CHAPTER12
SOME LESSONS FROM CAPITAL
MARKET HISTORY
Learning Objectives
LO1
LO2
LO3
LO4
How to calculate the return on investment.
The historical returns on various important types of investments.
The historical risks on various important types of investments.
The implications of market efficiency.
Answers to Concepts Review and Critical Thinking Questions
1.
(LO4) They all wish they had! Since they didn’t, it must have been the case that the stellar performance
was not foreseeable, at least not by most.
2.
(LO4) As in the previous question, it’s easy to see after the fact that the investment was terrible, but it
probably wasn’t so easy ahead of time.
3.
(LO2, 3) No, stocks are riskier. Some investors are highly risk averse, and the extra possible return
doesn’t attract them relative to the extra risk.
4.
(LO4) On average, the only return that is earned is the required return—investors buy assets with
returns in excess of the required return (positive NPV), bidding up the price and thus causing the return
to fall to the required return (zero NPV); investors sell assets with returns less than the required return
(negative NPV), driving the price lower and thus the causing the return to rise to the required return
(zero NPV).
5.
(LO4) The market is not weak form efficient.
6.
(LO4) Yes, historical information is also public information; weak form efficiency is a subset of semistrong form efficiency.
7.
(LO4) Ignoring trading costs, on average, such investors merely earn what the market offers; the trades
all have zero NPV. If trading costs exist, then these investors lose by the amount of the costs.
8.
(LO4) Unlike gambling, the stock market is a positive sum game; everybody can win. Also, speculators
provide liquidity to markets and thus help to promote efficiency.
9.
(LO4) The EMH only says, within the bounds of increasingly strong assumptions about the information
processing of investors, that assets are fairly priced. An implication of this is that, on average, the
typical market participant cannot earn excessive profits from a particular trading strategy. However,
that does not mean that a few particular investors cannot outperform the market over a particular
investment horizon. Certain investors who do well for a period of time get a lot of attention from the
financial press, but the scores of investors who do not do well over the same period of time generally
get considerably less attention from the financial press.
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10. (LO4)
a.
If the market is not weak form efficient, then this information could be acted on and a profit
earned from following the price trend. Under 2, 3, and 4, this information is fully impounded in
the current price and no abnormal profit opportunity exists.
b.
Under 2, if the market is not semi-strong form efficient, then this information could be used to
buy the stock “cheap” before the rest of the market discovers the financial statement anomaly.
Since 2 is stronger than 1, both imply that a profit opportunity exists; under 3 and 4, this
information is fully impounded in the current price and no profit opportunity exists.
c.
Under 3, if the market is not strong form efficient, then this information could be used as a
profitable trading strategy, by noting the buying activity of the insiders as a signal that the stock is
underpriced or that good news is imminent. Since 1 and 2 are weaker than 3, all three imply that a
profit opportunity exists. Under 4, this information does not signal any profit opportunity for
traders; any pertinent information the manager-insiders may have is fully reflected in the current
share price.
Solutions to Questions and Problems
NOTE: All end of chapter problems were solved using a spreadsheet. Many problems require multiple steps.
Due to space and readability constraints, when these intermediate steps are included in this solutions
manual, rounding may appear to have occurred. However, the final answer for each problem is found
without rounding during any step in the problem.
Basic
1.
(LO1) 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 = [($102 – 91) + 2.40] / $91 = .1472 or 14.73%
2.
(LO1) The dividend yield is the dividend divided by price at the beginning of the period price, so:
Dividend yield = $2.40 / $91 = .0263 or 2.637%
And the capital gains yield is the increase in price divided by the initial price, so:
Capital gains yield = ($102 – 91) / $91= .1209 or 12.09%
3.
(LO1) Using the equation for total return, we find:
R = [($83 – 91) + 2.40] / $91 = –.0615 or –6.15%
And the dividend yield and capital gains yield are:
Dividend yield = $2.40 / $91 = .02637 or 2.637%
Capital gains yield = ($83 – 91) / $91 = –.08791 or –8.791%
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.
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4.
