Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

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