Econ 210A PS1 Answer Key
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Jehle ehle-R -Ren eny y
1.2 (a) In order order to prove prove this, we must must show show that every every eleme element nt in is in the set . Let x, y ∈ X s.t. x y. Therefore, Therefore, x, y ∈. By definition, x y and x, y ∈. Therefore ⊂ (b) In order to prove this, we must show every element in ∼ is an element in . Let x, y ∈ X s.t. x ∼ y. If x ∼ y then, we have by definition x yANDy x . Sinc Sincee x y , then x, y ∈ and ∼⊂ (c) In order to prove prove this we must must show containment containment both ways. ways. That is, ( ∪ ∼) ⊂ and ⊂ ( ∪ ∼). Star Startin ting g with ( ∪ ∼) ⊂: Let x, y ∈ ( ∪ ∼). Then by definition, x, y are either a member of OR ∼. Therefore, Therefore, by definition definition either, either, x y AND NOT (y x) OR x y AND y x. In eit eithe herr case case,, x y and x, y ∈. Proving the other containment: let x, y ∈ s.t. X y. There are two possibilities between the relationship between y and x. Either Either,, y x or NOT(y NOT(y x) (these are complementary scenarios). scenarios). Since we have x y AND (y (y x OR NOT ( NOT (y x)), x, y ∈ ( ∪ ∼) by definition and ⊂ ( ∪ ∼). Sinc Sincee we hav have shown containment both ways, = ( ∪ ∼). (d) To show the set is empty empty, we do this by contradictio contradiction. n. Assume Assume that there exists x, y ∈ ( ∩ ∼). Then, (x (x y AND NOT y x AND y x) AND (x (x y AND y x). But, But, one cannot cannot have have NOT(y NOT(y x) and y x. Therefore, there ∃ no x, y ∈ ( ∩ ∼) 1.3 1.3 (a) (a) For For any any x, y ∈ X where x ∼ y we can have neither x y nor y x. For any x, y ∈ X where x y, we can have neither x ∼ y nor y ∼ x. (b) Suppose x1 x2 . Then NOTx2 x1 . This means means that that NOTx2 x1 OR x1 ∼ x2 . The case of x2 x1 can be shown by symmetry. try. Suppose x1 ∼ x2 . Then Then it cannot cannot be true true that that x1 x2 or x2 x1 , since a necessary condition for either relation is that NOTx1 ∼ x2 . 1.4 (a) (a) : Let Let x1 , x2 , x3 ∈ X and assume that x1 x2 and x2 x3 . Show that x1 x3 .
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This implies that x1 x2 and x2 x3 and by transitivity of , x1 x3 . It is also true that x2 N OT x1 and x3 N OT x2 (∗) . Assume that x3 x1 . This is necessary if x3 x1 . This implies that x3 x2 by transitivity. But this contradicts the earlier statement. (b) Suppose x ∼ y and y ∼ z. x ∼ y implies that x y and y x. y ∼ z implies that y z and z y. By transitivity of , x y and y z implies that x z. Also, because z y and y x, z x. Since x z and z x, it must be that x ∼ z. (c) Finally, show that if x1 ∼ x2 x3 , then x1 x3 . Assume x1 ∼ x2 and x2 x3 . If x1 ∼ x2 ⇔ x1 x2 and x2 x1 . Since x1 x2 and x2 x3 , by transitivity of , then x1 x3 . 1.5 (b) We must again show containment both ways. Let y ∈ (x0 ). Then, by definition y x0 . Further, there are two possibilities either x0 y OR NOT(x0 y) because these are complementary qualities. Thus, by definition y ∈∼ (x0 ) ∪ (x0 ). Let y ∈∼ (x0 ) ∪ (x0 ). Then y x AND x y OR y x AND NOT x y. But we already have that y x, so y ∈ (x0 ) and we have containment both directions and the sets are equal. (c) By contradiction. Suppose y ∈∼ (x0 ) ∩ (x0 ). Then y x AND x y AND y x AND NOT x y. Since x y AND NOT x y can’t be both true it is a contradiction. (g) Contradiction. Suppose x ∈ X,, but x ∈ / y ∈∼ (x0 ) ∪ (x0 ) ∪ (x0 ) . Then neither x x0 , x ∼ x0 , and x x0 . By definition, neither x x0 nor x x0 and this violates the completeness of the preference relationship. 1.6 Many examples, but something like when consumer goods are indivisible i.e. cars and refrigerators. 1.7 Let represent convex preferences. Let x0 , x1 and x2 ∈ X where x1 x2 x0 . Then x1 , x2 ∈ x0 . For any t ∈ [0, 1], consider the bundle tx1 + (1 − t) x2 . By convexity of , tx1 +(1 − t) x2 x0 , which is true iff tx1 + (1 − t) x2 ∈ x0 . Therefore, x0 is a convex set for any x0 ∈ X .
