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

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is selected and the amount of gas (therms) used

during the month of January is determined for

each house. The resulting observations are 103,

156, 118, 89, 125, 147, 122, 109, 138, 99. Let m

denote the average gas usage during January by

all houses in this area. Compute a point esti-

mate of m.

b. Suppose there are 10,000 houses in this area

that use natural gas for heating. Let t denote

the total amount of gas used by all of these

houses during January. Estimate t using the

data of part (a). What estimator did you use

in computing your estimate?

c. Use the data in part (a) to estimate p, the pro-

portion of all houses that used at least 100

therms.

d. Give a point estimate of the population median

usage (the middle value in the population of all

houses) based on the sample of part (a). What

estimator did you use?

8. In a random sample of 80 components of a certain

type, 12 are found to be defective.

a. Give a point estimate of the proportion of all

such components that are not defective.

b. A system is to be constructed by randomly

selecting two of these components and con-

necting them in series, as shown here.

The series connection implies that the system

will function if and only if neither component is

defective (i.e., both components work properly).

Estimate the proportion of all such systems that

work properly. [Hint:Ifp denotes the probabil-

ity that a component works properly, how can

P(system works) be expressed in terms of p?]

c. Let

p be the sample proportion of successes.

an unbiased estimator for p

?[Hint:

For any rv Y, E( Y

) ¼ V(Y) + [E(Y)]

9. Each of 150 newly manufactured items is exam-

ined and the number of scratches per item is

recorded (the items are supposed to be free of

scratches), yielding the following data:

Number of

scratches per item

01234567

Observed

frequency

18 37 42 30 13 7 2 1

Let X ¼ the number of scratches on a randomly

chosen item, and assume that X has a Poisson

distribution with parameter l.

a. Find an unbiased estimator of l and compute

the estimate for the data. [Hint: E(X) ¼ l for X

Poisson, so E(

X ¼ ?)]

b. What is the standard deviation (standard error)

of your estimator? Compute the estimated stan-

dard error. [Hint: s

¼ l for X Poisson.]

10. Using a long rod that has length m, you are going

to lay out a square plot in which the length of each

side is m. Thus the area of the plot will be m

However, you do not know the value of m,so

you decide to make n independent measurements

, X

,...X

of the length. Assume that each X

has mean m (unbiased measurements) and vari-

ance s

a. Show that

is not an unbiased estimator

for m

.[Hint: For any rv Y, E(Y

) ¼

V(Y) + [E(Y)]

. Apply this with Y ¼ X.]

b. For what value of k is the estimator

 kS

unbiased for m

?[Hint: Compute E(X

 kS

).]

11. Of n

randomly selected male smokers, X

smoked

filter cigarettes, whereas of n

randomly selected

female smokers, X

smoked filter cigarettes. Let

and p

denote the probabilities that a randomly

selected male and female, respectively, smoke

filter cigarettes.

a. Show that (X

) – (X

) is an unbiased

estimator for p

–p

.[Hint: E(X

) ¼ n

for

i ¼ 1, 2.]

b. What is the standard error of the estimator in

part (a)?

c. How would you use the observed values x

and x

to estimate the standard error of your estimator?

d. If n

¼ n

¼ 200, x

¼ 127, and x

¼ 176,

use the estimator of part (a) to obtain an esti-

mate of p

–p

e. Use the result of part (c) and the data of part (d)

to estimate the standard error of the estimator.

12. Suppose a certain type of fertilizer has an expected

yield per acre of m

with variance s

, whereas the

expected yield for a second type of fertilizer is

with the same variance s

. Let S

and S

denote

the sample variances of yields based on sample

sizes n

and n

, respectively, of the two fertilizers.

Show that the pooled (combined) estimator

ðn

 1ÞS

þðn

 1ÞS

þ n

 2

is an unbiased estimator of s

13. Consider a random sample X

,...,X

from the pdf

f ðx; yÞ¼:5ð1 þ yxÞ1  x  1

348

CHAPTER 7 Point Estimation

where 1  y  1 (this distribution arises in

particle physics). Show that

y ¼ 3

X is an unbiased

estimator of y.[Hint: First determine

m ¼ EðXÞ¼Eð

XÞ.]

14. A sample of n captured Pandemonium jet fighters

results in serial numbers x

, x

,...,x

. The CIA

knows that the aircraft were numbered consecu-

tively at the factory starting with a and ending

with b, so that the total number of planes manu-

factured is b – a + 1 (e.g., if a ¼ 17 and b ¼ 29,

then 2917 + 1 ¼ 13 planes having serial num-

bers 17, 18, 19, . . ., 28, 29 were manufactured).

