hw2 solution

# hw2 solution - UNIVERSITY OF TORONTO Joseph L Rotman School...

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Unformatted text preview: UNIVERSITY OF TORONTO Joseph L. Rotman School of Management RSM332 PROBLEM SET #2 SOLUTIONS 1. (a) Expected returns are: E [ R A ] = 0 . 3 × . 07 + 0 . 4 × . 06 + 0 . 3 × (- . 08) = 0 . 021 = 2 . 1% , E [ R B ] = 0 . 3 × . 14 + 0 . 4 × (- . 04) + 0 . 3 × . 08 = 0 . 05 = 5% . Variances are: σ 2 A = 0 . 3 × (0 . 07) 2 + 0 . 4 × (0 . 06) 2 + 0 . 3 × (0 . 08) 2- (0 . 021) 2 = 0 . 004389 , σ 2 B = 0 . 3 × (0 . 14) 2 + 0 . 4 × (0 . 04) 2 + 0 . 3 × (0 . 08) 2- (0 . 05) 2 = 0 . 00594 . Standard deviations are: σ A = √ . 004389 = 6 . 625% , σ B = √ . 00594 = 7 . 707% . Covariance is: σ AB = 0 . 3 × . 07 × . 14 + 0 . 4 × . 06 × (- . 04) + 0 . 3 × (- . 08) × . 08- . 021 × . 05 =- . 00099 . Correlation is: ρ AB = σ AB σ A σ B =- . 00099 . 06625 × . 07707 =- . 19389 . (b) We can use the following first order condition to figure out the weights of assets A and B in the market portfolio: E [ R A ]- R F σ AM = E [ R B ]- R F σ BM ⇒ E [ R A ]- R F w A σ 2 A + (1- w A ) σ AB = E [ R B ]- R F w A σ AB + (1- w A ) σ 2 B ⇒ . 021- . 02 w A × . 004389 + (1- w A ) × (- . 00099) = . 05- . 02 w A × (- . 00099) + (1- w A )0 . 00594 . 1 This can be solved to obtain w A = 0 . 2118 and thus w B = 1- w A = 0 . 7882. Expected return and standard deviation of the market portfolio are: E [ R M ] = 0 . 2118 × . 021 + 0 . 7882 × . 05 = 0 . 04386 = 4 . 386% σ M = √ . 2118 2 × . 004389 + 0 . 7882 2 × . 00594- 2 × . 2118 × . 7882 × . 00099 = 0 . 05964 = 5 . 964% (c) Betas of stock A and B can be found from the CAPM equation (or alternatively you can calculate σ AM and σ BM and use the definition of beta): . 021 = 0 . 02 + β A (0 . 04386- . 02) . 05 = 0 . 02 + β B (0 . 04386- . 02) Solving this gives β A = 0 . 042 and β B = 1 . 257. The portfolio this investor wants to create has to have a beta of 1 . 5: 1 . 5 = w A × . 042 + (1- w A ) × 1 . 257 Solving this gives w A =- . 2 and thus w B = 1 . 2. So the investor needs to short-sell \$200 of asset A and invest the \$1,200 in asset B . The expected return on this portfolio is E [ R p ] = R F + β p ( E [ R M ]- R F ) = 0 . 02 + 1 . 5 × (0 . 04386- . 02) = 0 . 0558, or E [ R p ] = w A E [ R A ] + w B E [ R B ] =- . 2 × . 021 + 1 . 2 × . 05 = 0 . 0558 . (d) There will now be two tangency portfolios: the first one (let’s call it T b ) will be the one we calculated in part (b) above using the risk-free borrowing rate. Recall that this portfolio has w b A = 0 . 2118, w b B = 0 . 7882, E [ R T b ] = 4 . 386% and σ T b = 5 . 964%. The second one ( T l ) will be at the tangency portfolio using the risk-free lending rate. We can calculate its expected return and standard deviation as we did in part (b) to get: w l A = 0 . 3955, w l B = 0 . 6045, E [ R T l...
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hw2 solution - UNIVERSITY OF TORONTO Joseph L Rotman School...

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