10.5.∗General Rank Scores 609As Exercise 10.5.4 shows,s^2 a/n≈1. SinceEH 0 (Wφ)=0,wehaveVarH 0 (Wφ)=EH 0 (Wφ^2 )=∑n^2j=1∑n^2j′=1EH 0 [aφ(R(Yj))aφ(R(Yj′))]=∑n^2j=1EH 0 [a^2 φ(R(Yj))] +∑∑j =j′EH 0 [aφ(R(Yj))aφ(R(Yj′))]=n 2
ns^2 a−n 2 (n 2 −1)
n(n−1)s^2 a (10.5.7)=
n 1 n 2
n(n−1)s^2 a; (10.5.8)see Exercise 10.5.2 for the derivation of the second term in expression (10.5.7). In
more advanced books, it is shown thatWφ is asymptotically normal underH 0.
Hence the corresponding asymptotic decision rule of levelαis
RejectH 0 in favor ofH 1 ifz=q Wφ
VarH 0 (Wφ)≥zα. (10.5.9)To answer the questions posed in the first paragraph of this section, the efficacy
of the test statisticWφis needed. To proceed along the lines of the last section,
define the process
Wφ(Δ) =∑n^2j=1aφ(R(Yj−Δ)), (10.5.10)whereR(Yj−Δ) denotes the rank ofYj−ΔamongX 1 ,...,Xn 1 ,Y 1 −Δ,...,Yn 2 −Δ.
In the last section, the process for the MWW statistic was also written in terms of
counts of the differencesYj−Xi. We are not as fortunate here, but as the next
theorem shows, this general process is a simple decreasing step function of Δ.
Theorem 10.5.1. The processWφ(Δ)is a decreasing step function ofΔwhich
steps down at each differenceYj−Xi,i=1,...,n 1 andj=1,...,n 2. Its maximum
and minimum values are
∑n
j=n 1 +1aφ(j)≥^0 and∑n 2
j=1aφ(j)≤^0 , respectively.Proof: Suppose Δ 1 <Δ 2 andWφ(Δ 1 ) =Wφ(Δ 2 ). Hence the assignment of the
ranks among theXiandYj−ΔmustdifferatΔ 1 and Δ 2 ; that is, then there must
be ajand anisuch thatYj−Δ 2 <XiandYj−Δ 1 >Xi. Thisimpliesthat
Δ 1 <Yj−Xi<Δ 2 .ThusWφ(Δ) changes values at the differencesYj−Xi.To
show it is decreasing, suppose Δ 1 <Yj−Xi<Δ 2 and there are no other differences
between Δ 1 and Δ 2 .ThenYj−Δ 1 andXimust have adjacent ranks; otherwise,
there would be more than one difference between Δ 1 and Δ 2 .SinceYj−Δ 1 >Xi
andYj−Δ 2 <Xi,wehave
R(Yj−Δ 1 )=R(Xi)+1 andR(Yj−Δ 2 )=R(Xi)− 1.Also, in the expression forWφ(Δ), only the rank of theYjterm has changed in the