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CHAPTER 10
RISK AND RETURN: LESSONS FROM
MARKET HISTORY
Solutions to Questions and Problems
1. The return of any asset is the increase in price, plus any dividends or cash flows, all divided by the initial price. The return of this stock is: R = [($86 – 75) + 1.20] / $ R = .1627, or 16.27% 2. The dividend yield is the dividend divided by price at the beginning of the period, so:
Dividend yield = $1.20 / $ Dividend yield = .0160, or 1.60% And the capital gains yield is the increase in price divided by the initial price, so: Capital gains yield = ($86 – 75) / $ Capital gains yield = .1467, or 14.67%
3. Using the equation for total return, we find:
R = [($67 – 75) + 1.20] / $ R = –.0907, or –9.07% And the dividend yield and capital gains yield are: Dividend yield = $1.20 / $ Dividend yield = .0160, or 1.60% Capital gains yield = ($67 – 75) / $ Capital gains yield = –.1067, or –10.67% Here’s a question for you: Can the dividend yield ever be negative? No, that would mean you were paying the company for the privilege of owning the stock. It has happened on bonds.
4. The total dollar return is the change in price plus the coupon payment, so:
Total dollar return = $1,063 – 1,040 + 60 Total dollar return = $ The total nominal percentage return of the bond is:
R = [($1,063 – 1,040) + 60] / $1,
R = .0798, or 7.98% Notice here that we could have simply used the total dollar return of $83 in the numerator of this equation. Using the Fisher equation, the real return was: (1 + R ) = (1 + r )(1 + h ) r = (1.0798 / 1.030) – 1 r = .0484, or 4.84%
5. The nominal return is the stated return, which is 11.80 percent. Using the Fisher equation, the real return was: (1 + R ) = (1 + r )(1 + h ) r = (1.1180)/(1.031) – 1 r = .0844, or 8.44% 6. Using the Fisher equation, the real returns for government and corporate bonds were:
(1 + R ) = (1 + r )(1 + h ) r G = 1.061/1.031 – 1 r G = .0291, or 2.91% r C = 1.064/1.031 – 1 r C = .0320, or 3.20%
7. The average return is the sum of the returns, divided by the number of returns. The average return for each stock was:
[ ] (^) .0620,or6.20% 5
1
⎥ = + − + +^ =
∑
X x N
N i i
[ ] (^) .0980,or9.80% 5
1
⎥ = + − + +^ =
∑
Y y N
N i i
We calculate the variance of each stock as:
Using the equation for variance, we find the variance for T-bills over this period was: Variance = 1/5[(.0729 – .0655) 2 + (.0799 – .0655) 2 + (.0587 – .0655) 2 + (.0507 – .0655)^2 + (.0545 – .0655) 2 + (.0764 – .0655)^2 ] Variance = 0. And the standard deviation for T-bills over this period was: Standard deviation = (0.000153)1/ Standard deviation = 0.0124 or 1.24% c. The average observed risk premium over this period was: Average observed risk premium = –19.90% / 6 Average observed risk premium = –3.32% The variance of the observed risk premium was: Variance = 1/5[(–.2198 – (–.0332)) 2 + (–.3446 – (–.0332)) 2 + (.3136 – (–.0332))^2 + (.1886 – (–.0332)) 2 + (–.1261 – (–.0332))^2 + (–.0107 – (–.0332)) 2 ] Variance = 0. And the standard deviation of the observed risk premium was: Standard deviation = (0.06278)1/ Standard deviation = 0.2492 or 24.92%
9. a. To find the average return, we sum all the returns and divide by the number of returns, so:
Arithmetic average return = (.27 +.13 + .18 – .14 + .09)/ Arithmetic average return = .1060, or 10.60% b. Using the equation to calculate variance, we find: Variance = 1/4[(.27 – .106)^2 + (.13 – .106) 2 + (.18 – .106)^2 + (–.14 – .106)^2 + (.09 – .106)^2 ] Variance = 0. So, the standard deviation is: Standard deviation = (0.023430)1/ Standard deviation = 0.1531, or 15.31%
