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

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When the pmf p(x) specifies a mathematical model for the distribution of

population values, both s

and s measure the spread of values in the population;

is the population variance, and s is the population standard deviation.

A Shortcut Formula for s

The number of arithmetic operations necessary to compute s

can be reduced by

using an alternative computing formula.

PROPOSITION

VðXÞ¼s

 pðxÞ

 m

¼ EðX

Þ½EðXÞ

In using this formula, E(X

) is computed first without any subtraction; then

E(X) is computed, squared, and subtracted (once) from E(X

Example 3.24 Referring back to the Apgar score scenario of Examples 3.16 and 3.23,

EðX

Þ¼

x¼0

 pðxÞ¼ð0

Þð:002Þþð1

Þð:001Þþ::: þð10

Þð:01Þ¼52:704

Thus s

¼ 52:704  7:15ðÞ

¼ 1:5815 as before:

■

Proof of the Shortcut Formu la Expand (x  m)

in the definition of s

obtain x

 2mx + m

, and then carry S through to each of the three terms:

 pðxÞ2m 

x  pðxÞþm

pðxÞ

¼ EðX

Þ2m  m þ m

¼ EðX

Þm

■

Rules of Variance

The variance of h(X) is the expected value of the squared difference between h(X)

and its expected value:

V½hðXÞ ¼ s

hðXÞ

fhðxÞE½hðxÞg

 pðxÞð3:13Þ

When h(x) is a linear function, V[h(X)] is easily related to V(X) (Exercise 40).

PROPOSITION

VðaX þ bÞ¼s

aXþb

¼ a

 s

and s

aXþb

¼jajs

118 CHAPTER 3 Discrete Random Variables and Probability Distributions

This result says that the addition of the constant b does not affect the

variance, which is intuitive, because the addition of b changes the location (mean

value) but not the spread of values. In particular,

¼ a

 s

¼jajs

ð3:14Þ

2. s

Xþb

¼ s

The reason for the absolute value in s

is that a may be negative, whereas a

standard deviation cannot be negative; a

results when a is brought outside the term

being squared in Equation 3.13.

Example 3.25 In the computer sales scenario of Example 3.22, E(X) ¼ 2 and

EðX

Þ¼ð0

Þð:1Þþð1

Þð:2Þþð2

Þð:3Þþð3

Þð:4Þ¼5

so V(X) ¼ 5  (2)

¼ 1. The profit function h(X) ¼ 800X  900 then has variance

(800)

·V(X) ¼ (640,000)(1) ¼ 640,000 and standard deviation 800. ■

Exercises Section 3.3 (28–43)

28. The pmf for X ¼ the number of major defects on

a randomly selected appliance of a certain type is

x 0 1234

p(x) .08 .15 .45 .27 .05

Compute the following:

a. E(X)

b. V(X) directly from the definition

c. The standard deviation of X

d. V(X) using the shortcut formula

29. An individual who has automobile insurance

from a company is randomly selected. Let Y be

the number of moving violations for which the

individual was cited during the last 3 years. The

pmf of Y is

y 0 123

p(y) .60 .25 .10 .05

a. Compute E(Y).

b. Suppose an individual with Y violations

incurs a surcharge of $100Y

. Calculate the

expected amount of the surcharge.

30. Refer to Exercise 12 and calculate V(Y)

and s

. Then determine the probability that

Y is within 1 standard deviation of its mean

value.

31. An appliance dealer sells three different

models of upright freezers having 13.5, 15.9,

and 19.1 cubic feet of storage space, respec-

tively. Let X ¼ the amount of storage space pur-

chased by the next customer to buy a freezer.

Suppose that X has pmf

x 13.5 15.9 19.1

p(x).2 .5 .3

a. Compute E(X), E(X

), and V(X).

b. If the price of a freezer having capacity X cubic

feet is 25X  8.5, what is the expected price

paid by the next customer to buy a freezer?

c. What is the variance of the price 25X  8.5

paid by the next customer?

d. Suppose that although the rated capacity of a

freezer is X, the actual capacity is h(X) ¼ X 

.01X

. What is the expected actual capacity of

the freezer purchased by the next customer?