However, the CIA does not know the values of

a or b. A CIA statistician suggests using the esti-

mator max(X

) – min(X

) + 1 to estimate the total

number of planes manufactured.

a. If n ¼ 5, x

¼ 237, x

¼ 375, x

¼ 202,

¼ 525, and x

¼ 418, what is the

corresponding estimate?

b. Under what conditions on the sample will the

value of the estimate be exactly equal to the

true total number of planes? Will the estimate

ever be larger than the true total? Do you think

the estimator is unbiased for estimating b –

a + 1? Explain in one or two sentences.

(A similar method was used to estimate German

tank production in World War II.)

15. Let X

, X

,...,X

represent a random sample from

a Rayleigh distribution with pdf

f ðx; yÞ¼

x

=ð2yÞ

x>0

a. It can be shown that E(X

) ¼ 2y. Use this fact

to construct an unbiased estimator of y based

(and use rules of expected value to

show that it is unbiased).

b. Estimate y from the following measurements

of blood plasma beta concentration (in pmol/L)

for n ¼ 10 men.

16.88 10.23 4.59 6.66 13.68

14.23 19.87 9.40 6.51 10.95

16. Suppose the true average growth m of one type

of plant during a 1-year period is identical to that

of a second type, but the variance of growth for

the first type is s

, whereas for the second type,

the variance is 4s

. Let X

,...,X

be m indepen-

dent growth observations on the first type [so

E(X

) ¼ m, V(X

) ¼ s

], and let Y

,...,Y

n independent growth observations on the

second type [E(Y

) ¼ m, V( Y

) ¼ 4s

]. Let c be a

numerical constant and consider the estimator

m ¼ c

X þð1  cÞY. For any c between 0 and 1

this is a weighted average of the two sample

means, e.g., :7

X þ :3Y

a. Show that for any c the estimator is unbiased.

b. For fixed m and n, what value c minimizes

Vð

mÞ?[Hint: The estimator is a linear combi-

nation of the two sample means and these

means are independent. Once you have an

expression for the variance, differentiate with

respect to c.]

17. In Chapter 3, we defined a negative binomial rv as

the number of failures that occur before the rth

success in a sequence of independent and identical

success/failure trials. The probability mass func-

tion (pmf) of X is

nbðx; r; pÞ

x þ r  1

ð1  pÞ

x ¼ 0; 1; 2; ...

0 otherwise

a. Suppose that r  2. Show that

p ¼ðr  1Þ=ðX þ r  1Þ

is an unbiased estimator for p.[Hint: Write out

Eð

pÞ and cancel x+r–1 inside the sum.]

b. A reporter wishing to interview five indivi-

duals who support a certain candidate begins

asking people whether (S) or not (F) they sup-

port the candidate. If the sequence of responses

is SFFSFFFSSS, estimate p ¼ the true propor-

tion who support the candidate.

18. Let X

, X

,...,X

be a random sample from a pdf

f(x) that is symmetric about m, so that

X is an

unbiased estimator of m.Ifn is large, it can be

shown that Vð

XÞ1=f4n½f ðmÞ

g. When the

underlying pdf is Cauchy (see Example 7.8),

Vð

XÞ¼1,soX is a terrible estimator. What is

Vð

XÞ in this case when n is large?

19. An investigator wishes to estimate the proportion

of students at a certain university who have vio-

lated the honor code. Having obtained a random

sample of n students, she realizes that asking each,

“Have you violated the honor code?” will proba-

bly result in some untruthful responses. Consider

the following scheme, called a randomized

response technique. The investigator makes up a

deck of 100 cards, of which 50 are of type I and 50

are of type II.

7.1 General Concepts and Criteria 349

Type I: Have you violated the honor code (yes or

no)?

Type II: Is the last digit of your telephone number

a 0, 1, or 2 (yes or no)?

Each student in the random sample is asked to mix

the deck, draw a card, and answer the resulting

question truthfully. Because of the irrelevant ques-

tion on type II cards, a yes response no longer

stigmatizes the respondent, so we assume that

responses are truthful. Let p denote the proportion

of honor-code violators (i.e., the probability of a

randomly selected student being a violator), and

let l ¼ P(yes response). Then l and p are related

by l ¼ .5p+(.5)(.3).

a. Let Y denote the number of yes responses, so

Y~Bin(n, l). Thus Y/n is an unbiased estimator

of l. Derive an estimator for p based on Y.If

n ¼ 80 and y ¼ 20, what is your estimate?

