Devore J.L., Berk K.N. Modern Mathematical Statistics with Applications

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negative product. Thus for a strong negative relationship, Cov(X , Y) should be quite

negative. If X and Y are not strongly related, positive and negative products

will tend to cancel each other, yielding a covariance near 0. Figure 5.4 illustrates

the different possibilities. The covariance depends on both the set of possible pairs

and the probabilities. In Figure 5.4, the probabilities could be change d without

altering the set of possible pairs, and this could drastically change the value of

Cov(X, Y).

Example 5.14 The joint and marginal pmf’s for X ¼ automobile policy deductible amount and

Y ¼ homeowner policy deductible amount in Example 5.1 were

p(x, y)

0 100 200 x 100 250 y 0 100 200

100 .20 .10 .20 p

(x).5 .5p

(y) .25 .25 .50

250 .05 .15 .30

from which m

¼ Sxp

ðxÞ¼175 and m

¼ 125. Therefore,

CovðX;YÞ¼

ðx;

yÞ

ðx 175Þðy 125Þpðx;yÞ

¼ð100 175Þð0 125 Þð: 20Þþþð250 175Þð200 125Þð:30Þ

¼1875

■

The following shortcut formula for Cov(X, Y) simplifies the computations.

PROPOSITION

CovðX; YÞ¼EXYðÞm

 m

According to this formula, no intermediate subtractions are necessary; only at the

end of the computation is m

· m

subtracted from E(XY). The proof involves

expanding ðX  m

ÞðY  m

Þ and then taking the expected value of each term

separately. Note that Cov ðX; XÞ¼EðX

Þm

¼ VðXÞ.

abc

Figure 5.4 p(x,y) ¼

for each of ten pairs corresponding to indicated points;

(a) positive covariance; (b) negative covariance; (c) covariance near zero

248

CHAPTER 5 Joint Probability Distributions

Example 5.15

(Example 5.5

continued)

The joint and marginal pdf’s of X ¼ amount of almonds and Y ¼amount of cashews

were

f ð x; yÞ¼

24xy 0  x  1; 0  y  1; x þ y  1

0 otherwise

(

ðxÞ¼

12xð1  xÞ

0  x  1

0 otherwise



with f

(y) obta ined by replacing x by y in f

(x). It is easily verified that

¼ m

, and

EðXYÞ¼

1

xyf ðx; yÞdx dy ¼

1x

xy  24xy dy dx

¼ 8

ð1  xÞ

dx ¼

Thus CovðX; YÞ¼







¼

. A negative covariance is reason-

able here because more almonds in the can implies fewer cashews.

■

The covariance satisfies a useful linearity prope rty (Exercise 33).

PROPOSITION

If X, Y, and Z are rv’s and a and b are constants then

Cov aX þ bY; ZðÞ¼a Cov X; ZðÞþb Cov Y; ZðÞ

It would appear that the relationship in the insurance example is quite strong

since Cov(X, Y) ¼ 1875, whereas in the nut example CovðX; YÞ¼

would seem

to imply quite a weak relationship. Unfortunately, the covariance has a serious

defect that makes it imposs ible to interpret a computed value of the covariance. In

the insurance example, suppose we had expressed the deductible amount in cents

rather than in dollars. Then 100X would replace X, 100Y would replace Y, and

the resulting covariance would be Cov(100X, 100Y) ¼ (100)(100)Cov(X, Y) ¼

18,750,000. If, on the other hand, the deductible amount had been expressed in

hundreds of dollars, the computed covariance would have been (.01)(.01)(1875) ¼

.1875. The defect of covariance is that its computed value depends critically on the

units of measurement. Ideally, the choice of units should have no effect on a

measure of strength of relationship. This is achieved by scaling the covariance.