32. Let X be a Bernoulli rv with pmf as in Example

3.17.

a. Compute E(X

b. Show that V(X) ¼ p(1  p).

c. Compute E( X

3.3 Expected Values of Discrete Random Variables 119

33. Suppose that the number of plants of a particular

type found in a rectangular region (called a quad-

rat by ecologists) in a certain geographic area is

an rv X with pmf

pðxÞ¼

c=x



x ¼ 1; 2; 3; ...

otherwise

Is E(X) finite? Justify your answer (this is

another distribution that statisticians would call

heavy-tailed).

34. A small market orders copies of a certain

magazine for its magazine rack each week. Let

X ¼ demand for the magazine, with pmf

x 1 23456

p(x)

Suppose the store owner actually pays $2.00 for

each copy of the magazine and the price to cus-

tomers is $4.00. If magazines left at the end of

the week have no salvage value, is it better to

order three or four copies of the magazine?

[Hint: For both three and four copies ordered,

express net revenue as a function of demand X,

and then compute the expected revenue.]

35. Let X be the damage incurred (in $) in a certain

type of accident during a given year. Possible X

values are 0, 1000, 5000, and 10,000, with prob-

abilities .8, .1, .08, and .02, respectively. A partic-

ular company offers a $500 deductible policy. If

the company wishes its expected profit to be $100,

what premium amount should it charge?

36. The n candidates for a job have been ranked 1, 2,

3, ... , n. Let X ¼ the rank of a randomly

selected candidate, so that X has pmf

pðxÞ¼

1=n

x ¼ 1; 2; 3; ... ; n

otherwise



(this is called the discrete uniform distribution).

Compute E(X) and V(X) using the shortcut for-

mula. [Hint: The sum of the first n positive

integers is n(n+1)/2, whereas the sum of their

squares is n(n+1)(2n+1)/6.]

37. Let X ¼ the outcome when a fair die is rolled once.

If before the die is rolled you are offered either

(1/3.5) dollars or h(X) ¼ 1/X dollars, would you

accept the guaranteed amount or would you gamble?

[

Note: It is not generally true that 1/E(X) ¼ E(1/X).]

38. A chemical supply company currently has in stock

100 lb of a chemical, which it sells to customers in

5-lb containers. Let X ¼ the number of containers

ordered by a randomly chosen customer, and sup-

pose that X has pmf

x 1234

p(x).2 .4 .3 .1

Compute E(X) and V(X). Then compute the

expected number of pounds left after the next

customer’s order is shipped and the variance of

the number of pounds left. [Hint: The number of

pounds left is a linear function of X.]

39. a. Draw a line graph of the pmf of X in Exercise

34. Then determine the pmf of X and draw its

line graph. From these two pictures, what can

you say about V(X) and V(X)?

b. Use the proposition involving V(aX + b)to

establish a general relationship between V(X)

and V(X).

40. Use the definition in Expression (3.13) to prove

that VðaX þ bÞ¼a

.[Hint: With h(X) ¼

aX + b, E[h(X)] ¼ am +bwhere m ¼ E(X).]

41. Suppose E(X) ¼ 5 and E[X(X  1)] ¼ 27.5.

What is

a. E(X

)? [Hint: E[X(X  1)] ¼ E[X

 X] ¼

E(X

)  E(X).]

b. V(X)?

c. The general relationship among the quantities

E(X), E[X(X–1)], and V(X)?

42. Write a general rule for E(X  c) where c is a

constant. What happens when you let c ¼ m, the

expected value of X?

