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

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That is, independence implies that the conditional distribution of Y is the same as

the unconditional distribution. The implication works in the other direct ion, too. If

YjX

ðyjxÞ¼p

ðyÞ

then

pðx; yÞ

ðxÞ

¼ p

ðyÞ

px; yðÞ¼p

ðxÞp

ðyÞ

and therefore X and Y are independent. Is this intuitively reasonable? Yes,

because independence means that knowing X does not change our probabilities

for Y.

In Example 5.7 we said that indep endence necessitates the region of positive

density being a rectangle (possibly infinite in extent). In terms of conditional

distribution this region tells us the domain of Y for each X. For independence we

need to have the domain of Y not be dependent on X. That is, the conditional

distributions must all be the same, so the interval of positive density must be the

same for each x, implying a rectangular region.

The Bivariate Normal Distribution

Perhaps the most useful example of a joint distribution is the bivariate normal.

Although the formula may seem rather messy, it is based on a simple quadratic

expression in the standardized variables (subtract the mean and then divide by the

standard deviation). The bivariate normal density is

f ð x; yÞ¼

2ps

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

1  r

f½ðxm

Þ=s



2rðxm

Þðym

Þ=s

þ½ðym

Þ=s



g=½2ð1r

Þ

There are five parameters, including the mean m

and the standard deviation s

of X

and the mean m

and the standard deviation s

of Y. The fifth parameter r is the

correlation between X and Y. The integration required to do bivariate normal

probability calculations is quite difficult. Computer code is available for calculat-

ing P(X < x, Y < y) approximately using numerical integration, and some statistical

software packages (e.g., R, SAS, Stata) include this feature.

What does the density look like when plotted as a function of x and y?Ifwe

set f(x, y) to a constant to investigate the contours, this is setting the exponent to a

constant, and it will give ellipses centered at (x, y) ¼ (m

, m

). That is, all of the

contours are concentric ellipses. The plot in three dimensions looks like a mountain

with elliptical cross-sections. The vertical cross-sections are all proportional to

normal densities. See Figure 5.6.

258 CHAPTER 5 Joint Probability Distributions

If r ¼ 0, then fx; yðÞ¼f

ðxÞ f

ðyÞ, where X is normal with mean m

and

standard deviation s

, and Y is normal with mean m

and standard deviation s

.That

is, X and Y have independent normal distributions. In this case the plot in three

dimensions has elliptical contours that reduce to circles. Recall that in Section 5.2

we emphasized that independence of X and Y implies r ¼ 0 but, in general, r ¼

0 does not imply independence. However, we have just seen that when X and Y are

bivariate normal r ¼0 does imply independence. Therefore, in the bivariate norm al

case r ¼ 0 if and only if the two rv’s are independent.

What do we get for the marginal distributions? As you might guess, the

marginal distribution f

(x) is just a normal distribution with mean m

and standard

deviation s

ðxÞ¼

ﬃﬃﬃﬃﬃﬃ

f½ðxm

Þ=s



g=2

The integration to show this [integ rating f(x,y)ony from 1 to 1 ] is rather messy.

More generally, any linear combination of the form aX + bY, where a and b are

constants, is normally distributed.

We get the conditional density by dividing the marginal density of X into

f(x,y). Unfortunately, the algebra is again a mess, but the result is fairly simple.

The conditional density f

Y|X

(y|x) is a normal density with mean and variance

given by

jX¼x

¼ EYjX ¼ xðÞ¼m

þ rs

x  m

YjX¼x

¼ VYjX ¼ xðÞ¼s

ð1  r

Notice that the conditional mean is a linear function of x and the conditional

variance doesn’ t depend on x at all. When r ¼ 0, the conditional mean is the

mean of Y and the conditional variance is just the variance of Y. In other words,

if r ¼ 0, then the conditional distribution of Y is the same as the unconditional

distribution of Y. This says that if r ¼ 0 then X and Y are independent, but we

already saw that previously in terms of the factorization of f(x,y) into the product of

the marginal densities.

