AnswersMidterm10

AnswersMidterm10 - Answers to Econ 210A Midterm October...

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Answers to Econ 210A Midterm, October 2010 Question 1. f ( x 1 ,x 2 ) = p max { x 1 ,x 2 } A. The function f is homogeneous of degree 1/2. To see this, note that for all t > 0 and all ( x 1 ,x 2 ) f ( tx 1 ,x 2 ) = p max { tx 1 ,tx 2 } = p t max { x 1 ,x 2 } = t 1 / 2 p max { x 1 ,x 2 } (1) = t 1 / 2 f ( x 1 ,x 2 ) B. The function f is neither concave, nor quasi-concave. One should be able to see this quickly by drawing an indifference curve and seeing that the at-least- as-good set is not convex. Therefore the function is not quasi-concave. Since a concave function must be quasi-concave, f cannot be concave either. One can show that the function is not quasi-concave by means of a single example. Let x = (1 , 0) and x 0 = (0 , 1). Then f ( x ) = 1 f ( x 0 ) = 1. Now consider the convex combination tx + (1 - t ) x 0 = ( t,t ) where 0 < t < 1. If f is quasi-concave, then it must be that f ( tx + (1 - t ) x 0 ) f ( x 0 ). But f ( tx + (1 - t ) x 0 ) = t < f ( x 0 ) = 1. So f cannot be quasi-concave. We also note from this same example that tf ( x ) + (1 - t ) f ( x 0 ) = 1 > f ( tx + (1 - t ) x 0 ) = t . But if f is a concave function tf ( x )+(1 - t ) f ( x 0 ) f ( tx +(1 - t ) x 0 ), so x is not a concave function. C The function f is not a convex function. The square root sign should be a tipoff. Square root is a strictly concave function. To show that f is not a convex function, consider the two points x = (1 , 0) and x 0 = (4 , 0). Then f ( x ) = 1 and f ( x 0 ) = 2. Therefore 1 2 f ( x ) + 1 2 f ( x 0 ) = 2 . 5 . But f ( 1 2 x + 1 2 x 0 ) = f (2 . 5 , 0) = 2 . 5 < 1 2 f ( x ) + 1 2 f ( x 0 ) The function f is a quasi-convex function. A quick way to see that this must be true is to look at an indifference curve and see that it looks like the “worse-than” sets must be convex. To show this formally, suppose
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AnswersMidterm10 - Answers to Econ 210A Midterm October...

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