43. A result called Chebyshev’s inequality states that

for any probability distribution of an rv X and any

number k that is at least 1, Pð X  m

 ksÞ

1/k

. In words, the probability that the value of X

lies at least k standard deviations from its mean is

at most 1/k

a. What is the value of the upper bound for

k ¼ 2? k ¼ 3? k ¼ 4? k ¼ 5? k ¼ 10?

b. Compute m and s for the distribution of

Exercise 13. Then evaluate Pð X  m

 ksÞ

for the values of k given in part (a). What

does this suggest about the upper bound rela-

tive to the corresponding probability?

c. Let X have three possible values, 1, 0, and 1,

with probabilities 1=18 , 8=9, and 1=18 respec-

tively. What is Pð X  m

 3sÞ, and how does

it compare to the corresponding bound?

d. Give a distribution for which

Pð X  m

 5sÞ¼:04.

120

CHAPTER 3 Discrete Random Variables and Probability Distributions

3.4

Moments and Moment Generating

Functions

Sometimes the expected values of integer powers of X and X  m are called

moments, terminology borrowed from physics. Expected values of powers of X

are called moments about 0 and powers of X  m are called moments about the

mean. For example, E(X

) is the second moment about 0, and E[(X  m)

] is the third

moment about the mean. Moments about 0 are sometimes simply called moments.

Example 3.26 Suppose the pmf of X, the number of points earned on a short quiz, is given by

x 0 123

p(x).1 .2.3.4

The first moment about 0 is the mean

m ¼ EðXÞ¼

x2D

xpðxÞ¼0ð:1Þþ1ð:2Þþ2ð:3Þþ3ð:4Þ¼2

The second moment about the mean is the variance

¼ E½ðX  mÞ

¼

x2D

ðx  mÞ

pðxÞ

¼ð0  2Þ

ð:1Þþð1  2Þ

ð:2Þþð2  2Þ

ð:3Þþð3  2Þ

ð:4Þ¼1

The third moment about the mean is also important.

E½ðX  mÞ

¼

x2D

ðx  mÞ

pðxÞ

¼ð0  2 Þ

ð:1Þþð1  2Þ

ð:2Þþð2  2Þ

ð:3Þþð3  2Þ

ð:4Þ¼:6

We would like to use this as a measure of lack of symmetry, but E [(X  m)

]

depends on the scale of measurement. That is, if X is measured in feet, the value

is different from what would be obtained if X were measured in inches.

Scale independence results from dividing the third moment about the mean by s

E½ðX  mÞ



¼ E

X  m



This is our measure of departure from symmetry, called the skewness.

For a symmetric distribution the third moment about the mean would be 0, so

the skewness in that case is 0. However, in the present example the skewness is

E[(X  m)

]/s

¼.6/1 ¼.6. When the skewness is negative, as it is here, we

say that the distribution is negatively skewed or that it is skewed to the left.

Generally speaking, it means that the distribution stretches farther to the left of

the mean than to the right.

If the skewness were positive then we would say that the distribution is

positively skewed or that it is skewed to the right. For example, reverse the order

of the probabilities in the p(x) table above, so the probabilities of the values 0, 1, 2,

and 3 are now .4, .3, .2, and .1, respectively (a much harder quiz). This changes the

sign but not the magnitude of the skewness, so it becomes .6 and the distribution is

skewed right (see Exercise 57).

■

3.4 Moments and Moment Generating Functions 121

Moments are not always easy to obtain, as shown by the calculation of E (X)

in Example 3.18. We now introduce the moment generating function, which will

help in the calculation of moments and the understanding of statistical distributions.

We have already discussed the expected value of a function, E[h(X)]. In particular,

let e denote the base of the natural logarithms, with approximate value 2.71828.

Then we may wish to calculate E(e

) ¼ Se

p(x), E(e

3.75X

), or E(e

2.56X

). That is,

for any particular number t, the expected value E (e

) is meaningful. When we

consider this expected value as a function of t, the result is called the moment

generating function.

DEFINITION

The moment generating function (mgf) of a discrete random variable X is

defined to be

ðtÞ¼Eðe

Þ¼

x 2D

pðxÞ

where D is the set of possible X values. We will say that the moment

generating function exists if M

(t) is defined for an interval of numbers that

includes zero as well as positive and negative values of t (an interv al

including 0 in its interior).