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1.8 Case 1: Take two points, xa and xb along the indifference curve. To show that the preferences are convex, it is seen that for any t ∈ [0, 1], txa + (1 − t) xb xa . We have shown Axiom 5 in this case. We can take this a step further and say that txa + (1 − t) xb xb and txa + (1 − t) xb ∈∼ (xa ). Because any two points on the line, txa + (1 − t) NOTxb xa ⇒the preferences are not strictly convex. We have shown how Axiom 5’ does not apply. Case 2: Take two points, xa and xb such that xa xb and let t ∈ (0, 1). Then by construction, xa lies to the northeast to xb . Since t > 0, that implies txa + (1 − t) xb lies to the northeast of xb . Therefore, txa + (1 − t) xb xb and we have shown Axiom 5.
1.9 Strict Monotonicity Axiom 4: ”Preferences increase northeasterly” translates to if xa xb ,then xa xb . The fact that ”indifference sets are parallel right angles that ’kink’ on the line x1 = x2 ” along with with ”preferences increase northeasterly” implies that if xa xb . These facts assert that satisfy strict monotonicity. Strict Convexity Axiom 5: Pick any two point along a ”leg” of an indifference curve such as (α, α) and (α, β ), where α < β . The point α, 12 (α + β ) lies in
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between these two and on the same indifference curve as (α, α) and (α, β ). Therefore, we cannot have α, 12 (α + β ) (α, α), illustrating that these preferences are not strictly convex. Convexity Axiom 5’: Consider any x, y ∈ X ⊂ R2 such that x ∼ y. Given the nature of these preferences, it must be true then that min [x1 , x2 ] = min[y1 , y2 ] . For any t ∈ [0, 1] consider the point tx + (1 − t) y. If we can show that min [tx1 + (1 − t) y1 , tx2 + (1 − t) y2 ] ≥ min[x1 , x2 ] = min[y1 , y2 ] , then we have shown that these preferences are convex.
A.1.2
a Let x ∈ S then we must show x ∈ S ∪ T . S ∪ T contains all elements in S and T . Therefore, since x ∈ S , x ∈ S ∪ T . b Same proof c Let x ∈ S ∩ T . That implies that x is in both S and T . Therefore x ∈ T c Same proof
A.1.5 Let A = [a1 , a2 ] and B = [b1 , b2 ], where a2 < b1 . Since ta2 +(1 − t) b1 ∈ / A ∪ B for t ∈ (0, 1) , A ∪ B is not convex. A.1.7
a This set is not convex. For example, (0, 1) and (1, e) ∈ {(x, y) |y = ex }, ∈ but 12 , e+1 / {(x, y) |y = ex } . 2
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b This set is convex. Let (x1 , y1 ) , (x2 , y2 ) ∈ S = {(x, y) |y ≥ ex } . Since y = ex is a continuous function, it is sufficient to show that (tx1 + (1 − t) x2 , ty1 + (1 − t) y2 ) ∈ S for any particular t ∈ (0, 1). Set t = 12 . Our task is to show that 12 (x1 + x2 ) , 12 (y1 + y2 ) ∈ S . 12 (y1 + y2 ) ≥ 12 (ex + ex ),
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1
2
since yi ≥ ex for i = 1, 2. Also, 12 (ex + ex ) ≥ e (x +x ) = e · e ⇔ ex + ex ≥ 2e · e ⇔ ex − 2e · e + ex ≥ 0 ⇔ (ex − ex )2 ≥ 0. 1
i
x1 2
x2 2
1
1
x1 2
2
x2 2
x1 2
1
2
x2 2
1
2
2
2
c This set is not convex.