When r is close to 1 or 1 the conditional variance will be much smaller

than V(Y), which says that knowledge of X will be very helpful in predicting Y.

(x, y)

Figure 5.6 A graph of the bivariate normal pdf

5.3 Conditional Distributions 259

If r is near 0 then X and Y are nearly independen t and knowledge of X is not very

useful in predicting Y.

Example 5.22 Let X be mother’s height and Y be daughter’s height. A similar situation was one

of the first applications of the bivariate normal distribution, by Francis Galton

in 1886, and the data was found to fit the distribution very well. Suppose a bivariate

normal distribution with mean m

¼ 64 in. and standard deviation s

¼ 3 in. for X

and mean m

¼ 65 in. and standard deviation s

¼ 3 in. for Y . Here m

> m

, which

is in accord with the increase in height from one generation to the next. Assume

r ¼ .4. Then

YjX¼x

¼ m

þ rs

x  m

¼ 65 þ :4ð3Þ

x  64

¼ 65 þ :4ðx  64Þ¼:4x þ 39:4

YjX¼x

¼ VYjX ¼ xðÞ¼s

ð1  r

Þ¼9 ð1  :4

Þ¼7:56 and s

YjX¼x

¼ 2:75

Notice that the conditional variance is 16% less than the variance of Y.

Squaring the correlation gives the percentage by which the conditional variance

is reduced relative to the variance of Y.

■

Regression to the Mean

The formula for the conditional mean can be re-expressed as

YjX¼x

 m

¼ r 

x  m

In words, when the formula is expressed in terms of standardized variables, the

standardized conditional mean is just r times the standardized x. In particular,

for the example of heights,

YjX¼x

 65

¼ :4 

x  64

If the mother is 5 in. above the mean of 64 in. for mothers, then the daughter’s

conditional expected height is just 2 in. above the mean for daughters. In this

example, with equal standard deviations for Y and X, the daughter’s conditional

expected height is always closer to its mean than the mother’s height is to its mean.

In general, the conditional expected Y is closer when it is measured in terms of

standard deviations. One can think of the conditional expectation as being pulled

back toward the mean, and that is why Galton called this regression to the mean.

Regression to the mean occurs in many contexts. For example, let X be a

baseball player’s average for the first half of the season and let Y be the average for

the second half. Most of the players with a high X (above .300) will not have such a

high Y. The same kind of reasoning applies to the “sophomore jinx,” which says

that if a player has a very good first season, then the player is unlikely to do as well

in the second season.

260 CHAPTER 5 Joint Probability Distributions

The Mean and Variance Via the Conditional

Mean and Variance

From the conditional mean we can obtain the mean of Y. From the conditional mean

and the conditional variance, the variance of Y can be obtained. The following

theorem uses the idea that the conditional mean and variance are themselves

random variables, as illustrated in the tables of Example 5.20.

THEOREM

a. EðYÞ¼EE YjXðÞ½

b. VðYÞ¼VEYjXðÞ½þEV YjXðÞ½

The result in (a) says that E(Y) is a weighted average of the conditional means

E(Y|X ¼ x), where the weights are given by the pmf or pdf of X. We give the

proof of just part (a) in the discrete case:

E½EðYjXÞ¼

x2D

EðYjX ¼xÞp

ðxÞ¼

x2D

y2D

YjX

ðyjxÞp

ðxÞ

x2D

y2D

pðx;yÞ

ðxÞ

ðxÞ¼

y2D

x2D

pðx;yÞ¼

y2D

ðyÞ¼EðYÞ

Example 5.23 To try to get a feel for the theorem, let’s apply it to Example 5.20. Here again is the

table for the conditional mean and variance of Y given X.

xP(X ¼ x) E(Y|X ¼ x) V(Y|X ¼ x)

100 .5 100 8000

250 .5 150 4500

Compute

EE YjXðÞ½¼EYjX ¼ 100ðÞPX¼ 100ðÞþEYj X ¼ 250ðÞPX¼ 250ðÞ

¼ 100 :5ðÞþ150 :5ðÞ¼125

Compare this with E(Y) computed directly:

EðYÞ¼0PY¼ 0ðÞþ100PY¼ 100ðÞþ200PY¼ 200ðÞ

¼ 0 :25ðÞþ100 :25ðÞþ200 :5ðÞ¼125

For the variance first compute the mean of the conditi onal variance:

EV YjXðÞ½¼VYjX ¼ 100ðÞPX¼ 100ðÞþVYjX ¼ 250ðÞPX¼ 250ðÞ

4500 :5ðÞþ8000 :5ðÞ¼6250

Then comes the variance of the conditional mean. We have already computed

the mean of this random variable to be 125. The variance is

VEYjXðÞ½¼:5 100  125ðÞ

þ :5 150  125ðÞ

¼ 625

5.3 Conditional Distributions 261

Finally, do the sum in part (b) of the theorem:

VðYÞ¼VEYjXðÞ½þEV YjXðÞ½¼625 þ 6250 ¼ 6875

To compare this with V(Y) calculated from the pmf of Y, compute first



¼ 0

PY¼ 0ðÞþ100

PY¼ 100ðÞþ200

PY¼ 200ðÞ

¼ 0 :25ðÞþ10; 000 :25ðÞþ40; 000 :5ðÞ¼22; 500

Thus, VðYÞ¼EY

ðÞEðYÞ½

¼ 22; 500  125

¼ 6875, in agreement with the

calculation based on the theorem.

■

Here is an example where the theorem is helpful in finding the mean and

variance of a random variable that is neither discrete nor continuous.

Example 5.24 The probability of a claim being filed on an insurance policy is .1, and

only one claim can be filed. If a claim is filed, the amount is exponentially

distributed with mean $1000. Recall from Section 4.4 that the mean and standard

deviation of the exponential distribution are the same, so the variance is the

square of this value. We want to find the mean and variance of the amount

paid. Let X be the number of claims (0 or 1) and let Y be the payment. We know

that E(Y| X ¼ 0) ¼ 0 and E(Y| X ¼ 1) ¼ 1000. Also, V(Y| X ¼ 0) ¼ 0 and V(Y|X ¼ 1)

¼ 1000

¼ 1,000,000. Here is a table for the distribution of E(Y|X ¼ x) and

V(Y|X ¼ x):

xP(X ¼ x) E(Y|X ¼ x) V(Y|X ¼ x)

0.9 0 0

1 .1 1000 1,000,000

Therefore,

EðYÞ¼EE YjXðÞ½¼EYjX ¼ 0ðÞPX¼ 0ðÞþEYjX ¼ 1ðÞPX¼ 1ðÞ

¼ 0 :9ðÞþ1000 :1ðÞ¼100

The variance of the conditional mean is

VEYjXðÞ½¼:90100ðÞ

þ :11000 100ðÞ

¼ 90; 000

The expected value of the conditional variance is

EV YjXðÞ½¼:9ð 0Þþ:11; 000; 000ðÞ¼100; 000

Finally, use part (b) of the theorem to get V(Y):

VðYÞ¼VEYjXðÞ½þEV YjXðÞ½¼90; 000 þ 100; 000 ¼ 190; 000

Taking the square root gives the standard deviation, s

¼ $435.89.

Suppose that we want to compute the mean and variance of Y directly.

Notice that X is discrete, but the conditional distribution of Y given X ¼ 1

is continuous. The random variable Y itself is neither discrete nor continuous, because

it has probability .9 of being 0, but the other .1 of its probability is spread out from

0to1. Such “mixed” distributions may require a little extra effort to evaluate means

and variances, although it is not especially hard in this case. Compute

262 CHAPTER 5 Joint Probability Distributions

¼ EðYÞ¼ð:1Þ

1000

y=1000

dy ¼ð:1Þð1000Þ¼100

EðY

Þ¼ð:1Þ

1000

y=1000

dy ¼ð:1Þ2ð1000

Þ¼200;000

VðYÞ¼EðY

Þ½EðYÞ

¼ 200;000  10;000 ¼ 190;000

These agree with what we found using the theorem.