If the mgf exists, it will be defined on a symmetric interval of the form

(t

, t

), where t

> 0, because t

can be chosen small enough so the symmetric

interval is contained in the interval of the definition.

When t ¼ 0, for any random variable X

ð0Þ¼Eðe

Þ¼

x 2D

pðxÞ¼

x 2D

1pðxÞ¼1

That is, M

(0) is the sum of all the probabilities, so it must always be 1. However, in

order for the mgf to be useful in generating moments, it will need to be defined for

an interval of values of t including 0 in its interior, and that is why we do not bother

with the mgf otherwise. As you might guess, the moment generating function fails

to exist in cases when moments themselves fail to exist, as in Example 3.19. See

Example 3.30 below.

The simplest example of an mgf is for a Bernoulli distribution, where only the

X values 0 and 1 receive positive probability.

Example 3.27 Let X be a Bernoulli random variable with pð0Þ¼

and p 1ðÞ¼

. Then

ðtÞ¼Eðe

Þ¼

x 2D

pðxÞ¼e

t0

þ e

t1

þ e

It should be clear that a Bernoulli random variable will always have an mgf of the

form p(0) +p(1)e

. This mgf exists because it is defined for all t. ■

122 CHAPTER 3 Discrete Random Variables and Probability Distributions

The idea of the mgf is to have an alternate view of the distribution based on an

infinite number of values of t. That is, the mgf for X is a function of t, and we get a

different functi on for each different distribution. When the function is of the form

of one constant plus another constant times e

, we know that it corresponds to a

Bernoulli random variable, and the constants tell us the probabilities. This is an

example of the followi ng “uniqueness property.”

PROPOSITION

If the mgf exists and is the same for two distributions, then the two distribu-

tions are the same. That is, the moment generating function uniquely speci-

fies the probability distribution; there is a one-to-one correspondence

between distributions and mgf’s.

Example 3.28 Let X be the number of claims in a year by someone holding an automobile

insurance policy with a company. The mgf for X is M

(t) ¼ .7 + .2e

+ .1e

Then we can say that the pmf of X is given by

x 012

p(x).7.2.1

Why? If we compute E( e

) based on this table, we get the correct mgf. Because X

and the random variable described by the table have the same mgf, the uniqueness

property requires them to have the same distribution. Therefor e, X has the given

pmf.

■

Example 3.29 This is a continuation of Example 3.18, except that here we do not consider the

number of births needed to produce a male child. Instead we are looking for a

person whose blood type is Rh+. Set p ¼ .85, which is the approximate probability

that a random person has blood type Rh+.IfX is the number of people we need to

check until we find someone who is Rh+, then p(x) ¼ p(1  p)

x1

¼ .85(.15)

x1

for x ¼ 1, 2, 3, ... .

Determination of the moment generating function here requires using the

formula for the sum of a geometric series:

a þ ar þ ar

þ¼

1  r

where a is the first term, r is the ratio of successive terms, and |r| < 1. The moment

generating function is

ðtÞ¼Eðe

Þ¼

x¼1

:85 ð:15Þ

x1

¼ :85e

x¼1

tðx1Þ

ð:15Þ

x1

¼ :85e

x¼1

½e

ð:15Þ

x1

:85e

1  :15e

3.4 Moments and Moment Generating Functions 123

The condition on r requires |.15e

| < 1. Dividing by .15 and taking logs, this gives

t < ln(.15)  1.90. The result is an interval of values that includes 0 in its

interior, so the mgf exists.

What about the value of the mgf at 0? Recall that M

(0) ¼ 1 always, because

the value at 0 amounts to summing the probabilities. As a check, after computing

an mgf we should make sure that this condition is satisfied. Here M

(0) ¼

.85/(1  .15) ¼ 1.