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 = (1.1060/1.042) – 1 r (^) = .0614, or 6.14%
b. The average risk premium is simply the average return of the asset, minus the average real risk-free rate, so, the average risk premium for this asset would be:
RP = R– (^) Rf RP = .1060 –. RP = .0550, or 5.50%
11. We can find the average real risk-free rate using the Fisher equation. The average real risk-free 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 risk-free rate from the average real return. So, the average real risk premium was: rp = r – r (^) f= 6.14% – 0.86% rp = 5.28%
12. Apply the five-year holding-period return formula to calculate the total return of the stock over the five-year period, we find: 5-year holding-period return = [(1 + R 1 )(1 + R 2 )(1 + R 3 )(1 + R 4 )(1 + R 5 )] – 1 5-year holding-period return = [(1 + .1612)(1 + .1211)(1 + .0583)(1 + .2614)(1 – .1319)] – 1 5-year holding-period return = 0.5086, or 50.86% 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 24 years to maturity. Using semiannual compounding, the price today is: P 1 = $1,000/1.045 48 P 1 = $120. There are no intermediate cash flows on a zero coupon bond, so the return is the capital gains, or: R = ($120.90 – 109.83) / $109. R = .1008, or 10.08% 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 $4, so the return for the year was: R = ($96.12 – 94.89 + 4.00) / $94. R = .0551, or 5.51%
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 ± 2σ = 6.4% ± 2(8.4%) = –10.40% to 23.20%
18. Looking at the large-company stock return history in Table 10.2, we see that the mean return was 11. percent, with a standard deviation of 20.3 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 ± 1σ = 11.8% ± 20.3% = –8.50% to 32.10% 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 ± 2σ = 11.8% ± 2(20.3%) = –28.80% to 52.40% 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: 5(.11) = .12 – .21 + .09 + .32 + 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[(.12 – .11)^2 + (–.21 – .11) 2 + (.09 – .11)^2 + (.32 – .11)^2 + (.23 – .11)^2 ] Variance = 0. And the standard deviation is: Standard deviation = (0.04035)1/ Standard deviation = 0.2009, or 20.09% 20. The arithmetic average return is the sum of the known returns divided by the number of returns, so:
Arithmetic average return = (.27 + .12 + .32 –.12 + .19 –.31) / 6 Arithmetic average return = .0783, or 7.83% Using the equation for the geometric return, we find: Geometric average return = [(1 + R 1 ) × (1 + R 2 ) × … × (1 + R (^) T )] 1/ T^ – 1 Geometric average return = [(1 + .27)(1 + .12)(1 + .32)(1 – .12)(1 + .19)(1 – .31)] (1/6)^ – 1 Geometric average return = .0522, or 5.22% 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: R 1 = ($64.83 – 61.18 + 0.72) / $61.18 = .0714, or 7.14% R 2 = ($72.18 – 64.83 + 0.78) / $64.83 = .1254, or 12.54% R 3 = ($63.12 – 72.18 + 0.86) / $72.18 = –.1136, or –11.36% R 4 = ($69.27 – 63.12 + 0.95)/ $63.12 = .1125, or 11.25% R 5 = ($76.93 – 69.27 + 1.08) / $69.27 = .1262, or 12.62% The arithmetic average return was: R A = (0.0714 + 0.1254 – 0.1136 + 0.1125 + 0.1262)/ R A = 0.0644, or 6.44% And the geometric average return was: R G = [(1 + .0714)(1 + .1254)(1 – .1136)(1 + .1125)(1 + .1262)] 1/5^ – 1 R G = 0.0601, or 6.01% 22. To find the real return we need to use the Fisher equation. Re-writing the Fisher equation to solve for the real return, we get: r = [(1 + R )/(1 + h )] – 1 So, the real return each year was: Year T-bill return Inflation Real return 1973 0.0729 0.
