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200a_f08_ps_3_ak

200a_f08_ps_3_ak - University of California Davis ARE/ECN...

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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 0 1 (~ x 1 ( p; w )) = ° ( p; w ) p 1 ; ::: u 0 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 > 0 for at least one j 2 f 1 ; :::; L g : Di/erentiating both sides of the equality u 0 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 < 0 : Hence, for any k 2 f 1 ; :::; L g ; u 00 k @ ~ x k @w = @w p k < 0 ; i:e:; @ ~ x k @w > 0 ; k = 1 ; :::; L: 2. 1
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Because the solution is interior, and by local nonsatiation, the solution is characterized by the system of three equations in three unknowns: 0 ( 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 0 ° p 1 0 ° 1 x 2 2 ° p 2 p 1 p 2 0 3 7 7 5 We compute: det J = ° ° ° ° ° ° ° ° 00 0 ° p 1 0 ° 1 x 2 2 ° p 2 p 1 p 2 0 ° ° ° ° ° ° ° ° = 00 ( p 2 ) 2 ° p 1 ± p 1 x 2 2 ² < 0 : 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 0 ° ° 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 0 ° ° ° ° ° ° ° ° ° ° ° 0 ° p 1 p 2 0 ° ° ° ° ° ° ° ° ° ° 0 ° p 1 ° 1 x 2 2 ° p 2 ° ° ° ° ° ° ° ° ° ° ° 0 ° p 2 p 1 0 ° ° ° ° ° ° ° ° ° ° 00 ° p 1 p 1 0 ° ° ° ° ° ° ° ° ° ° ° 00 ° p 1 0 ° p 2 ° ° ° ° ° ° ° ° ° ° 0 ° 1 x 2 2 p 1 p 2 ° ° ° ° ° ° ° ° ° ° ° 00 0 p 1 p 2 ° ° ° ° ° ° ° ° ° ° 00 0 0 ° 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
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i.e., 2 6 4 @ ~ x 1 @p 2 @ ~ x 2 @p 2 @p 2 3 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
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