COMPOSITE AGAINST COMPOSITE

In this section we consider testing a composite null hypothesis against a composite alternative hypothesis. As noted earlier, this situation is the most realistic. Let the null and alternative hypotheses be H0: 0 E 0O and Hi – 0 Є 0Ь where 0O and 0] are subsets of the parameter space 0. Here we define the concept of the UMP test as follows:

DEFINITION 9.5.1 A test R is the uniformly most powerful test of size (level) ct if sup0eeo P(R I 0) = (<) a and for any other test R such that sup9eeo P(R I 0) = (^) a we have P(R | 0) > P(Ri | 0) for any 0 Є 0j.

For the present situation we define the likelihood ratio test as follows.

definition 9.5.2 Let L(x | 0) be the likelihood function. Then the likelihood ratio test of H0 against Hi is defined by the critical region

sup L(0)

(9.5.1) A = ——– < c,

sup Д0)

OoUe,

where c is chosen to satisfy sup0oP(A < c | 0) = a for a certain specified value of a.

The following are examples of the likelihood ratio test.

Подпись: a.Подпись: /

image425
Подпись: (9.5.2)

EXAMPLE 9.5.1 Consider the same model as in Example 9.4.2, but here test H0: p < p0 against H. p > p0. If x/n ^ p0, A = 1; therefore, accept H0. Henceforth suppose x/n > p0. Since the numerator of the likelihood ratio attains its maximum зір = p0, A is the same as in (9.4.6). Therefore the critical region is again given by (9.4.8). Next we must determine d so as to satisfy

But since P(X/n > d I p) can be shown to be a monotonically increasing function of p, we have

(9.5.3)

sup P

(X, n>d

>

p

= p

(X „>dp0

P^Po

K.

/

)

Therefore the value of d is also the same as in Example 9.4.2.

This test is UMP. To see this, let R be the test defined above and let /ц be some other test such that suppSpoP(Ri p) ^ a. Then it follows that P(R I po) ^ a. But since R is the UMP test of H0: p = p0 against Hi ■p> p0, we have P(Ri p) — P(R p) for all p > p0 by the result of Example 9.4.2.

2 2

EXAMPLE 9.5.2 Let the sample be X, ~ N{|x, a ) with unknown a, і =

о

1, 2, . . . , n. We are to test H0: |x = p,0 and 0 < a < °° against Hp. p, > p-o and 0 < cr2 < °°.

о

Denoting (p., a ) by 0, we have

(9.5.4) Подпись: Z Oi ~ P-)2 • 2a2 i= L(0) = (2тг) n/2(u2) n/2exp

Therefore

(9.5.5) Подпись: ~ ^ ~ Bo)2 2ст2 І= і Подпись:sup L(0) = (2tt) n^u2) n/2exp во

/о — n/%-2 — я/2

= (2О (a ) exp

where a2 = n X’Lr-=i(xl — рь0)2. If x/n ^ po, A = 1; therefore, accept H0. Henceforth suppose x/n > p0. Then we have

(9.5.6) sup Ц0) = (2тт)-"/2(<т2)-п/,2ехр — >

eouei L

where d2 = n l’£/=i(x1 — x)2. Therefore the critical region is

(9.5.7)

Подпись: (9.5.8) image431

(crVd2) n/2 < c for some c, which can be equivalently written as

where a2 is the unbiased estimator of a2 defined by (n — 1)_1£”=1(хг – — x)2. Since the left-hand side of (9.5.8) is distributed as Student’s t with n — 1 degrees of freedom, k can be computed or read from the appropriate table. Note that since P{R HQ) is uniquely determined in this example in spite of the composite null hypothesis, there is no need to compute the supremum.

If the alternative hypothesis specifies p Ф p0, the critical region (9.5.8) should be modified by putting the absolute value sign around the left – hand side. In this case the same test can be performed using the confi­dence interval defined in Example 8.2.3.