■

Exercises Section 5.3 (36–57)

36. According to an article in the August 30, 2002

issue of the Chronicle of Higher Education,

30% of first-year college students are liberals,

20% are conservatives, and 50% characterize

themselves as middle-of-the-road. Choose two

students at random, let X be the number of liber-

als, and let Y be the number of conservatives.

a. Using the multinomial distribution from

Section 5.1, give the joint probability mass func-

tion p(x, y)ofX and Y. Give the joint probability

table showing all nine values, of which three

should be 0.

b. Determine the marginal probability mass func-

tions by summing p(x, y)numerically.How

could these be obtained directly? [Hint:What

are the univariate distributions of X and Y?]

c. Determine the conditional probability mass

function of Y given X ¼ x for x ¼ 0, 1, 2.

Compare with the Bin[2x, .2/(.2 + .5)] distri-

bution. Why should this work?

d. Are X and Y independent? Explain.

e. Find E(Y|X ¼ x) for x ¼ 0, 1, 2. Do this numer-

ically and then compare with the use of the

formula for the binomial mean, using the bino-

mial distribution given in part (c). Is E(Y|X ¼x)

a linear function of x?

f. Determine V(Y

|X ¼ x) for x ¼ 0, 1, 2. Do this

numerically and then compare with the use of

the formula for the binomial variance, using the

binomial distribution given in part (c).

37. Teresa and Allison each have arrival times uni-

formly distributed between 12:00 and 1:00. Their

times do not influence each other. If Y is the first

of the two times and X is the second, on a scale of

0–1, then the joint pdf of X and Y is f(x, y) ¼ 2 for

0 < y < x < 1.

a. Determine the marginal density of X.

b. Determine the conditional density of Y given

X ¼ x.

c. Determine the conditional probability that Y is

between 0 and .3, given that X is .5.

d. Are X and Y independent? Explain.

e. Determine the conditional mean of Y given

X ¼ x.IsE(Y|X ¼ x) a linear function of x?

f. Determine the conditional variance of Y given

X ¼ x.

38. Refer back to Exercise 37.

a. Determine the marginal density of Y.

b. Determine the conditional density of X given

Y ¼ y.

c. Determine the conditional mean of X given

Y ¼ y.IsE(X|Y ¼ y) a linear function of y?

d. Determine the conditional variance of X given

Y ¼ y.

39. A pizza place has two phones. On each phone the

waiting time until the first call is exponentially

distributed with mean one minute. Each phone is

not influenced by the other. Let X be the shorter of

the two waiting times and let Y be the longer.

It can be shown that the joint pdf of X and Y is

fx; yðÞ¼2e

ðxþyÞ

; 0 < x < y < 1

a. Determine the marginal density of X.

b. Determine the conditional density of Y given

X ¼ x.

c. Determine the probability that Y is greater

than 2, given that X ¼ 1.

d. Are X and Y independent? Explain.

e. Determine the conditional mean of Y given

X ¼ x.IsE(Y|X ¼ x) a linear function of x?

f. Determine the conditional variance of Y given

X ¼ x.

40. A class has 10 mathematics majors, 6 computer

science majors, and 4 statistics majors. A committee

of two is selected at random to work on a problem.

Let X be the number of mathematics majors and let Y

be the number of computer science majors chosen.

a. Determine the joint probability mass function

p(x,y). This generalizes the hypergeometric

distribution studied in Section 3.6. Give the

joint probability table showing all nine values,

of which three should be 0.

5.3 Conditional Distributions 263

b. Determine the marginal probability mass

functions by summing numerically. How could

these be obtained directly? [Hint: What are the

univariate distributions of X and Y?]

c. Determine the conditional probability mass

function of Y given X ¼ x for x ¼ 0, 1, 2.

Compare with the h(y;2x, 6, 10) distribution.