Correlation

DEFINITION

The correlation coefficient of X and Y, denoted by Corr(X, Y), or r

X,Y

or just r, is defined by

X;Y

CovðX; YÞ

 s

5.2 Expected Values, Covariance, and Correlation 249

Example 5.16 It is easily verified that in the insurance problem of Example 5.14, EðX

Þ¼

36; 250; s

¼ 36; 250  175ðÞ

¼ 5625; s

¼ 75; EðY

Þ¼22; 500; s

¼ 6875;

and s

¼ 82:92: This gives

r ¼

1875

ð75Þð82:92Þ

¼ :301

■

The following proposition show s that r remedies the defect of Cov(X, Y) and also

suggests how to recognize the existence of a strong (linear) relations hip.

PROPOSITION

1. If a and c are either both positive or both negative,

Corr aX þ b; cY þ dðÞ¼Corr X; YðÞ

2. For any two rv’s X and Y; 1  Corr X ; YðÞ1

Statement 1 says precisely that the correlation coefficient is not affected by a linear

change in the units of measurement (if, say, X ¼ temperature in



C, then 9X/5 +

32 ¼ temperature in



F). According to Statement 2, the strongest possible positive

relationship is evidenced by r ¼ +1, whereas the strongest possible negative

relationship corresponds to r ¼1. The proof of the first statement is sketched

in Exercise 31, and that of the second appears in Exercise 35 and also Supplemen-

tary Exercise 76 at the end of the next chapter. For descriptive purposes, the

relationship will be described as strong if |r|  .8, moderate if .5 < |r| <.8, and

weak if |r|  .5.

If we think of p(x, y)orf(x, y) as prescribing a mathematical model for how

the two numerical variables X and Y are distributed in some population (height and

weight, verbal SAT score and quantitative SAT score, etc.), then r is a population

characteristic or parameter that measures how strongly X and Y are related in

the population. In Chapter 12, we will consider taking a sample of pairs

; y

ðÞ; :::; x

; y

ðÞfrom the population. The sample correlation coefficient r

will then be defined and used to make inferences about r.

The correlation coefficient r is actually not a completely general measure of

the strength of a relationship.

PROPOSITION

1. If X and Y are independent, then r ¼ 0, but r ¼ 0 does not imply

independence.

2. r ¼ 1or1 iff Y ¼ aX + b for some numbers a and b with a 6¼ 0.

Exercise 29 and Example 5.17 relate to Property 1, and Property 2 is investigated

in Exercises 32 and 35.

This propositi on says that r isameasureofthedegreeoflinear

relationship between X and Y, and only when the two variables are perfectly

250 CHAPTER 5 Joint Probability Distributions

related in a linear manner will r be as positive or negative as it can be. A r

less than 1 in absolute value indicates only that the relationship is not

completely linear, but there m ay still be a very strong nonlinear relation. Also,

r ¼ 0 doe s not im pl y t hat X and Y are independent, but only that there is complete

absence of a linear r elationship. When r ¼ 0, X and Y ar e said to be uncorrelated.

Two variables could be uncorrelated yet highly dependent because of a strong

nonlinear relationship, so be careful not to conclude too much from knowing

that r ¼ 0.

Example 5.17 Let X and Y be discrete rv’s with joint pmf

pðx; yÞ¼

ðx; yÞ¼ð4; 1Þ; ð4; 1Þ; ð2 ; 2Þ ; ð2; 2Þ

0 otherwise

The points that receive positive probability mass are identified on the (x, y)

coordinate system in Figure 5.5. It is evident from the figure that the value of X

is completely determined by the value of Y and vice versa, so the two variables

are completely dependent. However, by symmetry m

¼ m

¼ 0 and EðXYÞ¼

ð4Þ

þð4Þ

þð4Þ

¼ 0 so Cov X; YðÞ¼EXYðÞm

 m

¼ 0 and thus

X,Y

¼ 0. Although there is perfect dependence, there is also complete absence of

any linear relationship!

A value of r near 1 does not necessarily imply that increasing the value of

X causes Y to increase. It implies only that large X values are associated with

large Y values. For example, in the population of children, vocabulary size and

number of cavities are quite positively correlated, but it is certainly not true

that cavities cause vocabulary to grow. Instead, the values of both these variables

tend to increase as the value of age, a third variable, increases. For children of a

fixed age, there is probably a very low correlation between number of cavities

and vocabulary size. In summary, association (a high correlation) is not the same

as causation.