1 1 9 1 For example, 10 , 2 , 1 10 , 2 ∈ S = 1 1 1 1
(x, y) |y ≥ 2x − x2; x > 0, y > 0 9 1 1 2
2
. However, 1, 2 =
2
10 , 2
+
1 10 , 2 ∈ / S .
(d) This set is convex. Consider any (x1 , y1 ) , (x2 , y2 ) ∈ S = {(x, y) |xy > 1, x,y > 0}. For any t ∈ [0, 1], (tx1 + (1 − t) x2 ) (ty1 + (1 − t) y2 ) = t2 x1 y1 +t (1 − t) (x1 y2 + x2 y1 )+ (1 − t)2 x2 y2 > t2 + (1 − t)2 + t (1 − t) (x1 y2 + x2 y1 ), since xi yi > 1. = 1 + 2t2 − 2t + t (1 − t) (x1 y2 + x2 y1 ) = 1 + 2t (t − 1) + t (1 − t) (x1 y2 + x2 y1 ) = 1 + t (1 − t) (x1 y2 + x2 y1 − 2) ≥ 1 iff x1 y2 + x2 y1 ≥ 0 x1 y2 + x2 y1 = x1 y1 yy + x2 y2 yy − 2 > yy + yy − 2 ≥ 0 y1 − 2y1 y2 + y2 ≥ 0 (y1 − y2 )2 ≥ 0, which is always true and therefore, (tx1 + (1 − t) x2 , ty1 + (1 − t) y2 ) ∈ S which is convex. e S is convex ⇐
⇔
1 2
1 2
⇔ (x1 x2 )1/2 ≤
1
2
1
1
2
1
2
ln (x1 ) + ln (x2 ) ≤ ln
ln (x1 x2 ) ≤ ln
1
2
1
2 x1
1 2 x2
+
1 2 x1 + 2 x2
1
1 2 x1 + 2 x2
⇔ x1 − 2 (x1 x2 )1/2 + x2 ≥ 0
1/2
⇔ x1
1/2 2
+ x2
≥ 0, which is always true.
A.1.8 R is not complete because there can be no R relation between any two people who do not know each other. R is not transitive, the obvious counter example being man R wife,wife Rwife’s mom, but man not R wife’s mom. That is although a man 5
may love his wife, and wife may love her mom, the man may not love her mother in law. A.1.9 Suppose there exists x such that x ∈ B but f (x) ∈ / f (A). Since x ∈ B and B ⊂ A, then x ∈ A. Then, f (x) ∈ f (A) which contradicts our initial assumption. A.1.10 Suppose there exists x such that x ∈ B f −1 (x) ∈ / f −1 (A) . Since x ∈ A and A ⊂ B, then x ∈ B. Then, f −1 (x) ∈ f −1 (A) which contradicts our initial assumption. A.1.16 (a) Let −x1 , −x2 ∈ −S . Then x1 , x2 ∈ S. Convexity of S implies that for any t ∈ [0, 1], tx1 + (1 − t) x2 ∈ S, which implies that − [tx1 + (1 − t) x2 ] = t (−x1 ) + (1 − t) (−x2 ) ∈ −S . Therefore, −S is convex. (b) Let x1 , x2 ∈ S − T . Then there are s1 , s2 ∈ S and t1 , t2 ∈ T , such that x1 = s1 − t1 and x2 = s2 − t2 . Since S and T are convex, λs1 + (1 − λ) s2 ∈ S and λt1 + (1 − λ) t2 ∈ T for any λ ∈ [0, 1]. Given this and the fact that λx1 + (1 − λ) x2 = λ (s1 − t1 ) + (1 − λ) (s2 − t2 ) = [λs1 + (1 − λ) s2 ] − [λt1 + (1 − λ) t2 ], λx1 + (1 − λ) x2 ∈ S − T , illustrating that S − T is convex. A.1.17 (a) Part 1-prove 2 convex sets intersect and form a convex set. See theorem A1.1 in Jehle and Reny page 414. Part 2-Show that additional sets formed from the intersection of convex sets is also a convex set. Define A12 = A1 ∩ A2 which is convex. A12 ∩ A3 must also be convex from part 1. This can be done for An = ∩ni=1 Ai and An ∩ An+1 . (b) Suppose x ∈ XAi , x = (x1, x2 ,...,xn ) . Suppose that y ∈ XAi , y = (y1 , y2 ,...,yn ). The convex combination of x and y is then: z = (tx1 + (1 − t) y1 , tx2 + (1 − t) y2 ,...,txn + (1 − t) yn ) . Since n A and X n A is convex. txi + (1 − t) yi ∈ Ai ∀i, Z ∈ X i=1 i i=1 i (c) Proof: If x ∈ ΣAi , ∃x1 ∈ A1 , x2 ∈ A2 ,...,xn ∈ An , s.t. Σxi = x and y ∈ ΣAi , ∃y1 ∈ A1 , y2 ∈ A2 ,...,yn ∈ An , s.t. Σyi = y. Therefore, tx1 + (1 − t) y1 ∈ A1 , tx2 + (1 − t) y2 ∈ A2 ,...,txn + (1 − t) yn ∈ An and Σ(tx1 + (1 − t) y1 ) ∈ ΣAi 6