Intuitively, why should this work?

d. Are X and Y independent? Explain.

e. Determine E(Y|X ¼ x), x ¼ 0, 1, 2. Do this

numerically and then compare with the use of

the formula for the hypergeometric mean,

using the hypergeometric distribution given in

part (c). Is E(Y|X ¼ x) a linear function of x?

f. Determine V(Y|X ¼ x), x ¼ 0, 1, 2. Do this

numerically and then compare with the use of

the formula for the hypergeometric variance,

using the hypergeometric distribution given in

part (c).

41. A stick is one foot long. You break it at a point X

(measured from the left end) chosen randomly

uniformly along its length. Then you break the

left part at a point Y chosen randomly uniformly

along its length. In other words, X is uniformly

distributed between 0 and 1 and, given X ¼ x, Y is

uniformly distributed between 0 and x.

Determine E(Y|X ¼ x) and then V(Y|X ¼ x). Is

E(Y|X ¼ x) a linear function of x?

b. Determine f(x,y) using f

(x) and f

Y|X

(y|x).

c. Determine f

(y).

d. Use f

(y) from (c) to get E(Y) and V(Y).

e. Use (a) and the theorem of this section to get

E(Y) and V(Y).

42. A system consisting of two components will con-

tinue to operate only as long as both components

function. Suppose the joint pdf of the lifetimes

(months) of the two components in a system

is given by fx; yðÞ¼c 10  x þ yðÞ½for x > 0;

y > 0; x þ y < 10

a. If the first component functions for exactly

3 months, what is the probability that the sec-

ond functions for more than 2 months?

b. Suppose the system will continue to work only

as long as both components function. Among 20

of these systems that operate independently of

each other, what is the probability that at least

half work for more than 3 months?

43. Refer to Exercise 1 and answer the following

questions:

a. Given that X ¼ 1, determine the conditional

pmf of Ythat is, p

Y|X

(0|1), p

Y|X

(1|1), and

Y|X

(2|1).

b. Given that two hoses are in use at the

self-service island, what is the conditional

pmf of the number of hoses in use on the full-

service island?

c. Use the result of part (b) to calculate the con-

ditional probability P(Y  1|X ¼ 2).

d. Given that two hoses are in use at the full-

service island, what is the conditional pmf of

the number in use at the self-service island?

44. The joint pdf of pressures for right and left front

tires is given in Exercise 9.

a. Determine the conditional pdf of Y given that

X ¼ x and the conditional pdf of X given that

Y ¼ y.

b. If the pressure in the right tire is found to be 22

psi, what is the probability that the left tire has

a pressure of at least 25 psi? Compare this to

P(Y  25).

c. If the pressure in the right tire is found to be

22 psi, what is the expected pressure in the left

tire, and what is the standard deviation of pres-

sure in this tire?

45. Suppose that X is uniformly distributed between

0 and 1. Given X ¼ x, Y is uniformly distributed

between 0 and x

a. Determine E(Y|X ¼ x) and then V(Y|X ¼ x).

Is E(Y|X ¼ x) a linear function of x?

b. Determine f(x,y) using f

(x) and f

Y|X

(y|x).

c. Determine f

(y).

46. This is a continuation of the previous exercise.

a. Use f

(y) from Exercise 45(c) to get E(Y) and

V(Y).

b. Use Exercise 45(a) and the theorem of this

section to get E(Y) and V(Y).

47. David and Peter independently choose at random a

number from 1, 2, 3, with each possibility equally

likely. Let X be the larger of the two numbers, and

let Y be the smaller.

a. Determine p( x, y).

b. Determine p

(x), x ¼ 1, 2, 3.

c. Determine p

Y|X

(y|x).

d. Determine E(Y|X ¼ x). Is this a linear function

of x?

e. Determine V(Y|X ¼ x).

48. In Exercise 47 find

a. E(X).

b. p

(y).

c. E(Y) using p

(y).

d. E(Y) using E(Y|X).

e. E(X)+E(Y). Intuitively, why should this be 4?

264

CHAPTER 5 Joint Probability Distributions

49. In Exercise 47 find

a. p

X|Y

(x|y).

b. E(X|Y ¼ y). Is this a linear function of y ?

c. V(X|Y ¼ y).