In Section 9.3 we gave a Bayesian interpretation of the classical method for the case of testing a simple null hypothesis against a simple alternative hypothesis. Here we shall do the same for the composite against composite case, and we shall see that the classical theory of hypothesis testing be­comes more problematic. Let us first see how the Bayesian would solve the problem of testing H0: 0 ^ 0O against Hj: 0 > 0O. Let 1^(0) be the loss incurred by choosing H0, and /^(0) by choosing HA. Then the Bayesian rejects H0 if

Подпись:‘со foo

A(0)/(0 I x)d0 < T2(0)/(0 I x)d0,

J — 00 J — 00 where / (0 I x) is the posterior density of 0. Suppose, for simplicity, that Li(0) and (0) are simple step functions defined by

(9.5.10) Lj(0) = 0 for 0 > 0O

= ух for 0 < 0O

and

1^(0) =0 for 0 < 0o = y2 for 0 — 00-

In this case the losses are as given in Table 9.2; therefore (9.5.9), as can be seen in (9.3.8), is reduced to

Подпись: (9.5.11)h(x І Я]) tjo L(x І Щ)

Recall that (9.5.11) is the basis for interpreting the Neyman-Pearson test. Here, in addition to the problem of not being able to evaluate тпо/ііі, the

classical statistician faces the additional problem of not being able to make sense of L(x I Hi) and L(x | H0).

The likelihood ratio test is essentially equivalent to rejecting H0 if

sup L(x I 0)

Подпись: (9.5.12)Подпись: > c.0>eo

sup L(x I 0)

0£во

A problem here is that the left-hand side of (9.5.12) may not be a good substitute for the left-hand side of (9.5.11).

Sometimes a statistical decision problem we face in practice need not and/or cannot be phrased as the problem of testing a hypothesis on a parameter. For example, consider the problem of deciding whether or not we should approve a certain drug on the basis of observing x cures in n independent trials. Let p be the probability of a cure when the drug is administered to a patient, and assume that the net benefit to society of approving the drug can be represented by a function U(p), nondecreasing in p. According to the Bayesian principle, we should approve the drug if

Подпись: (9.5.13)U(p)f(p I x)dp > 0,

where f(p I x) is the posterior density of p given x. Note that in this de­cision problem, hypothesis testing on the parameter p is not explicitly considered. The decision rule (9.5.13) is essentially the same kind as (9.5.9), however.

Next we try to express (9.5.13) more explicitly as an inequality concern­ing x, assuming for simplicity that f(p | x) is derived from a uniform prior density: that is, from (8.3.7), (9.5.14) f(p I x) = (n + 1)С*У(1 – p)n x.

image437 Подпись: = 1.

Now suppose у > x. Then f(p x) and f(p y) cross only once, except possibly at p = 0 or 1. To see this, put f(p | x) = f(p у). Hp Ф 0 or 1, this equality can be written as

The left-hand side of (9.5.15) is 0 if p = 1 and is monotonically increasing as p decreases to 0. Let p* be the solution to (9.5.15) such that p* Ф 0 or

1, and define h{p) = f(p y) — f(p x) and k(p) = f(p x) — f(p y). Then

Подпись: (9.5.16)
image440

we have

because the left-hand side is greater than U(p*), whereas the right-hand side is smaller than U{p*). But (9.5.16) is equivalent to

Подпись:[u(p)f{p I y)dp > flU(p)f(p I x)dp,

Jo Jo which establishes the result that the left-hand side of (9.5.13) is an increas­ing function in x. Therefore (9.5.13) is equivalent to

(9.5.18) x> c,

where c is determined by (9.5.13).

The classical statistician facing this decision problem will, first, para­phrase the problem into that of testing hypothesis H0: p > p0 versus Hx: p < po for a certain constant p0 and then use the likelihood ratio test. Her decision rule is of the same form as (9.5.18), except that she will deter­mine c so as to conform to a preassigned size a. If the classical statistician were to approximate the Bayesian decision, she would have to engage in a rather intricate thought process in order to let her p0 and a reflect the utility consideration.

Leave a reply

You may use these HTML tags and attributes: <a href="" title=""> <abbr title=""> <acronym title=""> <b> <blockquote cite=""> <cite> <code> <del datetime=""> <em> <i> <q cite=""> <s> <strike> <strong>