601-integ - x 1 = Max p 1 e ( p 1 ,p 2 ,u ) p 2 x 2 p 1 =...

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E frqrplfv 601 1 Integrability: An example If a system of demand equations satis f es the properties that follow from the preference based approach, can one f nd a utility function that rationalizes the demand system? MWG argue that under quite general conditions, the answer is yes. This note provides a simple example. As MWG explain, the f r s ts tepi stou sedua l i ty and the envelope theorem to integrate backwards from the demand equations to the expenditure function. The next step is to use the expenditure function to f nd the equation of a typical indiference curve. In a two-good demand system suppose x 1 ( p 1 ,p 2 ,w )= aw p 1 , 0 <a< 1 Find the equation for a typical indi f erence curve u 0 = u ( x 1 ,x 2 ) . Step 1 Use the envelope theorem and duality to write e ( p 1 ,p 0 2 ,u 0 ) p 1 = h ± p 1 ,p 0 2 ,u 0 ² = x 1 ± p 1 ,p 2 ,e ± p 1 ,p 0 2 ,u 0 ²² = ae ( p 1 ,p 0 2 ,u 0 ) p 1 The solution to this di f erential equation is e ± p 1 ,p 0 2 ,u 0 ² = kp a 1 ,forsome k> 0 . Step 2 Fix x 2 = x 0 2 .W ew a n tt h e x 1 that is consistent with
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Unformatted text preview: x 1 = Max p 1 e ( p 1 ,p 2 ,u ) p 2 x 2 p 1 = Max p 1 kp a 1 p 2 x 2 p 1 = Max p 1 kp a 1 1 p 2 x 2 p 1 1 Max p 1 f ( p 1 ) E frqrplfv 601 2 So f ( p 1 ) = 0 implies ( a 1) kp a 2 1 + p 2 x 2 p 2 1 = 0 or kp a 1 = p 2 x 2 1 a or p 1 = # p 2 x 2 k (1 a ) $ 1 /a Thus, at the optimal value of p 1 , x 1 = kp a 1 p 2 x 2 p 1 = p 2 x 2 1 a p 2 x 2 p 2 x 2 k (1 a ) 1 /a or p 2 x 2 k (1 a ) ( x 1 ) a = # ap 2 x 2 1 a $ a or ( x 1 ) a x 2 1 a = ka a (1 a ) 1 a p 2 a 1 = a constant. So we started with a Cobb-Douglas demand system and we have worked backwards to obtain the equation for a Cobb-Douglas indi f erence curve....
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601-integ - x 1 = Max p 1 e ( p 1 ,p 2 ,u ) p 2 x 2 p 1 =...

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