1234−1

−1

−2

−3−4

Figure 5.5 The population of pairs for Example 5.17 ■

5.2 Expected Values, Covariance, and Correlation 251

Exercises Section 5.2 (18–35)

18. An instructor has given a short quiz consisting of

two parts. For a randomly selected student, let X ¼

the number of points earned on the first part and

Y ¼ the number of points earned on the second

part. Suppose that the joint pmf of X and Y is given

in the accompanying table.

p(x, y)051015

0 .02 .06 .02 .10

5 .04 .15 .20 .10

10 .01 .15 .14 .01

a. If the score recorded in the grade book is the

total number of points earned on the two parts,

what is the expected recorded score EðX þ YÞ?

b. If the maximum of the two scores is recorded,

what is the expected recorded score?

19. The difference between the number of customers

in line at the express checkout and the number in

line at the superexpress checkout in Exercise 3 is

 X

. Calculate the expected difference.

20. Six individuals, including A and B, take seats

around a circular table in a completely random

fashion. Suppose the seats are numbered 1, ...,6.

Let X ¼ A’s seat number and Y ¼ B’s seat num-

ber. If A sends a written message around the table

to B in the direction in which they are closest, how

many individuals (including A and B) would you

expect to handle the message?

21. A surveyor wishes to lay out a square region

with each side having length L. However, because

of measurement error, he instead lays out a rect-

angle in which the north–south sides both have

length X and the east–west sides both have length

Y. Suppose that X and Y are independent and that

each is uniformly distributed on the interval

½L  A; L þ A (where 0 < A < L). What is the

expected area of the resulting rectangle?

22. Consider a small ferry that can accommodate cars

and buses. The toll for cars is $3, and the toll for

buses is $10. Let X and Y denote the number of

cars and buses, respectively, carried on a single

trip. Suppose the joint distribution of X and Y is as

given in the table of Exercise 7. Compute the

expected revenue from a single trip.

23. Annie and Alvie have agreed to meet for lunch

between noon (0:00 p.m.) and 1:00 p.m. Denote

Annie’s arrival time by X, Alvie’s by Y, and sup-

pose X and Y are independent with pdf’s

ðxÞ¼

0  x  1

0 otherwise



ðyÞ¼



0  y  1

otherwise

What is the expected amount of time that the one

who arrives first must wait for the other person?

[Hint: hðX; YÞ¼jX  Yj:]

24. Suppose that X and Y are independent rv’s with

moment generating functions M

(t) and M

(t),

respectively. If Z ¼ X þ Y, show that

ðtÞ¼M

ðtÞM

ðtÞ.[Hint: Use the proposition

on the expected value of a product.]

25. Compute the correlation coefficient r for X and Y

of Example 5.15 (the covariance has already been

computed).

26. a. Compute the covariance for X and Y in

Exercise 18.

b. Compute r for X and Y in the same exercise.

27. a. Compute the covariance between X and Y in

Exercise 9.

b. Compute the correlation coefficient r for this X

and Y.

28. Reconsider the computer component lifetimes

X and Y as described in Exercise 12. Determine

EXYðÞ. What can be said about CovðX; YÞ and r?

29. Use the proposition on the expected product to

show that when X and Y are independent,

CovðX; YÞ¼ CorrðX; YÞ¼0

30. a. Recalling the definition of s

for a single rv

X, write a formula that would be appropriate

for computing the variance of a function h(X, Y)

of two random variables. [Hint: Remember that

variance is just a special expected value.]

b. Use this formula to compute the variance of the

recorded score h(X, Y)[¼ max(X, Y)] in part

(b) of Exercise 18.

31. a. Use the rules of expected value to show that

CovðaX þ b; cY þ dÞ¼acCovðX; YÞ:

b. Use part (a) along with the rules of variance

and standard deviation to show that

CorrðaX þ b; cY þ dÞ¼ CorrðX; YÞ when a

and c have the same sign.

c. What happens if a and c have opposite signs?

