Devore J.L., Berk K.N. Modern Mathematical Statistics with Applications
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the straight line passing through the value of the fixed variable, and finally
integrating over all possible values of the fixed variable. Thus
ð
1
1
ð
1
1
f ð x; yÞdy dx ¼
ðð
D
f ð x; yÞdy dx ¼
ð
1
0
ð
1x
0
24xy dy
dx
¼
ð
1
0
24x
y
2
2
y¼1x
y¼0
()
dx ¼
ð
1
0
12xð1 xÞ
2
dx ¼ 1
To compute the probability that the two types of nuts together make up at most
50% of the can, let A ¼fðx; yÞ : 0 x 1; 0 y 1; and x þ y :5g; as shown
in Figure 5.3. Then
P½ðX; YÞ2AÞ¼
ðð
A
f ðx; yÞdx dy ¼
ð
:5
0
ð
:5x
0
24xy dy dx ¼ :0625
The marginal pdf for almonds is obtained by holding X fixed at x and integrating
f(x, y) along the vertical line through x:
f
X
ðxÞ¼
ð
1
1
f ð x; yÞdy ¼
Ð
1x
0
24xy dy ¼ 12xð1 xÞ
2
0 x 1
0 otherwise
By symmetry of f(x, y) and the region D, the marginal pdf of Y is obtained by
replacing x and X in f
X
(x)byy and Y, respectively. ■
Independent Random Variables
In many situations, information about the observed value of one of the two vari-
ables X and Y gives information about the value of the other variable. In Example 5.1,
the marginal probability of X at x ¼ 250 was .5, as was the probability that X ¼ 100.
If, however, we are told that the selected individual had Y ¼ 0, then X ¼ 10 0 is
fourtimesaslikelyasX ¼ 250. Thus there is a dependence between the two
variables.
x 1.50
1
.5
0
y = .5 − x
A = Shaded region
x
+y = 1
x
+y = .5
Figure 5.3 Computing
P
[(
X
,
Y
) 2
A
] for Example 5.5
238
CHAPTER 5 Joint Probability Distributions
In Chapter 2 we pointed out that one way of defining independence of two
events is to say that A and B are independent if PðA \ BÞ¼PðAÞPðBÞ . Here is an
analogous definition for the independence of two rv’s.
DEFINITION
Two random variables X and Y are said to be independent if for every pair of
x and y values,
px; yðÞ¼p
X
ðxÞp
Y
ðyÞ when X and Y are discrete
or (5.1)
fx; y
ðÞ
¼ f
X
ðxÞf
Y
ðyÞ when X and Y are continuous
If (5.1) is not satisfied for all (x, y), then X and Y are said to be dependent.
The definition says that two variables are independent if their joint pmf or pdf is the
product of the two marginal pmf’s or pdf’s.
Example 5.6 In the insurance situation of Examples 5.1 and 5.2,
p 100; 100ðÞ¼:10 6¼ :5ðÞ:25ðÞ¼p
X
100ðÞp
Y
100ðÞ
so X and Y are not independent. Independence of X and Y requires that every entry
in the joint probability table be the product of the corresponding row and column
marginal probabilities.
■
Example 5.7
(Example 5.5
continued)
Because f(x, y) in the nut scenario has the form of a product, X and Y would appear to
be independent. However, although f
X
3
4
¼ f
Y
3
4
¼
9
16
; f
3
4
;
3
4
¼ 0 6¼
9
16
9
16
so the
variables are not in fact independent. To be independent, f(x, y) must have the form
g(x) ·h(y) and the region of positive density must be a rectangle whose sides are
parallel to the coordinate axes.
