Lecture8 - Slutsky_Equation - Lecture 8 Slutsky Equation...

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Unformatted text preview: Lecture 8 Slutsky Equation Econ 101A: Microeconomic Theory UC Berkeley Spring 2011 Prof. Cristian Santesteban 6 Slutskys Equation 6.1 Review Expenditure function: e ( p 1 ,p 2 ,u ) = min x 1 ,x 2 p 1 x 1 + p 2 x 2 s.t. U ( x 1 ,x 2 ) = u = p 1 x c 1 ( p 1 ,p 2 ,u ) + p 2 x c 2 ( p 1 ,p 2 ,u ) , where x c 1 and x c 2 are the compensated demands, the cheapest choices that enable one to achieve utility level u at prices ( p 1 ,p 2 ). The Lagrangian for the E-min problem is L ( x 1 ,x 2 , ) = p 1 x 1 + p 2 x 2- ( U ( x 1 ,x 2 )- u ) . The FONC are: p 1- U x 1 ( x 1 ,x 2 ) = 0 , p 2- U x 2 ( x 1 ,x 2 ) = 0 , U ( x 1 ,x 2 ) = u . As for the derivatives of the expenditure function with respect to prices, e ( p 1 ,p 2 ,u ) p 1 = x c 1 ( p 1 ,p 2 ,u ) + p 1 x c 1 ( p 1 ,p 2 ,u ) p 1 + p 2 x c 2 ( p 1 ,p 2 ,u ) p 1 . ( ) In Appendix ?? , we discussed the Envelope Theorem, which says the second and third terms on the RHS cancel. Proof: Recall that U ( x c 1 ( p 1 ,p 2 ,u ) ,x c 2 ( p 1 ,p 2 ,u )) = u . Differentiate both sides with respect to p 1 : U x 1 x c 1 p 1 + U x 2 x c 2 p 1 = 0 . But U x 1 = p 1 / and U x 2 = p 2 / by the FONC. It follows by substitution that p 1 x c 1 p 1 + p 2 x c 2 p 1 = 0 , which means p 1 x c 1 p 1 + p 2 x c 2 p 1 = 0 . Thus we have e ( p 1 ,p 2 ,u ) p 1 = x c 1 ( p 1 ,p 2 ,u ) . There is a story we tell to go along with this. If you initially are minimizing expenditure, and the price of good 1 rises, what do you do? Your first order response is simply to continue buying the 33 old bundlethis increases your spending by...
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This note was uploaded on 01/21/2012 for the course ECON 101a taught by Professor Staff during the Spring '08 term at University of California, Berkeley.

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Lecture8 - Slutsky_Equation - Lecture 8 Slutsky Equation...

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