Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 32 / 5 / ii / Solution

Use the usual two mutually independent samples, and rank their $N=n+m$ pooled observations increasingly. Continuity gives no ties with probability one. Let $R_i$ be the pooled rank of $X_i$ and define the Wilcoxon rank sum test statisticThe last identity counts the ranks contributed by the other sample after subtracting the within-$X$ ranks.

When $F_X\ge F_Y$, the $X$ observations are stochastically smaller, so reject for small $W_X$, equivalently small $U_X$. For instance,The inequality follows from $F_X\ge F_Y$ and $\int F_{Y}d{F}_Y=1/2$ for continuous $F_Y$. A common-uniform quantile coupling also makes this direction transparent: $F_X^{-1}(u)\le F_Y^{-1}(u)$, and $U_X$ is nondecreasing in each $X_i$.

Under the common-distribution null hypothesis, the pooled observations are exchangeable. Every selection of $n$ ranks for the $X$ labels has probability ${\left(\genfrac{}{}{0ex}{}{N}{n}\right)}^{-1}$. Thus the exact null law isChoose a lower-tail critical value with probability at most the desired size, optionally randomizing at its boundary to attain the size exactly. This law is independent of the unknown common distribution. Common increasing transformations leave the pooled ranks of observations unchanged, so the statistic is a function of the maximal invariant from part (i). Although a rank sum is not itself maximal, it is an invariant test statistic. These facts justify the test within nonparametric statistics.

Sampling ranks without replacement givesso ${\mathbb{E}}_0{U}_X=nm/2$ with the same variance. For $n,m\to \infty$ with $n/N\to \rho \in (0,1)$, the normal approximation for a rank sum statesEquivalently replace $W_X-n(N+1)/2$ by $U_X-nm/2$. The large-sample one-sided test rejects when this standardized statistic is below the lower $\alpha$ normal distribution quantile. This is the two-sample rank-sum test, rather than the paired Wilcoxon signed-rank test.