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135. The reaction time (in seconds) to a stimulus is a

continuous random variable with pdf

f ðxÞ¼

1  x  3

0 otherwise

(

a. Obtain the cdf.

b. What is the probability that reaction time is at

most 2.5 s? Between 1.5 and 2.5 s?

c. Compute the expected reaction time.

d. Compute the standard deviation of reaction

time.

e. If an individual takes more than 1.5 s to react,

a light comes on and stays on either until one

further second has elapsed or until the person

reacts (whichever happens first). Determine

the expected amount of time that the light

remains lit. [Hint: Let h( X) ¼ the time that

the light is on as a function of reaction time

X.]

136. Let X denote the temperature at which a certain

chemical reaction takes place. Suppose that X

has pdf

f ðxÞ¼

ð4  x

Þ1  x  2

0 otherwise



a. Sketch the graph of f(x).

b. Determine the cdf and sketch it.

c. Is 0 the median temperature at which the

reaction takes place? If not, is the median

temperature smaller or larger than 0?

d. Suppose this reaction is independently carried

out once in each of ten different labs and that

the pdf of reaction time in each lab is as given.

Let Y ¼ the number among the ten labs at

which the temperature exceeds 1. What kind

of distribution does Y have? (Give the name

and values of any parameters.)

137. The article “Determination of the MTF of Posi-

tive Photoresists Using the Monte Carlo Method”

(Photographic Sci. Engrg., 1983: 254–260) pro-

poses the exponential distribution with parameter

l ¼ .93 as a model for the distribution of a

photon’s free path length (mm) under certain

circumstances. Suppose this is the correct model.

a. What is the expected path length, and what is

the standard deviation of path length?

b. What is the probability that path length

exceeds 3.0? What is the probability that

path length is between 1.0 and 3.0?

c. What value is exceeded by only 10% of all

path lengths?

138. The article “The Prediction of Corrosion by

Statistical Analysis of Corrosion Profiles” (Cor-

rosion Sci., 1985: 305–315) suggests the follow-

ing cdf for the depth X of the deepest pit in an

experiment involving the exposure of carbon

manganese steel to acidified seawater.

Fðx; a; bÞ¼e

e

ðxaÞ=b

1< x < 1

The authors propose the values a ¼ 150 and

b ¼ 90. Assume this to be the correct model.

a. What is the probability that the depth of the

deepest pit is at most 150? At most 300?

Between 150 and 300?

b. Below what value will the depth of the maxi-

mum pit be observed in 90% of all such

experiments?

c. What is the density function of X?

d. The density function can be shown to be

unimodal (a single peak). Above what value

on the measurement axis does this peak

occur? (This value is the mode.)

e. It can be shown that E(X)  .5772b + a.

What is the mean for the given values of a

and b, and how does it compare to the median

and mode? Sketch the graph of the density

function. [Note: This is called the largest

extreme value distribution.]

139. Let t ¼ the amount of sales tax a retailer owes the

government for a certain period. The article “Sta-

tistical Sampling in Tax Audits” (Statistics and

the Law, 2008: 320–343) proposes modeling the

uncertainty in t by regarding it as a normally

distributed random variable with mean value m

and standard deviation s (in the article, these

two parameters are estimated from the results of

a tax audit involving n sampled transactions). If a

represents the amount the retailer is assessed, then

an underassessment results if t > a and an over-

assessment if a > t. We can express this in terms

of a loss function, a function that shows zero loss

if t ¼ a but increases as the gap between t and a

increases. The proposed loss function is L(a,t) ¼

t  a

if t > a and ¼ k(a  t)ift  a (k > 1is

suggested to incorporate the idea that overassess-

ment is more serious than underassessment).

a. Show that a



¼ m þ sF

1

1=ðk þ 1ÞðÞis the

value of a that minimizes the expected loss,

where F

1

is the inverse function of the stan-

dard normal cdf.

b. If k ¼ 2 (suggested in the article),

m ¼ $100,000, and s ¼ $10,000, what is the

optimal value of a, and what is the resulting

probability of overassessment?

