

r

dr

=

d

( t ) ( t ) + 0.06 r p11 −0.02 r p10

() p11 t

dr

r

=

0.02 r

() () p10 − 0.16 r p11

p11

0

t

t

give us approximations p10

0.5

−

0

p10

0.5

0.5

0.02 0 p10 + 0.06 0 p11,

≈ −

−

p11

0.5

≈

0.02 0 p10

−

0.16 0 p11 ,

or 0.5

But

0

p10

p10

=

0.99 0 p10

≈

0,

p11

0

=

+

0.03 0 p11,

0.5

p11

≈

0.010 p10

+

0.92 0 p11 .

1, so that

0.5

p10

≈

0.99 0 p10

+

0.03 0 p11

=

0.99 0 + 0.03 1 = 0.03,

0.5

p11

≈

0.010 p10

+

0.92 0 p11

=

0.01 0 + 0.92 1 = 0.92.

+

0.03 0.5 p11

⋅

⋅

⋅

⋅

We conclude that 1

p10

≈

0.99 0.5 p10

≈

0.99 0.03 + 0.03 0.92 ⋅

⋅

≈

0.0573.

Answer D.


For a whole life insurance of 1000 with semi-annual premiums on (80), you are given: (i) A gross premium of 60 is payable every 6 months starting at age 80. (ii) Commissions of 10% are paid each time a premium is paid. (iii) Death benefits are paid at the end of the quarter of death. (iv) t V denotes the gross premium reserve at time t . (v) 10.75V 753.72. (vi) =

(vii) i ( ) Calculate 4

=

A. 680

t

l90+t

0 0.25 0.50 0.75

1000 898 800 706

0.08. V .

10.25

B. 690

C. 700

D. 710

E. 730

Solution. We can use the recursive reserve formula twice, in quarterly steps. Note that the effective quarterly interest rate is 2%, and that when going from policy duration 10.25 to 10.75, we only have a premium payment at policy duration 10.50. Thus Copyright © 2012 by Krzysztof Ostaszewski. All rights reserved. No reproduction in any form is permitted without explicit permission of the copyright owner.

( (

V + 0 ) ⋅ 1.02 =

10.25

0.25

q90.25 ⋅ 1000 +

V + 60 (1 − 0.10 )) ⋅ 1.02 =

10.50

0.25

p90.25 ⋅ 10.50V ,

0.25

q90.50 ⋅ 1000 +

0.25

p90.50 ⋅ 10.75V .

Now we substitute the following known data: V

10.75

753.72,

=

q 0.25 90.25

0.25

=

p90.25

q 0.25 90.50

0.25

=

=

p90.50

=

0.25

d 90.25

l90.25 l90.50 l90.25 0.25

l90.25

898

l90.50 l90.50

l90.25 − l90.50

800 =

d 90.50

l90.75

=

=

l90.50

800

898 − 800 898

98 =

898

,

,

l90.50 − l90.75

706 =

=

=

800 − 706 800

94 =

800

,

,

we obtain V ⋅ 1.02

10.25

(

98

=

V + 54

10.50

898

⋅ 1000 +

) ⋅1.02 =

94 800

800 898

⋅ 10.50V ,

⋅1000 +

706 800

⋅ 753.72.

From the second equation 94 V =

800

⋅

1000 +

10.50

706 800

⋅

753.72 −

1.02

54

≈

753.3117

Using this in the first equation, we obtain 98 V

10.25

≈

898

⋅

1000 +

800 898

⋅

753.3117

1.02

≈

729.9984.

Answer E.


For a special fully discrete whole life insurance on (40), you are given: (i) The death benefit is 1000 during the first 11 years and 5000 thereafter. (ii) Expenses, payable at the beginning of the year, are 100 in year 1 and 10 in years 2 and later. (iii) Π is the level annual premium, determined using the equivalence principle. (iv) G 1.02Π is the level annual gross premium. (v) Mortality follows the Illustrative Life Table. (vi) i = 0.06. (vii) 11 E 40 0.50330. Calculate the gross premium reserve at the end of year 1 for this insurance. =

=

A. −70

B. −60

C. −50

D. −40

E. −30


Solution. We begin by finding 1000 A40

+

4

Π.

The actuarial present value of benefits at issue is

E 40 1000 A51

⋅

⋅

11

= 161.32 +

4 0.50330 259.61 ⋅

⋅

683.9669.

