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

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Since the pdf factors and the region of positive density is rectangular, we have

demonstrated that

1. Th e distribution of system lifetime X

+ X

is gamma(a ¼ 2, b ¼ 1/l)

2. Th e distribution of the proportion of system lifetime during which the original

component functions is uniform on (0, 1)

3. Y

¼ X

+ X

and Y

¼ X

/(X

+ X

) are independent of each other ■

In the foregoing example, because the joint pdf factored into one pdf

involving y

alone and another pdf involving y

alone, the individual (i. e. marginal)

pdf’s of the two new variables were obtained from the joint pdf without any further

effort. Often this will not be the case – that is, Y

and Y

will not be independent.

Then to obtain the marginal pdf of Y

, the joint pdf must be integrated over all

values of the second variable. In fact, in many applications an investigator wishes to

obtain the distribution of a single function u

, X

) of the origina l variables. To

accomplish this, a second function u

, X

) is selected, the joint pdf is obtained,

and then y

is integrated out. There are of course many ways to select the second

function. The choice should be made so that the transformation can be easily

inverted and the integration in the last step is straightforward.

Example 5.26 Consider a rectangular coordinate system with a horizontal x

axis and a vertical x

axis as shown in Figure 5.7(a). First a point (X

, X

) is randomly select ed, where the

joint pdf of X

, X

f ðx

; x

Þ¼

þ x

0 < x

< 1; 0 < x

< 1

0 otherwise

(

Then a rectangle with vertices (0, 0), (X

, 0), (0, X

), and (X

, X

) is formed. What is

the distribution of X

, the area of this rectangle? To answer this question, let

¼ X

¼ u

; x

ðÞ¼x

¼ u

; x

ðÞ¼x

Then

¼ n

ðy

; y

Þ¼

¼ n

ðy

; y

Þ¼y

Notice that because x

(¼y

) is between 0 and 1 and y

is the product of the two x

’s,

it must be the case that 0 < y

< y

. The region of positive density for the new

variables is then

T ¼ y

; y

ðÞ: 0 < y

< y

; 0 < y

< 1

which is the triangular region shown in Figure 5.7(b).

268 CHAPTER 5 Joint Probability Distributions

Since ∂v

/∂y

¼ 0, the product of the two off-diagonal elements in the matrix

M will be 0, so only the two diagonal elements contribute to the Jacobian:

M ¼

jdetðMÞj ¼

The joint pdf of the two new variables is now

gðy

; y

Þ¼f

; y



 detðMÞ

þ y





0 < y

< y

; 0 < y

< 1

0 otherwise

To obtain the marginal pdf of Y

alone, we must now fix y

at some arbitrary

value between 0 and 1, and integrate out y

. Figure 5.7b shows that we must

integrate along the vertical line segment passing through y

whose lower limit is

and whose upper limit is 1:

ðy

Þ¼

þ y





¼ 2ð1  y

Þ 0 < y

< 1

This marginal pdf can now be integrated to obtain any desired probability involving

the area. For example, integrating from 0 to .5 gives P(area < .5) ¼ .75.

■

The Joint Distribution of More than Two New Variables

Consider now starting with three rando m variables X

, X

, and X

, and forming

three new variables Y

, Y

, and Y

. Suppose again that the transformation can be

inverted to express the original variables in terms of the new ones:

¼ v

; y

ðÞ; x

¼ v

; y

ðÞ; x

¼ v

; y

ðÞ

Then the foregoing theorem can be extended to this new situation. The Jacobian

matrix has dimension 3 3, with the entry in the ith row and jth column being ∂v

/∂y

The joint pdf of the new variables results from replacing each x

in the original pdf f(·)

by its expression in terms of the y

’s and multiplying by the absolute value of the

Jacobian.

A possible

rectangle

For (X

) For (Y

)

Figure 5.7 Regions of positive density for Example 5.26

5.4 Transformations of Random Variables 269

Example 5.27 Consider n ¼ 3 identical components with independent lifetimes X

, X

, each

having an exponential distribution with parameter l. If the first component is used

until it fails, replaced by the second one which remains in service until it fails, and

finally the third component is used until failure, then the total lifetime of these

components is Y

¼ X

+ X

. To find the distribution of total lifetime, let’s first

define two other new variables: Y

¼ X

and Y

¼ X

+ X

(so that Y

< Y

After finding the joint pdf of all three variabl es, we integrate out the first two

variables to obtain the desired information. Solving for the old variables in terms of

the new gives

¼ y

 y

¼ y

 y

It is obvious by inspection of these expressions that the three diagonal elements of

the Jacobian matrix are all 1’s and that the elements above the diagonal are all 0’s,

so the determinant is 1, the product of the diagonal elements. Since

f ð x

; x

Þ¼l

lðx

þx

> 0; x

> 0

by substitution,

gðy

; y

Þ¼l

ly

0 < y

< y

Integrating this joint pdf first with respect to y

between 0 and y

and then with

respect to y

between 0 and y

(try it!) gives

ðy

Þ¼

ly

> 0

This is a gamma pdf. The result is easily extended to n components. It can also be

obtained (more easily) by using a moment generating function argument.