■
Independence of two random variables is most useful when the description of
the experiment under study tells us that X and Y have no effect on each other. Then
once the marginal pmf’s or pdf’s have been specified, the joint pmf or pdf is simply
the product of the two marginal functions. It follows that
Pa X b; c Y dðÞ¼Pa X bðÞPc Y dðÞ
Example 5.8 Sup pose that the lifet imes of two components are independent of each other and
that the first lifetime, X
1
, has an exponential distribution with parameter l
1
whereas
the second, X
2
, has an exponential distribution with parameter l
2
. Then the joint
pdf is
f ð x
1
; x
2
Þ¼f
X
1
ðx
1
Þf
X
2
ðx
2
Þ
¼
l
1
e
l
1
x
1
l
2
e
l
2
x
2
¼ l
1
l
2
e
l
1
x
1
l
2
x
2
x
1
> 0; x
2
> 0
0 otherwise
(
5.1 Jointly Distributed Random Variables 239
Let l
1
¼ 1/1000 and l
2
¼ 1/1200, so that the expected lifetimes are 1000 h and
1200 h, respectively. The probability that both component lifetimes are at least
1500 h is
Pð1500 X
1
; 1500 X
2
Þ¼Pð1500 X
1
ÞPð1500 X
2
Þ
¼ e
l
1
ð1500Þ
e
l
2
ð1500Þ
¼ð:2231Þð:2865Þ¼:0639
■
More than Two Random Variables
To model the joint behavior of more than two random variables, we extend the
concept of a joint distribution of two variables.
DEFINITION
If X
1
, X
2
, ..., X
n
are all discrete random variables, the joint pmf of the
variables is the function
px
1
; x
2
; ...; x
n
ðÞ¼PX
1
¼ x
1
; X
2
¼ x
2
; ...; X
n
¼ x
n
ðÞ
If the variables are continuo us, the joint pdf of X
1
; X
2
; ...; X
n
is the function
f ð x
1
; x
2
; ...; x
n
Þ such that for any n intervals ½a
1
; b
1
; ...;½a
n
; b
n
;
Pða
1
X
1
b
1
; ...; a
n
X
n
b
n
Þ¼
ð
b
1
a
1
...
ð
b
n
a
n
f ð x
1
; ...; x
n
Þdx
n
...dx
1
In a binomial experiment, each trial could result in one of only two possible
outcomes. Consider now an experiment consisting of n independent and identical
trials, in which each trial can result in any one of r possible outcomes. Let p
i
¼
P(outcome i on any particular trial), and define random variables by X
i
¼ the
number of trials resulting in outcome i (i ¼ 1, ..., r). Such an experiment is called a
multinomial experiment, and the joint pmf of X
1
, ..., X
r
is called the multinomial
distribution. By using a counting argument analogous to the one used in deriving
the binomial distribution, the joint pmf of X
1
, ..., X
r
can be shown to be
pðx
1
;...;x
r
Þ
¼
n!
ðx
1
!Þðx
2
!Þðx
r
!Þ
p
x
1
1
p
x
r
r
; x
i
¼0;1;2; ...; with x
1
þþx
r
¼n
0 otherwise
8
>
<
>
:
The case r ¼ 2 gives the binomial distribution, with X
1
¼ number of successes and
X
2
¼ n X
1
¼ number of failures.
In the case r ¼ 3, the leading part of the expression for the joint pmf comes
from the number of ways of choosing x
1
of the n trials to be outcomes of the
first type and then x
2
of the remaining n x
1
trials to be outcomes of the second
type:
n
x
1
n x
1
x
2
¼
n!
x
1
!ðn x
1
Þ!
ðn x
1
Þ!
x
2
!ðn x
1
x
2
Þ!
¼
n!
x
1
!x
2
!ðn x
1
x
2
Þ!
¼
n!
x
1
!x
2
!x
3
!
:
240 CHAPTER 5 Joint Probability Distributions
Example 5.9 If the allele of each of ten independently obtained pea sections is determined and p
1
¼P(AA), p
2
¼P(Aa), p
3
¼P(aa) , X
1
¼number of AA’s, X
2
¼number of Aa’s, and
X
3
¼ number of aa’s, then
pðx
1
;x
2
;x
3
Þ¼
10!
ðx
1
!Þðx
2
!Þðx
3
!Þ
p
x
1
1
p
x
2
2
p
x
3
3
; x
i
¼0;1;2;... and x
1
þx
2
þx
3
¼10
If p
1
¼ p
3
¼ .25, p
2
¼ .5, then
PðX
1
¼ 2; X
2
¼ 5; X
3
¼ 3Þ¼pð2; 5; 3Þ¼
10!
2!5!3!
