45487306-Jehle-and-Reny-Solutions

45487306-Jehle-and-Reny-Solutions - Solutions to selected...

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Unformatted text preview: Solutions to selected exercises from Jehle and Reny (2001): Advanced Microeconomic Theory Thomas Herzfeld September 2010 Contents 1 Mathematical Appendix 2 1.1 Chapter A1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 1.2 Chapter A2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 2 Consumer Theory 12 2.1 Preferences and Utility . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 2.2 The Consumer’s Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 2.3 Indirect Utility and Expenditure . . . . . . . . . . . . . . . . . . . . . . . 16 2.4 Properties of Consumer Demand . . . . . . . . . . . . . . . . . . . . . . 18 2.5 Equilibrium and Welfare . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 3 Producer Theory 23 3.1 Production . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 3.2 Cost . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 3.3 Duality in production . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 3.4 The competitive firm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 1 1 Mathematical Appendix 1 Mathematical Appendix 1.1 Chapter A1 A1.7 Graph each of the following sets. If the set is convex, give a proof. If it is not convex, give a counterexample. Answer (a) ( x,y ) | y = e x This set is not convex. Any combination of points would be outside the set. For example, (0 , 1) and (1 ,e ) ∈ ( x,y ) | y = e x , but combination of the two vectors with t = 1 2 not: ( 1 2 , e +1 2 ) / ∈ ( x,y ) | y = e x . (b) ( x,y ) | y ≥ e x This set is convex. Proof: Let ( x 1 ,y 1 ), ( x 2 ,y 2 ) ∈ S = ( x,y ) | y ≥ e x . Since y = e x is a continuous function, it is sufficient to show that ( tx 1 + (1- t ) x 2 , ty 1 + (1- t ) y 2 ) ∈ S for any particular t ∈ (0 , 1). Set t = 1 2 . Our task is to show that ( 1 2 ( x 1 + x 2 ) , 1 2 ( y 1 + y 2 ) ) ∈ S . 1 2 ( y 1 + y 2 ) ≥ 1 2 ( e x 1 + e x 2 ), since y i ≥ e x 1 for i = 1 , 2. Also, 1 2 ( e x 1 + e x 2 ) ≥ e 1 2 ( x 1 + x 2 = e x 1 2 · e x 2 2 ⇔ e x 1 + e x 2 ≥ 2 e x 1 2 · e x 2 2 ⇔ e x 1- 2 e x 1 2 · e x 2 2 + e x 2 ≥ ⇔ ( e x 1- e x 2 ) 2 ≥ . (c) ( x,y ) | y ≥ 2 x- x 2 ; x > ,y > This set is not convex. For example, ( 1 10 , 1 2 ) , ( 1 9 10 , 1 2 ) ∈ S = ( x,y ) | y ≥ 2 x- x 2 ; x > ,y > 0. However, ( 1 , 1 2 ) = 1 2 ( 1 10 , 1 2 ) + 1 2 ( 1 9 10 , 1 2 ) / ∈ S (d) ( x,y ) | xy ≥ 1; x > ,y > This set is convex. Proof: Consider any ( x 1 ,y 1 ), ( x 2 ,y 2 ) ∈ S = ( x,y ) | xy ≥ 1; x > ,y > 0. For any t ∈ [0 , 1], ( tx 1 + (1- t ) x 2 )( ty 1 + (1- t ) y 2 ) = t 2 x 1 y 1 + t (1- t )( x 1 y 2 + x 2 y 1 ) + (1- t ) 2 x 2 y 2 > t 2 + (1- t ) 2 + t (1- t )( x 1 y 2 + x 2 y 1 ), since x i y i > 1 ....
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45487306-Jehle-and-Reny-Solutions - Solutions to selected...

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