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econ11_08_ps4_sol

# econ11_08_ps4_sol - Eco11 Fall 2008 Simon Board Economics...

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Eco11, Fall 2008 Simon Board Economics 11: Practice Problems 4 (Week 5) October 23, 2008 1. Expenditure Minimisation with CES demand A consumer has the utility u ( x 1 , x 2 ) = x 1 / 2 1 + x 1 / 2 2 . a) Find the Hicksian demand for x 1 and x 2 b) Find the expenditure function. c) Find the Marshallian demand for x 1 and x 2 . d) Find the indirect utility function. Depending on how you approach this question, you may find it useful to note that: p 2 p 1 1 / 2 + p 1 p 2 1 / 2 = p 1 + p 2 ( p 1 p 2 ) 1 / 2 e) Show that the Slutsky equation holds. Solution a) The Lagrangian is L = p 1 x 1 + p 2 x 2 + λ [ u - x 1 / 2 1 - x 1 / 2 2 ] The FOCs are p 1 = 1 2 λx - 1 / 2 1 p 2 = 1 2 λx - 1 / 2 2 Dividing we get, p 1 p 2 = x 1 / 2 2 x 1 / 2 1 1

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Eco11, Fall 2008 Simon Board This states that the MRS equals the price ratio. Substituting into the constraint, the Hicksian demands are given by h 1 = u 2 p 2 p 1 + p 2 2 h 2 = u 2 p 1 p 1 + p 2 2 b) The expenditure function is e = p 1 x * 1 + p 2 x * 2 = u 2 p 1 p 2 p 1 + p 2 c) The Lagrangian is L = x 1 / 2 1 + x 1 / 2 2 + λ [ m - p 1 x 1 - p 2 x 2 ] The FOCs are 1 2 x - 1 / 2 1 = λp 1 1 2 x - 1 / 2 2 = λp 2 Dividing we get, x 1 / 2 2 x 1 / 2 1 = p 1 p 2 This states that the MRS equals the price ratio. Substituting into the constraint, x * 1 = m p 1 p 2 p 1 + p 2 x * 2 = m p 2 p 1 p 1 + p 2 e) The indirect utility function is v = ( x * 1 ) 1 / 2 + ( x * 2 ) 1 / 2 = m p 1 + p 2 p 1 p 2 1 / 2 This can be derived by substituting the Marshallian demands into the utility function, or by inverting the expenditure function. 2
Eco11, Fall 2008 Simon Board d) The Slutsky equation states that dx * 1 dp 1 = dh 1 dp 1 - x * i dx * 1 dm That is, - m p 2 (2 p 1 + p 2 ) p 2 1 ( p 1 + p 2 ) 2 = - 2 u 2 p 2 2 ( p 1 + p 2 ) 3 - m p 2 2 p 2 1 ( p 1 + p 2 ) 2 Using the indirect utility function, u 2 = m p 1 + p 2 p 1 p 2 Substituting, - m p 2 (2 p 1 + p 2 ) p 2 1 ( p 1 + p 2 ) 2 = - 2 m p 1 + p 2 p 1 p 2 p 2 2 ( p 1 + p 2 ) 3 - m p 2 2 p 2 1 ( p 1 + p 2 ) 2 Simplifying, - m p 2 (2 p 1 + p 2 ) p 2 1 ( p 1 + p 2 ) 2 = - m 2 p

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