:25
2
:50
5
:25
3
¼ :0769
■
Example 5.10 When a certain method is used to collect a fixed volume of rock samples in a
region, there are four resulting rock types. Let X
1
, X
2
, and X
3
denote the proportion
by volume of rock types 1, 2, and 3 in a randomly selected sample (the proportion
of rock type 4 is 1 X
1
X
2
X
3
, so a variable X
4
would be redundant). If the joint
pdf of X
1
, X
2
, X
3
is
f ð x
1
;x
2
;x
3
Þ¼
kx
1
x
2
ð1 x
3
Þ
0
8
>
>
<
>
>
:
0 x
1
1; 0 x
2
1; 0 x
3
1;
x
1
þx
2
þx
3
1
otherwise
then k is determined by
1 ¼
ð
1
1
ð
1
1
ð
1
1
f ð x
1
; x
2
; x
3
Þdx
3
dx
2
dx
1
¼
ð
1
0
ð
1x
1
0
ð
1x
1
x
2
0
kx
1
x
2
ð1 x
3
Þdx
3
dx
2
dx
1
This iterated integral has value k/144, so k ¼ 144. The probability that rocks of
types 1 and 2 toge ther account for at most 50% of the sample is
PðX
1
þ X
2
:5Þ¼
ððð
0 x
i
1 for i ¼ 1; 2; 3
x
1
þ x
2
þ x
3
1; x
1
þ x
2
:5
f ð x
1
; x
2
; x
3
Þdx
3
dx
2
dx
1
¼
ð
:5
0
ð
:5x
1
0
ð
1x
1
x
2
0
144x
1
x
2
ð1 x
3
Þdx
3
dx
2
dx
1
¼ :6066 ■
The notion of independence of more than two random variables is similar to
the notion of independen ce of more than two events.
DEFINITION
The random variables X
1
; X
2
; :::; X
n
are said to be independent if for every
subset X
i
1
; X
i
2
; ...; X
i
k
of the variables (each pair, each triple, and so on), the joint
pmf or pdf of the subset is equal to the product of the marginal pmf’s or pdf’s.
5.1 Jointly Distributed Random Variables 241
Thus if the variables are independent with n ¼ 4, then the joint pmf or pdf of any two
variables is the product of the two marginals, and similarly for any three variables
and all four variables together. Most important, once we are told that n variables are
independent, then the joint pmf or pdf is the product of the n marginals.
Example 5.11 If X
1
, ..., X
n
represent the lifetimes of n components, the components operate
independently of each other, and each lifetime is expone ntially distributed with
parameter l, then
f ð x
1
; x
2
; ...; x
n
Þ¼ le
lx
1
le
lx
2
le
lx
n
¼
l
n
e
lSx
i
x
1
0; x
2
0; ...; x
n
0
0 otherwise
(
If these n components are connected in series, so that the system will fail
as soon as a single component fails, then the probability that the system lasts past
time t is
PðX
1
> t; ...; X
n
> tÞ¼
ð
1
t
...
ð
1
t
f ð x
1
; ...; x
n
Þdx
1
...dx
n
¼
ð
1
t
le
lx
1
dx
1
ð
1
t
le
lx
n
dx
n
¼ e
lt
n
¼ e
nlt
Therefore,
Pðsystem lifetime tÞ¼1 e
nlt
for t 0
which shows that system lifetime has an exponential distribution with parameter nl;
the expected value of system lifetime is 1/ nl .
■
In many experimental situations to be consi dered in this book, independence
is a reasonable assumption, so that specifying the joint distribution reduces to
deciding on appropriate marginal distributions.
Exercises Section 5.1 (1–17)
1. A service station has both self-service and full-
service islands. On each island, there is a single
regular unleaded pump with two hoses. Let X
denote the number of hoses being used on the
self-service island at a particular time, and let Y
denote the number of hoses on the full-service
island in use at that time. The joint pmf of X and
Y appears in the accompanying tabulation.
y
p(x, y)012
0 .10 .04 .02
x 1 .08 .20 .06
2 .06 .14 .30
a. What is PðX ¼ 1 and Y ¼ 1Þ?
b. Compute P ð X 1 and Y 1Þ:
c. Give a word description of the event
fX 6¼ 0 and Y 6¼ 0g; and compute the proba-
bility of this event.
d. Compute the marginal pmf of X and of Y.
Using p
X
(x), what is PðX 1Þ?
e. Are X and Y independent rv’s? Explain.