t (Σxi ) + ( 1 − t) (Σyi ) ∈ ΣAi and tx + (1 − t) y ∈ ΣAi and ΣAi is convex. (d) See part (c). A1.18 Proof: Suppose that Ω is not convex. The there exists x1 , x2 ∈ Ω, x1 = x2 and t ∈ (0, 1) for which tx1 + (1 − t) x2 ∈ / Ω. Thus for some j j j ∈ {1,...,m}, f (tx1 + (1 − t) x2 ) < 0 ⇒ a [tx1 + (1 − t) x2 ]+b j < 0. But this contradicts the fact that, since f j (x1 ) = a j x1 + b j ≥ 0 and f j (x2 ) = a j x1 + b j ≥ 0 ⇔ t a j x1 + b j ≥ 0 and (1 − t) a j x2 + b j ≥ 0 ⇔ t a j x1 + b j + (1 − t) a j x2 + b j = a j [tx1 + (1 − t) x2 ] + b j ≥ 0. Hence, Ω must be convex.
2
Rubinstein
Problem 1 Show satisfies property (1), which is saying that indifference curves do not cross. Consider any x, y ∈ X such that I (x) = I (y). Suppose that I (x) ∩ I (y) = 0, so that there is some z ∈ I (x) ∩ I (y). Then z ∈ I (x) and z ∈ I (y), implying that z ∼ y and z ∼ x. From transitivity it follows that z1 ∼ z2 for any z ∈ I (x) , z ∈ I (y), both z1 and z2 ∈ I (x) ∩ I (y). Thus, I (x) = I (y) ,contradicting the premise that I (x) = I (y). Therefore, I (x) ∩ I (y) = 0, if I (x) = I (y). Now suppose that I (x) = I (y). Then x, y ∈ I (x) ∩ I (y) = 0. Show satisfies property (2), indifference sets are non-empty. Proof: For any x ∈ X , x ∈ I (x). So for y = x, x ∈ I (y) . Problem 2 Solved in class Problem 4 Induction has three steps. First what are we inducting on? The size of the set X. Step 1 Base Case: i = 2. Let X have two elements x and y. Because of the definition of asymmetry, WLOG, let xP y. Then we have established an ordering and it is complete because both x and y are described. Step 2 Assume for cases up to i = n, that there is a complete ordering. Step 3 Prove for case i = n + 1 that there is a complete ordering. Let x1 , x2 ,...,xn+1 ∈ X of size n+1. We know that sets of size n have a complete ordering. Lets take elements x1 ,....,xn and form a 7
new set. WLOG, we know that we can rank x1 P x2 P...xn . Now construct a new set with xn+1 and x1 . By our base case, we know that either x1 P xn+1 or xn+1 P x1 . If, xn+1 P x1 , then we are done and we have our ordering xn+1 P x1 P x2 P...xn . If not, we make a new set with x2 and xn+1 which we know we can rank. If, xn+1 P x2 , then we are done and we have our ordering x1 P xn+1 P x2 P...xn , we repeat this process n − 2, a finite number, more times. If xn+1 does not outrank any of them we have our complete ordering x1 P x2 P...xn P xn+1 , otherwise as specified we have our complete ordering.
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