204PS6Solution2009 (1)

# 204PS6Solution2009 (1) - Econ 204 Problem Set 6 Solutions...

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Econ 204 Problem Set 6 Solutions Exercise 1 We know that there are solutions to the system 0 = F ( x;y;w;z ) = f ( x;y ) ( w;z ) . DF ( x;y;w;z ) = Df ( x;y ) D ( w;z ) = D x f 1 ( x;y ) D y f 1 ( x;y ) 1 0 D x f 2 ( x;y ) D y f 2 ( x;y ) 0 1 ± , where f i ( x;y ) is the i th component of f: DF is clearly of rank 2 for all ( x;y;w;z ) and since f 2 C 3 it follow that F 2 C 3 . By Transversality Theorem we know that there is a subset B of R 2 such that B c is of measure zero, and for all ( x;y;w;z ) satisfying F ( x;y;w;z ) = 0 where ( w;z ) 2 B , we have rank ( D ( x;y ) F ( x;y;w;z )) = 2 . Hence, for these ( x;y;w;z ) we satisfy the hypotheses of the Implicit Function Exercise 2 Check directly. Since 0 6 = 1 = 2 , x = 0 only chance we have is if x = 1 = ( x + 1) if and only if x ( x + 1) = 1 if and only if x 1 ; 2 = 1 ± p 1+4 2 . Thus f [0 ; 1] . f is a set function (correspondence) so Brower±s Thm. does not apply. Kakutani±s does not apply since f is not convex valued. For the next function, x = [0 ; 1] is compact (closed and bounded), convex (an interval), and non-empty, and the correspondence f is: convex valued since image of every point is an interval; closed valued since that way; upper-hemicontinuous since f has a closed graph and the set

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## This note was uploaded on 10/04/2010 for the course ECON 204 taught by Professor Anderson during the Summer '08 term at Berkeley.

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204PS6Solution2009 (1) - Econ 204 Problem Set 6 Solutions...

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