2. When an automobile is stopped by a roving safety
patrol, each tire is checked for tire wear, and each
headlight is checked to see whether it is properly
aimed. Let X denote the number of headlights that
need adjustment, and let Y denote the number of
defective tires.
a. If X and Y are independent with p
X
ð0Þ¼:5;
p
X
1ðÞ¼:3; p
X
2ðÞ¼:2; and p
Y
ð0Þ¼:6; p
Y
1ðÞ
¼ :1; p
Y
2ðÞ¼p
Y
3ðÞ¼:05; p
Y
4ðÞ¼:2; display
the joint pmf of (X, Y) in a joint probability table.
242
CHAPTER 5 Joint Probability Distributions
b. Compute PðX 1 and Y 1Þ from the joint
probability table, and verify that it equals the
product PðX 1ÞPðY 1Þ
c. What is PðX þ Y ¼ 0Þ (the probability of no
violations)?
d. Compute P ð X þ Y 1Þ
3. A market has both an express checkout line and a
superexpress checkout line. Let X
1
denote the num-
ber of customers in line at the express checkout at a
particular time of day, and let X
2
denote the number
of customers in line at the superexpress checkout at
the same time. Suppose the joint pmf of X
1
and X
2
is as given in the accompanying table.
x
2
0123
0 .08 .07 .04 .00
1 .06 .15 .05 .04
x
1
2 .05 .04 .10 .06
3 .00 .03 .04 .07
4 .00 .01 .05 .06
a. What is PðX
1
¼1; X
2
¼1Þ; that is, the probabil-
ity that there is exactly one customer in each line?
b. What is PðX
1
¼ X
2
Þ; that is, the probability
that the numbers of customers in the two lines
are identical?
c. Let A denote the event that there are at least two
more customers in one line than in the other
line. Express A in terms of X
1
and X
2
, and
calculate the probability of this event.
d. What is the probability that the total number of
customers in the two lines is exactly four? At
least four?
e. Determine the marginal pmf of X
1
, and then
calculate the expected number of customers in
line at the express checkout.
f. Determine the marginal pmf of X
2
.
g. By inspection of the probabilities PðX
1
¼ 4Þ;
PðX
2
¼ 0Þ; and PðX
1
¼ 4; X
2
¼ 0Þ; are X
1
and X
2
independent random variables? Explain.
4. According to the Mars Candy Company, the long-
run percentages of various colors of M&M milk
chocolate candies are as follows:
Blue:
24%
Orange:
20%
Green:
16%
Yellow:
14%
Red:
13%
Brown:
13%
a. In a random sample of 12 candies, what is the
probability that there are exactly two of each
color?
b. In a random sample of 6 candies, what is the
probability that at least one color is not included?
c. In a random sample of 10 candies, what is the
probability that there are exactly 3 blue candies
and exactly 2 orange candies?
d. In a random sample of 10 candies, what is the
probability that there are at most 3 orange
candies? [Hint: Think of an orange candy as a
success and any other color as a failure.]
e. In a random sample of 10 candies, what is the
probability that at least 7 are either blue,
orange, or green?
5. The number of customers waiting for gift-wrap ser-
vice at a department store is an rv X with possible
values 0, 1, 2, 3, 4 and corresponding probabilities
.1, .2, .3, .25, .15. A randomly selected customer will
have 1, 2, or 3 packages for wrapping with prob-
abilities .6, .3, and .1, respectively. Let Y ¼ the total
number of packages to be wrapped for the customers
waiting in line (assume that the number of packages
submitted by one customer is independent of the
number submitted by any other customer).
a. Determine PðX ¼ 3; Y ¼ 3Þ; that is, p(3, 3).
b. Determine p(4, 11).
6. Let X denote the number of Canon digital cameras
sold during a particular week by a certain store.
The pmf of X is
x 0123 4
p
X
(x) .1 .2 .3 .25 .15
Sixty percent of all customers who purchase these
cameras also buy an extended warranty. Let Y
denote the number of purchasers during this
week who buy an extended warranty.
a. What is PðX ¼ 4; Y ¼ 2Þ? [Hint: This proba-
bility equals PðY ¼ 2jX ¼ 4ÞPðX ¼ 4Þ; now
think of the four purchases as four trials of a
binomial experiment, with success on a trial
corresponding to buying an extended war-
ranty.]
b. Calculate PðX ¼ YÞ
c. Determine the joint pmf of X and Y and then the
marginal pmf of Y.