■

Exercises Section 5.4 (58–64)

58. Consider two components whose lifetimes X

and

are independent and exponentially distributed

with parameters l

and l

, respectively. Obtain

the joint pdf of total lifetime X

+ X

and the

proportion of total lifetime X

/(X

+ X

) during

which the first component operates.

59. Let X

denote the time (hr) it takes to perform a

first task and X

denote the time it takes to perform

a second one. The second task always takes at

least as long to perform as the first task. The

joint pdf of these variables is

f ðx

; x

Þ¼

2ðx

þ x

Þ 0  x

 x

 1

0 otherwise



a. Obtain the pdf of the total completion time for

the two tasks.

b. Obtain the pdf of the difference X

X

between the longer completion time and the

shorter time.

60. An exam consists of a problem section and a short-

answer section. Let X

denote the amount of time

(hr) that a student spends on the problem section

and X

represent the amount of time the same

student spends on the short-answer section.

Suppose the joint pdf of these two times is

f ðx

; x

Þ¼

< x

; 0 < x

< 1

0 otherwise

a. What is the value of c ?

b. If the student spends exactly .25 h on the short-

answer section, what is the probability that at

most .60 h was spent on the problem section?

[Hint: First obtain the relevant conditional dis-

tribution.]

c. What is the probability that the amount of time

spent on the problem part of the exam exceeds

the amount of time spent on the short-answer

part by at least .5 hr?

270

CHAPTER 5 Joint Probability Distributions

d. Obtain the joint distribution of Y

¼ X

, the

ratio of the two times, and Y

¼ X

. Then

obtain the marginal distribution of the ratio.

61. Consider randomly selecting a point (X

, X

)in

the unit cube {(x

, x

): 0 < x

< 1, 0 < x

< 1,

0 < x

< 1}according to the joint pdf

f ðx

; x

0 < x

< 1; 0 < x

< 1

0 otherwise

(

(so the three variables are independent). Then

form a rectangular solid whose vertices are (0, 0, 0),

,0,0),(0,X

,0),(X

, X

,0),(0,0,X

), (X

,0,X

(0, X

, X

), and (X

, X

). The volume of this cube

is Y

¼X

. Obtain the pdf of this volume. [Hint:

Let Y

¼ X

and Y

¼ X

62. Let X

and X

be independent, each having a

standard normal distribution. The pair (X

, X

)

corresponds to a point in a two-dimensional coor-

dinate system. Consider now changing to polar

coordinates via the transformation,

¼ X

þ X

arctan



> 0; X

 0

arctan



þ 2p X

> 0; X

< 0

arctan



þ p X

< 0

0 X

¼ 0

from which X

ﬃﬃﬃﬃﬃ

cosðY

Þ; X

ﬃﬃﬃﬃﬃ

sinðY

Þ.

Obtain the joint pdf of the new variables and then

the marginal distribution of each one. [Note:It

would be nice if we could simply let Y

¼ arctan

), but in order to insure invertibility of the

arctan function, it is defined to take on values

onl y between  p/2 and p/2. Our specification

of Y

allows it to assume any value between

0and2p .]

63. The result of the previous exercise suggests how

observed values of two independent standard nor-

mal variables can be generated by first generating

their polar coordinates with an exponential rv with

l ¼

and an independent uniform(0, 2p) rv: Let

and U

be independent uniform(0, 1) rv’s, and

then let

¼2ln U

ðÞY

¼ 2pU

ﬃﬃﬃﬃﬃ

cosðY

Þ Z

ﬃﬃﬃﬃﬃ

sinðY

Show that the Z

’s are independent standard nor-

mal. [Note: This is called the Box-Muller transfor-

mation after the two individuals who discovered

it. Now that statistical software packages will

generate almost instantaneously observations

from a normal distribution with any mean and

variance, it is thankfully no longer necessary for

people like you and us to carry out the transforma-

tions just described – let the software do it!]

