Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

610 Nonparametric and Robust Statistics

interval [Δ 1 ,Δ 2 ]. Therefore, since the scores are nondecreasing,

Wφ(Δ 1 )−Wφ(Δ 2 )=

∑

k =j

aφ(R(Yk−Δ 1 )) +aφ(R(Yj−Δ 1 ))

−

⎡

⎣

∑

k =j

aφ(R(Yk−Δ 2 )) +aφ(R(Yj−Δ 2 ))

⎤

⎦

= aφ(R(Xi)+1))−aφ(R(Xi)−1))≥ 0.

BecauseWφ(Δ) is a decreasing step function and steps only at the differencesYj−
Xi, its maximum value occurs when Δ<Yj−Xi, for alli, j, i.e., whenXi<Yj−Δ,
for alli, j. Hence, in this case, the variablesYj−Δ must get all the high ranks, so

max

Δ

Wφ(Δ) =

∑n

j=n 1 +1

aφ(j).

Note that this maximum value must be nonnegative. For suppose it was strictly
negative, then at least oneaφ(j)<0forj=n 1 +1,...,n. Because the scores are
nondecreasing,aφ(i)<0 for alli=1,...,n 1. This leads to the contradiction

0 >

∑n

j=n 1 +1

aφ(j)≥

∑n

j=n 1 +1

aφ(j)+

∑n^1

j=1

aφ(j)=0.

The results for the minimum value are obtained in the same way; see Exercise 10.5.6.

As Exercise 10.5.7 shows, the translation property, Lemma 10.2.1, holds for the
processWφ(Δ). Using this result and the last theorem, we can show that the power
function of the test statisticWφfor the hypotheses (10.5.1) is nondecreasing. Hence
the test is unbiased.

10.5.1 Efficacy

We next sketch the derivation of the efficacy of the test based onWφ. Our arguments

can be made rigorous; see advanced texts. Consider the statistic given by the average

Wφ(0) =

1

n

Wφ(0). (10.5.11)

Based on (10.5.5) and (10.5.8), we have

μφ(0) =E 0 (Wφ(0)) = 0 and σ^2 φ=Var 0 (Wφ(0)) =n(nn^1 n−^2 1)n−^2 s^2 a. (10.5.12)

Notice from Exercise 10.5.4 that the variance ofWφ(0) is of the orderO(n−^2 ). We
have

μφ(Δ) =EΔ[Wφ(0)] =E 0 [Wφ(−Δ)] =

1

n

∑n^2

j=1

E 0 [aφ(R(Yj+ Δ))]. (10.5.13)