7. The joint probability distribution of the number X
of cars and the number Y of buses per signal cycle
at a proposed left-turn lane is displayed in the
accompanying joint probability table.
y
p(x, y) 0 1 2
x
0 .025 .015 .010
1 .050 .030 .020
2 .125 .075 .050
3 .150 .090 .060
4 .100 .060 .040
5 .050 .030 .020
5.1 Jointly Distributed Random Variables 243
a. What is the probability that there is exactly one
car and exactly one bus during a cycle?
b. What is the probability that there is at most one
car and at most one bus during a cycle?
c. What is the probability that there is exactly one
car during a cycle? Exactly one bus?
d. Suppose the left-turn lane is to have a capacity
of five cars, and one bus is equivalent to three
cars. What is the probability of an overflow
during a cycle?
e. Are X and Y independent rv’s? Explain.
8. A stockroom currently has 30 components of a
certain type, of which 8 were provided by supplier
1, 10 by supplier 2, and 12 by supplier 3. Six of
these are to be randomly selected for a particular
assembly. Let X ¼ the number of supplier 1’s
components selected, Y ¼ the number of supplier
2’s components selected, and p(x, y) denote the
joint pmf of X and Y.
a. What is p(3, 2)? [Hint: Each sample of size 6 is
equally likely to be selected. Therefore, p(3, 2)
¼ (number of outcomes with X ¼ 3 and Y ¼ 2)/
(total number of outcomes). Now use the prod-
uct rule for counting to obtain the numerator
and denominator.]
b. Using the logic of part (a), obtain p(x, y). (This
can be thought of as a multivariate hypergeo-
metric distribution – sampling without replace-
ment from a finite population consisting of
more than two categories.)
9. Each front tire of a vehicle is supposed to be filled
to a pressure of 26 psi. Suppose the actual air
pressure in each tire is a random variable X
for the right tire and Y for the left tire, with
joint pdf
f ðx; yÞ¼
Kðx
2
þ y
2
Þ 20 x 30; 20 y 30
0 otherwise
(
a. What is the value of K?
b. What is the probability that both tires are
underfilled?
c. What is the probability that the difference in air
pressure between the two tires is at most 2 psi?
d. Determine the (marginal) distribution of air
pressure in the right tire alone.
e. Are X and Y independent rv’s?
10. Annie and Alvie have agreed to meet between
5:00 p.m. and 6:00 p.m. for dinner at a local
health-food restaurant. Let X ¼ Annie’s arrival
time and Y ¼ Alvie’s arrival time. Suppose X
and Y are independent with each uniformly
distributed on the interval [5, 6].
a. What is the joint pdf of X and Y?
b. What is the probability that they both arrive
between 5:15 and 5:45?
c. If the first one to arrive will wait only 10 min
before leaving to eat elsewhere, what is the
probability that they have dinner at the health-
food restaurant? [Hint: The event of interest is
A ¼ðx; yÞ : jx yj
1
6
.]
11. Two different professors have just submitted final
exams for duplication. Let X denote the number of
typographical errors on the first professor’s exam
and Y denote the number of such errors on the
second exam. Suppose X has a Poisson distribu-
tion with parameter l, Y has a Poisson distribution
with parameter y, and X and Y are independent.
a. What is the joint pmf of X and Y?
b. What is the probability that at most one error is
made on both exams combined?
c. Obtain a general expression for the probability
that the total number of errors in the two exams
is m (where m is a nonnegative integer). [Hint:
A ¼fðx; yÞ : x þy ¼ mg¼fðm; 0Þ; ðm 1; 1Þ;
:::; ð1;m 1Þ; ð0; mÞg. Now sum the joint pmf
over (x, y) 2 A and use the binomial theorem,
which says that
X
m
k¼0
m
k
a
k
b
mk
¼ða þ bÞ
m
for any a, b.]
12. Two components of a computer have the follow-
ing joint pdf for their useful lifetimes X and Y:
f ðx; yÞ¼
xe
xð1þyÞ
x 0 and y 0
0 otherwise
a. What is the probability that the lifetime X of
the first component exceeds 3?
b. What are the marginal pdf’s of X and Y? Are
the two lifetimes independent? Explain.
c. What is the probability that the lifetime of at
least one component exceeds 3?
