Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

(Jacob Rumans) #1
8.2. Uniformly Most Powerful Tests 483

maxθ≤θ′γ(θ), whereγ(θ) is the power function of the test. That is, the significance
level is the maximum probability of Type I error.
LetX′=(X 1 ,...,Xn) be a random sample with common pdf (or pmf)f(x;θ),
θ∈Ω, and, hence with the likelihood function


L(θ,x)=

∏n

i=1

f(xi;θ), x′=(x 1 ,...,xn).

We consider the family of pdfs that has monotone likelihood ratio as defined next.


Definition 8.2.2. We say that the likelihoodL(θ,x)hasmonotone likelihood
ratio(mlr) in the statisticy=u(x)if, forθ 1 <θ 2 ,theratio


L(θ 1 ,x)
L(θ 2 ,x)

(8.2.2)

is a monotone function ofy=u(x).
Assume then that our likelihood functionL(θ,x) has a monotone decreasing
likelihood ratio in the statisticy=u(x). Then the ratio in (8.2.2) is equal tog(y),
wheregis a decreasing function. The case where the likelihood function has a mono-
tone increasing likelihood ratio (i.e.,gis an increasing function) follows similarly
by changing the sense of the inequalities below. Letαdenote the significance level.
Then we claim that the following test is UMP levelαfor the hypotheses (8.2.1):


RejectH 0 ifY≥cY, (8.2.3)
wherecYis determined byα=Pθ′[Y≥cY]. To show this claim, first consider the
simple null hypothesisH 0 ′ : θ=θ′.Letθ′′>θ′be arbitrary but fixed. LetC
denote the most powerful critical region forθ′versusθ′′. By the Neyman–Pearson
Theorem,Cis defined by


L(θ′,X)
L(θ′′,X)
≤kif and only ifX∈C,

wherekis determined byα=Pθ′[X∈C]. But by Definition 8.2.2, becauseθ′′>θ′,


L(θ′,X)
L(θ′′,X)

=g(Y)≤k⇔Y≥g−^1 (k),

whereg−^1 (k)satisfiesα=Pθ′[Y≥g−^1 (k)]; i.e.,cY=g−^1 (k). Hence the Neyman–
Pearson test is equivalent to the test defined by (8.2.3). Furthermore, the test is
UMP forθ′versusθ′′>θ′because the test only depends onθ′′>θ′andg−^1 (k)is
uniquely determined underθ′.
LetγY(θ) denote the power function of the test (8.2.3). To finish, we need to
show that maxθ≤θ′γY(θ)=α. But this follows immediately if we can show that
γY(θ) is a nondecreasing function. To see this, letθ 1 <θ 2 .Notethatsinceθ 1 <θ 2 ,
the test (8.2.3) is the most powerful test for testingθ 1 versusθ 2 with the level
γY(θ 1 ). By Corollary 8.1.1, the power of the test atθ 2 must not be below the level;
i.e.,γY(θ 2 )≥γY(θ 1 ). HenceγY(θ) is a nondecreasing function. Since the power
function is nondecreasing, it follows from Definition 8.1.2 that the mlr tests are
unbiased tests for the hypotheses (8.2.1); see Exercise 8.2.14.

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