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

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Example 7.30 Suppose that the number of major defects on a randomly selected new vehicle of a

certain type has a Poisson distribution with parameter l, Consider estimating e

l

the probability that a vehicle has no such defects, based on a random sample of n

vehicles. Let’s start with the estimator U ¼ I(X

¼ 0), the indicator function of the

event that the first vehicle in the sample has no defects. That is,

U ¼

1ifX

¼ 0

0ifX

> 0



Then

EðUÞ¼1  PX

¼ 0ðÞþ0  PX

> 0ðÞ¼PX

¼ 0ðÞ¼e

l

 l

=0! ¼ e

l

Our estimator is therefore unbiased for estimating the probability of no defects. The

sufficient statistic here is T ¼ ∑X

, so of course the estimator U is not a function of

T. The improved estimator is U* ¼ E(U | ∑X

) ¼ P(X

¼ 0|∑X

). Let’s consider

P(X

¼ 0|∑X

¼ t) where t is some non-negative integer. The event that X

¼ 0

and ∑X

¼ t is identical to the event that the first vehicle has no defects and the

total number of defects on the last n1 vehicles is t. Thus

PðX

¼ 0 jS

i¼1

¼ tÞ¼

¼ 0

i¼1

¼ t



i¼1

¼ t



¼ 0

i¼2

¼ t



i¼1

¼ t



A moment generating function argument shows that the sum of all nX

’s has a

Poisson distribution with parameter nl and the sum of the last n  1 X

’s has a

Poisson distribution with parameter (n  1)l. Furthermore, X

is independent of

the other n  1 X

’s so it is independe nt of their sum, from which

PðX

¼ 0 jS

i¼1

¼ tÞ¼

l



ðn1Þl

½ðn  1Þl

nl

ðnlÞ

n  1



The improved unbiased estimator is then U* ¼ (11/n)

. If, for example, there are a

total of 15 defects among 10 randomly selected vehicles, then the estimate is

ð1 

¼ :206. For this sample,

l ¼



x ¼ 1:5, so the maximum likelihood esti-

mate of e

l

is e

1.5

¼ .223. Here as in some other situations the principles of

unbiasedness and maximum likelihood are in conflict. However, if n is large, the

improved estimate is ð1  1=nÞ

¼½ð1  1 =n Þ





 e





, which is the mle. That is,

the unbiased and maximum likelihood estimators are “asymptotically equivalent.”

■

We have emphasized that in general there will not be a unique sufficient

statistic. Suppose there are two different sufficient statistics T

and T

such that the

first one is not a one-to-one function of the second (e.g., we are not considering

¼ ∑X

and T

¼ X). Then it would be distressing if we started with an unbiased

368 CHAPTER 7 Point Estimation

estimator U and found that E(U | T

) 6¼ E(U | T

), so our improved estimator

depended on which sufficient statistic we used. Fortunately there are general

conditions under which, starting with a minimal sufficient statistic T, the improved

estimator is the MVUE (minimum variance unbiased estimator). That is, the new

estimator is unbiased and has smaller variance than any other unbiased estimator.

Please consult one of the chapter references for more detail.

Further Comments

Maximum likelihood is by far the most popular method for obtaining point esti-

mates, so it would be disappointing if maximum likelihood estimators did not make

full use of sample information. Fortunately the mle’s do not suffer from this defect.

If T

,...,T

are jointly sufficient statistics for parameters y

,...,y

, then the joint

pmf or pdf factors as follows:

; :::; x

; y

; :::; y

ðÞ¼gt

; :::; t

; y

; :::; y

ðÞhx

; :::; x

ðÞ

The maximum likelihood estimates result from maximizing f() with respect to the

’s. Because the h() factor does not involve the parameters, this is equivalent to

maximizing the g() factor with respect to the y

’s. The resulting

’s will involve

the data only through the t

’s. Thus it is always possible to find a maximum

likelihood estimator that is a function of just the sufficient statistic(s). There are

contrived examples of situations where the mle is not unique, in which case an mle

that is not a function of the sufficient statistics can be constructed—but there is also

one that is a function of the sufficient statistics.

