*The value of $1 invested at the beginning of the sample period (1/1/2001).
132
P A R T I I
Portfolio Theory and Practice
The larger the swings in rates of return, the greater the discrepancy between the arith-
metic and geometric averages, that is, between the compound rate earned over the sample
period and the average of the annual returns. If returns come from a normal distribution,
the expected difference is exactly half the variance of the distribution, that is,
E
3Geometric average4 5 E3Arithmetic average4 2 ½s
2
(5.15)
(A warning: To use Equation 5.15, you must express returns as decimals, not percentages.)
When returns are approximately normal, Equation 5.15 will be a good approximation.
9
Variance and Standard Deviation
When thinking about risk, we are interested in the likelihood of deviations from the
expected return. In practice, we usually cannot directly observe expectations, so we esti-
mate the variance by averaging squared deviations from our estimate of the expected
The geometric average in Example 5.6 (.54%) is substantially less than the arithmetic
average (2.10%). This discrepancy sometimes is a source of confusion. It arises from the
asymmetric effect of positive and negative rates of returns on the terminal value of the
portfolio.
Observe the returns in years 2002 ( 2 .2210) and 2003 (.2869). The arithmetic aver-
age return over the 2 years is ( 2 .2210 1 .2869)/2 5 .03295 (3.295%). However, if
you had invested $100 at the beginning of 2002, you would have only $77.90 at the
end of the year. In order to simply break even, you would then have needed to earn
$21.10 in 2003, which would amount to a whopping return of 27.09% (21.10/77.90).
Why is such a high rate necessary to break even, rather than the 22.10% you lost in
2002? The value of your investment in 2003 was much smaller than $100; the lower
base means that it takes a greater subsequent percentage gain to just break even.
Even a rate as high as the 28.69% realized in 2003 yields a portfolio value in 2003 of
$77.90 3 1.2869 5 $100.25, barely greater than $100. This implies a 2-year annually
compounded rate (the geometric average) of only .12%, significantly less than the arith-
metic average of 3.295%.
Example 5.7
Geometric versus Arithmetic Average
You invest $1 million at the beginning of 2018 in an S&P 500 stock-index fund. Given the rate of return for
2018, 2 40%, what rate of return in 2019 will be necessary for your portfolio to recover to its original value?
CONCEPT CHECK
5.4
9
We are told to measure historical performance over a particular sample period by the
geometric average but to
estimate expected future performance from that same sample as the arithmetic average. The question naturally
arises: If the same sample were to recur in the future, performance would be measured by the geometric aver-
age, so isn’t that the best estimator of expected return? Surprisingly, the answer is no. Future results will always
contain both positive and negative surprises compared to prior expectations. When compounded, a run of positive
surprises has greater impact on final wealth than a run of equal-sized negative ones. Because of this asymmetry,
the sample geometric average is actually a downward-biased estimate of the future average return drawn from
the same distribution. The bias turns out to be half the variance, and thus using the arithmetic average corrects
for this bias.
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C H A P T E R
5
Risk, Return, and the Historical Record
133
return, the arithmetic average, r . Adapting Equation 5.12 for historic data, we again use
equal probabilities for each observation, and use the sample average in place of the unob-
servable E ( r ).
Variance 5 Expected value of squared deviations
s
2
5 a
p(
s)3
r(
s) 2
E(
r)4
2
Using historical data with
n observations, we could
estimate variance as
s
^
2
5
1
n a
n
s51
3r(s) 2 r 4
2
(5.16)
where s
^ replaces s to denote that it is an estimate.
Take another look at Spreadsheet 5.2 . Column D shows the square deviations from
the arithmetic average, and cell D12 gives the standard deviation as .1774, which is
the square root of the sum of products of the (equal) probabilities times the squared
deviations.
Example 5.8
Variance and Standard Deviation
The variance estimate from Equation 5.16 is biased downward, however. The reason
is that we have taken deviations from the sample arithmetic average, r , instead of the
unknown, true expected value, E ( r ), and so have introduced an estimation error. Its effect
on the estimated variance is sometimes called a degrees of freedom bias. We can elimi-
nate the bias by multiplying the arithmetic average of squared deviations by the factor
n /( n 2 1). The variance and standard deviation then become
s
^
2
5
a
n
n 2 1
b 3
1
n a
n
s51
3r(s) 2 r4
2
5
1
n 2 1 a
n
s51
3r(s) 2 r 4
2
s
^ 5
Å
1
n 2 1 a
n
s51
3r(s) 2 r4
2
(5.17)
Cell D13 shows the unbiased estimate of standard deviation, .1983, which is higher than
the .1774 value obtained in cell D12. For large samples, n /( n 2 1) is close to 1, and the
adjustment for degrees of freedom becomes trivially small.
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