13. You have two lightbulbs for a particular lamp. Let
X ¼ the lifetime of the first bulb and Y ¼ the
lifetime of the second bulb (both in 1000’s of
hours). Suppose that X and Y are independent
and that each has an exponential distribution
with parameter l ¼ 1.
a. What is the joint pdf of X and Y?
b. What is the probability that each bulb lasts at
most 1000 h (i.e., X 1 and Y 1)?
c. What is the probability that the total lifetime
of the two bulbs is at most 2? [Hint: Draw a
picture of the region A ¼fðx; yÞ : x 0;
y 0; x þ y 2g before integrating.]
244
CHAPTER 5 Joint Probability Distributions
d. What is the probability that the total lifetime is
between 1 and 2?
14. Suppose that you have ten lightbulbs, that the
lifetime of each is independent of all the other
lifetimes, and that each lifetime has an exponen-
tial distribution with parameter l.
a. What is the probability that all ten bulbs fail
before time t?
b. What is the probability that exactly k of the ten
bulbs fail before time t?
c. Suppose that nine of the bulbs have lifetimes
that are exponentially distributed with parame-
ter l and that the remaining bulb has a lifetime
that is exponentially distributed with parameter
y (it is made by another manufacturer). What is
the probability that exactly five of the ten bulbs
fail before time t?
15. Consider a system consisting of three components
as pictured. The system will continue to function
as long as the first component functions and
either component 2 or component 3 functions.
Let X
1
, X
2
, and X
3
denote the lifetimes of compo-
nents 1, 2, and 3, respectively. Suppose the
X
i
’s are independent of each other and each
X
i
has an exponential distribution with
parameter l.
1
3
2
a. Let Y denote the system lifetime. Obtain
the cumulative distribution function of Y
and differentiate to obtain the pdf. [Hint:
FðyÞ¼PðY yÞ; express the event fY yg
in terms of unions and/or intersections of
the three events fX
1
yg; fX
2
yg; and
fX
3
yg:]
b. Compute the expected system lifetime.
16. a. For f ðx
1
; x
2
; x
3
Þ as given in Example 5.10,
compute the joint marginal density function
of X
1
and X
3
alone (by integrating over x
2
).
b. What is the probability that rocks of types 1
and 3 together make up at most 50% of the
sample? [Hint: Use the result of part (a).]
c. Compute the marginal pdf of X
1
alone. [Hint:
Use the result of part (a).]
17. An ecologist selects a point inside a circular sam-
pling region according to a uniform distribution.
Let X ¼ the x coordinate of the point selected and
Y ¼ the y coordinate of the point selected. If the
circle is centered at (0, 0) and has radius R, then
the joint pdf of X and Y is
f ðx; yÞ¼
1
pR
2
x
2
þ y
2
R
2
0 otherwise
8
<
:
a. What is the probability that the selected point is
within R/2 of the center of the circular region?
[Hint: Draw a picture of the region of positive
density D. Because f(x,y) is constant on D,
computing a probability reduces to computing
an area.]
b. What is the probability that both X and Y differ
from 0 by at most R/2?
c. Answer part (b) for R=
ffiffiffi
2
p
replacing R=2
d. What is the marginal pdf of X?OfY? Are X and
Y independent?
5.2
Expected Values, Covariance,
and Correlation
We previously saw that any function h(X) of a single rv X is itself a random
variable. However, to compute E[h(X)], it was not necessary to obtain the proba-
bility distribution of h(X); instead, E[h(X )] was computed as a weighted average of
h(X) values, where the weight function was the pmf p(x ) or pdf f(x)ofX. A similar
result holds for a function h(X, Y) of two jointly distributed random variables.
PROPOSITION
Let X and Y be jointly distributed rv’s with pmf p(x, y) or pdf f(x, y) according
to whether the variables are discrete or continuous. Then the expected value
of a function h(X, Y), denoted by E½hðX; YÞ or m
h
ðX;YÞ
is given by
5.2 Expected Values, Covariance, and Correlation 245
E½hðX; YÞ ¼
X
x
X
y
hðx; yÞpðx; yÞ if X and Y are discrete
ð
1
1
ð
1
1
hðx; yÞf ðx; yÞdx dy if X and Y are continuous
8
>
>
<
>
>
:
Example 5.12 Five friends have purchased tickets to a concert. If the tickets are for seats 1–5 in a
particular row and the tickets are randomly distributed among the five, what is the
expected number of seats separating any particular two of the five? Let X and Y
denote the seat numbers of the first and secon d indivi duals, respectively. Possible
(X, Y) pairs are 1; 2ðÞ; 1; 3ðÞ; :::; 5; 4ðÞfg; and the joint pmf of (X, Y)is
pðx; yÞ¼
1
20
x ¼ 1; ...; 5; y ¼ 1; ...; 5; x 6¼ y
0 otherwise
8
>
<
>
:
The number of seats separating the two individuals is hðX; YÞ¼jX Yj1:
.
