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Unformatted text preview: )(1 – .48)](1/6) – 1 Geometric average return = .0245 or 2.45% Remember, the geometric average return will always be less than the arithmetic average return if the returns have any variation. 21. To calculate the arithmetic and geometric average returns, we must first calculate the return for each year. The return for each year is: R1 = ($55.83 – 49.62 + 0.68) / $49.62 = .1389 or 13.89% R2 = ($57.03 – 55.83 + 0.73) / $55.83 = .0346 or 3.46% R3 = ($50.25 – 57.03 + 0.84) / $57.03 = –.1042 or –10.42% R4 = ($53.82 – 50.25 + 0.91)/ $50.25 = .0892 or 8.92% R5 = ($64.18 – 53.82 + 1.02) / $53.82 = .2114 or 21.14% The arithmetic average return was: RA = (0.1389 + 0.0346 – 0.1042 + 0.0892 + 0.2114)/5 RA = 0.0740 or 7.40% And the geometric average return was: RG = [(1 + .1389)(1 + .0346)(1 – .1042)(1 + .0892)(1 + .2114)]1/5 – 1 RG = 0.0685 or 6.85% 22. To find the real return we need to use the Fisher equation. Rewriting the Fisher equation to solve for the real return, we get: r = [(1 + R)/(1 + h)] – 1 264 So, the real return each year was: Year 1973 1974 1975 1976 1977 1978 1979 1980 Tbill return 0.0729 0.0799 0.0587 0.0507 0.0545 0.0764 0.1056 0.1210 0.6197 Inflation 0.0871 0.1234 0.0694 0.0486 0.0670 0.0902 0.1329 0.1252 0.7438 Real return –0.0131 –0.0387 –0.0100 0.0020 –0.0117 –0.0127 –0.0241 –0.0037 –0.1120 a. The average return for Tbills over this period was: Average return = 0.6197 / 8 Average return = .0775 or 7.75% And the average inflation rate was: W Y b. Average inflation = 0.7438 / 8 X Average inflation = .0930 or 9.30% Using the equation for variance, we find the variance for Tbills over this period was: Variance = 1/7[(.0729 – .0775)2 + (.0799 – .0775)2 + (.0587 – .0775)2 + (.0507 – .0775)2 + (.0545 – .0775)2 + (.0764 – .0775)2 + (.1056 – .0775)2 + (.1210 − .0775)2] Variance = 0.000616 And the standard deviation for Tbills was: Standard deviation = (0.000616)1/2 Standard deviation = 0.0248 or 2.48% The variance of inflation over this period was: Variance = 1/7[(.0871 – .0930)2 + (.1234 – .0930)2 + (.0694 – .0930)2 + (.0486 – .0930)2 + (.0670 – .0930)2 + (.0902 – .0930)2 + (.1329 – .0930)2 + (.1252 − .0930)2] Variance = 0.000971 And the standard deviation of inflation was: Standard deviation = (0.000971)1/2 Standard deviation = 0.0312 or 3.12% c. The average observed real return over this period was: Average observed real return = –.1122 / 8 265 d. Average observed real return = –.0140 or –1.40% The statement that Tbills have no risk refers to the fact that there is only an extremely small chance of the government defaulting, so there is little default risk. Since Tbills are short term, there is also very limited interest rate risk. However, as this example shows, there is inflation risk, i.e. the purchasing power of the investment can actually decline over time even if the investor is earning a positive return. 23. To find the return on the coupon bond, we first need to find the price of the bond today. Since one year has elapsed, the bond now has six years to maturity, so the price today is: P1 = $70(PVIFA8%,6) + $1,000/1.086 P1 = $953.77 You received the coupon payments on the bond, so the nominal return was: R = ($953.77 – 943.82 + 70) / $943.82 R = .0847 or 8.47% And using the Fisher equation to find the real return, we get: r = (1.0847 / 1.048) – 1 r = .0350 or 3.50% 24. Looking at the longterm government bond return history in Table 10.2, we see that the mean return was 6.1 percent, with a standard deviation of 9.4 percent. In the normal probability distribution, approximately 2/3 of the observations are within one standard deviation of the mean. This means that 1/3 of the observations are outside one standard deviation away from the mean. Or: Pr(R< –3.3 or R>15.5) ≈ 1/3 But we are only interested in one tail here, that is, returns less than –3.3 percent, so: Pr(R< –3.3) ≈ 1/6 You can use the zstatistic and the cumulative normal distribution table to find the answer as well. Doing so, we find: z = (X – µ)/σ z = (–3.3% – 6.1)/9.4% = –1.00 Looking at the ztable, this gives a probability of 15.87%, or: Pr(R< –3.3) ≈ .1587 or 15.87% The range of returns you would expect to see 95 percent of the time is the mean plus or minus 2 standard deviations, or: 95% level: R∈ µ n ± 2σ = 6.1% ± 2(9.4%) = –12.70% to 24.90% 266 The range of returns you would expect to see 99 percent of the time is the mean plus or minus 3 standard deviations, or: 99% level: R∈ µ ± 3σ = 6.1% ± 3(9.4%) = –22.10% to 34.30% 25. The mean return for small company stocks was 16.4 percent, with a standard deviation of 33.0 percent. Doubling your money is a 100% return, so if the return distribution is normal, we can use the zstatistic. So: z = (X – µ)/σ z = (100% – 16.4%)/33.0% = 2.533 standard deviations above the mean This corresponds to a probability of ≈ 0.565%, or about once every 200 years. Tripling your money would be: z = (200% – 16.4%)/33.0% = 5.5...
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 Spring '10
 eshmalwi
 Finance, Corporate Finance

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