Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

6.3. Maximum Likelihood Tests 379

Then Λ≤cis equivalent to−2log Λ≥−2logc. However,

−2log Λ=

(

X−θ 0

σ/

√

n

) 2

,

which has aχ^2 (1) distribution underH 0. Thus, the likelihood ratio test with
significance levelαstates that we rejectH 0 and acceptH 1 when

−2log Λ =

(

X−θ 0

σ/

√

n

) 2

≥χ^2 α(1). (6.3.6)

Note that this test is the same as thez-test for a normal mean discussed in Chapter

4withsreplaced byσ. Hence, the power function for this test is given in expression

(4.6.5).

Other examples are given in the exercises. In these examples the likelihood ratio
tests simplify and we are able to get the test in closed form. Often, though, this
is impossible. In such cases, similarly to Example 6.2.7, we can obtain the mle by
iterative routines and, hence, also the test statistic Λ. In Example 6.3.2,−2log Λ
had an exactχ^2 (1) null distribution. While not true in general, as the following
theorem shows, under regularity conditions, the asymptotic null distribution of
−2logΛ isχ^2 with one degree of freedom. Hence in all cases an asymptotic test
can be constructed.

Theorem 6.3.1. Assume the same regularity conditions as for Theorem 6.2.2.

Under the null hypothesis,H 0 : θ=θ 0 ,

−2logΛ

D

→χ^2 (1). (6.3.7)

Proof:Expand the functionl(θ) into a Taylor series aboutθ 0 of order 1 and evaluate

it at the mle,θ̂. This results in

l(θ̂)=l(θ 0 )+(θ̂−θ 0 )l′(θ 0 )+

1

2

(̂θ−θ 0 )^2 l′′(θ∗n), (6.3.8)

whereθn∗is betweenθ̂andθ 0. Becausêθ→P θ 0 , it follows thatθ∗n→P θ 0 .This,in
addition to the fact that the functionl′′(θ) is continuous, and equation (6.2.22) of
Theorem 6.2.2 imply that

−

1

n

l′′(θ∗n)→P I(θ 0 ). (6.3.9)

By Corollary 6.2.3,

1

√

n

l′(θ 0 )=

√

n(̂θ−θ 0 )I(θ 0 )+Rn, (6.3.10)

whereRn→0, in probability. If we substitute (6.3.9) and (6.3.10) into expression
(6.3.8) and do some simplification, we have

−2 log Λ = 2(l(θ̂)−l(θ 0 )) ={

√

nI(θ 0 )(θ̂−θ 0 )}^2 +R∗n, (6.3.11)