200a_f08_ps_1_ak

200a_f08_ps_1_ak - University of California, Davis ARE/ECN...

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University of California, Davis ARE/ECN 200A Fall 2008 PROBLEM SET 1 ANSWER KEY 1. From Textbook 1.B.3 Let x 2 X and y 2 X: Since u ( : ) represents ; x y if and only if u ( x ) ± u ( y ) : Since f ( : ) is strictly increasing, u ( x ) ± u ( y ) if and only if v ( x ) ± v ( y ) : Hence, x y if and only if v ( x ) ± v ( y ) : Therefore, v ( : ) represents : 1.B.4 Suppose ±rst x y: If, furthermore, y x; then x ² y and hence u ( x ) = u ( y ) : If, on the contrary, we do not have y x; then x ³ y: Hence, u ( x ) > u ( y ) : Thus, if x y; then u ( x ) ± u ( y ) : Suppose conversely that u ( x ) ± u ( y ) : If, furthermore, u ( x ) = u ( y ) ; then x ² y and hence x y: If, on the contrary, u ( x ) > u ( y ) ; then x ³ y; and hence x y: Thus, if u ( x ) ± u ( y ) ; then x y: So u ( : ) represents : 1.B.5 Lemma . Let be a transitive preference relation, and let N be a positive integer ± 3 : Then x 1 x 2 x 3 ::: x N 1 x N ) x 1 x N : Proof: Because x 1 x 2 and x 2 x 3 ; transitivity implies that x 1 x 3 ; which together with x 3 x 4 ; implies x 1 x 4 : Iterating the argument, we obtain x 1 x N : Now, turn to the question: (1) We consider ±rst the case where there is no indi²erence, i.e., x i x j , x i ³ x j : (1) Thus, by the completeness of the preference relation, given x i ;x j in X; either x i ³ x j or x j ³ x i : (2) Let Y be a nonempty subset of X; and let N be the number of elements of Y: We want to show that there is an element of Y that is preferred to all other elements of Y: The proof is by 1
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contradiction. Suppose not. Then given x 2 Y; there is an x 0 2 Y such that x is not preferred to x 0 : By (1) and (2), we must have that x 0 x;i:e:; given x 2 Y; there is an x 0 2 Y such that x 0 x: (3) Arbitrarily choose an x 2 Y: By (3), there is an x 0 2 Y such that x 0 x: Applying (3) again, there must exist an x 00 2 Y such that x 00 x 0 : We can apply (3) a total of N times to get x ( N ) x ( N 1) ::: x (3) x 00 x 0 x: (4) for some x ( N ) ;x ( N 1) ;:::;x (3) ;x 00 ;x 0 ;x in Y: But there are only N elements in
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This note was uploaded on 07/01/2009 for the course ARE 200a taught by Professor Silvestre,j during the Winter '08 term at UC Davis.

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200a_f08_ps_1_ak - University of California, Davis ARE/ECN...

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