(LO1) The total dollar return is the increase in price plus the coupon payment, so:
Total dollar return = $1070 – 1040 + 70 = $100
The total percentage return of the bond is:
R = [($1070 – 1040) + 70] / $1040 = .09615 or 9.615%
Notice here that we could have simply used the total dollar return of $100 in the numerator of this
equation.
Using the Fisher equation, the real return was:
(1 + R) = (1 + r)(1 + h)
r = (1.09615 / 1.04) – 1 = .0539 or 5.4%
5.
(LO2) The nominal return is the stated return, which is 10.45 percent from Table 12.4. Using the Fisher
equation, the real return was:
(1 + R) = (1 + r)(1 + h)
r = (1.1045)/(1.0395) – 1 = .0625 or 6.25%
6.
(LO2) The nominal return is the stated return, which is 8.74 percent from Table 12.4. Using the Fisher
equation, the real return was:
(1 + R) = (1 + r)(1 + h)
r = (1.0874)/(1.0395) – 1 = .04607 or 4.608%
7.
(LO1) The average return is the sum of the returns, divided by the number of returns. The average return for
each stock was:
.08  .21  .17  .16  .09  7.80%
N 
X   x i  N 
5
 i 1 
.16  .38.  14  .22  .26  14.40%
N 
Y   y i  N 
5
 i 1 
Remembering back to “sadistics,” we calculate the variance of each stock as:


N
 X 2     x i  x 2 
X2
Y 2
N  1
 i 1

1
.08  .0782  .21  .0782  .17  .0782   .16  .0782  .09  .0782  .02067

5 1
1
.16  .1442  .38  .1442  .14  .1442   .22  .1442  .26  .1442  .05048

5 1




The standard deviation is the square root of the variance, so the standard deviation of each stock is:
X = (.002067)1/2 = .1437 or 14.37%
Y = (.05048)1/2 = .22467 or 22.467%
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8.
(LO2, 3)
Year
1970
1971
1972
1973
1974
1975
Large co. stock return
– 3.57%
8.01
27.37
0.27
–25.93
18.48
24.63
T-bill return
6.89%
3.86
3.43
4.78
7.68
7.05
33.69
a.
Large company stocks: average return = 24.63 / 6 = 4.105%
T-bills: average return = 33.69 / 6 = 5.615%
b.
Large company stocks:
Risk premium
10.46%
4.15
23.94
–4.51
–33.61
11.43
–9.06
variance = 1/5[(–.0357 – .04105)2 + (.0801 – .04105)2 + (.2737 – .04105)2 + (.0027 – .04105)2 +
(–.2593 – .04105)2 + (.1848 – .04105)2] = 0.034777
standard deviation = (0.034777)1/2 = 0.186486 = 18.65%
T-bills:
variance = 1/5[(.0689 – .05615)2 + (.0386–.05615)2 + (.0343–.05615)2 + (.0478–.05615)2 +
(.0768 – . 05615)2 + (.0705 – . 05615)2] = 0.00033001
standard deviation = (0.00033001)1/2 = 0.018166 = 1.82%
c.
Average observed risk premium = –9.06 / 6 = –1.51%
variance = 1/5[(–.1046 + .0151)2 + (.0415 + .0151)2 + (.2394 + .0151)2 +
(–.0451 + .0151)2 + (–.3361 + .0151)2 + (.1143 + .0151)2] = 0.03933388
standard deviation = (0.03933388)1/2 = 0.1983277 = 19.83%
d.
9.
Before the fact, for most assets the risk premium will be positive; investors demand compensation over and above the risk-free return to invest their money in the risky asset. After the fact,
the observed risk premium can be negative if the asset’s nominal return is unexpectedly low, the
risk-free return is unexpectedly high, or if some combination of these two events occurs.
(LO1)
a.
To find the average return, we sum all the returns and divide by the number of returns, so:
Average return = (.07 –.12 +.11 +.38 +.14)/5 = .116 or 11.6%
b.
Using the equation to calculate variance, we find:
Variance = 1/4[(.07 – .116)2 + (–.12 – .116)2 + (.11 – .116)2 + (.38 – .116)2 +
(.14 – .116)2]
Variance = 0.03203
So, the standard deviation is:
Standard deviation = (0.03203)1/2 = 0.1789 or 17.89%
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10. (LO1)
a.