50. For a Calculus I class, the final exam score Y and

the average of the four earlier tests X are bivariate

normal with mean m

¼ 73, standard deviation

¼ 12, mean m

¼70, standard deviation s

15. The correlation is r ¼.71. Determine

a. m

Y|X¼x

b. s

YjX¼x

c. s

Y|X¼x

d. P(Y > 90|X ¼ 80), i.e., the probability that the

final exam score exceeds 90 given that the

average of the four earlier tests is 80

51. Let X and Y, reaction times (sec) to two different

stimuli, have a bivariate normal distribution with

mean m

¼ 20 and standard deviation s

¼ 2 for X

and mean m

¼30 and standard deviation s

¼ 5

for Y. Assume r ¼.8. Determine

a. m

Y|X¼x

b. s

YjX¼x

c. s

Y|X¼x

d. P(Y > 46|X ¼ 25)

52. Consider three ping pong balls numbered 1, 2, and 3.

Two balls are randomly selected with replacement.

If the sum of the two resulting numbers exceeds 4,

two balls are again selected. This process continues

until the sum is at most 4. Let X and Y denote the

last two numbers selected. Possible (X, Y) pairs are

{(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1)}.

a. Determine p

X,Y

(x,y).

b. Determine p

Y|X

(y|x).

c. Determine E(Y|X ¼ x). Is this a linear function

of x?

d. Determine E(X|Y ¼ y). What special property

of p(x, y) allows us to get this from (c)?

e. Determine V(Y|X ¼ x).

53. Let X be a random digit (0, 1, 2, ..., 9 are equally

likely) and let Y be a random digit not equal to X.

That is, the nine digits other than X are equally

likely for Y.

a. Determine p

(x), p

Y|X

(y|x), p

X,Y

(x,y).

b. Determine a formula for E(Y|X ¼ x). Is this a

linear function of x?

54. In our discussion of the bivariate normal, there is

an expression for E(Y|X ¼ x).

a. By reversing the roles of X and Y give a similar

formula for E(X|Y ¼ y).

b. Both E(Y|X ¼ x) and E(X|Y ¼ y) are linear

functions. Show that the product of the two

slopes is r

55. This week the number X of claims coming into an

insurance office is Poisson with mean 100. The

probability that any particular claim relates to

automobile insurance is .6, independent of any

other claim. If Y is the number of automobile

claims, then Y is binomial with X trials, each

with “success” probability .6.

a. Determine E( Y|X ¼ x) and V(Y|X ¼ x).

b. Use part (a) to find E(Y).

c. Use part (a) to find V(Y).

56. In Exercise 55 show that the distribution of Y is

Poisson with mean 60. You will need to recognize

the Maclaurin series expansion for the exponential

function. Use the knowledge that Y is Poisson with

mean 60 to find E(Y) and V(Y).

57. Let X and Y be the times for a randomly selected

individual to complete two different tasks, and

assume that (X, Y) has a bivariate normal distribution

with m

¼ 100, s

¼ 50, m

¼ 25, s

¼ 5, r ¼ .5.

From statistical software we obtain P(X < 100,

Y < 25) ¼ .3333, P(X < 50, Y < 20) ¼ .0625,

P(X < 50, Y < 25) ¼ .1274, and P(X < 100, Y <

20) ¼ .1274.

(a) Determine P(50 < X < 100, 20 < Y < 25).

(b) Leave the other parameters the same but

change the correlation to r ¼ 0 (indepen-

dence). Now recompute the answer to part

(a). Intuitively, why should the answer to

part (a) be larger?

5.4

Transformations of Random Variables

In the previous chapter we discussed the problem of starting with a single random

variable X, forming some function of X, such as X

or e

, to obtain a new random

variable Y ¼ h(X), and investigating the distribution of this new random variable.

