Mann–Whitney U test ($U_X=W_X-m(m+1)/2$)

For two independent samples of sizes $m,n$, rank their pooled observations and let $W_X$ be the sum of the ranks of the first sample. With continuous observations, the null hypothesis of a common distribution makes every allocation of $m$ ranks to that sample equally probable. A one-sided alternative that the first sample is stochastically smaller rejects for small $W_X$, equivalently small $U_X$. In a common-shape location model this tests a negative location difference. Without the common-shape assumption it is not simply a test of medians. The shifted statistic $U_X$ counts cross-sample pairs for which the first observation exceeds the second. Ties use average ranks and a conditional permutation test or a tie-adjusted normal approximation.

With $N=m+n$ and no ties, sampling $m$ ranks without replacement gives the displayed mean and variance. For observed rank sum below its mean, a continuity-corrected lower-tail statistic is $Z=[W_X-m(N+1)/2+1/2]/\sqrt{mn(N+1)/12}$. Ties of sizes $t_g$ replace the variance by $mn[(N+1)-{\sum }_g(t_g^3-t_g)/(N(N-1))]/12$. These formulas concern the raw rank sum; the Mann–Whitney U test statistic subtracts $m(m+1)/2$ from both the observation and its mean.

Under the continuous common-distribution null of the Wilcoxon rank sum test, pooled labels are exchangeable. The product records every choice of ranks: choosing $z{t}^k$ assigns rank $k$ to the first sample. Its indicated coefficient counts subsets of size $m$ and rank sum $w$. Divide by $\left(\genfrac{}{}{0ex}{}{m+n}{m}\right)$ and sum the lower tail for the one-sided P-value. Equivalently the counts satisfy $a_{k,h,w}=a_{k-1,h,w}+a_{k-1,h-1,w-k}$ with $a_{0,0,0}=1$, allowing an exact integer computation.

For two independent continuous samples of positive sizes $m,n$ under a common-distribution null, each $m$-subset of ranks $1,\dots ,m+n$ is equally likely. The smallest rank sum is $m(m+1)/2$, attained only by $\{1,\dots ,m\}$. The only subset with rank sum one larger is $\{1,\dots ,m-1,m+1\}$. Hence the exact lower tail at that threshold is $2/\left(\genfrac{}{}{0ex}{}{m+n}{m}\right)$. This follows directly by observing that any other change increases the sum by at least two.