[Hint:Solvel ¼ .5p+.15 for p and then sub-

stitute Y/n for l.]

b. Use the fact that E(Y/n) ¼ l to show that your

estimator

p is unbiased.

c. If there were 70 type I and 30 type II cards,

what would be your estimator for p?

20. Return to the problem of estimating the population

proportion p and consider another adjusted esti-

mator, namely

p ¼

X þ

ﬃﬃﬃﬃﬃﬃﬃﬃ

n=4

n þ

ﬃﬃﬃ

The justification for this estimator comes from the

Bayesian approach to point estimation to be intro-

duced in Section 14.4.

a. Determine the mean squared error of this esti-

mator. What do you find interesting about this

MSE?

b. Compare the MSE of this estimator to the

MSE of the usual estimator (the sample

proportion).

7.2

Methods of Point Estimation

So far the point estimators we have introduced were obtained via intuition and/or

educated guesswork. We now discuss two “constructive” methods for obtaining

point estimators: the method of moments and the method of maximum likelihood.

By constructive we mean that the general definition of each type of estimator

suggests explicitly how to obtain the estimator in any specific problem. Although

maximum likelihood estimators are generally preferable to moment estimators

because of certain efficiency properties, they often require significantly more

computation than do moment estimators. It is sometimes the case that these

methods yield unbiased estimators.

The Method of Moments

The basic idea of this method is to equate certain sample characteristics, such as the

mean, to the corresponding population expected values. Then solving these equa-

tions for unknown parame ter values yields the estimators.

DEFINITION

Let X

,...,X

be a random sample from a pmf or pdf f(x). For k ¼ 1, 2, 3, . . . ,

the kth population moment,orkth moment of the distribution f(x), is E(X

The kth sample moment is ð1= nÞ

i¼1

Thus the first population moment is E(X) ¼ m and the first sample moment is

=n ¼ X: The second population and sample moments are E(X

) and

=n, respectively. The population moments will be functions of any unknown

parameters y

, y

,....

350 CHAPTER 7 Point Estimation

DEFINITION

Let X

, X

,...,X

be a random sample from a distribution with pmf or pdf

f(x; y

,...,y

), where y

,...,y

are parameters whose values are unknown.

Then the moment estimators

; ...;

are obtained by equating the first m

sample moments to the corresponding first m population moments and

solving for y

,...,y

If, for example, m ¼ 2, E(X) and E(X

) will be functions of y

and y

. Setting

EðX Þ¼ð1=nÞ

ð¼ XÞ and EX

ðÞ¼ð1=nÞ

gives two equations in y

and

. The solution then defines the estimators. For estimating a population mean m,

the method gives m ¼

X, so the estimator is the sample mean.

Example 7.13 Let X

,...,X

represent a random sample of service times of n customers at a

certain facility, where the underlying distribution is assumed exponential with

parameter l. Since there is only one parameter to be estimated, the estimator is

obtained by equating E(X)to

X. Since E(X) ¼ 1/l for an exponential distribution,

this gives 1=l ¼

X or l ¼ 1=X. The moment estimator of l is then

l ¼ 1=X. ■

Example 7.14 Let X

,...,X

be a random sample from a gamma distribution with parameters a

and b. From Section 4.4 , E(X) ¼ ab and E(X

) ¼ b

G(a + 2)/G(a) ¼ b

(a +1)a.

The moment estimators of a and b are obtained by solving

X ¼ ab

¼ aða þ 1Þb

Since aða þ 1Þb

¼ a

þ ab

and the first equation implies a

¼ XðÞ

, the

second equatio n becomes

¼ XðÞ

þ ab

Now dividing each side of this second equation by the corresponding side of the

first equation and substituting back gives the estimators

a ¼

XðÞ

 XðÞ

b ¼

 XðÞ

To illustrate, the survival time data mentioned in Example 4.28 is

152 115 109 94 88 137 152 77 160 165

125 40 128 123 136 101 62 153 83 69

with



x ¼ 113:5 and ð1=20Þ

¼ 14; 087:8. The estimates are

a ¼

113:5ðÞ

14; 087:8  113:5ðÞ

¼ 10:7

b ¼

14; 087:8  113: 5ðÞ

113:5

¼ 10:6

These estimates of a and b differ from the values suggested by Gross and Clark

because they used a different estimation technique.