252

CHAPTER 5 Joint Probability Distributions

32. Show that if Y ¼aX þb ða 6¼0Þ,thenCorrðX;YÞ¼

þ1or1. Under what conditions will r ¼þ1?

33. Show that if X, Y, and Z are rv’s and a and b are

constants, then Cov aX þ bY; ZðÞ¼a Cov X; ZðÞþ

b Cov Y; ZðÞ

34. Let Z

be the standardized X; Z

¼ðX  m

Þ=s

and let Z

be the standardized Y; Z

ðY  m

Þ=s

. Use Exercise 31 to show that

Corr X; YðÞ¼Cov Z

; Z

ðÞ¼EZ

ðÞ

35. Let Z

be the standardized X, Z

¼ (X  m

)/s

and let Z

be the standardized Y, Z

¼ (Y  m

)/s

a. Show with the help of Exercise 34 that

Ef½ðZ

 rZ

Þ

g¼1  r

b. Use part (a) to show that 1  r  1.

c. Use part (a) to show that r ¼ 1 implies that

Y ¼ aX þ b where a > 0, and r ¼1 implies

that Y ¼ aX þ b where a < 0.

5.3

Conditional Distributions

The distribution of Y can depend strongly on the value of another variable X. For

example, if X is height and Y is weight, the distribution of weight for men who are

6 ft tall is very different from the distribution of weight for short men. The

conditional distribution of Y given X ¼ x describes for each possible x how

probability is distributed over the set of possible y values. We define the condi tional

distribution of Y given X, but the conditional distribution of X given Y can be

obtained by just reversing the roles of X and Y. Both definitions are analogous to

that of the conditional probability P(A |B) as the ratio PðA \ BÞ=PðBÞ.

DEFINITION

Let X and Y be two discrete random variables with joint pmf p(x,y) and

marginal X pmf p

(x). Then for any x valu e such that p

(x) > 0, the

conditional probability mass function of Y given X ¼ x is

YjX

ðyjxÞ¼

pðx; yÞ

ðxÞ

An analogous formula holds in the continuous case. Let X and Y be two

continuous random variables with joint pdf f(x,y) and marginal X pdf f

(x).

Then for any x valu e such that f

(x) > 0, the conditional probability density

function of Y given X ¼ x is

YjX

ðyjxÞ¼

f ðx; yÞ

ðxÞ

Example 5.18 For a discrete example, reconsider Example 5.1, where X represents the deductible

amount on an automobile policy and Y represents the deductible amount on a

homeowner’s policy. Here is the joint distribution again.

p(x, y) 0 100 200

100 .20 .10 .20

250 .05 .15 .30

5.3 Conditional Distributions 253

The distribution of Y depends on X. In particular, let’s find the conditional

probability that Y is 200, given that X is 250, using the definition of conditional

probability from Section2.4.

PðY ¼ 200jX ¼ 250Þ¼

PðY ¼ 200 and X ¼ 250Þ

PðX ¼ 250Þ

:05 þ :15 þ :3

¼ :6

With our new definition we obtain the same result

YjX

ð200j250Þ¼

pð250; 200Þ

ð250Þ

:05 þ :15 þ :3

¼ :6

The conditional probabilities for the two other possible values of Y are

YjX

ð0j250Þ¼

pð250; 0Þ

ð250Þ

:05

:05 þ :15 þ :3

¼ :1

YjX

ð100j250Þ¼

pð250; 100Þ

ð250Þ

:15

:05 þ :15 þ :3

¼ :3

Thus, p

YjX

0j250ðÞþp

YjX

100j250ðÞþp

YjX

200j250ðÞ¼:1 þ :3 þ :6 ¼ 1. This is no

coincidence; conditional probabilities satisfy the properties of ordinary probabil-

ities. They are nonnegative and they sum to 1. Essentially, the denominator in the

definition of conditional probability is designed to make the total be 1.

