Walsh average ($(X_i+X_j)/2$)

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.