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Econ 210A PS1 Answer Key

1 Jehle-Reny

1.2 (a) In order to prove this, we must show that every element in isin the set . Let x, y X s.t. x y. Therefore, x, y . Bydenition, x y and x, y . Therefore

(b) In order to prove this, we must show every element in is anelement in . Let x, y X s.t. x y. If x y then, we haveby denition x yANDy x . Since x y, then x, y and

(c) In order to prove this we must show containment both ways. Thatis, ( ) and ( ). Starting with ( ) :Let x, y ( ). Then by denition, x, y are either a memberof OR . Therefore, by denition either, x y AND NOT(y x) OR x y AND y x. In either case, x y andx, y . Proving the other containment: let x, y s.t. X y.There are two possibilities between the relationship between yand x. Either, y x or NOT( y x) (these are complementaryscenarios). Since we have x y AND (y x OR NOT (y x)),x, y ( ) by denition and ( ). Since we haveshown containment both ways, = ( ).

(d) To show the set is empty, we do this by contradiction. Assumethat there exists x, y ( ). Then, ( x y AND NOT y xAND y x) AND (x y AND y x). But, one cannot haveNOT( y x) and y x. Therefore, there no x, y ( )

1.3 (a) For any x, y X where x y we can have neither x y nory x. For any x, y X where x y, we can have neither x ynor y x.

(b) Suppose x1 x2. Then NOTx 2 x1. This means that NOTx 2

x1 OR x1 x2. The case of x2 x1 can be shown by symme-try. Suppose x1 x2. Then it cannot be true that x1 x2 orx2 x1, since a necessary condition for either relation is thatNOTx 1 x2.

1.4 (a) : 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 x2NOT x1 and x3NOT x2 () .Assume that x3 x1. This is necessary if x3 x1. This impliesthat x3 x2 by transitivity. But this contradicts the earlierstatement.

(b) Suppose x y and y z.x y implies that x y and y x. y z implies that y zand 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 x

2x

3, then x

1x

3.

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 denition y x0. Further, there are two possibilitieseither x0 y OR NOT( x0 y) because these are complementaryqualities. Thus, by denition y (x0) (x0). Let y (x0) (x0). Then y x AND x y OR y x AND NOTx y. But we already have that y x, so y (x0) and wehave containment both directions and the sets are equal.

(c) By contradiction. Suppose y (x0) (x0). Then y xAND x y AND y x AND NOT x y. Since x y ANDNOT x y cant 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 denition,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 indivis-ible i.e. cars and refrigerators.

1.7 Let represent convex preferences. Let x0, x1 and x2 X wherex1 x2 x0. Then x1, x2 x0 . For any t [0, 1], consider thebundle tx 1 +(1 t) x2. By convexity of , tx 1 +(1 t) x2 x0, whichis true iff tx 1 + (1 t) x2 x0 . Therefore, x0 is a convex setfor 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],tx a + (1 t) xb xa . We have shown Axiom 5 in this case. Wecan take this a step further and say that tx a + (1 t) xb xb andtx a + (1 t) xb (xa ). Because any two points on the line, tx a +(1 t) NOTx b xa the preferences are not strictly convex. Wehave 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 tx a + (1 t) xb lies to the northeast of xb. Therefore,tx a + (1 t) xb xb and we have shown Axiom 5.

1.9Strict Monotonicity Axiom 4: Preferences increase northeasterly translates to if xa xb,then

xa xb. The fact that indifference sets are parallel right anglesthat kink on the line x1 = x2 along with with preferencesincrease northeasterly implies that if xa xb. These facts assertthat satisfy strict monotonicity.

Strict Convexity Axiom 5: Pick any two point along a leg of an indifference curve such as(, ) and ( , ), where < . The point , 12 ( + ) lies in

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between these two and on the same indifference curve as ( , )

and ( , ). Therefore, we cannot have , 12 ( + ) (, ),illustrating that these preferences are not strictly convex.

Convexity Axiom 5: Consider any x, y X R2 such that x y. Given the na-ture 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 [ tx 1 + (1 t) y1, tx 2 + (1 t) y2] min[x1, x2] =min[y1, y2] , then we have shown that these preferences are con-vex.

A.1.2 a Let x S then we must show x S T . S T contains allelements 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, a 2] and B = [b1, b2], where a2 < b1. Since ta 2 +(1 t) b1 /A B for t (0, 1) , AB 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+12 / {(x, y) |y = e

x } .

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b This set is convex.

Let ( x1, y1) , (x2, y2) S = {(x, y) |y ex } . Since y = ex is a con-tinuous function, it is sufficient to show that ( tx 1 + (1 t) x2, ty1 + (1 t) y2) S for any particular t (0, 1). Set t = 12 . Our task is to showthat 12 (x1 + x2) ,

12 (y1 + y2) S .