≈

The actuarial present value of expenses at issue is

(  − 1) = 100 + 10 ⋅ (14.8166 − 1) = 238.17.

100 + 10 a40

= 100 + 10 a40

All the values used in calculations above came from the Illustrative Life Table. The premium is paid as a life annuity due on (40), hence Π≈

683.97 + 238.17

40

=

683.97 + 238.17 14.8166

a

Therefore, obtain G

G

−

=

If we calculated the reserve retrospectively, we would

1.02Π ≈ 63.4812.

1000 q40 1.06 ⋅

1

−

−

100

63.4812 ≈

E 40

(1

1

But prospectively, we obtain (1000 A41 + 4 10 E 41 1000 A51 ) + ⋅

≈ 62.2365.

2.78 1.06 ⋅

0.00278

41 10 a

⋅

 

−

−

−



Actuarial present value of future expenses

Actuarial present value of future benefits

)

⋅

(

41.6056 )

−

−

(

⋅

61.2012 )

−

≈

−

1.06

100

41.6056.

≈ −

1

−

41 = 63.4812 a   Actuarial present value of future gross premiums

= 168.69 + 4 0.53499 259.61 + 10 14.6864 The difference of these two is ⋅

1

−

⋅

−

63.48 14.6864 ⋅

61.2012.

≈ −

19.60.

Note that the actuarial present value at policy inception of the profit portion of the gross premium is

40 ≈ 0.02 ⋅ 62.2365 ⋅ 14.8166 ≈ 18.4427. 0.02 ⋅ Π ⋅ a Its actuarial accumulated value at policy duration 1 is approximately 18.4427 E 40

1

18.4427 =

(1

−

0.00278 ) 1.06 ⋅

1

−

≈

19.60.

Let us write the actuarial equivalence for gross premium valuation at policy inception: Actuarial Present Value of Future Gross Premiums = = Actuarial Present Value of Future Benefits and Expenses + + Actuarial Present Value of Future Profits. Note that at policy inception, the actuarial present value of future benefits and expenses minus the actuarial present value of future gross premiums equals the opposite (i.e., minus) of the actuarial present value of future profits. Now c onsider a policy duration k, where k is an integer. Let us write APV for actuarial present value, and AAV for actuarial accumulated value. We have at policy inception: APV (Future Gross Premiums) = = APV (Future Gross Premiums up to time k ) + + APV (Future Gross Premiums from time k on) = APV (Future Benefits and Expenses) + APV (Future Profits) = = APV (Future Benefits and Expenses up to time k ) + + APV (Future Benefits and Expenses from time k on) + + APV (Future Profits). Now we accumulate all of these amounts to policy duration k Copyright © 2012 by Krzysztof Ostaszewski. All rights reserved. No reproduction in any form is permitted without explicit permission of the copyright owner.

AAV (Past Gross Premiums up to time k ) + + APV (Future Gross Premiums from time k on) = = AAV (Past Benefits and Expenses up to time k ) + + APV (Future Benefits and Expenses from time k on) + + Actuarial Value (Past and Future Profits). Therefore, at policy duration k, Retrospective Reserve = AAV (Past Gross Premiums) – AAV (Past Benefits and Expenses) = = APV (Future Benefits and Expenses) – APV (Future Gross Premiums) + + Actuarial Value (Past and Future Profits) = = Prospective Reserve + Actuarial Value (Past and Future Profits). If you recall that the purpose of the reserve is to provide funds for benefits and expenses payment that are not paid by remaining premiums, you now should be able to clearly see that the prospective reserve calculation is the correct one, and if you do the calculation retrospectively, you must subtraction the actuarial value of all profits at the moment of calculation from the retrospective calculation result. In this problem, the reserve at policy duration 1 is: –61.20 = –41.60 – 19.60. Interestingly, –40 was one of the answer choices, nice little trap waiting for the candidates taking the exam. The correct answer choice is –60, the one closest to –61.20. Answer B.


