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Appendix B
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Properties of the Mean and the Variance of a Random Variable
Appendix B: Properties of the Mean and the Variance of
a Random Variable, and the Covariance
Suppose a company that manufactures TV sets has a fixed production cost of $2 million per year.
The gross profit for each TV set sold, which is the price minus the unit variable production cost,
is $50. Historical sales records indicate that the number of TV sets sold per year, x, is a random
variable with a mean of mx 100,000 and a standard deviation of sx 10,000. Let y denote the
company’s annual profit from selling the TV set. Since this profit equals the gross profit associated with selling x TV sets, which is 50x, minus the fixed cost of $2,000,000, it follows that
y 2,000,000 50x
In order to find the mean, variance, and standard deviation of y, we can use the following result:
If x is a random variable and a and b are fixed numbers, then
m(a bx) a bmx
s2(a bx) b2s2x
and
In the TV set manufacturing example, we have seen that the company’s annual profit is
y 2,000,000 50x
We have also seen that mx 100,000 and sx 10,000. Therefore,
my m(2,000,000 50x) 2,000,000 50mx 2,000,000 50(100,000)
3,000,000
and
s2y s2( 2,000,000 50x) (50)2s2x 2,500(10,000)2 250,000,000,000
sy 2s2y 1250,000,000,000 500,000
Chebyshev’s Theorem tells us that the probability is at least 3兾4 that the annual profit from
selling the TV set will be between my 2sy $2,000,000 and my 2sy $4,000,000.
We next consider a result concerning the mean and variance of a sum of random variables.
Let x1, x2, . . . , xn be n random variables. Then:
1
2
m(x
1x2 xn)
mx mx mx
1
2
n
If x1, x2, . . . , xn are statistically independent (that is, the value taken by any one of these random
variables is in no way associated with the value taken by any other of these random variables), then
s2(x
1x2 xn)
s2x s2x s2x
1
2
n
For example, the time to set up a new production system in a particular company is denoted
by the random variable y and is the sum of the following three random variables:
1
2
3
x1, the time to purchase the production equipment and have it delivered, which has mean
mx1 30 days and standard deviation sx1 3 days.
x2, the time to assemble the equipment, which has mean mx2 20 days and standard
deviation sx2 2 days.
x3, the time to train the factory workers to use the equipment, which has mean mx3 14 days
and standard deviation sx3 2 days.
It follows that
my m(x1x2x3) mx1 mx2 mx3 30 20 14 64 days
Furthermore, although we cannot train the factory workers until we assemble the equipment, and although we cannot assemble the equipment until the equipment is purchased and
delivered, it is reasonable that the times to do these tasks (x1, x2, and x3) are statistically
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independent. Therefore,
s2y s2(x1x2x3) s2x1 s2x2 s2x3 (3)2 (2)2 (2)2 9 4 4 17
and
sy 2s2y 117 4.1231 (days)
Chebyshev’s Theorem tells us that the probability is at least 34 that the time to set up the
production system will be between my 2sy 55.75 days and my 2sy 72.25 days.
To conclude this appendix, we note that sometimes random variables are not independent,
and we can measure their dependence by using the covariance. For example, below we present
(1) the probability distribution of x, the yearly proportional return for stock A, (2) the probability
distribution of y, the yearly proportional return for stock B, and (3) the joint probability
distribution of (x, y), the joint yearly proportional returns for stocks A and B [note that we have
obtained the data below from Pfaffenberger and Patterson (1987)].
x
p( x)
y
p( y)
0.10
0.05
0.15
0.38
0.400
0.125
0.100
0.375
0.15
0.05
0.12
0.46
0.300
0.200
0.150
0.350
mx .124
my .124
s2x .0454
s2y .0681
sx .2131
sy .2610
Stock B
Return, y
0.15
0.05
0.12
0.46
Joint Distribution of (x, y)
Stock A Return, x
0.10
0.05
0.15
0.38
0.025
0.075
0.050
0.250
0.025
0.025
0.025
0.050
0.025
0.025
0.025
0.025
0.225
0.075
0.050
0.025
To explain the joint probability distribution, note that the probability of .250 enclosed in the rectangle is the probability that in a given year the return for stock A will be .10 and the return
for stock B will be .46. The probability of .225 enclosed in the oval is the probability that in a
given year the return for stock A will be .38 and the return for stock B will be .15. Intuitively,
these two rather large probabilities say that (1) a negative return x for stock A tends to be associated with a highly positive return y for stock B, and (2) a highly positive return x for stock A
tends to be associated with a negative return y for stock B. To further measure the association
between x and y, we can calculate the covariance between x and y. To do this, we calculate
(x mx)(y my) (x .124)(y .124) for each combination of values of x and y. Then, we
multiply each (x mx)(y my) value by the probability p(x, y) of the (x, y) combination of values and add up the quantities that we obtain. The resulting number is the covariance, denoted
S2xy. For example, for the combination of values x .10 and y .46, we calculate
(x mx)(y my) p(x, y) (.10 .124)(.46 .124)(.250) .0188
Doing this for all combinations of (x, y) values and adding up the resulting quantities, we find that
the covariance is .0318. In general, a negative covariance says that as x increases, y tends to
decrease in a linear fashion. A positive covariance says that as x increases, y tends to increase in
a linear fashion.