64. Let X

and X

be independent random variables,

each having a standard normal distribution. Show

that the pdf of the ratio Y ¼ X

is given by f(y)

¼ 1/[p(1 + y

)] for 1< y < 1 (this is called

the standard Cauchy distribution).

5.5

Order Statistics

Many statistical procedures involve ordering the sample observations from smallest

to largest and then manipulating these ordered values in various ways. For example,

the sample median is either the middle value in the ordered list or the average

of the two middle values depending on whe ther the sample size n is odd or even.

The sample range is the difference between the largest and smallest values. And a

trimmed mean results from deleting the same number of observations from each

end of the ordered list and averaging the remaining values.

Suppose that X

, X

, ..., X

is a random sample from a continuous distribu-

tion with cumulative distributio n function F(x) and density function f(x ). Because

of continuity, for any i, j with i 6¼ j, P(X

¼ X

) ¼ 0. This implies that with

probability 1, the n sample observations will all be different (of course, in practice

all measuring instruments have accuracy limitations, so tied values may in fact

result).

5.5 Order Statistics 271

DEFINITION

The order statistics from a random sample are the rando m variables Y

, ...

given by

¼ the smallest among X

, X

, ..., X

¼ the second smallest among X

, X

, ..., X

...

¼ the largest among X

, X

, ..., X

so that with probability 1, Y

< Y

< ... < Y

n  1

< Y

The sample median is then Y

(n + 1)/2

when n is odd, the sample range is Y

 Y

, and

for n ¼ 10 the 20% trimmed mean is

i¼3

=6. The order statistics are defined as

random variables (hence the use of uppercase letters); observed values are denoted

by y

, ..., y

The Distributions of Y

and Y

The key idea in obtaining the distribution of the largest order statistic is that Y

is at

most y if and only if every one of the X

’s is at most y. Similarly, the distribution of

is based on the fact that it will be at least y if and only if all X

’s are at least y.

Example 5.28 Consider 5 identical components connected in parallel, as illustrated in Figure 5.8(a).

Let X

denote the lifetime (hr) of the ith component (i ¼ 1, 2, 3, 4, 5). Suppose that

the X

’s are independent and that each has an exponential distribution with l ¼ .01,

so the expected lifetime of any particular component is 1/l ¼ 100 h. Because of the

parallel configuration, the system will continue to function as long as at lea st one

component is still working, and will fail as soon as the last component functioning

ceases to do so. That is, the system lifetime is just Y

, the largest order statistic in a

sample of size 5 from the specifi ed exponential distribution. Now Y

will be at most

y if and only if every one of the five X

’s is at most y. With G

(y) denoting the

cumulative distribution function of Y

ðyÞ¼PY

 yðÞ¼PX

 y; X

 y; :::; X

 yðÞ

¼ PX

 yðÞPX

 yð Þ  PX

 yðÞ¼FðyÞ½

¼ 1  e

:01y



The pdf of Y

can now be obtained by differentiating the cdf with respect to y.

Suppose instead that the five components are connected in series rather than

in parallel (Figure 5.8(b)). In this case the system lifetime will be Y

, the smallest of

the five order statistics, since the system will crash as soon as a sing le one of the

individual components fails. Note that system lifetime will exceed y hr if and only

if the lifetime of every component exceeds y hr. Thus

ðyÞ¼PY

 yðÞ¼1  PY

> yðÞ¼1  PX

> y; X

> y; :::; X

> yðÞ

¼ 1  PX

> yðÞPX

> yð Þ  PX

> yðÞ¼1  e

:01y



¼ 1  e

:05y

This is the form of an exponential cdf with parameter .05. More generally, if the

n components in a series connection have lifetimes that are independent, each

exponentially distributed with the same parameter l, then system lifetime will be

272 CHAPTER 5 Joint Probability Distributions

exponentially distributed with parameter nl. The expected system lifetime will then

be 1/nl, much smaller than the expected lifetime of an individual component.

■

An argument para llel to that of the previous example for a general sample

size n and an arbitrary pdf f(x) gives the following general results.

PROPOSITION

Let Y

and Y

denote the smallest and largest order statistics, respectively,

based on a random sample from a continuo us distribution with cdf F(x) and

pdf f(x). Then the cdf and pdf of Y

are

ðyÞ¼ FðyÞ½

ðyÞ¼nFðyÞ½

1

 f ðyÞ

The cdf and pdf of Y

are

ðyÞ¼1  1  FðyÞ

ðyÞ¼n

½

1  FðyÞ

½

1

 f ðyÞ

Example 5.29 Let X denote the contents of a one-gallon container, and suppose that its pdf is f(x)

¼ 2x for 0  x  1 (and 0 otherwise) with corresponding cdf F(x) ¼ x

in the

interval of positive density. Con sider a random sample of four such containers.