■

7.2 Methods of Point Estimation 351

Example 7.15 Let X

,...,X

be a random sample from a generalized negative binomial

distribution with parameters r and p (Section 3.6). Since E(X) ¼ r(1 – p)/p and

V(X) ¼ r(1 – p)/p

, E(X

) ¼ V(X)+[E(X)]

¼ r(1 – p)(r – rp + 1)/p

. Equating

E(X)to

X and E(X

)toð1=nÞ

eventually gives

p ¼

 XðÞ

r ¼

XðÞ

 XðÞ

 X

As an illustration, Reep, Pollard, and Benjamin (“Skill and Chance in Ball

Games,” J. Roy. Statist. Soc. Ser. A, 1971: 623–629) consider the negative bino-

mial distribution as a model for the number of goals per game scor ed by National

Hockey League teams. The data for 1966–1967 follows (420 games):

Goals 0 12345678910

Frequency 29 71 82 89 65 45 24 7 4 1 3

Then,



x ¼

=420 ¼½ð0Þð29Þþð1Þð71Þþþð10Þð3Þ=420 ¼ 2:98

and

=420 ¼½ð0Þ

ð29Þþð1Þ

ð71Þþþð10Þ

ð3Þ=420 ¼ 12:40

Thus,

p ¼

2:98

12:40  2:98ðÞ

¼ :85

r ¼

2:98ðÞ

12:40  2:98ðÞ

 2:98

¼ 16:5

Although r by definition must be positive, the denominator of

r could be negative,

indicating that the negative binomial distribution is not appropriate (or that the

moment estimator is flawed).

■

Maximum Likelihood Estimation

The method of maximum likelihood was first introduced by R. A. Fisher, a

geneticist and statistician, in the 1920s. Most statisticians recommend this method,

at least when the sample size is large, since the resulting estimators have certain

desirable efficiency properties (see the proposition on large sample behavior

toward the end of this section).

Example 7.16 A sample of ten new bike helmets manufactured by a company is obtained . Upon

testing, it is found that the first, third, and tenth helmets are flawed, whereas the

others are not. Let p ¼ P(flawed helmet) and define X

,...,X

by X

¼ 1 if the i th

helmet is flawed and zero otherwise. Then the observed x

’s are 1, 0, 1, 0, 0, 0, 0, 0,

0, 1, so the joint pmf of the sample is

f ð x

; x

; ...; x

; pÞ¼pð1  pÞp p ¼ p

ð1  pÞ

ð7:4Þ

352 CHAPTER 7 Point Estimation

We now ask, “For what value of p is the obser ved sample most likely to have

occurred?” That is, we wish to find the value of p that maximizes the pmf (7.4) or,

equivalently, maximizes the natural log of (7.4).

Since

ln½f ðx

; x

; ...; x

; pÞ ¼ 3lnðpÞþ7lnð1  pÞð7:5Þ

and this is a differentiable function of p, equating the derivative of (7.5) to zero

gives the maximizing value

ln½f ðx

; x

; ...; x

; pÞ ¼



1  p

¼ 0 ) p ¼

where x is the observed number of successe s (flawed helmets). The estimate

of p is now

p ¼

. It is called the maximum likelihood estimate because

for fixed x

,...,x

, it is the parameter value that maximizes the likelihood

(joint pmf) of the observed sample. The likelihood and log likelihoo d are

graphed in Figure 7.5. Of course, the maximum on both graphs occurs at the

same value, p ¼ .3.

Note that if we had been told only that among the ten helmets there were

three that were flawed, Equation (7.4) would be replaced by the binomial pmf



ð1  pÞ

, which is also maximized for

p ¼

Since ln[g(x)] is a monotonic function of g(x), finding x to maximize ln[g(x)] is equivalent to

maximizing g(x) itself. In statistics, taking the logarithm frequently changes a product to a sum, which

is easier to work with.

This conclusion requires checking the second derivative, but the details are omitted.

