200a_f08_ps_3_ak

200a_f08_ps_3_ak - University of California Davis ARE/ECN 200A Fall 2008 Joaquim Silvestre& Shaofeng Xu PROBLEM SET 3 ANSWER KEY 1 1.a The

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Unformatted text preview: University of California, Davis ARE/ECN 200A Fall 2008 Joaquim Silvestre & Shaofeng Xu PROBLEM SET 3 ANSWER KEY 1. 1.a The function u is continuous, and the budget set is compact. Hence, a maximum exists. Uniqueness follows from the strict convexity of preferences. 1.b By di¡erentiability, interiority and monotonicity, the solution ~ x ( p; w ) satis¢es, 8 ( p; w ) ; u 1 (~ x 1 ( p; w )) = & ( p; w ) p 1 ; ::: u L (~ x L ( p; w )) = & ( p; w ) p L ; X L j =1 p j ~ x j ( p; w ) = w: Di¡erentiating both sides of the last equality with respect to w yields X L j =1 p j @ ~ x j @w > 0 = 1 ; Hence, @ ~ x j @w > for at least one j 2 f 1 ; :::; L g : Di¡erentiating both sides of the equality u j (~ x j ( p; w )) = & ( p; w ) p j with respect to w yields, by the chain rule, u 00 j @ ~ x j @w = @& @w p j ; i:e:; @& @w < : Hence, for any k 2 f 1 ; :::; L g ; u 00 k @ ~ x k @w = @& @w p k < ; i:e:; @ ~ x k @w > ; k = 1 ; :::; L: 2. 1 Because the solution is interior, and by local nonsatiation, the solution is characterized by the system of three equations in three unknowns: ’ ( x 1 ) & &p 1 = 0 ; 1 x 2 & &p 2 = 0 ; p 1 x 1 + p 2 x 2 = w: The Jocabian of the system relative to the endogeneous variables ( x 1 ; x 2 ; & ) is J = 2 6 6 4 ’ 00 & p 1 & 1 x 2 2 & p 2 p 1 p 2 3 7 7 5 We compute: det J = & & & & & & & & ’ 00 & p 1 & 1 x 2 2 & p 2 p 1 p 2 & & & & & & & & = ’ 00 ( p 2 ) 2 & p 1 ¡ p 1 x 2 2 ¢ < : We want to ascertain the sign of @ ~ x 1 @p 2 : Hence, we are only interested in the parameter p 2 (not p 1 or w ). The Implicit Function Theorem guarantees that 2 6 4 @ ~ x 1 @p 2 @ ~ x 2 @p 2 @& @p 2 3 7 5 = & J & 1 2 6 4 & & x 2 3 7 5 : The next step is inverting the 3 ¡ 3 matrix J (see, e.g, Simon and Blume, section 9.2), i.e., J & 1 = 1 det J 2 6 6 6 6 6 6 6 6 6 4 & & & & & & 1 x 2 2 & p 2 p 2 & & & & & & & & & & & & p 1 p 2 & & & & & & & & & & & p 1 & 1 x 2 2 & p 2 & & & & & & & & & & & & p 2 p 1 & & & & & & & & & & ’ 00 & p 1 p 1 & & & & & & & & & & & ’ 00 & p 1 & p 2 & & & & & & & & & & & 1 x 2 2 p 1 p 2 & & & & & & & & & & & ’ 00 p 1 p 2 & & & & & & & & & & ’ 00 & 1 x 2 2 & & & & & 3 7 7 7 7 7 7 7 7 7 5 = 1 det J 2 6 6 4 p 2 2 & p 1 p 2 & p 1 x 2 2 & p 1 p 2 p 2 1 ’ 00 p 2 p 1 x 2 2 & ’ 00 p 2 & ’ 00 x 2 2 3 7 7 5 : 2 i.e., 2 6 4 @ ~ x 1 @p 2 @ ~ x 2 @p...
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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_3_ak - University of California Davis ARE/ECN 200A Fall 2008 Joaquim Silvestre& Shaofeng Xu PROBLEM SET 3 ANSWER KEY 1 1.a The

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