Let’s determine the expected value of Y

 Y

, the difference between the contents

of the most-filled container and the least-filled contain er; Y

 Y

is just the sample

range. The pdf’s of Y

and Y

are

ðyÞ¼4 y

ðÞ

 2y 0  y  1

ðyÞ¼41 y

ðÞ

 2y 0  y  1

The correspo nding density curves appear in Figure 5.9

Figure 5.8 Systems of components for Example 5.28: (a) parallel connection;

(b) series connection

5.5 Order Statistics 273

EðY

 Y

Þ¼EðY

ÞEðY

Þ¼

y  8y

dy

y  8yð1  y



384

945

¼ :889  :406 ¼ :483

If random samples of four containers were repeatedly selected and the sample

range of contents determined for each one, the long run average value of the range

would be .483.

■

The Joint Distribution of the n Order Statistics

We now develop the joint pdf of Y

, Y

, ..., Y

. Consider first a random sample

, X

of fuel efficiency measurements (mpg). The joint pdf of this random

sample is

; x

ðÞ¼fx

ðÞfx

ðÞ

The joint pdf of Y

, Y

will be positive only for values of y

, y

satisfying

< y

. What is this joint pdf at the values y

¼ 28.4, y

¼ 29.0, y

¼ 30.5?

There are six different ways to obtain these ordered values:

¼ 28.4 X

¼ 29.0 X

¼ 30.5

¼ 28.4 X

¼ 30.5 X

¼ 29.0

¼ 29.0 X

¼ 28.4 X

¼ 30.5

¼ 29.0 X

¼ 30.5 X

¼ 28.4

¼ 30.5 X

¼ 28.4 X

¼ 29.0

¼ 30.5 X

¼ 29.0 X

¼ 28.4

These six possibilities come from the 3! ways to order the three numerical observa-

tions once their values are fixed. Thus

g 28:4;29:0;30:5ðÞ¼f 28: 4ðÞf 29:0ðÞf 30:5ðÞþþf 30: 5ðÞf 29:0ðÞf 28:4ðÞ

¼3!f 28: 4ðÞf 29:0ðÞf 30:5ðÞ

Figure 5.9 Density curves for the order statistics (a) Y

and (b) Y

in Example 5.29

274

CHAPTER 5 Joint Probability Distributions

Analogous reasoning with a sample of size n yields the following result:

PROPOSITION

Let g(y

, y

, ..., y

) denote the joint pdf of the order statistics Y

, Y

, ..., Y

resulting from a random sample of X

’s from a pdf f(x). Then

gðy

; y

; ...; y

Þ¼

n!f ðy

Þf ðy

Þf ðy

Þ y

< y

<  < y

0 otherwise



For example, if we have a random sample of component lifetimes and the lifetime

distribution is exponential with parameter l, then the joint pdf of the order statistics is

gðy

; ...; y

Þ¼n!l

lðy

þþy

0 < y

< y

<  < y

Example 5.30 Suppose X

, X

, and X

are independent random variables, each uniformly

distributed on the interval from 0 to 1. The joint pdf of the four corresponding order

statistics Y

, Y

, and Y

is f(y

, y

) ¼ 4!∙1 for 0 < y

< y

< 1.

The probability that every pair of X

s is separated by more than .2 is the same as the

probability that Y

 Y

> .2, Y

 Y

> .2, and Y

 Y

> .2. This latter

probability results from integrating the joint pdf of the Y

s over the region .6 < y

< 1, .4 < y

< y

 .2, .2 < y

< y

 .2, 0 < y

< y

 .2:

PðY

Y

>:2; Y

Y

>:2; Y

Y

>:2Þ¼

:2

4!dy

The inner integration gives 4!(y

 .2), and this must then be integrated between .2

and y

.2. Making the change of variable z

¼y

.2, the integration of z

is from

0toy

 .4. The result of this integration is 4!∙(y

 .4)

/2. Continuing with the 3rd

and 4th integration, each time making an appropriate change of variable so that the

lower limit of each integration becomes 0, the result is

PðY

 Y

>:2; Y

 Y

>:2 ; Y

 Y

>:2Þ¼:4

¼ :0256

A more gener al multiple integration argument for n independe nt uniform (0, B)

rvs shows that the probability that at all values are separated by at least d is 0 if d 

