Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

472 Optimal Tests of Hypotheses

Theorem 8.1.1. Neyman–Pearson Theorem. LetX 1 ,X 2 ,...,Xn,wheren
is a fixed positive integer, denote a random sample from a distribution that has pdf
or pmff(x;θ). Then the likelihood ofX 1 ,X 2 ,...,Xnis

L(θ;x)=

∏n

i=1

f(xi;θ), forx′=(x 1 ,...,xn).

Letθ′andθ′′be distinct fixed values ofθso thatΩ={θ:θ=θ′,θ′′},andletkbe
a positive number. LetCbe a subset of the sample space such that

(a)

L(θ′;x)

L(θ′′;x)

≤k,foreachpointx∈C.

(b)

L(θ′;x)

L(θ′′;x)

≥k,foreachpointx∈Cc.

(c)α=PH 0 [X∈C].

ThenCis a best critical region of sizeαfor testing the simple hypothesisH 0 :θ=θ′
against the alternative simple hypothesisH 1 :θ=θ′′.
Proof: We shall give the proof when the random variables are of the continuous
type. IfC is the only critical region of sizeα, the theorem is proved. If there
is another critical region of size∫ α,denoteitbyA. For convenience, we shall let
···
R

∫

L(θ;x 1 ,...,xn)dx 1 ···dxnbe denoted by

∫
RL(θ). In this notation we wish
to show that ∫

C

L(θ′′)−

∫

A

L(θ′′)≥ 0.

SinceCis the union of the disjoint setsC∩AandC∩AcandAis the union of the
disjoint setsA∩CandA∩Cc,wehave
∫

C

L(θ′′)−

∫

A

L(θ′′)=

∫

C∩A

L(θ′′)+

∫

C∩Ac

L(θ′′)−

∫

A∩C

L(θ′′)−

∫

A∩Cc

L(θ′′)

=

∫

C∩Ac

L(θ′′)−

∫

A∩Cc

L(θ′′). (8.1.5)

However, by the hypothesis of the theorem,L(θ′′)≥(1/k)L(θ′) at each point ofC,
and hence at each point ofC∩Ac;thus,
∫

C∩Ac

L(θ′′)≥

1

k

∫

C∩Ac

L(θ′).

ButL(θ′′)≤(1/k)L(θ′) at each point ofCc, and hence at each point ofA∩Cc;
accordingly, ∫

A∩Cc

L(θ′′)≤

1

k

∫

A∩Cc

L(θ′).

These inequalities imply that
∫

C∩Ac

L(θ′′)−

∫

A∩Cc

L(θ′′)≥

1

k

∫

C∩Ac

L(θ′)−

1

k

∫

A∩Cc

L(θ′);