hw5soln - ORIE 3500/5500, Fall 10 HW 5 Solutions Assignment...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

This note was uploaded on 11/11/2010 for the course ORIE 3500 at Cornell University (Engineering School).

Page1 / 6

hw5soln - ORIE 3500/5500, Fall 10 HW 5 Solutions Assignment...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online