To calculate the average real return, we can use the average nominal return of the asset, and the
average inflation rate in the Fisher equation. Doing so, we find:
(1 + R) = (1 + r)(1 + h)
r = (1.116/1.035) – 1 = .0783 or 7.83%
b.
The average risk premium is simply the average nominal return of the asset, minus the average
risk-free rate, so, the average risk premium for this asset would be:
RP  R – R f = .116 – .042 = .074 or 7.40%
11. (LO1) 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 = (1.042/1.035) – 1 = .0068 or 0.68%
And to calculate the average real risk premium, we can subtract the average real risk free rate from the
average real return. So, the average real risk premium was:
rp  r – r f = 7.83% – 0.68% = 7.15%
12. (LO2) T-bill rates were highest in the early eighties. This was during a period of high inflation and is
consistent with the Fisher effect.
Intermediate
13. (LO1) To find the real return, we first need to find the nominal return, which means we need the
current price of the bond. Going back to the chapter on pricing bonds, we find the current price is:
P1 = $80(PVIFA7%,6) + $1,000(PVIF7%,6) = $1047,67
So the nominal return is:
R = [($1047.67 – 1030) + 80]/$1030 = .0948 or 9.48%
And, using the Fisher equation, we find the real return is:
1 + R = (1 + r)(1 + h)
r = (1.0948/1.042) – 1 = .0507 or 5.07%
14. (LO1) 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:
.525 = .07 – .12 + .18 + .19 + R
R = .205 or 20.5%
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The missing return has to be 20.5 percent. Now we can use the equation for the variance to find:
Variance = 1/4[(.07 – .105)2 + (–.12 – .105)2 + (.18 – .105)2 + (.19 – .105)2 + (.205 – .105)2]
Variance = 0.018675
And the standard deviation is:
Standard deviation = (0.018675)1/2 = 0.13675 or 13.67%
15.
(LO1) The arithmetic average return is the sum of the known returns divided by the number of returns,
so:
Arithmetic average return = (.03 + .38 + .21 – .15 + .29 – .13) / 6
Arithmetic average return = .105 or 10.5%
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 + .03)(1 + .38)(1 + .21)(1 – .15)(1 + .29)(1 – .13)](1/6) – 1
Geometric average return = .08602 or 8.602%
Remember, the geometric average return will always be less than the arithmetic average return if the
returns have any variation.
16. (LO1) To calculate the arithmetic and geometric average returns, we must first calculate the return for
each year. The return for each year is:
R1 = ($73.66 – 60.18 + 0.60) / $60.18 = .2339 or 23.39%
R2 = ($94.18 – 73.66 + 0.64) / $73.66 = .2873 or 28.73%
R3 = ($89.35 – 94.18 + 0.72) / $94.18 = –.0436 or –4.36%
R4 = ($78.49 – 89.35 + 0.80)/ $89.35 = -.1126 or -11.26%
R5 = ($95.05 – 78.49 + 1.20) / $78.49 = .2263 or 22.63%
The arithmetic average return was:
RA = (0.2339 + 0.2873 – 0.0436 - 0.1126 + 0.2263)/5 = 0.11825 or 11.825%
And the geometric average return was:
RG = [(1 + .2339)(1 + .2873)(1 – .0436)(1-.1126)(1 + .2263)]1/5 – 1 = 0.1058 or 10.58%
17. (LO3) Looking at the long-term corporate bond return history in Table 12.4, we see that the mean
return was 8.74 percent, with a standard deviation of 9.75percent.
We use the z-statistic and the cumulative normal distribution table to find the answer. Doing so, we
find:
z = (X – µ)/
z = (–2.2% – 8.74)/9.75% = –1.122
Looking at a z-table, a z-value of – 1.122 gives a probability of 13.09%, or:
Pr(R< –2.2)  .1309 or 13.09%
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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 
 = 8.74% ± 2(9.75%) = –10.76% to 28.24%
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 
 = 8.74% ± 3(9.75%) = –20.51% to 37.99%
18. (LO3) For small stocks:  = 13.71%;  = 26.65%.