For whole life annuities-due of 15 per month on each of 200 lives age 62 with independent future lifetimes, you are given: (i) i = 0.06. 2 (12 ) (12 ) (ii) A62 0.4075 and A62 0.2105. (iii) Π is the single premium to be paid by each of the 200 lives. (iv) S is the present value random variable at time 0 of total payments made to the 200 lives. Using the normal approximation, calculate Π such that Pr(200 Π > S ) = 0.90. =

=

A. 1850

B. 1860

C. 1870

D. 1880

E. 1890

Solution. Let X 1 , X 2 , …, X 200 , be the random present values of life annuities-due considered. These are independent identically distributed random variables, and S = X 1

+

X 2

+ ... + X 200 .

This implies that S is approximately normal and

( ) = E ( X + X

E S

1

2

+ ... + X 200

) = 200 E ( X ) 1

as well as

( ) = Var ( X

Var S

Recall that

(12 )

d

1

+

X 2

+ ... + X 200

) = 200Var ( X ). 1

⎛ 1 − (1 + i )− ⎞ . We have ⎜⎝ ⎟ ⎠ 1

= 12

12


( )

E X 1

(12 )

62 15 ⋅ 12 ⋅ a

=

=

15 ⋅ 12 ⋅

(12 )

1 − A62

=

(12 )

d

15 ⋅ 12 ⋅

1 − 0.4075 − ⎞ ⎛ 12 ⎜ 1 − 1.06 ⎟ ⎝ ⎠ 1

≈ 1834.7545,

12

and (12 ) A62

2

Var ( X 1 )

2

=

(15 ⋅12 ) ⋅

(

− ( A(

12 ) 62

(12) d

2

)

2

)

=

180

2

⋅

0.2105 − 0.4075

2 2

⎛ ⎛ − ⎞ ⎞ − 12 1 1.06 ⎜⎝ ⎜⎝ ⎟ ⎠ ⎟ ⎠ 1

≈ 426,176.9086.

12

Based on this,

( )

E S

200 E ( X 1 )

=

≈

200 1834.7545 ⋅

≈

366,936.12,

and Var ( S )

=

200Var ( X 1 )

≈

200 426,176.9086 ⋅

≈

85,235,381.71.

We have

⎛ 200Π − E ( S ) S − E ( S ) ⎞ 0.90 = Pr ( 200Π > S ) = Pr ⎜ > ⎟ . Var Var S S ( ) ( ) ⎝ ⎠ Since

( ) is approximately standard normal, Var ( S )

S − E S

( ) must be the 90-th Var ( S )

200Π − E S

percentile of the standard normal distribution, i.e., 1.282. This gives the equation 1.282

=

200Π − E ( S ) Var ( S )

=

200Π − 366,936.12 85,235,381.71

,

resulting in Π=

366,936.12 + 1.282 ⋅ 85,235,381.71 200

≈ 1893.8597.

Answer E.


Stuart, now age 65, purchased a 20-year deferred whole life annuity-due of 1 per year at age 45. You are given: (i) Equal annual premiums, determined using the equivalence principle, were paid at the beginning of each year during the deferral period. (ii) Mortality at ages 65 and older follows the Illustrative Life Table. (iii) i = 0.06. (iv) Y is the present value random variable at age 65 for Stuart’s annuity benefits. Calculate the probability that Y is less than the actuarial accumulated value of Stuart’s premiums. A. 0.40

B. 0.42

C. 0.44

D. 0.46

E. 0.48

Solution. Copyright © 2012 by Krzysztof Ostaszewski. All rights reserved. No reproduction in any form is permitted without explicit permission of the copyright owner.

Let us write P for the annual premium that Stuart has paid. Since it was calculated using the equivalence principle, we have 65 E 45 ⋅ a 20 . P 45:20 a =

Let us write K 65 for the curtate future lifetime of (65) (we could also write K (65), but we need to get used to the new notation, so this is a good time to practice). The probability we are looking for is (note that a65 9.8969 from the Illustrative Life Table) =

⎛ Pa Pr ⎜⎝ E

45:20

20

>

 a

20

K 65 +1

45

⎛ E ⋅ a a ⎞ ⋅ = Pr ⎜ ⎟ E ⎠ ⎝ a 45

65

45:20

20

45:20

=

(

65 Pr a

>

 a

K 65 +1

)

>

 a

K 65 +1

45

=

⎞ ⎟ = ⎠

(

 Pr 9.8969 > a

K 65 +1

).

