Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

6.3. Maximum Likelihood Tests 377

Recall that the likelihood function and its log are given by

L(θ)=

∏n

i=1

f(Xi;θ)

l(θ)=

∑n

i=1

logf(Xi;θ).

Letθ̂denote the maximum likelihood estimate ofθ.
To motivate the test, consider Theorem 6.1.1, which says that ifθ 0 is the true
value ofθ, then, asymptotically,L(θ 0 ) is the maximum value ofL(θ). Consider the
ratio of two likelihood functions, namely,

Λ=

L(θ 0 )

L(θ̂)

. (6.3.2)

Note that Λ≤1, but ifH 0 is true, Λ should be large (close to 1), while ifH 1 is true,
Λ should be smaller. For a specified significance levelα, this leads to the intuitive
decision rule
RejectH 0 in favor ofH 1 if Λ≤c, (6.3.3)
wherecis such thatα=Pθ 0 [Λ≤c]. We call it thelikelihood ratio test(LRT).
Theorem 6.3.1 derives the asymptotic distribution of Λ underH 0 , but first we look
at two examples.

Example 6.3.1 (Likelihood Ratio Test for the Exponential Distribution). Sup-
poseX 1 ,...,Xnare iid with pdff(x;θ)=θ−^1 exp{−x/θ},forx, θ >0. Let the
hypotheses be given by (6.3.1). The likelihood function simplifies to

L(θ)=θ−nexp{−(n/θ)X}.

From Example 4.1.1, the mle ofθisX. After some simplification, the likelihood
ratio test statistic simplifies to

Λ=en

(

X

θ 0

)n

exp{−nX/θ 0 }. (6.3.4)

The decision rule is to rejectH 0 if Λ≤c. But further simplification of the test is
possible. Other than the constanten, the test statistic is of the form

g(t)=tnexp{−nt},t> 0 ,

wheret=x/θ 0. Using differentiable calculus, it is easy to show thatg(t)has
a unique critical value at 1, i.e., g′(1) = 0, and further thatt=1providesa
maximum, becauseg′′(1)<0. As Figure 6.3.1 depicts,g(t)≤c if and only if
t≤c 1 ort≥c 2. This leads to

Λ≤c, if and only if,Xθ 0 ≤c 1 orXθ 0 ≥c 2.