The concept of sufficiency is very compelling when an inve stigator is sure

the underlying distributio n that generated the data is a member of some particular

family (normal, exponential , etc.). However, two different families of distributions

might each furnish plausible models for the data in a particular application, and yet

the sufficient statistics for these two families might be different (an analogous

comment applies to maximum likelihood estimation). For example, there are data

sets for which a gamma probability plot suggests that a member of the gamma

family would give a reasonable model and also a lognormal probability plot

(normal probability plot of the logs of the observations) indicates that lognormality

is plausible. Yet the jointly sufficient statistics for the parameters of the gamma

family are not the same as those for the parameters of the lognormal family. When

estimating some parameter y in such situations (e.g., the mean m or median

m), one

would look for a robust estimator that performs well for a wide variety of underly-

ing distributions, as discussed in Section 7.1. Please consult a more advanced

source for additional information.

Exercises Section 7.3 (32–41)

32. The long run proportion of vehicles that pass a

certain emissions test is p. Suppose that three

vehicles are independently selected for testing.

Let X

¼ 1 if the ith vehicle passes the test and

¼ 0 otherwise (i ¼ 1, 2, 3), and let X ¼ X

+ X

. Use the definition of sufficiency to

show that X is sufficient for p by obtaining the

conditional distribution of the X

’s given that

X ¼ x for each possible value x. Then generalize

by giving an analogous argument for the case of

n vehicles.

7.3 Sufﬁciency 369

33. Components of a certain type are shipped in

batches of size k. Suppose that whether or not

any particular component is satisfactory is inde-

pendent of the condition of any other component,

and that the long run proportion of satisfactory

components is p. Consider n batches, and let X

denote the number of satisfactory components in

the ith batch (i ¼ 1, 2, . . ., n). Statistician A is

provided with the values of all the X

’s, whereas

statistician B is given only the value of X ¼ ∑X

Use a conditional probability argument to decide

whether statistician A has more information

about p than does statistician B.

34. Let X

,...,X

be a random sample of component

lifetimes from an exponential distribution with

parameter l. Use the factorization theorem to

show that ∑X

is a sufficient statistic for l.

35. Identify a pair of jointly sufficient statistics for

the two parameters of a gamma distribution

based on a random sample of size n from that

distribution.

36. Suppose waiting time for delivery of an item is

uniform on the interval from y

to y

(so f(x; y

, y

)

¼ 1/(y

 y

) for y

< x < y

and is 0 other-

wise). Consider a random sample of n waiting

times, and use the factorization theorem to show

that min(X

), max(X

) is a pair of jointly sufficient

statistics for y

and y

.[Hint: Introduce an appro-

priate indicator function as we did in Example

7.27.]

37. For y > 0 consider a random sample from a

uniform distribution on the interval from y to

2y (pdf 1/y for y < x < 2y), and use the factori-

zation theorem to determine a sufficient statistic

for y.

38. Suppose that survival time X has a lognormal

distribution with parameters m and s (which are

the mean and standard deviation of ln(X), not of

X itself). Are ∑X

and

jointly sufficient for

the two parameters? If not, what is a pair of

jointly sufficient statistics?

39. The probability that any particular component of

a certain type works in a satisfactory manner is p.

If n of these components are independently

selected, then the statistic X, the number among

the selected components that perform in a satis-

factory manner, is sufficient for p. You must

purchase two of these components for a particu-

lar system. Obtain an unbiased statistic for the

probability that exactly one of your purchased

components will perform in a satisfactory man-

ner. [Hint: Start with the statistic U, the indicator

function of the event that exactly one of the first

two components in the sample of size n performs

as desired, and improve on it by conditioning on

the sufficient statistic.]

40. In Example 7.30, we started with U ¼ I(X

¼ 0)

and used a conditional expectation argument to

obtain an unbiased estimator of the zero-defect

probability based on the sufficient statistic. Con-

sider now starting with a different statistic: U ¼

[∑I(X

¼ 0)]/n. Show that the improved esti-

mator based on the sufficient statistic is identical

to the one obtained in the cited example.

[Hint: Use the general property E(Y+Z| T) ¼

E(Y | T)+E(Z | T).]

