Wilcoxon signed-rank test ($W_{+}={\sum }_{D_i>0}\mathrm{rank}(|D_i|)$)

For independent paired observations, form differences $D_i=U_i-V_i$. Assume the differences are identically distributed, continuous and symmetric about a location $\delta$. Under the zero-location null hypothesis, the signs are independent fair signs conditional on their absolute values. Rank the absolute differences and add ranks with positive signs. A negative-location alternative rejects for small $W_{+}$. Its exact conditional distribution is that of a sum of an independently selected subset of ranks, each of the $2^n$ subsets equally likely. Thus its mean is $n(n+1)/4$ and its variance is $n(n+1)(2n+1)/24$. Symmetry is essential: a zero median alone does not justify this signed-rank null law. Zero differences are omitted in the usual version, and tied absolute differences receive average ranks.

The Walsh averages of a sample are $(X_i+X_j)/2$ for $1\le i\le j\le n$, including the diagonal pairs. If the absolute observations are distinct and nonzero, order them as $|X_{(1)}|<\cdots <|X_{(n)}|$. Among the $r$ pairs whose larger absolute observation is $X_{(r)}$, every sum has the sign of $X_{(r)}$. Hence the number of positive Walsh averages is exactly ${\sum }_{r}r{1}_{X_{(r)}>0}$, the positive Wilcoxon signed-rank statistic. This proves equivalence of the corresponding tests; the Walsh averages themselves are generally dependent.

For a continuous sample, the statistic also equals ${\sum }_i{1}_{\{X_i>0\}}+{\sum }_{i<j}{1}_{\{X_i+X_j>0\}}$. In each pair the sign of the sum is the sign of the observation with larger absolute value, proving the identity by counting its smaller-rank partners. Under a continuous symmetric null, the pair kernel has projection $F(x)-1/2$ and projection variance $1/12$. The U-statistic central limit theorem yields asymptotic normality of $W_{+}$, with variance asymptotic to $n^3/12$.