.0010

.0005

.0020

.0015

.0025

0 .2.4.6.81.0 0 .2.4.6.81.0

Likelihood

−5

−10

−15

−20

−25

−30

−35

ln(likelihood)

Figure 7.5 Likelihood and log likelihood plotted against

■

7.2 Methods of Point Estimation 353

DEFINITION

Let X

,...,X

have joint pmf or pdf

f ð x

; x

; :::; x

; y

; :::; y

Þð7:6Þ

where the parameters y

,...,y

have unknown values. When x

,...,x

are the

observed sample values and (7.6) is regarded as a function of y

,...,y

,itis

called the likelihood function. The maximum likelihood estimates

; ...;

are those values of the y

’s that maximize the likelihood function, so that

f(x

, x

,. . ., x

;

; ...;

)  f(x

, x

,. . ., x

; y

,...,y

) for all y

,...,y

When the X

’s are substituted in place of the x

’s, the maximum likelihood

estimators (m le’s) result.

The likelihood function tells us how likely the observed sample is as a function

of the possible parameter values. Maximizing the likelihood gives the parameter

values for which the observed sample is most likely to have been generated, that is,

the parameter values that “agree most closely” with the observed data.

Example 7.17 Suppose X

,...,X

is a random sample from an exponential distribution with

parameter l. Because of independence, the likelihood function is a product of the

individual pdf’s:

f ð x

; ...; x

; lÞ¼ðle

lx

Þðle

lx

Þ¼l

lSx

The ln(likeliho od) is

ln f ðx

; ...; x

; lÞ½¼n lnðlÞl

Equating (d/dl)[ln(likelihood)] to zero results in n/l – Sx

¼ 0, or l ¼

n=Sx

¼ 1=



x. Thus the mle is

l ¼ 1=X; it is identical to the method of moments

estimator but it is not an unbiased estimator, since Eð1=

XÞ 6¼ 1=EðXÞ. ■

Example 7.18 Let X

,...,X

be a random sample from a normal distribution. The likelihood

function is

f ð x

; ...; x

; m; s

Þ¼

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

2ps

ðx

mÞ

=ð2s



ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

2ps

ðx

mÞ

=ð2s

2ps



n=2



ðx

mÞ

=ð2s

ln½f ðx

; ...; x

; m; s

Þ ¼ 

lnð2ps

Þ

ðx

 mÞ

To find the maximizing values of m and s

, we must take the partial derivatives of

ln( f ) with respect to m and s

, equate them to zero, and solve the resulting two

equations. Omitting the details, the resulting mle’s are

354 CHAPTER 7 Point Estimation

m ¼

ðX

 XÞ

The mle of s

is not the unbiased estimator, so two different principles of estimation

(unbiasedness and maximum likelihood) yield two different estimators. ■

Example 7.19 In Chapter 3, we discussed the use o f the Poisson distribution for modeling the

number of “events” that occur in a two-dimensional region. Assume that when the

region R being sampled has area a(R), the number X of events occurring in R has a

Poisson distribution with parameter la(R) (where l is the expected number of

events per unit area) and that nonoverlapping regions yield independent X’s.

Suppose an ecol ogist selects n nonoverlapping regions R

,...,R

and counts

the number of plants of a certain species found in each region. The joint pmf

(likelihood) is then

pðx

; ...; x

; lÞ¼

l  aðR

Þ½

laðR



l  aðR

Þ½

laðR

aðR

Þ½

 aðR

Þ½

 l

 e

lSaðR

! x

The ln(likelihood) is

ln½pðx

; ...; x

; lÞ ¼

 ln½aðR

Þ þ lnðlÞ

 l

aðR

Þ

lnðx

!Þ

Taking d/dl ln( p) and equating it to zero yields



aðR

Þ¼0

l ¼

aðR

The mle is then

l ¼

aðR

Þ. This is intuitively reasonable because l is the

true density (plants per unit area), whereas

l is the sample density since

aðR

Þ is

just the total area sampled. Because E(X

) ¼ l · a(R

), the estimator is unbiased.

Sometimes an alternative sampling procedure is used. Instead of fixing

regions to be sampled, the ecologist will select n points in the entire region of

interest and let y

¼ the distance from the ith point to the nearest plant. The

cumulative distribution function (cdf) of Y ¼ distance to the nearest plant is

ðyÞ¼PðY  yÞ¼1  PðY > yÞ¼1  P

no plants in a

circle of radius y



¼ 1 

lpy

lpy

ðÞ

¼ 1  e

lpy

Taking the derivative of F

(y) with respect to y yields

ðy; lÞ¼

2plye

lpy

y  0

0 otherwise

(

If we now form the likelihood f

; l)·····f

; l), differentiate ln(likelihood),

and so on, the resu lting mle is

7.2 Methods of Point Estimation 355

l ¼

number of plants observed

total area sampled

which is also a sample density. It can be shown that in a sparse environment (small l),

the distance method is in a certain sense better, whereas in a dense environment, the

first sampling method is better.