41. A particular quality characteristic of items pro-

duced using a certain process is known to be

normally distributed with mean m and standard

deviation 1. Let X denote the value of the charac-

teristic for a randomly selected item. An unbiased

estimator for the parameter y ¼ P(X  c), where

c is a critical threshold, is desired. The estimator

will be based on a random sample X

,...,X

a. Obtain a sufficient statistic for m.

b. Consider the estimator

y ¼ IðX

 cÞ. Obtain

an improved unbiased estimator based on the

sufficient statistic (it is actually the minimum

variance unbiased estimator). [Hint: You may

use the following facts: (1) The joint distribu-

tion of X

and X is bivariate normal with means

m and m, respectively, variances 1 and 1/n,

respectively, and correlation r (which you

should determine). (2) If Y

and Y

have a

bivariate normal distribution, then the condi-

tional distribution of Y

given that Y

¼ y

normal with mean m

+(rs

)(y

 m

) and

variance s

ð1  rÞ

370

CHAPTER 7 Point Estimation

7.4

Information and Efficiency

In this section we introduce the idea of Fisher information and two of its applica-

tions. The first application is to find the minimum possible variance for an unbiased

estimator. The second application is to show that the maximum likelihood estima-

tor is asymptotically unbiased and normal (that is, for large n it has expected value

approximately y and it has approximately a normal distribution) with the minimum

possible variance.

Here the notation f(x; y) will be used for a probability mass function or a

probability density function with unknown parameter y. The Fisher information is

intended to measure the precision in a single observation. Consider the random

variable U obtained by taking the partial derivative of ln[f(x;y)] with respect to y

and then replacing x by X: U ¼ @[ln[f(X;y)]/@y. For example, if the pdf is yx

y1

for 0 < x < 1(y > 0), then @[ln(yx

y1

)]/@y ¼ @[ln(y)+(y1)ln(x)]/@y ¼

1/y + ln(x), so U ¼ ln(X)+1/y.

DEFINITION

The Fisher information I(u) in a single observation from a pmf or pdf

f(x;u) is the variance of the random variable U ¼ @[ln[f(X;y)]/@y :

IðyÞ¼V

lnðf ðX; yÞÞ



ð7:7Þ

It may seem strange to differentiate the logarithm of the pmf or pdf, but

this is exactly what is often done in maximum likelihood estimation. In what

follows we will assume that f(x; y) is a pmf, but everything that we do will apply

also in the continuous case if appropriate assumptions are mad e. In particular, it is

important to assume that the set of possible x’s does not depend on the value of the

parameter.

When f(x; y) is a pmf , we know that 1 ¼

f ð x; yÞ. Therefore, differentiat-

ing both sides with respect to y and using the fact that [ln(f)]

¼ f

/f, we find that the

mean of U is 0:

0 ¼

f ð x; yÞ¼

f ðx; yÞ

½ln f ðx; yÞf ðx; yÞ¼E½

lnðf ðX; yÞÞ ¼ EðUÞ

ð7:8Þ

This involves interchanging the order of differentiation and summation, which

requires certain technical assumptions if the set of possible x values is infinite.

We will omit those assumptions here and elsewhere in this section, but we

emphasize that switching differentiation and summation (or integration) is not

allowed if the set of possible valu es depends on y. For example, if the summation

were from –y to y there would be additional variability, and therefore terms for the

limits of summation would be needed .

7.4 Information and Efﬁciency 371

There is an alternative expression for I(y ) that is sometimes easier to compute

than the variance in the definition:

IðyÞ¼E

lnðf ðX; yÞÞ



ð7:9Þ

This is a consequence of taking another derivative in (7.8 ):

0 ¼

ln f ðx; yÞ½f ðx;yÞþ

ln f ðx; yÞ½

½ln f ð x;yÞf ðx; yÞ

¼ E

½ln f ð X; y



þE

ln f ðX; y



()

ð7:10Þ

To complete the derivation of (7.9), recall that U has mean 0, so its variance is

IðyÞ¼V

½ln f ðX; yÞ



¼ E

ln f ðX; yÞ



()

¼E

½ln f ðX; yÞ



where Equation (7.10) is used in the last step.