B/(n  1) and

Pðall values are separated by more than dÞ

½1 ðn  1Þd=B

0  d  B=ðn  1Þ

0 d > B= ðn  1Þ



As an application, consider a year that has 365 days, and suppose that the birth time

of someone born in that year is uniformly distributed throughout the 365-day

period. Then in a group of 10 independently selected people born in that year, the

probability that all of their birth times are separated by more than 24 h (d ¼1 day)

is (1  9/365)

¼ .779. Thus the probability that at least two of the 10 birth times

are separated by at most 24 h is .221. As the group size n increases, it becomes more

likely that at least two people have birth times that are within 24 h of each other

5.5 Order Statistics 275

(but not necessarily on the same day). For n ¼ 16, this probability is .467, and for

n ¼ 17 it is .533. So with as few as 17 people in the group, it is more likely than not

that at least two of the people were born within 24 h of each other. Coincidences

such as this are not as surprising as one might think. The probability that at least

two people are born on the same day (assuming equally likely birthdays) is much

easier to calculate than what we have shown here; see Exercise 2.98.

■

The Distribution of a Single Order Statistic

We have already obtained the (marginal) distribution of the largest order statistic Y

and also that of the smallest order statistic Y

. Let’s now focus on an intermediate

order statistic Y

where 1 < i < n. For concreteness, consider a random sample X

, ..., X

of n ¼ 6 component lifetimes, and suppos e we wish the distribution of

the 3

smallest lifetime Y

. Now the joint pdf of all six order statistics is

gðy

; y

; ::: ; y

Þ¼6! fy

ðÞfy

ðÞ y

< y

To obtain the pdf of Y

alone, we must hold y

fixed in the joint pdf and integr ate

out all the other y

’s. One way to do this is to

1. Int egrate y

from 1 to y

, and then integrate y

from 1to y

2. Int egrate y

from y

to 1, then integrate y

from y

to 1, and finally integrate y

from y

to 1.

That is,

gðy

Þ¼

1

6!f ðy

Þf ðy

Þf ðy

Þdy

¼6!

1

f ðy

Þf ðy

Þdy





f ðy

Þf ðy

Þdy



f ðy

In these integrations we use the following general results:

½FðxÞ

f ð xÞdx ¼

k þ 1

½FðxÞ

kþ1

þ c ½let u ¼ Fð xÞ

½1  FðxÞ

f ð xÞdx ¼

k þ 1

½1  FðxÞ

kþ1

þ c ½let u ¼ 1  FðxÞ

Therefore

1

f ð y

Þf ðy

Þ dy

1

Fðy

Þf ðy

Þ dy

½Fðy

Þ

276 CHAPTER 5 Joint Probability Distributions

and

f ð y

Þf ðy

Þ dy

½1  Fðy

Þf ð y

Þf ðy

Þ dy

¼

½1  Fðy

Þ

f ð y

Þ dy

3  2

½1  Fðy

Þ

Thus

gðy

Þ¼

2!3!

½Fðy

Þ

½1  Fðy

Þ

f ð y

Þ1< y

< 1

A generalization of the foregoing argument gives the following expre ssion

for the pdf of any single order statistic.

PROPOSITION

The pdf of the ith smallest order statistic Y

gðy

Þ¼

ði  1Þ! ðn  iÞ!

½Fðy

Þ

i1

½1  Fðy

Þ

ni

f ð y

Þ1< y

< 1

Example 5.31 Suppose that component lifetime is exponentially distributed with parameter l. For

a random sample of n ¼ 5 components, the expected value of the sample median

lifetime is

EðY

Þ¼

y 

2!  2!

ð1  e

ly

ðe

ly

 le

ly

Expanding out the integrand and integrating term by term, the expected

value is .783/l. The median of the exponential distribution is, from solving

Fð

mÞ¼:5;

m ¼ :693=l . Thus if sample after sample of five components is

selected, the long run average value of the sample median will be somewhat larger

than the value of the lifetime population distribution median. This is because the

exponential distribution has a positive skew.

■

The Joint Distribution of Two Order Statistics

We now focus on the joint distribution of two order statistics Y

and Y

with i < j.

Consider first n ¼ 6 and the two order statistics Y

and Y

We must then take the

joint pdf of all six order statistics, hold y

and y

fixed, and integrate out y

, y

and y

.Thatis,

gðy

; y

Þ¼

1

6! f ðy

Þf ðy

Þdy

5.5 Order Statistics 277