PS2_solution

PS2_solution - UNIVERSITY OF NORTH CAROLINA KENAN-FLAGLER...

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Unformatted text preview: UNIVERSITY OF NORTH CAROLINA KENAN-FLAGLER BUSINESS SCHOOL BUSI 580: INVESTMENTS PART I: PORTFOLIO THEORY AND ASSET PRICING Prof. Günter Strobl Spring 2010 Solution to Problem Set 2 A. The Capital Asset Pricing Model 1. If the stock’s correlation coefficient with the market portfolio doubles (with all other variables such as variances unchanged), then beta, and therefore the risk premium, will also double. The current risk premium is 14% - 6% = 8%. The new risk premium will be 16% and the new discount rate for the security will be 16% + 6% = 22%. If the stock pays a constant perpetual dividend, then we know that the dividend (D) must satisfy the equation for the present value of a perpetuity: Price = Dividend / Discount rate → D = $50 × 14% = $7 At the new discount rate of 22%, the stock will be worth $7 / 0.22 = $31.82. 2. (a) σp2 = 0.752 × 0.42 + 0.252 × 0.32 + 2 × 0.75 × 0.25 × 0.02 = 0.103 (b) β1 = σ1,M / σM2 = 0.064 / 0.04 = 1.6 β2 = σ2,M / σM2 = 0.032 / 0.04 = 0.8 βp = 0.75 × β1,M + 0.25 × β2,M = 1.4 (c) R12 = β12 σM2 / σ12 = 1.62 × 0.04 / 0.16 = 0.64 R22 = β22 σM2 / σ22 = 0.82 × 0.04 / 0.09 = 0.284 Rp2 = βp2 σM2 / σp2 = 1.42 × 0.04 / 0.103 = 0.761 3. In a world with only two risky securities, we know that the weight of stock A in the optimal risky portfolio (tangency portfolio) is given by: 1 1 See the lecture notes on portfolio theory. 1 wA = ( ) e A Er ( ) ( ) ( ) ( ( ) ( )) e 2 e E rA σ B − E rB σ Aσ B ρ AB . 2 e 2 e e σ B + E rB σ A − E rA + E rB σ Aσ B ρ AB Since the weight of stock A in the market portfolio is wA = 2/3, we can now solve this equation for the correlation coefficient ρAB (or use the Solver in Excel to find the correlation coefficient ρAB that would result in an optimal weight of wA = 2/3): 2 ρAB = 0.188 The market risk premium is E(rM) – rf = 2/3 × 6% + 1/3 × 8% = 6.67% and the variance of rM is σM2 = (2/3)2 × (0.14)2 + (1/3)2 × (0.22)2 + 2 × 2/3 × 1/3 × 0.14 × 0.22 × 0.188 = 0.01667 = (12.91%)2. If the risk-free rate is 4%, the CML is given by: E (rp ) = 4% + 6.67% σp, 12.91% and the SML is given by: E(ri ) = 4% + β i × 6.67% . We can now use the SML to calculate the betas of the two stocks: βA = 6% = 0.9 6.67% βB = 8% = 1.2 6.67% B. Testing the CAPM 4. Please see the Excel spreadsheet PS2_solution.xls. All of the estimated betas are significantly positive, which indicates that the market return has explanatory power for the returns of the individual stocks. The betas of Merck, Coca Cola, and Eastman Kodak are significantly less than unity (pvalue < 5%), which suggests that these stocks have significantly less systematic risk than the market portfolio. 2 Note that wA has to be 2/3, since wA + wB = 1 and the market capitalization of A is twice that of B, that is, wA = 2 wB. 2 SML 1.4% 1.2% Average Excess Return 1.0% 0.8% 0.6% 0.4% 0.2% 0.0% 0.6 0.7 0.9 0.8 1.0 1.1 1.2 -0.2% -0.4% Beta Contrary to expectations, the SML has a negative slope. Some of the stocks in our sample (e.g., Merck and Coca Cola) have large returns and low betas, which indicate that these stocks are underpriced relative to the CAPM. Moreover, our sample of five stocks is not only very small, but may not be representative of the average firm in the stock market. Of the five stocks, only Coca Cola has an intercept coefficient (alpha) that is significantly different from zero (at the 5% level of significance). The alpha of Merck is marginally significant with a p-value of 5.7%. In both cases, the intercept is positive, which means that the two stocks are underpriced relative to the CAPM. This provides mixed evidence against the CAPM. Although we can reject the CAPM for one stock, it is not clear if we can reject the CAPM overall (at the same level of significance). To do that, we would need to test jointly that all five intercepts are equal to zero. However, with only five observations, this test would clearly be lacking in power (i.e., the ability to distinguish between the null and the alternative hypothesis). Of particular concern is the fact that the slope coefficient of the SML doesn’t even have the correct sign. C. Getting Better Betas 5. Please see the Excel spreadsheet PS2_solution.xls. Using the unconditional market model, we implicitly assume that the parameters of the model are constant over time. However, from comparing betas over various sub-periods, it is clear that beta is not constant. It follows that the estimated betas from the market model correspond to the average betas over the sample period under 3 consideration. There is a trade-off between using longer sample periods for more precise estimates (i.e., smaller standard errors) and shorter but more recent sample periods for more accurate estimates of current (and future) beta. One way to reconcile these requirements is to adjust historical betas to provide more accurate estimates of future betas. Here we assume that the current date is the end of December 2003. Letting the subscript “1” indicate the first time interval (January 1994 to December 1998) and the subscript “2” the second time interval (January 1999 to December 2003), we can capture the tendency of beta to “regress to the mean” with the following equation: ˆ ˆ β i ,(2 ) = c 0 + c1 β i ,(1) , where the regression coefficients are constrained to sum to one: c0 + c1 = 1. Estimating this equation via OLS yields c0 = 0.685 and c1 = 0.315. The predicted betas for the period from January 2004 to December 2008 (indicated by the subscript “3”) are given by: ˆ ˆ β i , (3 ) = c 0 + c 1 β i , ( 2 ) The actual betas of Merck and Kodak differ sharply from the second 5-year period (January 1999 to December 2003) to the third 5-year period (January 2004 to December 2008). None of the predicted betas captures the sharp increase in the actual beta for Kodak. Overall, Blume’s adjusted betas have lower forecast errors than historical betas. Clearly, predicting future betas is a difficult process. 4 ...
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This note was uploaded on 04/06/2010 for the course BUSI 580 taught by Professor Strobl during the Fall '09 term at UNC.

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