Example 7.31 Let X be a Bernoulli rv, so f(x; p) ¼ p

(1–p)

1–x

, x ¼ 0, 1. Th en

lnðf ðX; pÞÞ ¼

½Xln p þð1  XÞlnð1  pÞ ¼



1  X

1  p

X  p

pð1  pÞ

ð7:11Þ

This has mean 0, in accord with Equation (7.8), because E(X) ¼ p. Computing the

variance of the partial derivative, we get the Fisher information:

IðpÞ¼V

lnðf ðX; pÞÞ



VðX  p Þ

½pð1  pÞ

VðXÞ

½pð1  pÞ

pð1  pÞ

½pð1  pÞ

pð1  pÞ

ð7:12Þ

The alternative method uses Equation (7.9). Differentiating Equation (7.11) with

respect to p gives

lnðf ðX; pÞÞ ¼

X



1  X

ð1  pÞ

ð7:13Þ

Taking the negative of the expected value in Equation (7.13) gives the information

in an observation:

IðpÞ¼E

lnðf ðX; pÞÞ



1  p

ð1  pÞ

pð1  pÞ

ð7:14Þ

372 CHAPTER 7 Point Estimation

Both methods yield the answer I(p ) ¼ 1/[p(1 – p)], which says that the information

is the reciprocal of V(X). It is reasonable that the information is greatest when the

variance is smallest.

■

Information in a Random Sample

Now assume a random sample X

, X

, ..., X

from a distribution with pmf or pdf

f(x; y). Let f(X

, X

, ..., X

; y) ¼ f(X

; y)  f(X

; y)    f(X

; y) be the likelihood

function. The Fisher information I

(y) for the random sample is the variance of the

score funct ion

ln f ðX

; X

; ...; X

; yÞ¼

ln½f ðX

; yÞf ðX

; yÞf ðX

; yÞ

The log of a product is the sum of the logs, so the score function is a sum:

ln f ðX

; X

; ...; X

; yÞ¼

ln f ðX

; yÞþ

ln f ðX

; yÞþ

ln f ðX

; yÞð7:15Þ

This is a sum of terms for which the mean is zero, by Equation (7.8), and therefore

ln f ðX

; X

; ...; X

; yÞ



¼ 0 ð7:16Þ

The right-hand-side of Equation (7.15) is a sum of independent identically

distributed random variables, and each has variance I(y). Taking the variance of

both sides of Equation (7.15) gives the information I

(y) in the random sample

ðyÞ¼V

ln f ðX

; X

; ...; X

; yÞ



¼ nV

ln f ðX

; yÞ



¼ nIðyÞ: ð7:17Þ

Therefore, the Fisher information in a random sample is just n times the information

in a single observat ion. This should make sense intuitively, because it says that

twice as many observations yield twice as much information.

Example 7.32 Continuing with Example 7.31, let X

, X

, ..., X

be a random sample from the

Bernoulli distribution with f(x; p) ¼ p

(1 – p)

1–x

, x ¼ 0, 1. Suppose the purpose is

to estimate the proportion p of drivers who are wearing seat belts. We saw that the

information in a single observation is I(p) ¼ 1/[p(1 – p)], and therefore the Fisher

information in the random sample is I

(p) ¼ nI(p) ¼ n/[p(1 – p)]. ■

The Crame

r-Rao Inequality

We will use the concept of Fisher information to show that if t(X

, X

, ..., X

)isan

unbiased estimator of y, then its minimum possible variance is the reciprocal of

(y). Harald Crame

r in Sweden and C. R. Rao in India independently derived this

7.4 Information and Efﬁciency 373

inequality during World War II, but R. A. Fisher had some notion of it 20 years

previously.

THEOREM

(CRAME

R-

RAO

INEQUALITY)

Assume a random sample X

, X

, ..., X

from the distribution with pmf or pdf

f(x; y) such that the set of possible values does not depend on y. If the statist ic

T ¼ t(X

, X

, ..., X

) is an unbiased estimator for the parameter y, then

VðTÞ

½ln f ðX

; ...; X

; yÞ



nIðyÞ

ðyÞ

Proof The basic idea here is to consider the correlation r between T and the

score function, and the desired inequality will result from 1  r  1. If T ¼

t(X

, X

, ..., X

) is an unbiased estimator of y, then

y ¼ EðTÞ¼

;...;x

tðx

; ...; x

Þf ðx

; ...; x

; yÞ

Differentiating this with respect to y,

1 ¼

;...;x

tðx

; ...; x

Þf ðx

; ...; x

; yÞ¼

;...;x

tðx

; ...; x

f ð x

; ...; x

; yÞ

Multiplying and dividing the last term by the likelihood f(x

, ..., x

;y) gives

1 ¼

;...;x

tðx

; ...; x

f ð x

; ...; x

; yÞ

f ðx

; ...; x

; yÞ

f ð x

; ...; x

; yÞ

which is equivalent to

1 ¼

;...;x

tðx

; ...; x

½ln f ðx

; ...; x

; yÞf ðx

; ...; x

; yÞ

¼ EtðX

; :::; X

½lnf ðX

; :::; X

; yÞ



Therefore, becau se of Equation 7.16, the covariance of T with the score function is