Doubling your money is a 100% return, so if the return distribution is normal, z = (100 –
13.71)/26.65 = 3.238 standard deviations above the mean; this corresponds to a probability of
Pr(R=100%) = 1 – Pr(R=100%) = 1 – 0.9994 ≈ 0.0006%
Tripling your money would be z = (200 – 13.71)/26.65 = 6.99 standard deviations above the mean; this
corresponds to a probability that is very close to zero. In this case, you should never expect your
money to triple.
19. (LO3) It is impossible to lose more than 100 percent of your investment. Therefore, return distributions
are truncated on the lower tail at –100 percent.
20. (LO2, 3) 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
1973
1974
1975
1976
1977
1978
1979
1980
a.
T-bill return
Inflation
Real return
0.0478
0.0768
0.0705
0.0910
0.0764
0.0790
0.1101
0.1223
0.6739
0.0936
0.1230
0.0952
0.0587
0.0945
0.0844
0.0969
0.1120
0.7583
-0.0419
-0.0411
-0.0226
0.0305
-0.0165
-0.0050
0.0120
0.0093
-0.0753
The average return for T-bills over this period was:
Average return = 0.6739 / 8
Average return = .0842 or 8.42%
And the average inflation rate was:
Average inflation = 0.7583 / 8
Average inflation = .0948 or 9.48%
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b.
Using the equation for variance, we find the variance for T-bills over this period was:
Variance = 1/7[(.0478 – .0842)2 + (.0768 – .0842)2 + (.0705 – .0842)2 + (.0910 – .0842)2 +
(.0764 – .0842)2 + (.0790 – .0842)2 + (.1101 – .0842)2 + (.1223  .0842)2]
Variance = 0.000546
And the standard deviation for T-bills was:
Standard deviation = (0.000546)1/2
Standard deviation = 0.0234 or 2.34%
The variance of inflation over this period was:
Variance = 1/7[(.0936 – .0948)2 + (.1230 – .0948)2 + (.0952 – .0948)2 + (.0587 – .0948)2 +
(.0945 – .0948)2 + (.0844 – .0948)2 + (.0969 – .0948)2 + (.1120  .0948)2]
Variance = 0.000358
And the standard deviation of inflation was:
Standard deviation = (0.000358)1/2
Standard deviation = 0.0189 or 1.89%
c.
The average observed real return over this period was:
Average observed real return = –.0753 / 8
Average observed real return = –.0094 or –0.94%
d. The statement that T-bills 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 T-bills 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.
Challenge
21. (LO3) For this problem, the Z value should be rounded to the nearest value appearing in the cumulative
normal probability table. As an alternative the Excel function =NORMSDIST can be used to provide
greater accuracy.
z = (X – µ)/
z = (0 – 10.45)/16.93 = –0.6172;
Pr ( R< 0)  26.85%
22. (LO3) For this problem, the Z value should be rounded to the nearest value appearing in the cumulative
normal probability table. As an alternative the Excel function =NORMSDIST can be used to provide
greater accuracy.
For each of the questions asked here, we need to use the z-statistic, which is:
z = (X – µ)/
a.
The probability that long bond returns will be greater than 10 percent is:
z1 = (10% – 8.74)/9.75% = 0.1292
S12-8
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%) = 1 – .5514  44.86%
For a return less than 0 percent:
z2 = (0% – 8.74)/9.75 = –0.8964
Pr(R<0%)  18.50%
b.
The probability that T-bill returns will be greater than 10 percent is:
z3 = (10% – 6.15)/3.75% = 1.02667
Pr(R>10%) = 1 – Pr(R<10%) = 1 – .8477  15.23%
And the probability that T-bill returns will be less than 0 percent is:
z4 = (0% – 6.15)/3.75% = –1.64
Pr(R<0)  5.05%
c.
The probability that the return on long-term corporate bonds will be less than –2.09 percent is:
z5 = (–2.09% – 8.74)/9.75% = –1.1108
Pr(R<–2.09%)  13.33%
And the probability that T-bill returns will be greater than 19.11 percent is:
z6 = (19.11% – 6.15)/3.75% = 3.456
Pr(R<19.11%) = 1 – Pr(R<19.11%) = 1 – .99973  0.027%
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