200a_f08_ps_2_ak

200a_f08_ps_2_ak - University of California, Davis ARE/ECN...

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Unformatted text preview: University of California, Davis ARE/ECN 200A Fall 2008 Joaquim Silvestre & Shaofeng Xu PROBLEM SET 2 ANSWER KEY 1. An interesting demand function 1.1 YES. It is 0-homogeneous: ~ x ( &p;&w ) = & &w & 24 &p 2 &p 1 & &p 2 ; 24 &p 1 & &w &p 1 & &p 2 ¡ = & w & 24 p 2 p 1 & p 2 ; 24 p 1 & w p 1 & p 2 ¡ = ~ x ( p;w ) ; 8 & > : YES. It satis¡es WL: p T ~ x ( p;w ) = p 1 w & 24 p 2 p 1 & p 2 + p 2 24 p 1 & w p 1 & p 2 = w ( p 1 & p 2 ) p 1 & p 2 = w: YES. It satis¡es WARP: Method 1 : we use proposition 2.F.1 to prove it by showing that & p ¡ & x ¢ for all Slutsky- compensated price-wealth changes. Note ~ x 1 ( p;w ) + ~ x 2 ( p;w ) = 24 ; 8 ( p;w ) 2 P: Consider a Slutsky-compensated price-wealth change: ( p;w ) ! ( p ;w ) with w = p ¡ ~ x ( p;w ) : Then ~ x 1 ( p ;w ) = w & 24 p 2 p 1 & p 2 = p ¡ ~ x ( p;w ) & 24 p 2 p 1 & p 2 = p 1 ~ x 1 ( p;w ) + p 2 ~ x 2 ( p;w ) & 24 p 2 p 1 & p 2 = p 1 ~ x 1 ( p;w ) + p 2 (~ x 2 ( p;w ) & 24) p 1 & p 2 = p 1 ~ x 1 ( p;w ) & p 2 ~ x 1 ( p;w ) p 1 & p 2 (~ x 1 ( p;w ) + ~ x 2 ( p;w ) = 24) = ~ x 1 ( p;w ) Similarly, ~ x 2 ( p ;w ) = ~ x 2 ( p;w ) ; i:e:; & x = ~ x ( p ;w ) & ~ x ( p;w ) = 0 ; showing that & p ¡ & x = 0 for all Slutsky-compensated price-wealth changes. Thus, ~ x satis¡es WARP by proposition 2.F.1. Method 2 : we use the de¡nition of WARP to prove it by showing that 8 ( p ;w ) 2 P; ( p 1 ;w 1 ) 2 P;x £ ~ x ( p ;w ) ;x 1 £ ~ x ( p 1 ;w 1 ) ; we have that f p ¡ x 1 ¢ w ;x 6 = x 1 g ) 1 p 1 & x > w 1 : Note if p & x 1 ¡ w ;i:e:; then we have the following p 1 w 1 ¢ 24 p 1 2 p 1 1 ¢ p 1 2 + p 2 24 p 1 1 ¢ w 1 p 1 1 ¢ p 1 2 ¡ w , & p 1 ¢ p 2 ¡ w 1 ¢ 24 p 1 p 1 2 + 24 p 1 1 p 2 ¡ & p 1 1 ¢ p 1 2 ¡ w & p 1 1 > p 1 2 ¡ , & p 1 2 ¢ p 1 1 ¡ w ¢ 24 p 1 p 1 2 + 24 p 1 1 p 2 ¡ & p 2 ¢ p 1 ¡ w 1 , p 1 2 w ¢ 24 p 1 p 1 2 + 24 p 1 1 p 2 ¢ p 1 1 w ( p 1 ¢ p 2 ) ¡ ¢ w 1 & p 1 > p 2 ¡ , p 1 2 ( w ¢ 24 p 1 ) + p 1 1 (24 p 2 ¢ w ) ( p 1 ¢ p 2 ) ¡ ¢ w 1 , p 1 1 w ¢ 24 p 2 p 1 ¢ p 2 + p 1 2 24 p 1 ¢ w p 1 ¢ p 2 £ w 1 : i.e., p & x 1 ¡ w , p 1 & x £ w 1 and p & x 1 < w , p 1 & x > w 1 : Next we show f p & x 1 ¡ w ;x 6 = x 1 g ) p & x 1 < w : Since ~ x 1 ( p;w ) + ~ x 2 ( p;w ) = 24 ; 8 ( p;w ) 2 P; we have x & (1 ; 1) = 24 = x 1 & (1 ; 1) . Thus, we have ( x ¢ x 1 ) & (1 ; 1) = 0 ; i:e:;x 1 ¢ x 1 1 = x 1 2 ¢ x 2 . Hence, x 6 = x 1 shows that x 1 ¢ x 1 1 = x 1 2 ¢ x 2 6 = 0 : Consequently, p & x 1 ¢ w = p & x 1 ¢ p & x = p 1 ( x 1 1 ¢ x 1 ) + p 2 ( x 1 2 ¢ x 2 ) = p 1 ( x 1 1 ¢ x 1 ) + p 2 ( x 1 ¢ x 1 1 ) = ( p 2 ¢ p 1 ) ( x 1 ¢ x 1 1 ) 6 = 0 ;i:e:; p & x 1 6 = w : The inequality comes from the facts that x 1 ¢ x 1 1 6 = 0 and p 2 < p 1 : Thus, we conclude that f p x 1 ¡ w ;x 6 = x 1 g ) p & x 1 < w , p 1 & x > w 1 ; i.e., ~ x satis&es WARP by de&nition....
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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_2_ak - University of California, Davis ARE/ECN...

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