■

Example 3.30 Reconsider Example 3.19, where p(x) ¼ k/x

, x ¼ 1, 2, 3, ... . Recall that E(X)

does not exist, so there might be problems with the mgf, too:

ðtÞ¼Eðe

Þ¼

x¼1

With the help of tests for convergence such as the ratio test, we find that the series

converges if and only if e

 1, which means that t  0. Because zero is on the

boundary of this interval, not the interior of the interval (the interval must include

both positive and negative values), this mgf does not exis t. Of course, it could not

be useful for finding moments, because X does not have even a first moment

(mean).

■

How does the mgf produce moments? We will need various derivatives of

(t). For any positive integer r, let M

ðrÞ

ðtÞ denote the rth derivative of M

(t). By

computing this and then setting t ¼ 0, we get the rth moment about 0.

THEOREM

If the mgf exists,

EðX

Þ¼M

ðrÞ

ð0Þ

Proof We show that the theorem is true for r ¼ 1 and r ¼ 2. A proof by

mathematical induction can be used for general r. Differentiate

ðtÞ¼

x 2D

pðxÞ¼

x 2D

pðxÞ¼

x 2D

pðxÞ

where we have interchanged the order of summation and differentiation. This is

justified inside the interval of convergence, which includes 0 in its interior. Next we

set t ¼ 0 and get the first moment

ð0Þ¼M

ð1Þ

ð0Þ¼

x 2D

xpðxÞ¼EðXÞ

124 CHAPTER 3 Discrete Random Variables and Probability Distributions

Differentiate again:

ðtÞ¼

x 2D

pðxÞ¼

x 2D

pðxÞ¼

x 2D

pðxÞ

Set t ¼ 0 to get the second moment

ð0Þ¼M

ð2Þ

ð0Þ¼

x 2D

pðxÞ¼EðX

■

Example 3.31 This is a continuation of Example 3.28, where X represents the number of claims in

a year with pmf and mgf

x 012M

(t) ¼ .7 + .2e

+ .1e

p(x).7 .2 .1

First, find the derivatives

ðtÞ¼:2e

þ :1ð2Þe

ðtÞ¼:2e

þ :1ð2Þð2Þe

Setting t to 0 in the first derivative gives the first moment

EðXÞ¼M

ð0Þ¼M

ð1Þ

ð0Þ¼:2e

þ :1ð2Þe

2ð0Þ

¼ :2 þ :1ð2Þ¼:4

Setting t to 0 in the second derivative gives the second moment

EðX

Þ¼M

ð0Þ¼M

ð2Þ

ð0Þ¼:2e

þ :1ð2Þð2Þe

2ð0Þ

¼ :2 þ :1ð2Þð2Þ¼:6

To get the variance recall the shortcut formula from the previous section:

VðXÞ¼s

¼ EðX

Þ½EðXÞ

¼ :6  :4

¼ :6  :16 ¼ :44

Taking the square root gives s ¼ .66 approximately. Do a mean of .4 and

a standard deviation of .66 seem about right for a distribution concentrated mainly

on 0 and 1?

■

Example 3.32

(Example 3.29

continued)

Recall that p ¼ .85 is the probability of a person having Rh + blood and we keep

checking people until we find one with this blood type. If X is the number of people

we need to check, then p(x) ¼ .85(.15)

x1

, x ¼ 1, 2, 3, ..., and the mgf is

ðtÞ¼Eðe

Þ¼

:85e

1  :15e

Differentiating with the help of the quotient rule,

ðtÞ¼

:85e

ð1  :15e

3.4 Moments and Moment Generating Functions 125

Setting t ¼ 0,

m ¼ EðXÞ¼M

ð0Þ¼

:85

Recalling that .85 corresponds to p, we see that this agrees with Example 3.18.