Reversing the roles of X and Y, we find the conditional probabilities for X,

given that Y ¼ 0:

XjY

ð100j0Þ¼

pð100; 0Þ

ð0Þ

:20

:20 þ :05

¼ :8

XjY

ð250j0Þ¼

pð250; 0Þ

ð0Þ

:05

:20 þ :05

¼ :2

Again, the conditional probabilities add to 1.

■

Example 5.19 For a continuous example, recall Example 5.5, where X is the weight of almonds

and Y is the weight of cashe ws in a can of mixed nuts. The sum of X + Y is at most

one pound, the total weight of the can of nuts. The joint pdf of X and Y is

f ð x; yÞ¼

24xy



0  x  1; 0  y  1; x þ y  1

otherwise

In Example 5.5 it was shown that

ðxÞ¼

12xð1  xÞ

0  x  1

0 otherwise



The conditional pdf of Y given that X ¼ x is

YjX

ðyjxÞ¼

f ð x; yÞ

ðxÞ

24xy

12xð1  xÞ

ð1  xÞ

0  y  1  x

254 CHAPTER 5 Joint Probability Distributions

This can be used to get conditional probabiliti es for Y. For example,

PðY  :25jX ¼ :5Þ¼

:25

1

YjX

ðyj:5Þdy ¼

:25

ð1  :5Þ

dy ¼ 4y



:25

¼ :25

Recall that X is the weight of almonds and Y is the weight of cashews, so this says

that, given that the weight of almonds is .5 pound, the probability is .25 for the

weight of cashews to be less than .25 pound.

Just as in the discrete case, the conditional distribution assigns a total

probability of 1 to the set of all possible Y values. That is, integrating the condi-

tional density over its set of possible values should yield 1:

1

YjX

ðyjxÞdy ¼

1x

ð1  xÞ

dy ¼

ð1  xÞ



1x

¼ 1

Whenever you calculate a conditional density, we recommend doing this integra-

tion as a validity check.

■

Because the conditional distribution is a valid probability distribution, it

makes sense to define the conditional mean and variance.

DEFINITION

Let X and Y be two discrete random variables with conditional probability

mass function p

Y|X

(y|x). Then the conditional mean or expected value of Y

given that X ¼ x is

YjX¼x

¼ EðYjX ¼ xÞ¼

y 2D

YjX

ðyjxÞ

An analogous formula holds in the continuous case. Let X and Y be

two continuous random variables with conditional probability density func-

tion f

Y|X

(y|x). Then

YjX¼x

¼ EðYjX ¼ xÞ¼

1

YjX

ðyjxÞdy

The conditional mean of any function g(Y) can be obtained similarly. In

the discrete case,

EðgðYÞjX ¼ xÞ¼

y 2D

gðyÞp

YjX

ðyjxÞ

In the continuous case

EðgðYÞjX ¼ xÞ¼

1

gðyÞf

YjX

ðyjxÞdy

The conditional variance of Y given X ¼ x is

YjX¼x

¼ VYjX ¼ xðÞ¼EY EYjX ¼ xðÞ½

jX ¼ x

5.3 Conditional Distributions 255

There is a shortcut formula for the conditional variance analogous to that

for V(Y) itself:

YjX¼x

¼ VðYjX ¼ xÞ¼EðY

jX ¼ xÞm

YjX¼x

Example 5.20 Having found the conditional distribution of Y given X ¼ 250 in Example 5.18, we

compute the conditional mean and variance.

YjX¼250

¼ EYjX ¼ 250ðÞ¼0p

YjX

0j250ðÞþ100p

YjX

100j250ðÞ

þ 200p

YjX

200j250ðÞ¼0 :1ðÞþ100 :3ðÞþ200 :6ðÞ¼150:

Given that the possibilities for Y are 0, 100, and 200 and most of the probability is

on 100 and 200, it is reasonable that the conditional mean should be between 100

and 200.