We now generalize this scenario by starting with more than a single random

variable. Consider as an example a system having a component that can be replaced

just once before the system itself expires. Let X

denote the lifetime of the original

5.4 Transformations of Random Variables 265

component and X

the lifetime of the replacement component. Then any of the

following functions of X

and X

may be of interest to an investigator:

1. Th e total lifeti me X

+ X

2. Th e ratio of lifetimes X

; for example, if the value of this ratio is 2, the

original component lasted twice as long as its replacement

3. Th e ratio X

/(X

+ X

), which represents the proportion of system lifetime during

which the original component operated

The Joint Distribution of Two New Random Variables

Given two random variables X

and X

, consider forming two new random vari-

ables Y

¼ u

, X

) and Y

¼ u

, X

). We now focus on finding the joint

distribution of these two new variables. Since most applications assume that the

’s are continuous we restrict ourselves to that case. Some notation is needed

before a general result can be given. Let

f(x

, x

) ¼ the joint pdf of the two original variables

g(y

, y

) ¼ the joint pdf of the two new variables

The u

(·) and u

(·) functions express the new variables in terms of the original ones.

The general result presumes that these functions can be inverted to solve for the

original variables in terms of the new ones:

¼ v

; Y

ðÞ; X

¼ v

; Y

ðÞ

For example, if

¼ x

þ x

and y

þ x

then multiplying y

by y

gives an expression for x

, and then we can substitute this

into the expression for y

and solve for x

¼ y

¼ v

; y

ðÞx

¼ y

1  y

ðÞ¼v

; y

ðÞ

In a final burst of notation, let

S ¼ x

; x

ðÞ: fx

; x

ðÞ> 0

T ¼ y

; y

ðÞ: gy

; y

ðÞ> 0

That is, S is the region of positive density for the original variables and T is the

region of positive density for the new variables; T is the “image” of S under the

transformation.

THEOREM

Suppose that the partial derivative of each v

, y

) with respect to both y

and y

exists for every (y

, y

) pair in T and is continuous. Form the 2  2

matrix

M ¼

ðy

; y

ðy

; y

ðy

; y

ðy

; y

266 CHAPTER 5 Joint Probability Distributions

The determinant of this matrix, called the Jacobian,is

detðMÞ¼







The joint pdf for the new variables then results from taking the joint pdf

f(x

, x

) for the original variables, replacing x

and x

by their expressions in

terms of y

and y

, and finally multiplying this by the absolute value of the

Jacobian:

; y

ðÞ¼fv

; y

ðÞ; v

; y

ðÞ½jdetðMÞj y

; y

ðÞ2T

The theorem can be rewritten slightly by using the notation

detðMÞ¼

@ðx

; x

@ðy

; y



Then we have

gðy

; y

Þ¼f ðx

; x

Þ

@ðx

; x

@ðy

; y



which is the natural extension of the univariate result (transforming a single rv X to

obtain a single new rv Y) g(y) ¼ f(x)



|dx/dy| discussed in Chapter 4.

Example 5.25 Continuing with the component lifetime situation, suppose that X

and X

are

independent, each having an exponential distribution with parameter l. Let’s

determine the joint pdf of

¼ u

; X

ðÞ¼X

þ X

and Y

¼ u

; X

ðÞ¼

þ X

We have already inve rted this transformation:

¼ v

; y

ðÞ¼y

¼ v

; y

ðÞ¼y

1  y

ðÞ

The image of the transformation, i.e. the set of (y

, y

) pairs with positive density,

is 0 < y

and 0 < y

< 1. The four relevant partial derivatives are

¼ y

¼ 1  y

¼y

from which the Jacobian is  y

 y

1  y

ðÞ¼y

Since the joint pdf of X

and X

f ð x

; x

Þ¼le

lx

 le

lx

¼ l

lðx

þx

> 0; x

> 0

we have

gðy

; y

Þ¼l

ly

 y

¼ l

ly

 10< y

; 0 < y

< 1

The joint pdf thus factors into two parts. The first part is a gamma pdf with

parameters a ¼ 2 and b ¼ 1/l, and the second part is a uniform pdf on (0, 1).

5.4 Transformations of Random Variables 267