■

Example 7.20 Let X

,...,X

be a random sample from a Weibull pdf

f ð x; a; bÞ¼

 x

a1

 e

ðx=bÞ

x  0

0 otherwise

(

Writing the likelihood and ln(likelihood), then setting both ð@=@aÞ½lnðf Þ ¼ 0 and

ð@=@bÞ½lnðf Þ ¼ 0 yields the equations

a ¼

½x

 lnðx

Þ



lnðx



1

b ¼



1=a

These two equations cannot be solved explicitly to give general formulas for the

mle’s

a and

b. Instead, for each sample x

,...,x

, the equations must be solved

using an iterative numerical procedure. Even moment estimators of a and b are

somewhat complicated (see Exercise 22).

The iterative mle computations can be done on a computer, and they are available

in some statistical packages. MINITAB gives maximum likelihood estimates for both

the Weibull and the gamma distributions (under “Quality Tools”). Stata has a general

procedure that can be used for these and other distributions. For the data of Example

7.14 the maximum likelihood estimates for the Weibull distribution are

a ¼ 3:799

and

b ¼ 125:88. (The mle’s for the gamma distribution are

a ¼ 8:799 and

b ¼ 12:893,

a little different from the moment estimates in Example 7.14). Figure 7.6 shows the

Weibull log likelihood as a function of a and b. The surface near the top has a rounded

shape, allowing the maximum to be found easily, but for some distributions the surface

can be much more irregular, and the maximum may be hard to find.

Log likelihood

3.0

3.5

4.0

4.5

120

125

130

135

Figure 7.6 Weibull log likelihood for Example 7.20 ■

356

CHAPTER 7 Point Estimation

Some Properties of MLEs

In Exam ple 7.18, we obtained the mle of s

when the underlying distribution is

normal. The mle of s ¼

ﬃﬃﬃﬃﬃ

, as well as many other mle’s, can be easily derived

using the following proposition.

PROPOSITION

The Invariance Principle

Let

;

; ...;

be the mle’s of the parameters y

, y

,...,y

. Then the mle

of any function h (y

, y

,...,y

) of these parameters is the function

hð

;

; ...;

Þ, of the mle’s.

Proof For an intuitive idea of the proof, consider the special case m ¼ 1, with

¼ y, and assume that h(·) is a one-to-one function. On the graph of the likelihood

as a function of the parameter y, the highest point occurs where y ¼

y. Now consider

the graph of the likelihood as a function of h(y). In the new graph the same heights

occur, but the height that was previously plotted at y ¼ a is now plotted at

hðyÞ¼hðaÞ, and the highest point is now plotted at hðyÞ¼hð

yÞ. Thus, the maxi-

mum remains the same, but it now occurs at hð

yÞ. ■

Example 7.21

(Example 7.18

continued)

In the normal case, the mle’s of m and s

are

m ¼ X and

ðX

 XÞ

=n.To

obtain the mle of the function hðm; s

Þ¼

ﬃﬃﬃﬃﬃ

¼ s, substitute the mle’s into the

function:

s ¼

ﬃﬃﬃﬃﬃ

ðX

 XÞ



1=2

The mle of s is not the sample standard deviation S, although they are close unless n

is quite small. Similar ly, the mle of the population coefficient of variation 100m/ s

is 100

■

Example 7.22

(Example 7.20

continued)

The mean value of an rv X that has a Weibull distribution is

m ¼ b  Gð 1 þ 1=aÞ

The mle of m is therefore

m ¼

b  Gð1 þ 1=

aÞ, where

a and

b are the mle’s of a and

b. In particular,

X is not the mle of m, although it is an unbiased estimator. At least

for large n,

m is a better estimator than X. ■

Large-Sample Behavior of the MLE

Although the principle of maximum likelihood estimation has considerabl e intui-

tive appeal, the following proposition provides additional rationale for the use

of mle’s. (See Section 7.4 for more details.)

PROPOSITION

Under very general conditions on the joint distribution of the sample, when

the sample size is large, the maximum likelihood estimator of any parameter

y is close to y (consistency), is approximately unbiased [Eð

yÞy], and has

7.2 Methods of Point Estimation 357