1 ¼ Cov T;

½ln f ðX

; ...; X

; yÞ



ð7:18Þ

Recall from Section 5.2 that the correlation between two rv’s X and Y is r

X,Y

Cov(X, Y)/(s

), and that 1  r

X,Y

 1. Therefore,

Cov(X; YÞ

¼ r

X;Y

 s

374 CHAPTER 7 Point Estimation

Apply this to Equation 7.18:

1 ¼ Cov T;

½ln f ðX

; ...; X

; yÞ



 VðTÞV

½ln f ðX

; ...; X

; yÞ



ð7:19Þ

Dividing both sides by the variance of the score function and using the fact that this

variance equals nI(y), we obtain the desired result. ■

Because the variance of T must be at least 1/nI(y ), it is natural to call T an

efficient estimator of y if V(T) ¼ 1/[nI(y)].

DEFINITION

Let T be an unbiased estimator of y. The ratio of the lower bound to the

variance of T is its efficiency. Then T is said to be an efficient estimator if

T achieves the Crame

r–Rao lower bound (the efficiency is 1). An efficient

estimator is a minimum variance unbiased (MVUE) estimator, as discussed

in Section 7.1.

Example 7.33 Continuing with Example 7.32, let X

, X

, ..., X

be a random sample from the

Bernoulli distribution, where the purpose is to est imate the proportion p of drivers

who are wearing seat belts. We saw that the information in the sample is I

(p) ¼

n/[p(1 – p)], and therefore the Crame

r–Rao lower bound is 1/I

(p) ¼ p(1 – p)/n.Let

T(X

, X

, ..., X

) ¼

p ¼ X ¼

=n. Then EðTÞ¼Eð

Þ=n ¼ np=n ¼ p so T is

unbiased, and VðTÞ¼Vð

Þ=n

¼ npð1  pÞ= n

¼ pð1  pÞ=n. Because T is

unbiased and V(T) is equal to the lower bound, T has efficiency 1 and therefore it

is an efficient estimator.

■

Large Sample Properties of the MLE

As discusse d in Section 7.2, the maximum likelihood estimator

y has some nice

properties. First of all it is consistent, which means that it converges in probability

to the parameter y as the sample size increases. A verification of this is beyond the

level of this book, but we can use it as a basis for showing that the mle is

asymptotically normal with mean y (asymptotic unbiasedness) and variance equal

to the Crame

r–Rao lower bound.

THEOREM

Given a random sample X

, X

, ..., X

from a distribution with pmf or pdf

f(x; y), assume that the set of possible x values does not depend on y.

Then for large n the maximum likelihood estimator

y has approximately

a normal distribution with mean y and variance 1/[nI(y)]. More precisely,

the limiting distribution of

ﬃﬃﬃ

y  yÞ is normal with mean 0 and

variance 1/I(y).

7.4 Information and Efﬁciency 375

Proof Consider the score function

SðyÞ¼

ln f ðX

; X

; ...; X

; yÞ

Its derivative S

(y) at the true y is approximately equal to the difference quotient

ðyÞ¼

Sð

yÞSðyÞ

y  y

ð7:20Þ

and the error approaches zero asymptotically because

y approaches y (consistency).

Equation (7.20) connects the mle

y to the score function, so the asymptotic

behavior of the score function can be applied to

y. Because

y is the maximum

likelihood estimate, Sð

yÞ¼0, so in the limit,

y  y ¼

SðyÞ

S

ðyÞ

Multiplying both sides by

ﬃﬃﬃ

, then dividing numerator and denominator by

ﬃﬃﬃﬃﬃﬃﬃﬃ

IðyÞ

ﬃﬃﬃ

y  yÞ¼

ﬃﬃﬃ

=½n

ﬃﬃﬃﬃﬃﬃﬃﬃ

IðyÞ





SðyÞ

 1=½n

ﬃﬃﬃﬃﬃﬃﬃﬃ

IðyÞ





ðyÞ

SðyÞ=

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

nIðyÞ

ð1=nÞS

ðyÞ=

ﬃﬃﬃﬃﬃﬃﬃﬃ

IðyÞ

Now rewrite S(y) and S

(y) as sums using Equation 7.15:

ﬃﬃﬃ

y  yÞ¼

ln f ðX

; yÞ½ þþ

ln f ðX

; yÞ½



ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

IðyÞ=n





ln f ðX

; yÞ½ 

ln f ðX

; yÞ½

ﬃﬃﬃﬃﬃﬃﬃﬃ

IðyÞ



ð7:21Þ

The denominator braces contain a sum of independent iden tically distributed rv’s

each with mean

IðyÞ¼E

½ln f ðX; yÞ



by Equation (7.9). Therefore, by the law of large numbers, the denominator average

converges to I(y). Thus the denominator converges to

ﬃﬃﬃﬃﬃﬃﬃﬃ

IðyÞ

. The numerator

average

is the mean of independent identically distributed rv’s with mean 0 [by

Equation (7.8)] and variance I(y), so the numerator ratio is an average minus its

expected value, divided by its standard deviation. Therefore, by the Central Limit

Theorem it is approximately normal with mean 0 and standard deviation 1. Thus, the

ratio in Equation (7.21) has a numerator that is approximately N(0, 1) and a denomi-

nator that is approximately

ﬃﬃﬃﬃﬃﬃﬃﬃ

IðyÞ

, so the ratio is approximately N(0, 1/

ﬃﬃﬃﬃﬃﬃﬃﬃ

IðyÞ

) ¼

N(0, 1/I(y)).Thatis,

ﬃﬃﬃ

y  yÞ is approximately N(0, 1/I(y)), and it follows that

is approximately normal with mean y and variance 1/[nI(y)], the Crame

r–Rao

lower bound. ■

376 CHAPTER 7 Point Estimation

Example 7.34 Continuing with the previous example, let X

, X

, ..., X

be a random sample from

the Bernoulli distribution. The objective is to estimate the proportion p of drivers

who are wearing seat belts. The pmf is f(x; p) ¼ p

(1 – p)

1–x

, x ¼ 0, 1 so the

likelihood is

f ð x

; x

; ...; x

; pÞ¼p

þx

þ...þx

ð1  pÞ

nðx

þx

þ...þx

Then the log likelihood is

ln½f ðx

; x

; ...; x

; pÞ ¼

lnðpÞþðn 

Þlnð1  pÞ

and therefore its derivative, the score function, is

ln½f ðx

; x

; ...; x

; pÞ ¼



n 

1  p

 np

pð1  pÞ

Conclude that the maximum likelihood estimator is

p ¼

X ¼

=n. Recall from

Example 7.33 that this is unbiased and efficient with the minimum variance of the

Crame

r–Rao inequality. It is also asymptotically normal by the Central Limit

Theorem. These properties are in accord with the asymptotic distribution given

by the theorem,

p  Nðp; 1=½nIðpÞÞ.

■

Example 7.35 Let X

, X

, ..., X

be a random sample from the distribution with pdf f(x; y) ¼ yx

y1

for 0 < x < 1, assuming y > 0. Here X

, i ¼ 1, 2, ..., n, represents the fraction of a

perfect score assigned to the ith applicant by a recruiting team. The Fisher infor-

mation is the variance of

U ¼

ln½f ðX; yÞ ¼

½ln y þðy  1ÞlnðXÞ ¼

þ lnðXÞ

However, it is easier to use the alternative method of Equation (7.9):

IðyÞ¼E

ln½f ðX; yÞ



¼E

þ lnðXÞ



¼E

1



To obtain the maximum likelihood estimator, we first find the log likelihood:

ln½f ðx

; x

; ...; x

; yÞ ¼ lnðy

y1

Þ¼n lnðyÞþðy  1Þ

lnðx

Its derivative, the score function, is

ln½f ðx

; x

; ...; x

; yÞ ¼

lnðx

Setting this to 0, we find that the maximum likelihood estimate is

y ¼

1

lnðx

Þ=n

ð7:22Þ

The expected value of ln(X)is1/y, because E(U ) ¼ 0, so the denominator

of (7.22) converges in probability to 1/y by the law of large numbers. Therefore

converges in probability to y, which means that

y is consistent. We knew

this because the mle is always consistent, but it is also nice to show it directly.

By the theorem, the asymptotic distribution of

y is normal with mean y and

variance 1/[nI(y)] ¼ y

/n. ■

7.4 Information and Efﬁciency 377