a. The average return for T-bills over this period was: Average return = 0.6197 / 8 Average return = .0775, or 7.75%
r = (1.0596 / 1.032) – 1 r = .0268, or 2.68%
24. Looking at the long-term government bond return history in Table 10.2, we see that the mean return was 6.1 percent, with a standard deviation of 9.8 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 < –3.7 or R > 15.9) ≈ 1 / 3 But we are only interested in one tail here, that is, returns less than –3.7 percent, so: Pr( R < –3.7) ≈ 1 / 6 You can use the z-statistic and the cumulative normal distribution table to find the answer as well. Doing so, we find: z = (X – μ) / σ z = (–3.7% – 6.1) / 9.8% = –1. Looking at the z-table, 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 ± 2σ = 6.1% ± 2(9.8%) = –13.50% to 25.70% 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(9.8%) = –23.30% to 35.50% 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 z-statistic. So: z = (X – μ) / σ z = (100% – 16.5%) / 32.5% = 2.569 standard deviations above the mean This corresponds to a probability of ≈ 0.510%, or about once every 200 years. Tripling your money would be: z = (200% – 16.5%) / 32.5% = 5.646 standard deviations above the mean. This corresponds to a probability of (much) less than 0.5%. The actual answer is ≈.00000082039%, or about once every 1 million years.
26. It is impossible to lose more than 100 percent of your investment. Therefore, return distributions are truncated on the lower tail at –100 percent. Challenge 27. Using the z-statistic, we find:
z = (X – μ) / σ z = (0% – 11.8%) / 20.3% = –0. Pr(R ≤ 0) ≈ 28.05%
28. For each of the questions asked here, we need to use the z-statistic, which is:
z = (X – μ) / σ a. z 1 = (10% – 6.4%) / 8.4% = 0. This z-statistic gives us the probability that the return is less than 10 percent, but we are looking for the probability the return is greater than 10 percent. Given that the total probability is 100 percent (or 1), the probability of a return greater than 10 percent is 1 minus the probability of a return less than 10 percent. Using the cumulative normal distribution table, we get: Pr( R ≥ 10%) = 1 – Pr( R ≤ 10%) = 33.41% For a return less than 0 percent: z 2 = (0% – 6.4%) / 8.4 = –0. Pr( R < 10%) = 1 – Pr( R > 0%) = 22.31% b. The probability that T-bill returns will be greater than 10 percent is: z 3 = (10% – 3.6%) / 3.1% = 2. Pr( R ≥ 10%) = 1 – Pr( R ≤ 10%) = 1 – .9805 ≈ 1.95% And the probability that T-bill returns will be less than 0 percent is: z 4 = (0% – 3.6%) / 3.1% = –1. Pr( R ≤ 0) ≈ 12.28% c. The probability that the return on long-term corporate bonds will be less than –4.18 percent is: z 5 = (–4.18% – 6.4%) / 8.4% = –1. Pr( R ≤ –4.18%) ≈ 10.39%
CHAPTER 11
RETURN AND RISK: THE CAPITAL
ASSET PRICING MODEL (CAPM)
1. The portfolio weight of an asset is total investment in that asset divided by the total portfolio value. First, we will find the portfolio value, which is:
Total value = 135($47) + 105($41) = $10,
The portfolio weight for each stock is:
X A = 135($47)/$10,650 =.
X B = 105($41)/$10,650 =.