If we use BA II Plus Pro, set BGN mode by pushing 2ND, BGN if BGN is not displayed, and enter 0 FV, –1 PMT, 9.8969 PV, 6 I/Y, CPT N, we obtain 14.0973895. This means that the probability sought is Pr

( K

65

+ 1 < 14.0973895

) = Pr ( K

65

= 1−

l79 l65

+ 1 ≤ 14

= 1−

) = 1−

4225163 7533964

14

p65

=

≈ 0.43917133.

For a moment, when you see the formula for P you might worry about working with mortality from before age 65, when we are told that mortality follows the Illustrative Life Table only from age 65 on, but as you can see, items related to mortality from before age 65 cancel nicely in the work involved in finding the probability sought. Answer C.

November 2012 Course MLC Examination, Problem No. 21 For a fully continuous whole life insurance issued on ( x) and ( y), you are given:

(i) The death benefit of 100 is payable at the second death. (ii) Premiums are payable until the first death. (iii) The future lifetimes of ( x) and ( y) are dependent. (iv)

t

p xy

(v) t p x (vi)

t

=

=

p y

e

=

1 4

e

−0.01t

−0.01t

e

,

−0.02 t

t ≥

,

3

+

4

e

−0.03t

,

t ≥

0.

0.

t ≥

0.

(vii) δ 0.05. Calculate the annual benefit premium rate for this insurance. =

A. 0.96

B. 1.43

C. 1.91

D. 2.39

E. 2.86

Solution. Note that the future lifetime of ( x) is ruled by constant force of mortality of 1%, and the future lifetime of ( y) is rules by constant force of mortality of 2%. Let us write P for the annual benefit premium rate sought. We have Copyright © 2012 by Krzysztof Ostaszewski. All rights reserved. No reproduction in any form is permitted without explicit permission of the copyright owner.

P

=

100 A x: y

.

a x: y

Also, ∞

+

a x: y

∫

e−0.05

=

t

0

1

1

0.01t

∞

+

e 4 ∫

=

− ⋅ ⎛ ⎜⎝ 4 e

−0.06 t

3

0

A x: y

=

Poohsticks

∞

+

e 4 ∫

dt +

−0.08 t

1

dt =

A x

+

A y

−

A x: y

=

A x

+

A y

−

(1

0.02

0.01 + 0.05

+

1

⋅

4 0.06

0

0.01 =

3 −0.03t ⎞ e dt = ⎟ ⎠ 4

+

0.02 + 0.05

−

−

(1

+

δ a x: y

−

3

⋅

1

4 0.08

)

≈ 13.5416667,

=

0.05 13.5416667 ) ⋅

≈

0.12946429.

Therefore, P

=

100 A x: y

a x: y

100 0.12946429 ⋅

≈

13.5416667

≈

0.95604396.

Answer A.


You are given the following information about a special fully discrete 2-payment, 2-year term insurance on (80): (i) Mortality follows the Illustrative Life Table. (ii) i = 0.0175. (iii) The death benefit is 1000 plus a return of all premiums paid without interest. (iv) Level premiums are calculated using the equivalence principle. Calculate the benefit premium for this special insurance. A. 82

B. 86

C. 90

D. 94

E. 98

Solution. Let us write P for the level annual premium sought. Based on the equivalence principle, we write

 Pa

=

80:2

1

1000 A80:2

+

1

( )80:2 ,

P ⋅ IA

so that 1

1000 A P

80:2

=

 a

80:2

1

− ( IA)80:2

.

We calculate

 a

80:2

= 1+

vp80 = 1 +

1

−

q80

1.0175

= 1+

1 0.0803 −

1.0175

≈

1.90388206,


⎛ q + p q ⎞ = ⎜⎝ 1.0175 1.0175 ⎟ ⎠ ⎛ 0.0803 + 0.9197 ⋅ 0.08764 ⎞ ≈ 156.772702, = 1000 ⎜ ⎟ ⎝ 1.0175 ⎠ 1.0175

1

1000 A80:2

80

= 1000

80

81

2

2

and q80

1

( IA )

80:2

=

+

1.0175

2 p80 q81 1.0175

2

0.0803 =

1.0175

2 0.9197 0.08764 +

⋅

⋅

1.0175

2

≈

0.23462648.

Therefore, 1

1000 A P

156.772702

80:2

=

≈

1

 a

−

80:2

( IA)

1.90388206

80:2

−

0.23462648

≈

93.9177342.

Answer D.