12 (y1 + y2)

12 (e

x 1 + ex 2 ),

since yi ex i for i = 1 , 2. Also, 12 (ex 1 + ex 2 ) e

12 (x 1 + x 2 ) =

ex 1

2 ex 2

2 ex 1 + ex 2 2ex 1

2 ex 2

2 ex 1 2ex 1

2 ex 2

2 + ex 2 0 (ex 1 ex 2 )2 0.

c This set is not convex. For example, 110 ,12 , 1

910 ,

12 S =

(x, y) |y 2x x2; x > 0, y > 0 . However, 1, 12 =12

110 ,

12 +

12 1

910 ,

12 /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],(tx 1 + (1 t ) x2) ( ty1 + (1 t) y2) = t2x1y1+ t (1 t) (x1y2 + x2y1)+(1 t)2 x2y2> t 2 + (1 t)2 + t (1 t) (x1y2 + x2y1), since x iyi > 1.= 1 + 2 t2 2t + t (1 t) (x1y2 + x2y1)= 1 + 2 t (t 1) + t (1 t) (x1y2 + x2y1)= 1 + t (1 t) (x1y2 + x2y1 2) 1 iff x 1y2 + x2y1 0

x1y2 + x2y1 = x1y1y2y1 + x2y2

y1y2 2 >

y2y1 +

y1y2 2 0y1 2y1y2 + y2 0

(y1 y2)2 0, which is always true and therefore, ( tx 1 + (1 t) x2, ty1 + (1 t) y2) S which is convex.

e S is convex 12 ln (x1) + ln ( x2) ln12 x1 +

12 x2

12 ln (x1x2) ln

12 x1 +

12 x2

(x1x2)1/ 2 12 x1 +12 x2

x1 2 (x1x2)1/ 2 + x2 0

x1/ 21 + x1/ 22

2 0, which is always true.

A.1.8 R is not complete because there can be no R relation between any twopeople who do not know each other.

R is not transitive, the obvious counter example being man R wife,wifeRwifes mom, but man not R wifes mom. That is although a man

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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 ourinitial assumption.

A.1.10 Suppose there exists x such that x B f 1 (x) /f 1 (A) . Since x Aand A B , then x B . Then, f 1 (x) f 1 (A) which contradictsour initial assumption.

A.1.16 (a) Let x1, x2 S . Then x1, x2 S. Convexity of S impliesthat for any t [0, 1], tx 1 + (1 t) x2 S, which implies that

[tx 1 + (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 areconvex, s 1 + (1 ) s2 S and t 1 + (1 ) t2 T for any [0, 1]. Given this and the fact thatx 1 + (1 ) x2 = (s1 t1) + (1 ) (s2 t2)= [s 1 + (1 ) s2] [t 1 + (1 ) t2], x 1 + (1 ) 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. Seetheorem 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.Dene A12 = A1 A2 which is convex. A12 A3 must alsobe convex from part 1 . This can be done for An = ni=1 Ai andAn An +1 .

(b) Suppose x XA i , x = ( x1,x2,...,x n ) . Suppose that y XA i ,y = ( y1, y2,...,yn ). The convex combination of x and y is then:z = ( tx 1 + (1 t) y1, tx 2 + (1 t) y2,...,tx n + (1 t) yn ) . Sincetx i + (1 t) yi Aii, Z X ni=1 Ai and X ni=1 Ai is convex.

(c) Proof: If x Ai , x1 A1, x2 A2,...,x n An , s.t. x i = xandy Ai , y1 A1, y2 A2,...,yn An , s.t. yi = y.Therefore, tx 1 + (1 t) y1 A1, tx 2 + (1 t) y2 A2,...,tx n +(1 t) yn An and ( tx 1 + (1 t ) y1) Ai

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t ( x i )+(1 t) ( yi ) Ai and tx + (1 t) y Ai and Ai is

convex.(d) See part (c).

A1.18 Proof: Suppose that is not convex. The there exists x1, x2 ,x1 = x2 and t (0, 1) for which tx 1 + (1 t) x2 /. Thus for some j {1,...,m }, f j (tx 1 + (1 t) x2) < 0 a j [tx 1 + (1 t) x2]+ b j < 0.But this contradicts the fact that, since f j (x1) = a j x1 + b j 0 andf 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 [tx 1 + (1 t) x2] + b j 0.Hence, must be convex.

2 RubinsteinProblem 1 Show satises 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 transitivityit follows that z1 z2 for any z I (x) , z I (y), both z1 andz2 I (x) I (y). Thus, I (x) = I (y) ,contradicting the premise thatI (x) = I (y). Therefore, I (x) I (y) = 0 , if I (x) = I (y). Nowsuppose that I (x) = I (y). Then x, y I (x) I (y) = 0.

Show satises 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 sizeof the set X.

Step 1 Base Case: i = 2. Let X have two elements x and y. Becauseof the denition of asymmetry, WLOG, let xP y . Then we haveestablished an ordering and it is complete because both x and yare 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,...,x n +1 X of size n+1. We know that sets of size n havea complete ordering. Lets take elements x1,....,x n and form a

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new set. WLOG, we know that we can rank x1P x 2P...x n . Now

construct a new set with xn +1 and x1. By our base case, weknow that either x1P x n +1 or xn +1 P x 1. If, xn +1 Px 1, then weare done and we have our ordering xn +1 Px 1P x 2P...x n . If not,we make a new set with x2 and xn +1 which we know we canrank. If, xn +1 P x 2, then we are done and we have our orderingx1P x n +1 P x 2P...x n , we repeat this process n 2, a nite number,more times. If xn +1 does not outrank any of them we have ourcomplete ordering x1P x 2P...x n P x n +1 , otherwise as specied wehave our complete ordering.

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