2. The expected return of a portfolio is the sum of the weight of each asset times the expected return of each asset. The total value of the portfolio is:
Total value = $2,100 + 3,200 = $5,
So, the expected return of this portfolio is:
E( R p ) = ($2,100/$5,300)(0.11) + ($3,200/$5,300)(.14) = .1281, or 12.81%
3. The expected return of a portfolio is the sum of the weight of each asset times the expected return of each asset. So, the expected return of the portfolio is:
E( R p ) = .25(.11) + .40(.17) + .35(.14) = .1445, or 14.45%
4. Here we are given the expected return of the portfolio and the expected return of each asset in the portfolio and are asked to find the weight of each asset. We can use the equation for the expected return of a portfolio to solve this problem. Since the total weight of a portfolio must equal 1 (100%), the weight of Stock Y must be one minus the weight of Stock X. Mathematically speaking, this means:
E( R p ) = .129 = .14 X X + .09(1 – X X)
We can now solve this equation for the weight of Stock X as:
.129 = .14 X X + .09 – .10 X X .039 = .04 X X X X =.
So, the dollar amount invested in Stock X is the weight of Stock X times the total portfolio value, or:
Investment in X = .7800($10,000) = $7,
And the dollar amount invested in Stock Y is:
Investment in Y = (1 – .7800)($10,000) = $2,
5. The expected return of an asset is the sum of the probability of each state occurring times the rate of return if that state occurs. So, the expected return of each stock asset is:
E( R A) = .20(.06) + .55(.07) + .25(.11) = .0780, or 7.80%
E( R B) = .20(–.20) + .55(.13) + .25(.33) = .1140, or 11.40%
To calculate the standard deviation, we first need to calculate the variance. To find the variance, we find the squared deviations from the expected return. We then multiply each possible squared deviation by its probability, and then add all of these up. The result is the variance. So, the variance and standard deviation of each stock are:
σA^2 =.20(.06 – .0780)^2 + .55(.07 – .0780) 2 + .25(.11 – .0780) 2 =.
σA = (.00036)1/2^ = .0189, or 1.89%
σB^2 =.20(–.20 – .1140)^2 + .55(.13 – .1140) 2 + .25(.33 – .1140) 2 =.
σB = (.03152)1/2^ = .1775, or 17.75%
b. This portfolio does not have an equal weight in each asset. We still need to find the return of the portfolio in each state of the economy. To do this, we will multiply the return of each asset by its portfolio weight and then sum the products to get the portfolio return in each state of the economy. Doing so, we get:
Boom: R p =.20(.07) +.20(.15) + .60(.33) =.2420 or 24.20% Bust: R p =.20(.13) +.20(.03) + .60(−.06) = –.0040 or –0.40%
And the expected return of the portfolio is:
E( R p ) = .65(.2420) + .35(−.004) = .1559, or 15.59%
To find the variance, we find the squared deviations from the expected return. We then multiply each possible squared deviation by its probability, and then add all of these up. The result is the variance. So, the variance of the portfolio is:
σp^2 = .65(.2420 – .1559) 2 + .35(−.0040 – .1559)^2 =.
9. a. This portfolio does not have an equal weight in each asset. We first need to find the return of the portfolio in each state of the economy. To do this, we will multiply the return of each asset by its portfolio weight and then sum the products to get the portfolio return in each state of the economy. Doing so, we get:
Boom: R p = .30(.24) + .40(.45) + .30(.33) = .3510, or 35.10% Good: R p = .30(.09) + .40(.10) + .30(.15) = .1120, or 11.20% Poor: R p = .30(.03) + .40(–.10) + .30(–.05) = –.0460, or –4.60% Bust: R p = .30(–.05) + .40(–.25) + .30(–.09) = –.1420, or –14.20%
And the expected return of the portfolio is:
E( R p ) = .20(.3510) + .35(.1120) + .30(–.0460) + .15(–.1420) = .0743, or 7.43%
b. To calculate the standard deviation, we first need to calculate the variance. To find the variance, we find the squared deviations from the expected return. We then multiply each possible squared deviation by its probability, and then add all of these up. The result is the variance. So, the variance and standard deviation the portfolio is:
σp^2 = .20(.3510 – .0743) 2 + .35(.1120 – .0743) 2 + .30(–.0460 – .0743) 2
- .15(–.1420 – .0743)^2 σp^2 =.
