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Lecture8 - Slutsky_Equation

Lecture8 - Slutsky_Equation - Lecture8 Slutsky Equation...

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Lecture 8 Slutsky Equation Econ 101A: Microeconomic Theory UC Berkeley Spring 2011 Prof. Cristian Santesteban

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6 Slutsky’s Equation 6.1 Review Expenditure function: e ( p 1 , p 2 , u 0 ) = min x 1 ,x 2 p 1 x 1 + p 2 x 2 s.t. U ( x 1 , x 2 ) = u 0 = p 1 x c 1 ( p 1 , p 2 , u 0 ) + p 2 x c 2 ( p 1 , p 2 , u 0 ) , where x c 1 and x c 2 are the compensated demands, the cheapest choices that enable one to achieve utility level u 0 at prices ( p 1 , p 2 ). The Lagrangian for the E-min problem is L ( x 1 , x 2 , μ ) = p 1 x 1 + p 2 x 2 - μ ( U ( x 1 , x 2 ) - u 0 ) . The FONC are: p 1 - μU x 1 ( x 1 , x 2 ) = 0 , p 2 - μU x 2 ( x 1 , x 2 ) = 0 , U ( x 1 , x 2 ) = u 0 . As for the derivatives of the expenditure function with respect to prices, ∂e ( p 1 , p 2 , u 0 ) ∂p 1 = x c 1 ( p 1 , p 2 , u 0 ) + p 1 ∂x c 1 ( p 1 , p 2 , u 0 ) ∂p 1 + p 2 ∂x c 2 ( p 1 , p 2 , u 0 ) ∂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 0 ) , x c 2 ( p 1 , p 2 , u 0 )) = u 0 . 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 μ ∂x c 1 ∂p 1 + p 2 μ ∂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 0 ) ∂p 1 = x c 1 ( p 1 , p 2 , u 0 ) . 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 bundle—this increases your spending by x c 1 × Δ p 1 . That is the first term on the RHS of ( ).

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Lecture8 - Slutsky_Equation - Lecture8 Slutsky Equation...

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