parenleftbigg p 2 p 1 parenrightbigg \u03b1 \u03b1 1 x \u03b1 1 yielding h 1 p u u

Parenleftbigg p 2 p 1 parenrightbigg α α 1 x α 1

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parenleftbigg p 2 p 1 parenrightbigg α/ ( α 1) x α 1 yielding h 1 ( p, u ) = u parenleftBigg p α/ ( α 1) 1 p α/ ( α 1) 1 + p α/ ( α 1) 2 parenrightBigg 1 h 2 ( p, u ) = u parenleftBigg p α/ ( α 1) 2 p α/ ( α 1) 1 + p α/ ( α 1) 2 parenrightBigg 1 so that e ( p, u ) = p 1 u parenleftBigg p α/ ( α 1) 1 p α/ ( α 1) 1 + p α/ ( α 1) 2 parenrightBigg 1 + p 2 u parenleftBigg p α/ ( α 1) 2 p α/ ( α 1) 1 + p α/ ( α 1) 2 parenrightBigg 1 or e ( p, u ) = u p 1+1 / ( α 1) 1 ( p α/ ( α 1) 1 + p α/ ( α 1) 2 ) 1 + u p 1+1 / ( α 1) 2 ( p α/ ( α 1) 1 + p α/ ( α 1) 2 ) 1 or e ( p, u ) = u p 1+1 / ( α 1) 1 + p 1+1 / ( α 1) 2 ( p α/ ( α 1) 1 + p α/ ( α 1) 2 ) 1 2
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or e ( p, u ) = u parenleftBig p α/ ( α 1) 1 + p α/ ( α 1) 2 parenrightBig 1 1 and letting α/ ( α 1) = σ , e ( p, u ) = u ( p σ 1 + p σ 2 ) 1 Taking derivatives with respect to p 1 and p 2 yields the Hicksian demands. ii. To get the indirect utility function, v ( p, w ) = w ( p σ 1 + p σ 2 ) 1 (iii) Use Roy’s identity on the indirect utility function to get the Walrasian demands. (iv) The Lagrangian for the UMP is L = x α 1 + x α 2 λ ( p 1 x 1 + p 2 x 2 w ) with FONC’s αx α 1 1 λp 1 = 0 αx α 1 2 λp 2 = 0 ( p 1 x 1 + p 2 x 2 w ) = 0 The first two equations imply x 1 = x 2 parenleftbigg p 1 p 2 parenrightbigg 1 / ( α 1) substituting into the constraint yields w = p 1 x 2 parenleftbigg p 1 p 2 parenrightbigg 1 / ( α 1) + p 2 x 2 wp 1 / ( α 1) 2 = p 1 x 2 p 1 / ( α 1) 1 + p 1 / ( α 1) 2 p 2 x 2 so that x 2 = wp 1 / ( α 1) 2 p 1+1 / ( α 1) 1 + p 1+1 / ( α 1) 2 x 1 = wp 1 / ( α 1) 1 p 1+1 / ( α 1) 1 + p 1+1 / ( α 1) 2 and the indirect utility function is v ( p, w ) = w parenleftBiggparenleftBigg p 1 / ( α 1) 1 p 1+1 / ( α 1) 1 + p 1+1 / ( α 1) 2 parenrightBigg α + parenleftBigg p 1 / ( α 1) 2 p 1+1 / ( α 1) 1 + p 1+1 / ( α 1) 2 parenrightBigg α parenrightBigg 1 v ( p, w ) = w p α/ ( α 1) 1 + p α/ ( α 1) 2 parenleftBig p α/ ( α 1) 1 + p α/ ( α 1) 2 parenrightBig α 1 v ( p, w ) = w parenleftBig p α/ ( α 1) 1 + p α/ ( α 1) 2 parenrightBig (1 α ) 3
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3. A linear functional f ( x ) is a mapping f : x R N R satisfying f ( x ) = α x = α · x = < x, α > depending on your notational preferences. Let K be a convex set in R N . The support functional of K is S ( α ) = sup x K α · x Note that S () is a functional mapping linear functionals, f ( x ) = α · x , into real numbers, or S : α R N R . The support functional S ( α ) asks you to find the x that maximizes α · x , taking α as given, so it is really a value function. (i) Graph the unit circle, K = { ( x 1 , x 2 ) : x 2 1 + x 2 2 1 } , and for α 1 = (1 , 0), α 2 = (1 , 1) and α 3 = ( 1 , 2), graph α · x = u for some different values of u and solve the constrained maximization problem sup x K α · x . (ii) Now, think of S ( α ) as the value function associated with the constrained maximization problem sup x K α · x where α is allowed to vary, in the case of the unit circle. Show that by varying α , we can recover K and x ( α ). Explain how this is also true for all convex sets K .
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