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answersmid11 - Midterm Examination Economics 210A October...

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Midterm Examination: Economics 210A October 2011 The exam has 6 questions. Answer as many as you can. Good luck. 1) A) Must every quasi-concave function must be concave? If so, prove it. If not, provide a counterexample. (In all answers where you provide a counterexample, you must show that your example is really a counterexample.) Answer: Not every quasi-concave function is concave. Here is a counterexample. Define the function F with domain + (the positive real numbers.) such that F ( x ) = x 2 . We show two things: 1) This function is quasi-concave. 1 To see this, note that F is a strictly increasing function on + . Therefore if F ( y ) F ( x ), it must be that y x and hence for any t [0 , 1], ty + (1 t ) x x . Since F is an increasing function, it follows that F ( ty + (1 t ) x ) F ( x ) . Therefore F is quasi-concave. 2) The function F is not concave. To see this, note that F (2) = 4 and F (0) = 0, but F ( 1 2 2 + 1 2 0) = F (1) = 1 1 2 F (2) + 1 2 F (0) = 2 . This cannot be the case if F is a concave function. B) Must every concave function be quasi-concave? If so, prove it. If not, provide a counterexample. Answer: Every concave function is quasi-concave. Proof. If f is concave, its domain is a convex set A . For all x and y in A , and t between 0 and 1, if f ( tx + (1 t ) y ) tf ( x ) + (1 t ) f ( y ) . (1) From the Expression ?? it follows that f ( tx + (1 t ) y ) f ( y ) + t ( f ( x ) f ( y )) . (2) If f ( x ) f ( y ), then it follows from Expression ?? that f ( tx + (1 t ) y ) f ( y ) . (3) which means that f is quasi-concave. 1 By the way, F ( x ) = x 2 defined on the entire real line would not be quasi-concave. Can you show this? 1
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2) Let F and G be real-valued concave functions with the same domain, A . Define the function H so that for all x A , H ( x ) = F ( x ) + G ( x ). Is H a concave function? If so, prove it. If not, provide a counterexample. Answer: Since F and G are concave functions with domain A , it must be that if x A and y A , then for all t between 0 and 1, F ( tx + (1 t ) y ) tF ( x ) + (1 t ) F ( y ) and G ( tx + (1 t ) y ) tG ( x ) + (1 t ) G ( y ) Given these two inequalities, we see that H ( tx + (1 t ) y ) = F ( tx + (1 t ) y ) + G ( tx + (1 t ) y ) tF ( x ) + (1 t ) F ( y ) + tG ( x ) + (1 t ) G ( y ) = t ( F ( x ) + G ( x )) + (1 t ) ( F ( y ) + G ( y )) = tH ( x ) + (1 t ) H ( y ) By definition, the left side of Expression ?? equals H ( tx + (1 t ) y ). The right side of Expression ?? equals tF ( x )+ tG ( x )+(1 t ) F ( y )+ (1 t ) G ( y ) = tH ( x ) + (1 t ) HY ). Therefore H ( tx + (1 t ) y tH ( x ) + (1 t ) H ( y ) for all t between 0 and 1, which means that H is a concave function.
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