228

CHAPTER 4 Continuous Random Variables and Probability Distributions

140. A mode of a continuous distribution is a value x*

that maximizes f(x).

a. What is the mode of a normal distribution

with parameters m and s?

b. Does the uniform distribution with para-

meters A and B have a single mode? Why or

why not?

c. What is the mode of an exponential distribu-

tion with parameter l? (Draw a picture.)

d. If X has a gamma distribution with parameters

a and b, and a > 1, find the mode. [Hint:

ln[f(x)] will be maximized if and only if f(x)

is, and it may be simpler to take the derivative

of ln[f(x)].]

e. What is the mode of a chi-squared distribution

having n degrees of freedom?

141. The article “Error Distribution in Navigation”

(J. Institut. Navigation, 1971: 429–442) suggests

that the frequency distribution of positive errors

(magnitudes of errors) is well approximated by

an exponential distribution. Let X ¼ the lateral

position error (nautical miles), which can be

either negative or positive. Suppose the pdf of

X is

f ðxÞ¼ð:1Þe

:2jxj

1< x < 1

a. Sketch a graph of f(x) and verify that f(x)isa

legitimate pdf (show that it integrates to 1).

b. Obtain the cdf of X and sketch it.

c. Compute P(X  0), P(X  2), P(1  X

 2), and the probability that an error of

more than 2 miles is made.

142. In some systems, a customer is allocated to one

of two service facilities. If the service time for a

customer served by facility i has an exponential

distribution with parameter l

(i ¼ 1, 2) and p is

the proportion of all customers served by facility 1,

then the pdf of X ¼ the service time of a randomly

selected customer is

f ðx; l

; l

; pÞ

l

þð1  pÞl

l

x  0

0 otherwise

(

This is often called the hyperexponential or

mixed exponential distribution. This distribution

is also proposed as a model for rainfall amount in

“Modeling Monsoon Affected Rainfall of Paki-

stan by Point Processes” (J. Water Resources

Planning Manag., 1992: 671–688).

a. Verify that f(x; l

, l

, p) is indeed a pdf.

b. What is the cdf F(x; l

, l

, p)?

c. If X has f(x; l

, l

, p) as its pdf, what is E(X)?

d. Using the fact that E(X

) ¼ 2/l

when X has

an exponential distribution with parameter l,

compute E(X

) when X has pdf f(x; l

, l

, p).

Then compute V(X).

e. The coefficient of variation of a random vari-

able (or distribution) is CV ¼ s/m. What is the

CV for an exponential rv? What can you say

about the value of CV when X has a hyper-

exponential distribution?

f. What is the CV for an Erlang distribution with

parameters l and n as defined in Exercise 76?

[Note: In applied work, the sample CV is used

to decide which of the three distributions

might be appropriate.]

143. Suppose a state allows individuals filing tax

returns to itemize deductions only if the total of

all itemized deductions is at least $5,000. Let X

(in 1,000’s of dollars) be the total of itemized

deductions on a randomly chosen form. Assume

that X has the pdf

f ðx; aÞ¼

k=x



x  5

otherwise

a. Find the value of k. What restriction on a is

necessary?

b. What is the cdf of X?

c. What is the expected total deduction on a

randomly chosen form? What restriction on

a is necessary for E(X) to be finite?

d. Show that ln(X/5) has an exponential distri-

bution with parameter a  1.

144. Let I

be the input current to a transistor and I

the output current. Then the current gain is pro-

portional to ln(I

). Suppose the constant of

proportionality is 1 (which amounts to choosing

a particular unit of measurement), so that current

gain ¼ X ¼ ln(I

). Assume X is normally

distributed with m ¼ 1 and s ¼ .05.

a. What type of distribution does the ratio I

have?

b. What is the probability that the output current

is more than twice the input current?

c. What are the expected value and variance of

the ratio of output to input current?

145. The article “Response of SiC

/Si

Composites

Under Static and Cyclic Loading—An Experi-

mental and Statistical Analysis” (J. Engrg. Mate-

rials Tech., 1997: 186–193) suggests that tensile

strength (MPa) of composites under specified

conditions can be modeled by a Weibull distri-

bution with a ¼ 9 and b ¼ 180.

Supplementary Exercises 229

a. Sketch a graph of the density function.

b. What is the probability that the strength of a

randomly selected specimen will exceed 175?

Will be between 150 and 175?

c. If two randomly selected specimens are cho-

sen and their strengths are independent of

each other, what is the probability that at

least one has strength between 150 and 175?

d. What strength value separates the weakest

10% of all specimens from the remaining

90%?

