Question

1. You have two independent samples, Xi,..., Xn and Yi,... , Ym drawn from populations with continuous distributions. Suppose the two samples are combined and the combined set of values are put in increasing order. Let Vr- 1 if the value with rank r in the combined sample is a Y and V,-0 if it is an X, for r-1, . . . ,N, where N-m+ n. Show that, if the two populations are the same then mn E(V) TES The general linear rank statistic is defined to be T-:-1 ary, where (ar) are given constants. Show that the mean and variance of the general linear rank statistic are mn E (T)- N 2arandVar (T) - Show that Wilcoxon's W is a linear rank statistic. Use the formulae above to derive its mean and variance, under the null hypothesis that the populations are the same. Wilcoxon's W is designed to test whether two populations differ in their location. How could W be adapted to test whether two populations with the same median differ in their variance?

Answer 1

$P(V_r=1)=\frac{\binom{m}{1}}{\binom{N}{1}}=\frac{m}{N}\\ Since~V_r=1\Rightarrow the~value~with~rank~r~in~the~combined~sample~is~a~Y\\ so~the~favourable~cases~of~event~\{V_r=1\}~is~to~choose~any~one~from~Y_1,\\ Y_2,...,Y_m~in~\binom{m}{1}~ways~whereas~total~no.~of~possible~cases~\binom{N}{1}~ways.\\ E(V_r)=P(V_r=1)=\frac{m}{N}\\ E(V_r^2)=P(V_r=1)=\frac{m}{N},~Var(V_r)=E(V_r^2)-E^2(V_r)=\frac{m}{N}-\frac{m^2}{N^2}\\=\frac{m}{N}\left(1-\frac{m}{N} \right )=\frac{mn}{N^2},~since~N=n+m\\$

$P(V_r=1,V_s=1)=\frac{\binom{m}{2}}{\binom{N}{2}}=\frac{m(m-1)}{N(N-1)}\\ Since~\{V_r=1,V_s=1\}\Rightarrow the~values~with~ranks~r~and~s~in~the~combined~sample~are~Y\\ so~the~favourable~cases~of~event~\{V_r=1,V_s=1\}~is~to~choose~any~two~from~Y_1,\\ Y_2,...,Y_m~in~\binom{m}{2}~ways~whereas~total~no.~of~possible~cases~\binom{N}{2}~ways.\\ E(V_rV_s)=P(V-r=1,V_s=1)=\frac{m(m-1)}{N(N-1)}\\ Cov(V_r,V_s)=E(V_rV_s)-E(V_r)E(V_s)=\frac{m}{N}\left(\frac{m-1}{N-1}-\frac{m}{N} \right )=-\frac{mn}{N^2(N-1)},\\ since~N=m+n$

$E(T)=\sum_{r=1}^Na_rE(V_r)=\frac{m}{N}\sum_{r=1}^Na_r\\ Var(T)=\sum_{r=1}^Na_r^2Var(V_r)+\sum_{r\neq s}^Na_ra_sCov(V_r,V_s)\\ =\frac{mn}{N^2}\sum_{r=1}^Na_r^2-\frac{mn}{N^2(N-1)}\sum_{r\neq s}^Na_ra_s=\frac{mn}{N^2(N-1)}\left((N-1)\sum_{r=1}^Na_r^2-\sum_{r=1}^Na_r\left(\sum_{s=1}^Na_s-a_r \right ) \right )\\ =\frac{mn}{N^2(N-1)}\left((N-1)\sum_{r=1}^Na_r^2-N^2\bar{a} +\sum_{r=1}^Na_r^2\right ),~where~\bar{a}=\frac{1}{N}\sum_{r=1}^Na_r\\$

$=\frac{mn}{N(N-1)}\left(\sum_{r=1}^Na_r^2-N\bar{a}^2 \right ) =\frac{mn}{N(N-1)}\sum_{r=1}^N(a_r-\bar{a})^2$

$W=sum~of~the~rank~of~Y's~in~combined~sample=\sum_{r=1}^NrV_r\\ i.e.~W=T~when~a_r=r,~r=1,2,...,N.\\ Hence~W~is~a~linear~rank~statistic.\\ E(W)=\frac{m}{N}\times \frac{N(N+1)}{2}=\frac{m(N+1)}{2}\\ Var(W)=\frac{mn}{N(N-1)}\left(\sum_{r=1}^Nr^2-\frac{1}{N}\left(\sum_{r=1}^Nr \right )^2 \right )\\ =\frac{mn}{N(N-1)}\left(\frac{N(N+1)(2N+1)}{6}-\frac{1}{N}\times \frac{N^2(N+1)^2}{4} \right )=\frac{mn(N+1)}{12}$