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Answers to some old prelim questions: 1. Ms Blue and Mr. Green A Ms Blue has same preferences as Mr. Green, since her utility function is a monotone increasing function of his U B = e U G B Preferences are homothetic if ( x,y ) I ( x 0 ,y 0 ) implies ( tx,ty ) I ( tx 0 ,ty 0 ) for all t > 0. His preferences are not homothetic. To see this, we note for example that U (0 , 1) = 3 / 2 = U (3 / 2 , 0). But U (0 , 2) = 2 6 = U (3 , 0) = 3. C y ( p,w ) = 2 - p if p 2: y ( p,w ) = 0 if p > 2. x ( p,w ) = w - py ( p,w ). D If p 2, v ( p,w ) = ln ( w - p (2 - p ) + 2(2 - p ) - (2 - p ) 2 / 2) = ln ( w + (2 - p ) 2 / 2) . If p > 2, v ( p,w ) = w . E Roy’s law requires that y ( p,w ) = - ∂v ( p,w ) ∂p ∂v ( p,w ) ∂w . In this instance if p < 2, - ∂v ( p,w ) ∂p ∂v ( p,w ) ∂w = - (2 - p )( w + (2 - p ) 2 / 2) w + (2 - p ) 2 / 2 = - y ( p,w ) . If p > 0 - ∂v ( p,w ) ∂p ∂v ( p,w ) ∂w = 0 = - y ( p,w ) . F Mr. Green’s expenditure function satisﬁes the equation u = v ( p,e ( p,u )) = ln ( e ( p,u ) + (2 - p ) 2 2 ) and hence e ( p,u ) = e u - (2 - p ) 2 2 . G If p < 2, h ( p,u ) = 2 - p . If p > 2, h ( p,u ) = 0. 1

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2. Utility Question A) The same preferences are represented by the utility function V where V ( x ) = U ( x ) 1 α + β + γ This is true since V = f ( U ) where f is a strictly increasing function. But V ( x ) = ( x 1 + b 1 ) α 0 ( x 2 + b 2 ) β 0 ( x 3 + b 3 ) γ 0 where α 0 + β 0 + γ 0 = 1. B) These preferences are not in general homothetic. Preferences are homothetic only if U ( x ) = U ( y ) implies that U ( kx ) = U ( ky ) for all k > 0. To show that U is not homothetic, all we need is an example that violates this necessary condition. Here is one such example. Let α = β = γ = 1. Let b 1 = 1, b 2 = 0, and b 3 = 1. Now let x = (1 , 1 , 0) and y = (0 , 2 , 0). Then U ( x ) = U ( y ) = 2. But U (2 x ) = 6 and U (2 y ) = 4 6 = U (2 x ). C) The same preferences are represented by the function V where V ( x ) = ln U ( x ) = α ln( x 1 + b 1 ) + β ln( x 2 + b 2 ) + γ ln( x 3 + b 3 ) which is of the additively separable form. D) At an interior solution we would have x 1 = α m + p 1 b 1 + p 2 b 2 + p 3 b 3 p 1 - b 1 . x 2 = β m + p 1 b 1 + p 2 b 2 + p 3 b 3 p 1 - b 2 . x 3 = γ m + p 1 b 1 + p 2 b 2 + p 3 b 3 p 1 - b 3 . The only equilibrium will be an interior solution if all three of the above expressions are positive. If this is not the case, we need to ﬁnd a corner solution. A corner solution in which x 3 = 0 and the other two quantities are positive would have x 1 = α α + β m + p 2 b 1 + p 2 b 2 p 1 - b 1 x 2 = β α + β m + p 2 b 1 + p 2 b 2 p 1 - b 2 . and it would also have to be true that when x 3 = 0, the ratio of marginal utility of good 3 to its price is less than the ratio of marginal utility to price for the other two goods. That is γ p 3 b 3 α + β m + p 3 αb 1 + p 2 βb 2 2
Similar conditions would apply to the cases where x 2 = 0 and x 3 = 0. There could also be corner equilibria in which only one good was purchased,

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