The covariance helps us in this situation to understand the importance of investment diversification. If we invest all of our money in stock A, we have seen that mx .124 and sx .2131. If
we invest all of our money in stock B, we have seen that my .124 and sy .2610. If we invest
half of our money in stock A and half of our money in stock B, the return for the portfolio is
P .5x .5y. The expected value of the portfolio return is
mP m(.5x.5y) m.5x m.5y .5mx .5my .5(.124) .5(.124) .124
To find the variance of the portfolio return, we must use a new rule. In general, if x and y have a
nonzero covariance s2xy, and a and b are constants, then
s2(axby) a2s2x b2s2y 2abs2xy
Therefore
s2P s2(.5x.5y) (.5)2s2x (.5)2s2y 2(.5)(.5)s2xy
(.5)2(.0454) (.5)2(.0681) 2(.5)(.5)(.0318) .012475
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and
sP 1.012475 .1117
Note that, since mP .124 equals mx .124 and my .124, the portfolio has the same expected
return as either stock A or B. However, since sP .1117 is less than sx .2131 and sy .2610,
the portfolio is a less risky investment. In other words, diversification can reduce risk. Note,
however, that the reason that sP is less than sx and sy is that s2xy .0318 is negative. Intuitively, this says that the two stocks tend to balance each other’s returns. However, if the covariance between the returns of two stocks is positive, sP can be larger than sx and/or sy. The reader
will demonstrate this in Exercise D.3.
Finally, note that a measure of linear association between x and y that is unitless and always
between 1 and 1 is the correlation coefficient, denoted r. We define r to be s 2xy divided by
(sx)(sy). For the stock return example, r equals (.0318)((.2131)(.2610)) .5717.
Exercises for Appendix B
B.1 The gross profit (price minus unit variable cost) for each computer sold by a company is $500. The
company’s fixed cost is $5,000,000 per year. The number of computers sold per year is a random
variable having mean 15,000 and standard deviation 2,000. Let y denote the company’s annual
profit. Find my and sy. Then, use Chebyshev’s Theorem to find an interval containing at least
75 percent of the annual profits that might be obtained.
B.2 A product is manufactured on three different assembly lines that operate independently. The mean
and standard deviation of the hourly production (in units produced) for each assembly line are as
follows: (1) for assembly line 1, m1 35 and s1 4; (2) for assembly line 2, m2 25 and s2 2;
(3) for assembly line 3, m3 40 and s3 4. Let T denote the total hourly production on all three
assembly lines.
a Find mT and sT.
b Assuming that total hourly production is normally distributed, find an interval that contains
99.73 percent of the possible hourly production totals.
c Each unit of the product requires two 34" bolts for assembly. Use your result of part b to
estimate the hourly supply of bolts needed in order to be very certain that the assembly lines
will not run short of bolts during the hour.
B.3 Let x be the yearly proportional return for stock C, and let y be the yearly proportional return for
stock D. If mx .11, my .09, sx .17, sy .17, and s2xy .0412, find the mean and standard
deviation of the portfolio return P .5x .5y. Discuss the risk of the portfolio.
B.4 Below we give what is called a joint probability table for two utility bonds where the random
variable x represents the percentage return for bond 1 and the random variable y represents the
percentage return for bond 2.
8
9
x
10
11
12
p( y)
8
9
10
11
12
.03
.04
.02
.00
.00
.04
.06
.08
.04
.00
.03
.06
.20
.06
.03
.00
.04
.08
.06
.04
.00
.00
.02
.04
.03
.10
.20
.40
.20
.10
p(x)
.09
.22
.38
.22
.09
y
Source: David K. Hildebrand and Lyman Ott, Statistical Thinking for Managers,
2nd edition (Boston, Ma: Duxbury Press, 1987), p. 101.
In this table, probabilities associated with values of x are given in the row labeled p(x) and probabilities associated with values of y are given in the column labeled p(y). For example, P(x 9) .22
and P(y 11) .20. The entries inside the body of the table are joint probabilities—for instance,
the probability that x equals 9 and y equals 10 is .08. Use the table to do the following:
a Calculate mx, sx, my, and sy.
b Calculate s2xy, the covariance between x and y.
c Calculate the variance and standard deviation of a portfolio in which 50 percent of the money
is used to buy bond 1 and 50 percent is used to buy bond 2. That is, find s2P and sP, where
P .5x .5y. How does the portfolio’s risk compare to the risk associated with investing only
in bond 1? Only in bond 2?
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