146. a. Suppose the lifetime X of a component, when

measured in hours, has a gamma distribution

with parameters a and b. Let Y ¼ lifetime

measured in minutes. Derive the pdf of Y.

b. If X has a gamma distribution with parameters

a and b, what is the probability distribution of

Y ¼ cX?

147. Based on data from a dart-throwing experiment,

the article “Shooting Darts” (Chance, Summer

1997: 16–19) proposed that the horizontal and

vertical errors from aiming at a point target

should be independent of each other, each with

a normal distribution having mean 0 and vari-

ance s

. It can then be shown that the pdf of the

distance V from the target to the landing point is

f ðnÞ¼

 e

n

=ð2s

n > 0

a. This pdf is a member of what family intro-

duced in this chapter?

b. If s ¼ 20 mm (close to the value suggested in

the paper), what is the probability that a dart

will land within 25 mm (roughly 1 in.) of the

target?

148. The article “Three Sisters Give Birth on the

Same Day”(Chance, Spring 2001: 23–25) used

the fact that three Utah sisters had all given

birth on March 11, 1998, as a basis for posing

some interesting questions regarding birth coin-

cidences.

a. Disregarding leap year and assuming that the

other 365 days are equally likely, what is the

probability that three randomly selected births

all occur on March 11? Be sure to indicate

what, if any, extra assumptions you are

making.

b. With the assumptions used in part (a), what is

the probability that three randomly selected

births all occur on the same day?

c. The author suggested that, based on extensive

data, the length of gestation (time between

conception and birth) could be modeled as

having a normal distribution with mean

value 280 days and standard deviation 19.88

days. The due dates for the three Utah sisters

were March 15, April 1, and April 4, respec-

tively. Assuming that all three due dates are at

the mean of the distribution, what is the prob-

ability that all births occurred on March 11?

[Hint: The deviation of birth date from due

date is normally distributed with mean 0.]

d. Explain how you would use the information

in part (c) to calculate the probability of a

common birth date.

149. Let X denote the lifetime of a component, with

f(x) and F(x) the pdf and cdf of X. The proba-

bility that the component fails in the interval

(x, x + Dx) is approximately f(x)·Dx. The condi-

tional probability that it fails in (x, x + Dx) given

that it has lasted at least x is f(x)·D x/[1  F(x)].

Dividing this by

Dx produces the failure rate

function:

rðxÞ¼

f ðxÞ

1  FðxÞ

An increasing failure rate function indicates

that older components are increasingly likely to

wear out, whereas a decreasing failure rate is

evidence of increasing reliability with age. In

practice, a “bathtub-shaped” failure is often

assumed.

a. If X is exponentially distributed, what is r(x)?

b. If X has a Weibull distribution with para-

meters a and b, what is r(x)? For what param-

eter values will r(x) be increasing? For what

parameter values will r(x) decrease with x?

c. Since r(x) ¼(d/dx)ln[1  F(x)],

ln[1  F(x)] ¼

r(x) dx. Suppose

rðxÞ¼

a 1 



0  x  b

0 otherwise

(

so that if a component lasts b hours, it will last

forever (while seemingly unreasonable, this

model can be used to study just “initial wear-

out”). What are the cdf and pdf of X?

150. Let U have a uniform distribution on the interval

[0, 1]. Then observed values having this distribu-

tion can be obtained from a computer’s random

number generator. Let X ¼(1/l)ln(1  U).

a. Show that X has an exponential distribution

with parameter l.

b. How would you use part (a) and a random

number generator to obtain observed values

230

CHAPTER 4 Continuous Random Variables and Probability Distributions

from an exponential distribution with param-

eter l ¼ 10?

151. If the voltage v across a medium is fixed but

current I is random, then resistance will also be

a random variable related to I by R ¼ v/I.If

¼ 20 and s

¼ .5, calculate approximations

to m

and s

152. A function g(x)isconvex if the chord connecting

any two points on the function’s graph lies above

the graph. When g(x) is differentiable, an equiv-

alent condition is that for every x, the tangent line

at x lies entirely on or below the graph. (See the

figure below.) How does g(m) ¼ g[E(X)] com-

pare to E[g(X)]? [Hint: The equation of the tan-

gent line at x ¼ m is y ¼ g(m)+g

(m)·(x  m).

