Answers-B-1

Answers-B-1 - John Riley 24 July 2008 ANSWERS TO EXERCISES...

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© John Riley 24 July 2008 Answers to Exercises in Appendix B page 1 ANSWERS TO EXERCISES IN APPENDIX B Section B.1 VECTORS ANS SETS Exercise B.1-1: Convex sets (a) Let 01 12 ,, x xX X , hence 1 , x and 2 , x . Since 1 X and 2 X are convex, 1 x X λ and 2 x X . Then x XX ∈∩ , which implies that the intersection of the two sets is convex. (b) Let { } 1 X a = and { } 2 X b = where , n ab \ and . You should convince yourself that both sets are convex. Then any convex combination between a and b , with << , does not belong to { } , X Xa b ∪= . (c) Take 1 , x and 23 2 , x , so 002 y xxY = +∈ and 113 y = . Since 1 X and 2 X are convex, 1 (1 ) x λλ −+ and 2 ) x , for all [ ] 0,1 . Hence () ( ) ( ) 11 1 y yy xx Y =− + = − + + − + , which shows that Y is convex. Using a similar argument is easy to show that Z is also convex. Section B.2. FUNCTIONS OF VECTORS Exercise B.2-1: Positive semi-definite quadratic form If qx is everywhere non-negative, then is everywhere non-positive. This means that necessary and sufficient conditions are 11 22 ,0 aa −≤ and 11 22 12 21 0 −− , that is 11 22 and ( )( ) ( )( ) 11 22 12 21 0 . Exercise B.2-2: Positive definite quadratic form
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© John Riley 24 July 2008 Answers to Exercises in Appendix B page 2 If () 0 qx > for all 0 x , then ( ) 0 −< for all 0 x . This is the case where 11 0 a −> and 11 22 12 21 0 aa −− > , which is the same as 11 0 a < and 11 22 12 21 0 . Exercise B.2-3: Concave and Quasi-concave functions Proof of Proposition B.2.5 : Let f and g be concave on the convex set n X \ . Then ()() () 01 1 f xf x f x λ λλ ≥− + and ( ) ( ) ( ) ( ) 1 gx + . Summing both expressions we have () () ( ) ( ) ( ) ( )( ) ( ) 00 11 1 f xg x f x f x g x +≥ + + + , which gives the desired result. Proof of Proposition B.2.6 : (a) Since f is concave we know ( ) ( ) ( ) ( ) 1 f x f x + . Since g is increasing ()() ( ) 1 gfx g fx + . Now, using the concavity of g , ( ) ( ) ( ) ( ) 0 1 gf x f x g f x g f x −+ + . Hence 1 hx + , which proves the statement. (b) By linearity of f , we have that ( ) ( ) ( ) 0 1 g f x x g f x =− + = + . Since g is concave ( ) ( ) ( ) 0 1 g f x + + .
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© John Riley 24 July 2008 Answers to Exercises in Appendix B page 3 Exercise B.2-4: Convex lower contour sets (a) The proof is almost identical to the related proof for quasi-concave functions. We demonstrate necessity (only if). Suppose the lower contour sets of
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This note was uploaded on 01/11/2012 for the course ECON 200 taught by Professor Riley during the Fall '10 term at UCLA.

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Answers-B-1 - John Riley 24 July 2008 ANSWERS TO EXERCISES...

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