The accompanying table gives h(x, y) for each possible (x, y) pair.
x
h(x, y) 1 2345
1– 0123
20 –012
y 31 0–01
42 10–0
53 210–
Thus
E½hðX; YÞ ¼
X
ðx;
X
yÞ
hðx; yÞpðx; yÞ¼
X
5
x¼1
X
5
y¼1
x6¼y
ðjx yj1Þ
1
20
¼ 1
■
Example 5.13 In Example 5.5, the joint pdf of the amount X of almonds and amount Y of cashews
in a 1 lb can of nuts was
f ð x; yÞ¼
24xy 0 x 1; 0 y 1; x þ y 1
0 otherwise
(
If 1 lb of almonds costs the company $2.00, 1 lb of cashews costs $3.00, and 1 lb of
peanuts costs $1.00, then the total cost of the contents of a can is
hðX; YÞ¼2X þ 3Y þ 1ð1 X YÞ¼1 þ X þ 2Y
(since 1 X Y of the weight consists of peanuts). The expected total cost is
E½hðX; YÞ ¼
ð
1
1
ð
1
1
hðx; yÞf ðx; yÞdx dy
¼
ð
1
0
ð
1x
0
ð1 þ x þ 2yÞ24xy dy dx ¼ $2:20
■
246 CHAPTER 5 Joint Probability Distributions
The method of computing the expected value of a function hðX
1
; :::; X
n
Þ of
n random variables is similar to that for two random variables. If the X
i
’s are
discrete, E½hðX
1
; :::; X
n
Þ is an n-dimensional sum; if the X
i
’s are continuous, it is an
n-dimensional integral.
When h(X, Y) is a product of a function of X and a function of Y, the expected
value simplifies in the case of independence. In particular, let X and Y be continu-
ous independent random variables and suppose h(X, Y) ¼ XY. Then
EðXYÞ¼
ð
1
1
ð
1
1
xyfðx; yÞdx dy ¼
ð
1
1
ð
1
1
xyf
X
ðxÞ f
Y
ðyÞdx dy
¼
ð
1
1
yf
Y
ðyÞ½
ð
1
1
xf
X
ðxÞdx dy ¼ EðXÞEðYÞ
The discrete case is similar. More generally, essentially the same derivation works
for several functions of random variables.
PROPOSITION
Let X
1
, X
2
, ..., X
n
be independent random variables and assume that the
expected valu es of h
1
X
1
ðÞ; h
2
X
2
ðÞ; ...; h
n
X
n
ðÞall exist. Then
Eh
1
X
1
ðÞh
2
X
2
ð Þh
n
X
n
ðÞ½¼Eh
1
X
1
ðÞ½Eh
2
X
2
ðÞ½ Eh
n
X
n
ðÞ½
Covariance
When two random variables X and Y are not independent, it is frequently of interest
to assess how strongly they are related to each other.
DEFINITION
The covariance between two rv’s X and Y is
CovðX;YÞ¼E½ðX m
X
ÞðY m
Y
Þ
¼
X
x
X
y
ðx m
X
Þðy m
Y
Þpðx;yÞ if X and Y are discrete
ð
1
1
ð
1
1
ðx m
X
Þðy m
Y
Þf ðx;yÞdx dy if X and Y are continuous
8
>
>
>
<
>
>
>
:
The rationale for the definition is as follows. Suppose X and Y have a strong
positive relations hip to each other, by which we mean that large values of X tend
to occur with large values of Y and small values of X with small values of Y.
Then most of the probability mass or density will be associated with (x m
X
) and
(y m
Y
) either both positive (both X and Y above their respective means) or both
negative, so the product (x m
X
)(y m
Y
) will tend to be positive. Thus for a strong
positive relationship, Cov(X, Y) should be quite positive. For a strong negative
relationship, the signs of (x m
X
) and (y m
Y
) will tend to be opposite, yielding a
5.2 Expected Values, Covariance, and Correlation 247