Solution to H5

# Solution to H5 - H AAS S CHOOL OF B USINESS U NIVERSITY OF...

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Unformatted text preview: H AAS S CHOOL OF B USINESS U NIVERSITY OF C ALIFORNIA AT B ERKELEY UGBA 103 A VINASH V ERMA S OLUTION TO H OMEWORK 5 1. Suppose you have an investment portfolio of \$4,500,000. Seven sixteenths of your portfolio consists of investment in Coca Cola (CC). \$900,000 of the portfolio is in risk-free US Treasury securities yielding a certain return of 4.86%. The remaining amount is allocated between General Electric (GE) and Microsoft (MS) in such a way that the amount invested in Microsoft is five times the amount in General Electric. The probability distribution of the returns on the securities in the portfolio is as follows: Event Probability ~ R CC ~ R GE MS R ~ ~ R P A 0.15-15% 4.5% 22.5% ? B 0.45 24% 25% 15.3% ? C ? 18% 14.4% 0% ? ( 29 ? ~ = CC R E ( 29 ? ~ = GE R E ( 29 ? ~ = MS R E ( 29 ? ~ = P R E Calculate the portfolio weights 1 and the return on the portfolio in each of the three events. Next, calculate the expected return on Coca Cola, on General Electric, on Microsoft, and finally on the entire portfolio. Make sure that your calculations are in agreement with the fact that the expected return on the portfolio is a weighted sum of the expected return on the securities in the portfolio where expected return on each security is weighted with the fraction invested in that security. Or, in mathematical terms, confirm that: E R x E R x E R x E R P n n ( ~ ) ( ~ ) ( ~ ) ( ~ ) = + + + 1 1 2 2 ........ . Let us denote CC as Security 1 , GE as Security 2 , MS as Security 3 , and the risk free asset as Security 4 , We are given that: x 1 = 7/16 = 0.4375; x 4 = \$900,000/4,500,000 = 0.2. Because our money is divided only among these four securities, we can say that: x 1 + x 2 + x 3 + x 4 = 1. Subtracting the values of x 1 = 0.4375 and x 4 = 0.2 from the equation above, we get: x 2 + x 3 = 0.3625 Equation [1] We are also given that: x 3 = 5* x 2 Equation [2] Solving two equations [1] and [2] in two unknowns by substituting Equation [2] into Equation [1], we get: 6 *x 2 = 0.3625 x 2 = 0.060417. And x 3 = 5* x 2 = 0.060417*5 = 0.302083. Denoting p A as the Probability of Event A, and so on for Events B and C, we are given that p A = 0.15 and p B = 0.45. From the fact that probabilities add up to one, we conclude that p C = 0.4. Since by definition, expected value of a random variable is a weighted average of its realizations where each realization is weighted by the probability of its occurrence, we compute: 1 The fraction of your wealth invested in a security is known as the portfolio weight on that security. Page 1 of 6 H AAS S CHOOL OF B USINESS U NIVERSITY OF C ALIFORNIA AT B ERKELEY UGBA 103 A VINASH V ERMA ( 29 ? ~ = CC R E = E 1 = 0.15*( –15%) + 0.45*(24%) + 0.4*(18%) =15.75% Similarly: ( 29 ?...
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## This note was uploaded on 02/12/2012 for the course UGBA 101A taught by Professor Mccullough during the Spring '08 term at Berkeley.

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Solution to H5 - H AAS S CHOOL OF B USINESS U NIVERSITY OF...

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