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prelims_Micro Prelim ANSWERS June 2005

# prelims_Micro Prelim ANSWERS June 2005 - Page 1 of 18...

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Unformatted text preview: Page 1 of 18 Microeconomics Prelim June 2005 Answer Keys 1. a . Proof of the property . Let x ∈ B 1 ⊂ B , and let x be chosen at ( B , s ). First, if x = s , then, by (1), u ( s ) > u ( x *( B )) - c ( s ). Because B 1 ⊂ B , u ( x *( B )) > u ( x *( B 1 )), and hence u ( s ) > u ( x *( B 1 )) - c ( s ) and, again by (1), s is chosen at ( B 1 , s ). Second, let x *( B ) ≠ s and let x = x *( B ) be chosen at ( B , s ). Then, by (2), u ( s ) < u ( x *( B )) - c ( s ). Because x = x *( B ) ∈ B 1 and because u ( x *( B )) > u ( x *( B 1 )), the uniqueness of the maximizers implies that x *( B 1 ) = x *( B ) = x . Thus, x is also chosen at ( B 1 , s ). Revealed preference interpretation. If x is chosen at ( B , s ), then x is revealed (weakly) preferred to any point in B when the status quo is s . The consistency of choices would then dictate that x be also revealed (weakly) preferred to any point in a subset B 1 of B , as long as x belongs to it and as long as the status quo is still s . This is precisely what the property says. b. Consider the function f : X × X →ℜ : f ( x , s ) = u ( x ) – g ( x , s ), where g ( s , s ) = 0 and, for x ≠ s , g ( x , s ) = c ( s ) > 0. When X = ℜ N + , f is clearly discontinuous at any point in its domain of the form ( s , s ), even when both u and c are continuous functions. We now show that the maximization of f on B is equivalent to the choice rules (1) and (2). First, let x be chosen at ( B , s ). If x = s , then, by (1), f ( s , s ) ≡ u ( s ) > u ( x *( B )) - c ( s ) > u ( x ) - c ( s ), for all x ∈ B , with f ( x , s ) ≡ u ( x ) - c ( s ), for all x ∈ B , x ≠ s , i. e., x = s maximizes f on B . (Trivially, f ( s , s ) > f ( s , s ).) If, on the contrary, x ≠ s , then, by (2), x = x *( B ) and f ( s , s ) ≡ u ( s ) < u ( x *( B )) - c ( s ) ≡ f ( x *( B ), s ). Moreover, if x ∈ B , then u ( x *( B )) > u ( x ), by the definition of x *( B ), and hence f ( x *( B ), s ) ≡ u ( x *( B )) - c ( s ) > u ( x ) - c ( s ) ≡ f ( x , s ), for all x ∈ B , x ≠ s . Summarizing, if x is chosen at ( B , s ), then x maximizes f ( x , s ) on B . Conversely, let x solve max x f ( x , s ) subject to x ∈ B . If x = s = x *( B ), then (1) is trivially satisfied. If x = s ≠ x *( B ), then, because of maximization, f ( s , s ) > f ( x , s ) for all x ∈ B , x ≠ s , i.e., u ( s ) > u ( x ) – c ( s ) for all x ∈ B , x ≠ s , or u ( s ) + c ( s ) > u ( x ) for all x ∈ B , x ≠ s . This together with the fact that s ≠ x *( B ) implies that u ( s ) + c ( s ) > u ( x *( B )) (otherwise there would be an x in B , x ≠ s, with u ( s ) + c ( s ) < u ( x ), a contradiction). Hence, (1) is satisfied. Last, let x ≠ s ....
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