Question

Suppose that X = (Xi, X2, . . . , Xn) and Y = (y,Y2, . . . ,Yn) are random samples from continuous distributions F and G, respectively. Wilcoxon's two-sample test statistic W = W(X,Y) is defined to be Σ-ngi Ri where Ri is the rank of in the combined sample.

Answer 1

$P(I_{i,j}=1)=P(X_i<Y_j)=\frac{1}{2}\\ ~under~F=G~there~are~two~possible~cases:~(X_i<Y_j)~or~(X_i>Y_j)~with~equal\\ chances.\\ P(I_{i,j}=1,I_{i,l}=1)=P(X_i<Y_j,~X_i<Y_l)\\ =P(X_i<Y_j<Y_l)&plus;P(X_i<Y_l<Y_j)=\frac{1}{6}&plus;\frac{1}{6}=\frac{1}{3}\\ ~under~F=G~there~are~six~possible~cases:~(X_i<Y_j<Y_l),~(X_i<Y_l<Y_j),\\ (Y_j<X_i<Y_l),~(Y_l<X_i<Y_j),~(Y_j<Y_l<X_i),~(Y_l<Y_j<X_i)~with~equal\\ chances.\\ P(I_{i,j}=1,I_{k,l}=1)=P(X_i<Y_j,~X_k<Y_l)\\ =P(X_i<Y_j)P(X_k<Y_l),~since~they~are~independent\\ =\frac{1}{4}\\ Hence~E(I_{i,j})=P(I_{i,j}=1)=\frac{1}{2},~E(I_{i,j}I_{i,l})=P(I_{i,j}=1,I_{i,l}=1)=\frac{1}{3}\\ E(I_{i,j}I_{k,l})=P(I_{i,j}=1,I_{k,l}=1)=\frac{1}{4}\\ Var(I_{i,j})=E(I_{i,j}^2)-E^2(I_{i,j})=\frac{1}{4},~Cov(I_{i,j},I_{i,k})=E(I_{i,j}I_{i,k})-E(I_{i,j})E(I_{i,k})\\ =\frac{1}{3}-\frac{1}{4}=\frac{1}{12},~Cov(I_{i,j},I_{l,k})=0~for~i\neq l,~j\neq l\\ E(U)=\sum_{i=1}^n\sum_{j=1}^mE(I_{ij})=\frac{mn}{2}\\$

$Similarly,~Cov(I_{i,j},I_{k,j})=\frac{1}{12},~for~i\neq k \\Var(U)=\sum_{i=1}^n\sum_{j=1}^mVar(I_{i,j})&plus;\sum_{i=1}^n\sum_{j=1}^m\sum_{k=1(k\neq j)}^mCov(I_{i,j},I_{i,k})&plus;\sum_{i=1}^n\sum_{j=1}^m\sum_{k=1(k\neq i)}^nCov(I_{i,j},I_{k,j})\\ =\frac{nm}{4}&plus;\frac{nm(m-1)}{12}&plus;\frac{nm(n-1)}{12}=\frac{nm(3&plus;m-1&plus;n-1)}{12}=\frac{mn(m&plus;n&plus;1)}{12} \\ E(W)=E(U)&plus;\frac{m(m&plus;1)}{2}=\frac{m(m&plus;n&plus;1)}{2}\\ Var(W)=Var(U)=\frac{mn(m&plus;n&plus;1)}{12}$