Use the condition of convexity, substitute X

for x, and take expected values. Note: Unless

g(x) is linear, the resulting inequality (usually

called Jensen’s inequality) is strict (< rather

than ); it is valid for both continuous and dis-

crete rv’s.]

Tangent

line

153. Let X have a Weibull distribution with parameters

a ¼ 2andb. Show that Y ¼ 2X

has a chi-

squared distribution with n ¼ 2.

154. Let X have the pdf f(x) ¼ 1/[p(1 + x

)] for 1

< x < 1 (a central Cauchy distribution), and

show that Y ¼ 1/X has the same distribution.

[Hint: Consider P(|Y|  y), the cdf of |Y|, then

obtain its pdf and show it is identical to the pdf

of |X|.]

155. A store will order q gallons of a liquid product to

meet demand during a particular time period.

This product can be dispensed to customers in

any amount desired, so demand during the period

is a continuous random variable X with cdf F(x).

There is a fixed cost c

for ordering the product

plus a cost of c

per gallon purchased. The per-

gallon sale price of the product is d. Liquid left

unsold at the end of the time period has a salvage

value of e per gallon. Finally, if demand exceeds

q, there will be a shortage cost for loss of good-

will and future business; this cost is f per gallon

of unfulfilled demand. Show that the value of q

that maximizes expected profit, denoted by q*,

satisfies

Pðsatisfying demand) ¼ FðqÞ ¼

d  c

þ f

d  e þ f

Then determine the value of F(q*) if d ¼ $35,

¼ $25, c

¼ $15, e ¼ $5, and f ¼ $25. [Hint:

Let x denote a particular value of X. Develop an

expression for profit when x  q and another

expression for profit when x > q. Now write an

integral expression for expected profit (as a func-

tion of q) and differentiate.]

156. An insurance company issues a policy covering

losses up to 5 (in thousands of dollars). The loss,

X, follows a distribution with density function:

f ðxÞ¼

x  1

0 x < 1

What is the expected value of the amount paid

under the policy?

Bibliography

Bury, Karl, Statistical Distributions in Engineering,

Cambridge University Press, Cambridge, England,

1999. A readable and informative survey of distri-

butions and their properties.

Johnson, Norman, Samuel Kotz, and N. Balakrishnan,

Continuous Univariate Distributions, vols. 1–2,

Wiley, New York, 1994. These two volumes

together present an exhaustive survey of various

continuous distributions.

Nelson, Wayne, Applied Life Data Analysis, Wiley,

New York, 1982. Gives a comprehensive discus-

sion of distributions and methods that are used in

the analysis of lifetime data.

Olkin, Ingram, Cyrus Derman, and Leon Gleser,

Probability Models and Applications (2nd ed.),

Macmillan, New York, 1994. Good coverage of

general properties and specific distributions.

Bibliography 231

CHAPTER FIVE

Joint Probability

Distributions

Introduction

In Chapters 3 and 4, we studied probability models for a single random variable.

Many problems in probability and statistics lead to models involving several

random variables simultaneously. In this chapter, we ﬁrst discuss pro bability

models for the joint behavior of several random variables, putting special emphasis

on the case in which the variables are independent of each other. We then

study expected values of functions of several random variables, including covari-

ance and correlation as measures of the degree of association between two

variables.

The third section considers conditional distributions, the distributions of

random variables given the values of other random variables. The next section is

about transformations of two or more random variables, generalizing the results of

Section 4.7. In the last section of this chap ter we discuss the distribution of order

statistics: the minimum, maximum, median, and other statistics that can be found

by arranging the observations in order.

J.L. Devore and K.N. Berk, Modern Mathematical Statistics with Applications, Springer Texts in Statistics,

DOI 10.1007/978-1-4614-0391-3_5,

Springer Science+Business Media, LLC 2012

232

5.1

Jointly Distributed Random Variables

There are many experimental situations in which more than one random variable

(rv) will be of interest to an investigator. We shall first consider joi nt probability

distributions for two discrete rv’s, then for two continuous variables, and finally for

more than two variables.

The Joint Probability Mass Function

for Two Discrete Random Variables

The probability mass function (pmf) of a single discrete rv X specifies how much

probability mass is placed on each possible X value. The joint pmf of two discrete

rv’s X and Y describes how much probability mass is placed on each possible pair

of values (x, y).

