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200a_f08_ps_2_ak

200a_f08_ps_2_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 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 ° > 0 : 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 ² 0 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 0 ; w 0 ) with w 0 = p 0 ± ~ x ( p; w ) : Then ~ x 1 ( p 0 ; w 0 ) = w 0 ° 24 p 0 2 p 0 1 ° p 0 2 = p 0 ± ~ x ( p; w ) ° 24 p 0 2 p 0 1 ° p 0 2 = p 0 1 ~ x 1 ( p; w ) + p 0 2 ~ x 2 ( p; w ) ° 24 p 0 2 p 0 1 ° p 0 2 = p 0 1 ~ x 1 ( p; w ) + p 0 2 (~ x 2 ( p; w ) ° 24) p 0 1 ° p 0 2 = p 0 1 ~ x 1 ( p; w ) ° p 0 2 ~ x 1 ( p; w ) p 0 1 ° p 0 2 (~ x 1 ( p; w ) + ~ x 2 ( p; w ) = 24) = ~ x 1 ( p; w ) Similarly, ~ x 2 ( p 0 ; w 0 ) = ~ x 2 ( p; w ) ; i:e:; ° x = ~ x ( p 0 ; w 0 ) ° ~ 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 0 ; w 0 ) 2 P; ( p 1 ; w 1 ) 2 P; x 0 ³ ~ x ( p 0 ; w 0 ) ; x 1 ³ ~ x ( p 1 ; w 1 ) ; we have that f p 0 ± x 1 ² w 0 ; x 0 6 = x 1 g ) 1
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p 1 ± x 0 > w 1 : Note if p 0 ± x 1 ² w 0 ; i:e:; then we have the following p 0 1 w 1 ° 24 p 1 2 p 1 1 ° p 1 2 + p 0 2 24 p 1 1 ° w 1 p 1 1 ° p 1 2 ² w 0 , ² p 0 1 ° p 0 2 ³ w 1 ° 24 p 0 1 p 1 2 + 24 p 1 1 p 0 2 ² ² p 1 1 ° p 1 2 ³ w 0 ² p 1 1 > p 1 2 ³ , ² p 1 2 ° p 1 1 ³ w 0 ° 24 p 0 1 p 1 2 + 24 p 1 1 p 0 2 ² ² p 0 2 ° p 0 1 ³ w 1 , p 1 2 w 0 ° 24 p 0 1 p 1 2 + 24 p 1 1 p 0 2 ° p 1 1 w 0 ( p 0 1 ° p 0 2 ) ² ° w 1 ² p 0 1 > p 0 2 ³ , p 1 2 ( w 0 ° 24 p 0 1 ) + p 1 1 (24 p 0 2 ° w 0 ) ( p 0 1 ° p 0 2 ) ² ° w 1 , p 1 1 w 0 ° 24 p 0 2 p 0 1 ° p 0 2 + p 1 2 24 p 0 1 ° w 0 p 0 1 ° p 0 2 ´ w 1 : i.e., p 0 ± x 1 ² w 0 , p 1 ± x 0 ´ w 1 and p 0 ± x 1 < w 0 , p 1 ± x 0 > w 1 : Next we show f p 0 ± x 1 ² w 0 ; x 0 6 = x 1 g ) p 0 ± x 1 < w 0 : Since ~ x 1 ( p; w ) + ~ x 2 ( p; w ) = 24 ; 8 ( p; w ) 2 P; we have x 0 ± (1 ; 1) = 24 = x 1 ± (1 ; 1) . Thus, we have ( x 0 ° x 1 ) ± (1 ; 1) = 0 ; i:e:; x 0 1 ° x 1 1 = x 1 2 ° x 0 2 . Hence, x 0 6 = x 1 shows that x 0 1 ° x 1 1 = x 1 2 ° x 0 2 6 = 0 : Consequently, p 0 ± x 1 ° w 0 = p 0 ± x 1 ° p 0 ± x 0 = p 0 1 ( x 1 1 ° x 0 1 ) + p 0 2 ( x 1 2 ° x 0 2 ) = p 0 1 ( x 1 1 ° x 0 1 ) + p 0 2 ( x 0 1 ° x 1 1 ) = ( p 0 2 ° p 0 1 ) ( x 0 1 ° x 1 1 ) 6 = 0 ; i:e:; p 0 ± x 1 6 = w 0 : The inequality comes from the facts that x 0 1 ° x 1 1 6 = 0 and p 0 2 < p 0 1 : Thus, we conclude that f p 0 x 1 ² w 0 ; x 0 6 = x 1 g ) p 0 ± x 1 < w 0 , p 1 ± x 0 > w 1 ; i.e., ~ x satis°es WARP by de°nition. 1.2 D ~ x ( p; w ) = " ° w +24 p 2 ( p 1 ° p 2 ) 2 ° 24 p 1 + w ( p 1 ° p 2 ) 2 1 p 1 ° p 2 ° 24 p 2 + w ( p 1 ° p 2 ) 2 24 p 1 ° w ( p 1 ° p 2 ) 2 ° 1 p 1 ° p 2 # : 1.3 Good 1 is normal because from 1.2: @ ~ x 1 ( p;w ) @w = 1 p 1 ° p 2 > 0; Good 1 is a luxury because " 1 w = @ ~ x 1 ( p;w ) @w w ~ x 1 ( p;w ) = w w ° 24 p 2 > 1; Good 1 is ordinary because from 1.2: @ ~ x 1 ( p;w ) @p 1 = ° w +24 p 2 ( p 1 ° p 2 ) 2 < 0 : 1.4 Good 2 is inferior because from 1.2: @ ~ x 2 ( p;w ) @w = ° 1 p 1 ° p
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