601-sampleq1a

# 601-sampleq1a - E 601 1 Answers to sample questions 1 1 MWG...

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E frqrplfv 601 1 Answers to sample questions 1 1. MWG, 1.B.3 Proof: We say that u ( · ) represents ± on choice set X if for all x, y X , x ± y if and only if u ( x ) u ( y ) (1) To complete the proof note that if f ( · ) is a strictly increasing function de f ned on the real numbers then u ( x ) u ( y ) if and only if f ( u ( x )) f ( u ( y )) . MWG, 1.B.4 Proof: Denote for all x, y X , x y if and only if u ( x )= u ( y ) (2) for all x, y X , x " y if and only if u ( x ) >u ( y ) (3) Given that (2) and (3) are true we need to prove that (1) is true. Note that x ± y implies either x " y in which case u ( x ) ( y ) or x y in which case u ( x u ( y ) ; combining the two options then x ± y implies u ( x ) u ( y ) . Conversely, u ( x ) u ( y ) implies either u ( x ) ( y ) in which case x " y or u ( x u ( y ) in which case x y . So combining the two options then u ( x ) u ( y ) implies x ± y . 2. (a) WARP holds in some choice structure ( B ,C ( · )) if: B,B ± B , { x, y } B, { x, y } B ± : x C ( B ) ,y C ( B ± ) x C ( B ± ) . The only way to contradict WARP here is for some problem to arise with budget sets that don’t contain d. But there is none of these with two or more elements in common, so the supposition in the de f nition of WARP is not satis f ed and thus WARP must be true. Remember, anything is true of the empty set. (b) No, because a " b " c but c " a, so transitivity doesn’t hold. (c) Try C ( { a, b, c } { a } ; but C ( { c, a } { c } so WARP is violated. Try C ( { a, b, c } { b } ; but C ( { a, b } { a } so WARP is violated. C ( { a, b, c } { c } ; but C ( { b, c } { b } so WARP is violated.

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E frqrplfv 601 2 So the only way to satisfy WARP is to set C ( { a, b, c } )= the empty set, which is not allowed. 3. (a) A utility function u : X R represents a preference relation ± on the choice set X if x, y X, x ± y u ( x ) u ( y ) . Completeness: x, y X , u ( x ) and u ( y ) are real numbers so either u ( x ) u ( y ) in which case x ± y or u ( y ) u ( x ) in which case y ± x. Transitivity: Suppose x, y, z X ,and x ± y and y ± z. We need to show x ± Since u represents ± ,u ( x ) u ( y ) and u ( y ) u ( z ) . Since these are real numbers u ( x ) u ( z ) x ± (b) Lexicographic preferences are complete and transitive but not continuous. Completeness: Consider any two distinct points in R 2 + ( x 1 1 ,x 1 2 ) (point a )and ( x 2 1 2 2 ) (point b ). If x 1 1 >x 2 1 ,a " b. If x 2 1 1 1 ,b " a. If x 1 1 = x 2 1 , since a and b are distinct, it must be that either x 1 2 2 2 in which case a " b or x 2 2 1 2
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601-sampleq1a - E 601 1 Answers to sample questions 1 1 MWG...

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