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Unformatted text preview: Problem Set 7 Solutions Economics 6170 TA: Christopher Handy November 14, 2008 Problem 1 The proofs of parts (a) and (b) are very similar, so lets construct and prove the more general result requested in part (b). Then part (a) will follow for the special case where the degree of homogeneity is one. Claim: Let f : R n ++ R ++ be a continuously differentiable function on R n ++ , and suppose that x f ( x ) = rf ( x ) for all x R n ++ . Then f is homogeneous of degree r on R n ++ . Proof: Let x be an arbitrary vector in R n ++ . We want to show that for all t > 0, we have f ( t x ) = t r f ( x ). Given this x R n ++ , define the function g : R ++ R ++ by g ( t ) = f ( t x 1 ,...,t x n ) for all t > Using the Chain Rule and the hypothesis x f ( x ) = rf ( x ), we have that for all t > 0, g ( t ) = n X i =1 D i f ( t x 1 ,...,t x n ) x i = x f ( t x ) = r t f ( t x ) = r t g ( t ) Rearranging, we have that tg ( t ) = rg ( t ) for all t > Now, consider g ( t ) t r . Differentiating with respect to t , we have that for all t > 0, t g ( t ) t r = g ( t ) t r g ( t ) rt r 1 t 2 r = t r 1 t 2 r ( tg ( t ) rg ( t ) ) = 0 This implies that for all t > 0, g ( t ) t r = c for some c R . Evaluating at t = 1, we have that c = g (1) = f ( x ), so g ( t ) t r = f ( x ). Since g ( t ) = f ( t x ), we have that f ( t x ) = t r f ( x ) for all t > 0, which is what we wanted to show. 1 Problem 2 Note that part (a) shows that all homogeneous functions are also homothetic functions, while part (b) shows that a homothetic function is not necessarily homogeneous. (a) Homogeneity of degree r of F requires F ( tx ) = t r F ( x ), which is equivalent to g ( tf ( x )) = t r g ( f ( x )) since f must be homogeneous of degree one. This suggests choosing g ( x ) = x r . That choice implies F ( x ) = ( f ( x )) r , which suggests choos ing f ( x ) = ( F ( x )) 1 r . Now we must formally use these choices to show that F is homothetic. Define f : R n + R + and g : R + R + by f ( x ) = ( F ( x )) 1 r for all x R n + , g ( x ) = x r for all x R + Note that g is increasing on R + because we are given that r > 0. Also, f is homoge neous of degree one on R n + because for any x R n + and any t > 0, we have f ( tx ) = ( F ( tx )) 1 r = ( t r F ( x )) 1 r = t ( F ( x )) 1 r = tf ( x ) Finally, verify that for all x R n + , g ( f ( x )) = g ( F ( x )) 1 r = ( F ( x )) 1 r r = F ( x ) By the given definition, then, F is homothetic on R n + ....
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This note was uploaded on 10/03/2009 for the course ECON 6170 taught by Professor Mitra during the Fall '08 term at Cornell University (Engineering School).
 Fall '08
 MITRA
 Economics

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