To get the second moment, differentiate again:

ðtÞ¼

:85e

ð1 þ :15e

ð1  :15e

Setting t ¼ 0,

EðX

Þ¼M

ð0Þ¼

1:15

:85

Now use the shortcut formula for the variance from the previous section:

VðXÞ¼s

¼ EðX

Þ½EðXÞ

1:15

:85



:85

:15

:85

¼ :2076

■

There is an alternate way of doing the differentiation that can sometimes

make the effort easier. Define R

(t) ¼ ln[M

(t)], where ln(u) is the natural log

of u. In Exercise 54 you are requested to verify that if the moment generating

function exists ,

m ¼ EðXÞ¼R

ð0Þ

¼ VðXÞ¼R

ð0Þ

Example 3.33 Here we apply R

(t) to Example 3.32. Using ln(e

) ¼ t,

ðtÞ¼ln½M

ðtÞ ¼ ln

:85e

1  :15e



¼ lnð:85Þþt  lnð1  :15e

The first derivative is

ðtÞ¼

1  :15e

and the second derivative is

ðtÞ¼

:15e

ð1  :15e

Setting t to 0 gives

m ¼ EðXÞ¼R

ð0Þ¼

:85

¼ VðXÞ¼R

ð0Þ¼

:15

:85

These are in agreement with the results of Example 3.32. ■

126 CHAPTER 3 Discrete Random Variables and Probability Distributions

As mentioned at the end of the previous section, it is common to

transform X using a linear function Y ¼ aX + b. What happens to the mgf when

we do this?

PROPOSITION

Let X have the mgf M

(t) and let Y ¼ aX + b. Then M

(t) ¼ e

(at).

Example 3.34 Let X be a Bernoulli random variable with pð0Þ¼

and pð1Þ¼

. Think of X

as the number of wins, 0 or 1, in a single play of roulette. If you play roulette at

an American casino and you bet red, then your chances of winning are

because

18 of the 38 possible outcomes are red. Then from Example 3.27

ðtÞ¼

þ e

. Let your bet be $5 and let Y be your winnings. If X ¼ 0 then

Y ¼5, and if X ¼ 1 then Y ¼ +5. The linear equation Y ¼ 10X  5 gives the

appropriate relationship.

The equation is of the form Y ¼ aX + b with a ¼ 10 and b ¼5, so by the

proposition

ðtÞ¼ e

ðatÞ¼e

5t

ð10tÞ

¼ e

5t

þ e

10t



¼ e

5t

þ e

From this we can read off the probabilities for Y: p(5) ¼

and p(5) ¼

. ■

Exercises Section 3.4 (44–57)

44. For a new car the number of defects X has the

distribution given by the accompanying table.

Find M

(t) and use it to find E(X) and V(X).

x 0 123456

p(x) .04 .20 .34 .20 .15 .04 .03

45. In flipping a fair coin let X be the number of

tosses to get the first head. Then p(x) ¼ .5

for

x ¼ 1, 2, 3, .... Find M

(t) and use it to get E(X)

and V(X).

46. Given M

(t) ¼ .2 + .3e

+ .5e

, find p(x), E(X),

V(X).

47. Using a calculation similar to the one in Example

3.29 show that, if X has the distribution of

Example 3.18, then its mgf is

ðtÞ¼

1 ð1  pÞe

If Y has mgf M

(t) ¼ .75e

/(1 .25e

), determine

the probability mass function p

(y) with the help

of the uniqueness property.

48. Let X have the moment generating function

of Example 3.29 and let Y ¼ X  1. Recall that

X is the number of people who need to be

checked to get someone who is Rh+,soY is the

number of people checked before the first Rh+

person is found. Find M

(t) using the second

proposition.

49. If M

ðtÞ¼e

5tþ2t

then find E(X) and V(X)by

differentiating

a. M

(t)

b. R

(t)

50. Prove the result in the second proposition, i.e.,

aX+b

(t) ¼ e

(at).

51. Let M

ðtÞ¼e

5tþ2t

and let Y ¼ (X  5)/2. Find

(t) and use it to find E(Y) and V(Y).

52. If you toss a fair die with outcome X, pðxÞ¼

for x ¼ 1, 2, 3, 4, 5, 6. Find M

(t).

53. If M

(t) ¼ 1/(1  t

), find E(X) and V(X)by

differentiating M

(t).

54. Prove that the mean and variance are obtainable

from R

(t) ¼ ln(M

(t)):

3.4 Moments and Moment Generating Functions 127