DEFINITION

Let X and Y be two discrete rv’s defined on the sample space S of an

experiment. The joint probability mass function p(x, y) is defined for each

pair of numbers (x, y )by

px; yðÞ¼PðX ¼ x and Y ¼ yÞ

Let A be any set consisting of pairs of (x, y) values. Then the probability that

the random pair (X, Y) lies in A is obtained by summing the joint pmf over

pairs in A:

P½ðX; YÞ2A¼

ðx;yÞ

pðx; yÞ

Example 5.1 A large insurance agency services a number of customers who have purchased both

a homeowner’s policy and an automobile policy from the agency. For each type of

policy, a deductible amount must be specified. For an automobile policy, the

choices are $100 and $250, whereas for a homeowner’s policy, the choices are 0,

$100, and $200. Suppose an individual with both types of policy is selected at

random from the agency’s files. Let X ¼ the deductible amount on the auto policy

and Y ¼ the deductible amount on the homeowner’s policy. Possible (X, Y) pairs are

then (100, 0), (100, 100), (100, 200), (250, 0), (250, 100), and (250, 200); the joint

pmf specifies the probability associated with each one of these pairs, with any other

pair having probability zero. Suppose the joint pmf is given in the accompanying

joint probability table:

p(x, y) 0 100 200

100 .20 .10 .20

250 .05 .15 .30

5.1 Jointly Distributed Random Variables 233

Then p(100, 100) ¼ P(X ¼ 100 and Y ¼100) ¼ P($100 deductible on both policies)

¼ .10. The probability P(Y  100) is computed by summing probabilities of all

(x, y) pairs for which y  100:

PðY 100Þ¼p 100;100ðÞþp 250;100ðÞþp 100;200ðÞþp 250;200ðÞ¼:75

■

A function p(x, y) can be used as a joint pmf provided that p( x, y)  0 for all x

and y and

pðx; yÞ¼1 :

The pmf of one of the variables alon e is obtained by summing p(x, y)

over values of the other variable. The result is called a marginal pmf because

when the p(x, y) values appear in a rectangular table, the sums are just marginal

(row or column) totals

DEFINITION

The marginal probability mass functions of X and of Y, denoted by p

(x)

and p

(y), respectively, are given by

ðxÞ¼

pðx; yÞ p

ðyÞ¼

pðx; yÞ

Thus to obtain the marginal pmf of X evaluated at, say, x ¼ 100, the probabilities

p(100, y) are added over all possible y values. Doing this for each possible X value

gives the marginal pmf of X alone (without reference to Y). From the marginal pmf’s,

probabilities of events involving only X or only Y can be computed.

Example 5.2

(Example 5.1

continued)

The possible X v alues are x ¼ 100 and x ¼ 250, so computing row totals in the joint

probability table yields

100ðÞ¼p 100; 0ðÞþp 100; 100ðÞþp 100; 200ðÞ¼:50

And

250ðÞ¼p 250; 0ðÞþp 250; 100ðÞþp 250; 200ðÞ¼:50

The marginal pmf of X is then

ðxÞ¼



x ¼ 100; 250

otherwise

Similarly, the marginal pmf of Y is obtained from column totals as

ðyÞ¼

:25

:50

y ¼ 0; 100

y ¼ 20

otherwise

so PðY  100Þ¼p

100ðÞþp

200ðÞ¼:75 as before. ■

234 CHAPTER 5 Joint Probability Distributions

The Joint Probability Density Function for Two

Continuous Random Variables

The probability that the observed value of a continuous rv X lies in a one-

dimensional set A (such as an interval) is obtained by integrating the pdf f(x) over

the set A. Similarly, the probability that the pair (X, Y) of continuous rv’s falls in a

two-dimensional set A (such as a rectangle) is obtained by integrating a function

called the joint density function.