σp = (.02717)1/2^ = .1648, or 16.48%
10. The beta of a portfolio is the sum of the weight of each asset times the beta of each asset. So, the beta of the portfolio is:
βp = .10(.75) + .35(1.90) + .20(1.38) + .35(1.16) = 1.
11. The beta of a portfolio is the sum of the weight of each asset times the beta of each asset. If the portfolio is as risky as the market it must have the same beta as the market. Since the beta of the market is one, we know the beta of our portfolio is one. We also need to remember that the beta of the risk-free asset is zero. It has to be zero since the asset has no risk. Setting up the equation for the beta of our portfolio, we get:
βp = 1.0 = 1 / 3 (0) + 1 / 3 (1.65) + 1 / 3 (βX)
Solving for the beta of Stock X, we get:
βX = 1.
12. CAPM states the relationship between the risk of an asset and its expected return. CAPM is:
E( R i ) = R f + [E( R M) – R f] × βi
Substituting the values we are given, we find:
E( R i ) = .05 + (.11 – .05)(1.15) = .1190, or 11.90%
13. We are given the values for the CAPM except for the β of the stock. We need to substitute these values into the CAPM, and solve for the β of the stock. One important thing we need to realize is that we are given the market risk premium. The market risk premium is the expected return of the market minus the risk-free rate. We must be careful not to use this value as the expected return of the market. Using the CAPM, we find:
X (^) R f = 1 – .4425 =.
c. We need to find the portfolio weights that result in a portfolio with an expected return of 10 percent. We also know the weight of the risk-free asset is one minus the weight of the stock since the portfolio weights must sum to one, or 100 percent. So:
E( R p ) = .10 = .121 X S + .05(1 – X S) .10 = .121 X S + .05 – .05 X S X S =.
So, the β of the portfolio will be:
βp = .7042(1.13) + (1 – .7042)(0) = 0.
d. Solving for the β of the portfolio as we did in part b , we find:
βp = 2.26 = X S(1.13) + (1 – X S)(0)
X S = 2.26/1.13 = 2
X (^) R f = 1 – 2 = –
The portfolio is invested 200% in the stock and –100% in the risk-free asset. This represents borrowing at the risk-free rate to buy more of the stock.
17. First, we need to find the β of the portfolio. The β of the risk-free asset is zero, and the weight of the risk-free asset is one minus the weight of the stock, so the β of the portfolio is:
ß (^) p = X W(1.3) + (1 – X W)(0) = 1.3 X W
So, to find the β of the portfolio for any weight of the stock, we simply multiply the weight of the stock times its β.
Even though we are solving for the β and expected return of a portfolio of one stock and the risk-free asset for different portfolio weights, we are really solving for the SML. Any combination of this stock and the risk-free asset will fall on the SML. For that matter, a portfolio of any stock and the risk-free asset, or any portfolio of stocks, will fall on the
SML. We know the slope of the SML line is the market risk premium, so using the CAPM and the information concerning this stock, the market risk premium is:
E( R W) = .123 = .04 + MRP(1.30) MRP = .083/1.3 = .0638, or 6.38%
So, now we know the CAPM equation for any stock is:
E( R p ) = .04 + .0638βp
The slope of the SML is equal to the market risk premium, which is 0.0638. Using these equations to fill in the table, we get the following results:
X W E( R p ) ß (^) p
18. There are two ways to correctly answer this question. We will work through both. First, we can use the CAPM. Substituting in the value we are given for each stock, we find:
E( R Y) = .045 + .073(1.35) = .1436, or 14.36%
It is given in the problem that the expected return of Stock Y is 14 percent, but according to the CAPM, the return of the stock based on its level of risk should be 14.36 percent. This means the stock return is too low, given its level of risk. Stock Y plots below the SML and is overvalued. In other words, its price must decrease to increase the expected return to 14.36 percent.
For Stock Z, we find:
E( R Z ) = .045 + .073(0.80) = .1034, or 10.34%
