382 Maximum Likelihood Methods
conditions (R0)–(R5), but the results below can be derived rigorously; see, for
example, Hettmansperger and McKean (2011). Consider testing the hypotheses
H 0 :θ=θ 0 versusH 1 : θ =θ 0 ,
whereθ 0 is specified. Here Ω = (−∞,∞)andω={θ 0 }. By Example 6.1.1, we
know that the mle ofθunder Ω isQ 2 =med{X,...,Xn}, the sample median. It
follows that
L(̂Ω) = 2−nexp
{
−
∑n
i=1
|xi−Q 2 |
}
,
while
L(ω̂)=2−nexp
{
−
∑n
i=1
|xi−θ 0 |
}
.
Hence the negative of twice the log of the likelihood ratio test statistic is
−2log Λ =2
[n
∑
i=1
|xi−θ 0 |−
∑n
i=1
|xi−Q 2 |
]
. (6.3.24)
Thus the sizeαasymptotic likelihood ratio test forH 0 versusH 1 rejectsH 0 in favor
ofH 1 if
2
[n
∑
i=1
|xi−θ 0 |−
∑n
i=1
|xi−Q 2 |
]
≥χ^2 α(1).
By (6.2.10), the Fisher information for this model isI(θ) = 1. Thus, the Wald-type
test statistic simplifies to
χ^2 W=[
√
n(Q 2 −θ 0 )]^2.
For the scores test, we have
∂logf(xi−θ)
∂θ
=
∂
∂θ
[
log
1
2
−|xi−θ|
]
=sgn(xi−θ).
Hence the score vector for this model isS(θ)=(sgn(X 1 −θ),...,sgn(Xn−θ))′.
From the above discussion [see equation (6.3.17)], the scores test statistic can be
written as
χ^2 R=(S∗)^2 /n,
where
S∗=
∑n
i=1
sgn(Xi−θ 0 ).
As Exercise 6.3.5 shows, underH 0 ,S∗is a linear function of a random variable with
ab(n, 1 /2) distribution.
Which of the three tests should we use? Based on the above discussion, all three
tests are asymptotically equivalent under the null hypothesis. Similarly to the con-
cept of asymptotic relative efficiency (ARE), we can derive an equivalent concept