A life insurance company issues fully discrete 20-year term insurance policies of 1000. You are given: (i) Expected mortality follows the Illustrative Life Table. (ii) Death is the only decrement. (iii) 3V , the reserve at the end of year 3, is 12.18. On January 1, 2009, the company sold 10,000 of these policies to lives all aged 45. You are also given: (i) During the first two years, there were 30 actual deaths from these policies. (ii) During 2011, there were 18 actual deaths from these policies. Calculate the company’s gain due to mortality for the year 2011. A. 28,100

B. 28,300

C. 28,500

D. 28,700

E. 28,900

Solution. The mortality gain equals ( Expected deaths − Actual deaths ) ⋅ Net Amount at Risk. In this case, the net amount at risk is 1000 – 3V = 1000 – 12.18 = 987.82. Expected deaths are q47 from the Illustrative Life Table, i.e., 0.00466, times the number of alive policyholders at the beginning of 2011, which is 10,000 – 30 = 9970, which gives 0.00466 9970 ⋅

≈

46.4602.

The actual number of deaths in 2011 is 18. Hence, the mortality gain equals

( 46.4602

−

18 ) 987.82 ⋅

≈

28,113.5548.

Answer A.


An insurance company is designing a special 2-year term insurance. Transitions are modeled as a four-state homogeneous Markov model with states:


(H) Healthy (Z) infected with virus “Zebra” (L) infected with virus “Lion” (D) Death The annual transition probability matrix is given by: H Z L D H 0.90 0.05 0.04 0.01 Z 0.10 0.20 0.00 0.70 L 0.20 0.00 0.20 0.60 D 0.00 0.00 0.00 1.00 You are given: (i) Transitions occur only once per year. (ii) 250 is payable at the end of the year in which you become infected with either virus. (iii) For lives infected with either virus, 1000 is payable a t the end of the year of death. (iv) The policy is issued only on healthy lives. (v) i = 0.05. Calculate the actuarial present value of the benefits at policy issue. A. 66

B. 75

C. 84

D. 93

E. 102

Solution. We list all possible cases of payouts, with their probabilities and actuarial present values (calculated as products of probabilities and discounted benefits) Possible Probability Discounted Actuarial transition benefit present value HàZ

0.05

HàL

0.04

HàZ

à

D

0.05 ⋅ 0.7

HàL

à

D

0.04 ⋅ 0.6

HàH

à

Z

0.9 ⋅ 0.05

HàH

à

Z

0.9 ⋅ 0.04

250

11.9047619

1.05 250

9.52380952

1.05 1000 1.05

2

1000 1.05

2

250 1.05

2

250 1.05

2

31.7460317 21.7687075 10.2040816 8.16326531

The total actuarial present value of the benefits at policy issue is the sum of all numbers in the last column, which is approximately 93.3106576. Answer D.

November 2012 Course MLC Examination, Problem No. 25 For a fully discrete 10-year term life insurance policy on ( x), you are given: Copyright © 2012 by Krzysztof Ostaszewski. All rights reserved. No reproduction in any form is permitted without explicit permission of the copyright owner.

(i) Death benefits are 100,000 plus the return of all gross premiums paid without interest. (ii) Expenses are 50% of the first year’s gross premium, 5% of rene wal gross premiums and 200 per policy expenses each year. (iii) Expenses are payable at the beginning of the year. (iv) A x1:10 0.17094. =

(v)

1

( IA)

(vi) a :10

0.96728.

=

x:10 =

6.8865.

x

Calculate the gross premium using the equivalence principle. A. 3200

B. 3300

C. 3400

D. 3500

E. 3600

Solution. Let us write G for the gross premium sought. Using the equivalence principle, we write:  Ga

x:10

=

100,000 A x1:10

  

+

APV of death benefit

1

( ) :10 + 0.45G + 0.05Ga :10

G ⋅ IA

x

 

x   

APV of return of premium benefit

% of premium expenses

200a x:10

+

.

   

Per policy expenses

We solve for G and obtain 1

G

=

100,000 A x:10

 x:10 0.95a

+

 x:10 200a

1

−

( IA) :10 x

−

0.45

=

100,000 0.17094 + 200 6.8865 ⋅

0.95 6.8865 ⋅

⋅

−

0.96728

−

0.45

≈

3604.2299.

Answer E.


Nov 2012 Mlc

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