DEFINITION

Let X and Y be continuous rv’s. Then f(x, y) is the joint probability density

function for X and Y if for any two-dimensional set A

P½ðX; YÞ2A¼

f ð x; yÞdx dy

In particular, if A is the two-dimensional rectangle fðx; yÞ :

a  x  b; c  y  dg; then

P½ðX; YÞ2A¼Pða  X  b; c  Y  dÞ¼

f ð x; yÞdy dx

For f(x, y) to be a candidate for a joint pdf, it must satisfy f(x, y)  0

and

1

f ðx; yÞdx dy ¼ 1: We can think of f(x, y) as specifying a surface at

height f(x, y) above the point (x, y ) in a three-dimensional coordinate system. Then

P[(X, Y) 2 A] is the volume underneath this surface and above the region A,

analogous to the area under a curve in the one-dimensional case. This is illustrated

in Figure 5.1.

Example 5.3 A bank operates both a drive-up facility and a walk-up window. On a randomly

selected day, let X ¼ the proportion of time that the drive-up facility is in use (at

least one customer is being served or waiting to be served) and Y ¼ the proportion

of time that the walk-up window is in use. Then the set of possible values for (X, Y)

f (x, y)

Surface f (x, y)

A = Shaded

rectangle

Figure 5.1

[(

) 2

] ¼ volume under density surface above

5.1 Jointly Distributed Random Variables 235

is the rectangle D ¼fðx; yÞ : 0  x  1; 0  y  1g: Suppose the joint pdf of

(X, Y) is given by

f ð x; yÞ¼

ðx þ y

Þ 0  x  1; 0  y  1

0 otherwise

To verify that this is a legitimate pdf, note that f (x, y)  0 and

1

f ð x; yÞdx dy ¼

ðx þ y

Þdx dy

xdx dyþ

dx dy

xdxþ

dy ¼

¼ 1

The probability that neither facility is busy more than one-quarter of the time is

P 0  X 

; 0  Y 



1=4

ðx þ y

Þdx dy

1=4

xdxdyþ

1=4

dx dy





x¼1=4

x¼0





y¼1=4

y¼0

640

¼ :0109

■

As with joint pmf’s, from the joint pdf of X and Y, each of the two marginal

density functions can be computed.

DEFINITION

The marginal probability density functions of X and Y, denoted by f

(x)

and f

(y), respectively, are given by

ðxÞ¼

1

f ð x; yÞdy for 1< x < 1

ðyÞ¼

1

f ð x; yÞdx for 1< y < 1

Example 5.4

(Example 5.3

continued)

The marginal pdf of X, which gives the probability distribution of busy time for the

drive-up facil ity without reference to the walk-up window, is

ðxÞ¼

1

f ð x; yÞdy ¼

ðx þ y

Þdy ¼

x þ

236 CHAPTER 5 Joint Probability Distributions

for 0  x  1 and 0 otherwise. The marginal pdf of Y is

ðyÞ¼

0  y  1

0 otherwise

Then

 Y 



3=4

1=4



dy ¼

¼ :4625:

■

In Example 5.3, the region of positive joint density was a rectangle, which

made computation of the marginal pdf’s relatively easy. Consider now an example

in which the region of positive density is a more complicated figure.

Example 5.5 A nut company markets cans of deluxe mixed nuts containing almonds, cashews, and

peanuts. Suppose the net weight of each can is exactly 1 lb, but the weight contribution

of each type of nut is random. Because the three weights sum to 1, a joint probability

model for any two gives all necessary information about the weight of the third

type. Let X ¼ the weight of almonds in a selected can and Y ¼ the weight of cashews.

Then the region of positive density is D ¼fðx; yÞ : 0  x  1; 0  y  1;

x þ y  1 g, the shaded region pictured in Figure 5.2.

Now let the joint pdf for (X, Y)be

f ð x; yÞ¼

24xy 0  x  1; 0  y  1; x þ y  1

0 otherwise

(

For any fixed x, f(x, y) increases with y; for fixed y, f(x, y) increases with x. This is

appropriate because the word deluxe implies that most of the can should consist of

almonds and cashews rather than peanuts, so that the density function should be

large near the upper boundary and small near the origin. The surface determined by

f(x, y) slopes upward from zero as (x

, y) moves away from either axis.

Clearly, f(x, y)  0. To verify the second condition on a joint pdf, recall that a

double integral is computed as an iterated integral by holding one variable fixed

(such as x as in Figure 5.2), integrating over values of the other variable lying along

(0, 1)

(1, 0)

(x, 1−x)

Figure 5.2 Region of positive density for Example 5.5

5.1 Jointly Distributed Random Variables 237