Let’s use the alternative formula for the conditional variance.

jX ¼ 250



¼ 0

YjX

0j250ðÞþ100

YjX

100j250ðÞþ200

YjX

200j250ðÞ

¼ 0

:1ðÞþ100

:3ðÞþ200

:6ðÞ¼27; 000:

Thus,

YjX¼250

¼VYjX ¼ 250ðÞ¼EY

jX ¼ 250



m

YjX¼250

¼27; 000 150

¼ 4500:

Taking the square root, we get s

YjX¼250

¼ 67:08, which is in the right ballpark when

we recall that the possible valu es of Y are 0, 100, and 200.

It is important to realize that E(Y|X ¼ x) is one particular possible value

of a random variable E(Y|X), which is a function of X. Similarly, the conditional

variance V(Y|X ¼ x) is a value of the rv V(Y|X). The value of X might be 100

or 250. So far, we have just E(Y|X ¼ 250) ¼ 150 and V(Y|X ¼ 250) ¼ 4500.

If the calculat ions are repeated for X ¼ 100, the results are E( Y|X ¼ 100) ¼ 100 and

V(Y|X ¼ 100) ¼ 8000. Here is a summary in the form of a table:

xP(X ¼ x) E(Y|X ¼ x) V(Y|X ¼ x)

100 .5 100 8000

250 .5 150 4500

Similarly, the conditional mean and variance of X can be computed for

specific Y. Taking the conditional probabilities from Example 5.18,

XjY¼0

¼ EXjY ¼ 0ðÞ¼100p

XjY

100j0ðÞþ250p

XjY

250j0ðÞ

¼ 100 :8ðÞþ250 :2ðÞ¼130

XjY¼0

¼ VXjY ¼ 0ðÞ¼EX EXjY ¼ 0ðÞ½

jY ¼ 0



¼ 100  130ðÞ

XjY

100j0ðÞþ250  130ðÞ

XjY

250j0ðÞ

¼ 30

:8ðÞþ120

:2ðÞ¼3600:

256 CHAPTER 5 Joint Probability Distributions

Similar calcul ations give the other entries in this table:

yP(Y ¼ y) E(X|Y ¼ y) V(X|Y ¼ y )

0 .25 130 3600

100 .25 190 5400

200 .50 190 5400

Again, the conditional mean and variance are random because they depend on the

random value of Y.

■

Example 5.21

(Example 5.19

continued)

For any given weight of almonds, let’s find the expected weight of cashews. Using

the definition of conditional mean,

YjX¼x

¼ EðYjX ¼ xÞ¼

1

YjX

ðyjxÞdy ¼

1x

ð1  xÞ

ð1  xÞ 0  x  1

The conditional mean is a linear decreasing function of x. When there are

more almonds, we expect less cashews. This is in accord with Figure 5.2, which

shows that for large X the domain of Y is restricted to small values. To get the

corresponding variance, compute first

EðY

jX ¼xÞ¼

1

YjX

ðyjxÞdy ¼

1x

ð1 xÞ

dy ¼

ð1 xÞ

0  x  1

Then the conditional variance is

YjX¼x

¼VYjX ¼xðÞ¼EY

jX ¼x



m

YjX¼x

ð1xÞ



4ð1xÞ

ð1xÞ

and the conditional standard deviation is

YjX¼x

1  x

ﬃﬃﬃﬃﬃ

This says that the variance gets smaller as the weight of almonds approaches 1.

Does this make sense? When the weight of almonds is 1, the weight of cashews is

guaranteed to be 0, implying that the variance is 0. This is clarified by Figure 5.2,

which shows that the set of y-values narrow s to 0 as x approaches 1.

■

Independence

Recall that in Section 5.1 two random variables were defined to be independent if

their joint pmf or pdf factors into the product of the marginal pmf’s or pdf’s. We

can understand this definition better with the help of conditional distributions. For

example, suppos e there is independence in the discrete case. Then

YjX

ðyjxÞ¼

pðx; yÞ

ðxÞ

ðxÞp

ðyÞ

ðxÞ

¼ p

ðyÞ

5.3 Conditional Distributions 257