Question

Suppose that X = (Xi, X2, , X.) and Y-X,,Y2, , Ym) are random samples from continuous distributions F and G, respectively.Wilcoxon's two-sample test statistic W - W(X, Y) is defined to beRi where R, is the rank of Y in the combined sample 1. Let T Σǐn i Zi where Zi,Z2, ,Zm are numbers sampled at random without replacement from the set {1,2,..., N} Show that E(Z) = (N + 1)/2 and hence E(T) m(N + 1)/2 Show that 3N2 +5N+2 E(Z) = and hence 12 ー1 Deduce that under the null hypothesis that F = G, the expectation and variance of Wilcoxon's two-sample test statistic are m(n+m+1)/2 and nm(n+m+1)/12, respectively

Answer 1

$1.~P(Z_i=j)=\frac{\binom{N-1}{m-1}}{\binom{N}{m}}=\frac{1}{N},~j=1,2,...,N\\ P(Z_i=j,~Z_k=j^\prime)=\frac{\binom{N-2}{m-2}}{\binom{N}{m}}=\frac{1}{N(N-1)},~j\neq j^\prime=1(1)N.\\ E(Z_i)=\sum_{j=1}^Nj\times \frac{1}{N}=\frac{1}{N}\times \frac{N(N+1)}{2}=\frac{N+1}{2}.\\ E(T)=E\left(\sum_{i=1}^mZ_i \right )=\sum_{i=1}^mE(Z_i)=\frac{m(N+1)}{2}\\ E(Z_i^2)=\sum_{j=1}^Nj^2\times \frac{1}{N}=\frac{1}{N}\times \frac{N(N+1)(2N+1)}{6}=\frac{(N+1)(2N+1)}{6}\\ E(Z_iZ_k)=\sum_{j\neq j^\prime}^N jj^{\prime}\times \frac{1}{N(N-1)}=\frac{1}{N(N-1)}\left(\sum_{j=1}^N j \left(\sum_{j^\prime=1}^Nj^\prime -j \right )\right )\\ =\frac{1}{N(N-1)}\left(\left(\sum_{j=1}^N j \right )\left(\sum_{j^\prime=1}^Nj^\prime\right) -\sum_{j=1}^Nj ^2\right )\\ =\frac{1}{N(N-1)}\left(\frac{N^2(N+1)^2}{4} -\frac{N(N+1)(2N+1)}{6}\right )=\frac{3N^3+2N^2-3N-2}{12(N-1)}\\ =\frac{3N^2+5N+2}{12}\\ Cov(Z_i,Z_k)=E(Z_iZ_k)-E(Z_i)E(Z_k)=-\frac{N+1}{12}\\ Var(Z_i)=E(Z_i^2)-E^2(Z_i)=\frac{N^2-1}{12}\\$

$Var(T)=\sum_{i=1}^mVar(Z_i)+\sum_{i\neq k}^mCov(Z_i,Z_k)=\frac{m(N^2-1)}{12}-\frac{m(m-1)(N+1)}{12}\\ =\frac{m(N-m)(N+1)}{12}\\ If~Z_i's~are~rank~of~Y_i's~in~the~combined~sample~then~R_i=Z_i~and~under~H_0\\ F=G~then~we~select~Z_i's,~i=1(1)m~randomly~from~\{1,2,...,N\}~without~replacement.\\ Here,~N=n+m~then~put~m+n~instead~of~N~in~E(T)~and~Var(T),~we~get~\\ expectation=m(n+m+1)/2,~variance=mn(m+n+1)/12.$