Unformatted text preview: )2] Variance = 0.041270 So, the standard deviation is: Standard deviation = (0.041270)1/2 Standard deviation = 0.2032 or 20.32% 10. a. To calculate the average real return, we can use the average return of the asset and the average inflation rate in the Fisher equation. Doing so, we find: (1 + R) = (1 + r)(1 + h)
r = (1.1120/1.042) – 1 r = .0672 or 6.72% b. The average risk premium is simply the average return of the asset, minus the average real riskfree rate, so, the average risk premium for this asset would be: RP = R – R f RP = .1120 – .0510 RP = .0610 or 6.10% 11. We can find the average real riskfree rate using the Fisher equation. The average real riskfree rate was: (1 + R) = (1 + r)(1 + h)
r f = (1.051/1.042) – 1 r f = .0086 or 0.86% And to calculate the average real risk premium, we can subtract the average riskfree rate from the average real return. So, the average real risk premium was: rp = r – r f = 6.72% – 0.86% rp = 5.85%
12. Apply the fiveyear holdingperiod return formula to calculate the total return of the stock over the fiveyear period, we find: 5year holdingperiod return = [(1 + R1)(1 + R2)(1 +R3)(1 +R4)(1 +R5)] – 1 5year holdingperiod return = [(1 + .1843)(1 + .1682)(1 + .0683)(1 + .3219)(1 – .1987)] – 1 5year holdingperiod return = 0.5655 or 56.55% 261 13. To find the return on the zero coupon bond, we first need to find the price of the bond today. Since one year has elapsed, the bond now has 29 years to maturity, so the price today is: P1 = $1,000/1.0929 P1 = $82.15 There are no intermediate cash flows on a zero coupon bond, so the return is the capital gains, or: R = ($82.15 – 77.81) / $77.81 R = .0558 or 5.58% 14. The return of any asset is the increase in price, plus any dividends or cash flows, all divided by the initial price. This preferred stock paid a dividend of $5, so the return for the year was: R = ($94.63 – 92.85 + 5.00) / $92.85 R = .0730 or 7.30% 15. The return of any asset is the increase in price, plus any dividends or cash flows, all divided by the initial price. This stock paid no dividend, so the return was: R = ($82.01 – 75.15) / $75.15 R = .0913 or 9.13% This is the return for three months, so the APR is: APR = 4(9.13%) APR = 36.51% And the EAR is: EAR = (1 + .0913)4 – 1 EAR = .4182 or 41.82% 16. To find the real return each year, we will use the Fisher equation, which is: 1 + R = (1 + r)(1 + h) Using this relationship for each year, we find: Tbills 0.0330 0.0315 0.0405 0.0447 0.0227 0.0115 0.0088 Inflation (0.0112) (0.0226) (0.0116) 0.0058 (0.0640) (0.0932) (0.1027) Real Return 0.0447 0.0554 0.0527 0.0387 0.0926 0.1155 0.1243 1926 1927 1928 1929 1930 1931 1932 262 So, the average real return was: Average = (.0447 + .0554 + .0527 + .0387 + .0926 + .1155 + .1243) / 7 Average = .0748 or 7.48% Notice the real return was higher than the nominal return during this period because of deflation, or negative inflation. 17. Looking at the longterm corporate bond return history in Table 10.2, we see that the mean return was 6.2 percent, with a standard deviation of 8.4 percent. The range of returns you would expect to see 68 percent of the time is the mean plus or minus 1 standard deviation, or: R∈ µ n ± 1σ = 6.2% ± 8.4% = –2.20% to 14.60% The range of returns you would expect to see 95 percent of the time is the mean plus or minus 2 standard deviations, or: R∈ µ n ± 2σ = 6.2% ± 2(8.4%) = –10.60% to 23.00% 18. Looking at the largecompany stock return history in Table 10.2, we see that the mean return was 11.7 percent, with a standard deviation of 20.6 percent. The range of returns you would expect to see 68 percent of the time is the mean plus or minus 1 standard deviation, or: R∈ µ n ± 1σ = 11.7% ± 20.6% = –8.90% to 32.30% The range of returns you would expect to see 95 percent of the time is the mean plus or minus 2 standard deviations, or: R∈ µ n ± 2σ = 11.7% ± 2(20.6%) = –29.50% to 52.90% Intermediate 19. Here we know the average stock return, and four of the five returns used to compute the average return. We can work the average return equation backward to find the missing return. The average return is calculated as: .55 = .19 – .27 + .06 + .34 + R R = .23 or 23% The missing return has to be 23 percent. Now we can use the equation for the variance to find: Variance = 1/4[(.19 – .11)2 + (–.27 – .11)2 + (.06 – .11)2 + (.34 – .11)2 + (.23 – .11)2] Variance = 0.05515 And the standard deviation is: Standard deviation = (0.05515)1/2 Standard deviation = 0.2348 or 23.48% 263 20. The arithmetic average return is the sum of the known returns divided by the number of returns, so: Arithmetic average return = (.34 + .18 + .29 –.06 + .16 –.48) / 6 Arithmetic average return = .0717 or 7.17% Using the equation for the geometric return, we find: Geometric average return = [(1 + R1) × (1 + R2) × … × (1 + RT)]1/T – 1 Geometric average return = [(1 + .34)(1 + .18)(1 + .29)(1 – .06)(1 + .16...
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This note was uploaded on 07/10/2010 for the course FIN 6301 taught by Professor Eshmalwi during the Spring '10 term at University of TexasTyler.
 Spring '10
 eshmalwi
 Finance, Corporate Finance

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