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Unformatted text preview: ORIE 3500/5500, Fall 10 HW 5 Solutions Assignment 5 Solutions Problem 1 Suppose the investor buys m shares of company A and K m shares of company B. So, the value of the resulting portfolio is mX A + ( K m ) X B . Now, V ar ( mX A + ( K m ) X B ) = m 2 2 A + ( K m ) 2 2 B + 2 m ( K m )Cov( X A ,X B ) . Differentiating with respect to m and setting that equal to 0 , we get V ar ( mX A + ( K m ) X B ) m = 0 2 m ( 2 A Cov( X A ,X B )) + 2( K m )( 2 B + Cov( X A ,X B )) = 0 m = K ( 2 B Cov( X A ,X B )) 2 A + 2 B 2Cov( X A ,X B ) . Differentiating twice with respect to m, we get 2 V ar ( mX A + ( K m ) X B ) m 2 = 2( 2 A + 2 B 2Cov( X A ,X B )) = 2 V ar ( X A X B ) . Hence, the variance is minimized for m = K ( 2 B Cov( X A ,X B )) 2 A + 2 B 2Cov( X A ,X B ) . Problem 2 Since the possible values of X are 1 , 2 , and 4, p 1 + p 2 + p 4 = 1. We also have Var( X ) = E ( X EX ) 2 = (1 2) 2 p 1 + (2 2) 2 p 2 + (4 2) 2 p 4 = p 1 + 4 p 4 = 2 2 p 2 . (a) To maximize Var( X ), we should choose p 2 = 0. Solving p 1 + p 4 = 1 , EX = p 1 + 4 p 4 = 2 for p 1 and p 4 , we get p 1 = 2 / 3 and p 4 = 1 / 3. (b) Var( X ) is minimized when p 2 = 1. Then p 1 + 1 + p 4 = 1 , EX = p 1 + 2 + 4 p 4 = 2 , yields p 1 = p 4 = 0....
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This note was uploaded on 11/11/2010 for the course ORIE 3500 at Cornell University (Engineering School).
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