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Discarding zero observations and resolving ties is unnecessary with probability one under the stated independent continuous model. Rank the absolute observations from $1$ to $n$, let $R_i$ be the rank of $|X_i|$, and form the positive Wilcoxon signed-rank statisticReject the null hypothesis for large $W^{+}$. Under symmetry about zero, each sign is a fair independent binary variable, independent of all the absolute values. To see this independence, for every measurable positive set $B$, symmetry gives $\mathrm{Pr}(X\in B)=\mathrm{Pr}(X\in -B)=\frac{1}{2}\mathrm{Pr}(|X|\in B)$. Independence of observations then gives independence of the whole sign vector from the absolute-value vector. Conditional on those absolute values, relabeling by rank therefore givesThe null distribution is consequently independent of the unknown symmetric shape. Its probability generating function is $2^{-n}{\prod }_{r=1}^n(1+z^r)$, so the coefficient of $z^w$ gives the exact probability of $W^{+}=w$. Choose an upper critical value with null tail probability at most the desired level, or randomize at the boundary to obtain that level exactly. The observed upper-tail probability is an exact one-sided p-value. A positive translation raises each Walsh average, as established below, so this upper-tail rule has the correct direction against positive centers for every fixed symmetric shape.

The independent-sign representation provesThe asymptotic null distribution, stated without proof, isThis supplies the usual upper normal distribution critical value when exact calculation is inconvenient; for the discrete upper tail $\mathrm{Pr}(W^{+}\ge c)$, use $c-1/2$ in the continuity correction.

For exact equivalence with the proposed statistic, reorder the observations so $|X_{(1)}|<\cdots <|X_{(n)}|$. Partition the unordered pairs, including diagonals, by the member with larger absolute value. The block belonging to $X_{(r)}$ consists of the $r-1$ pairs with an earlier member and its diagonal pair. For an earlier member $X_{(j)}$, the strict inequality $|X_{(j)}|<|X_{(r)}|$ means the sign of $X_{(j)}+X_{(r)}$ is the sign of $X_{(r)}$; the diagonal sum has that same sign. Thus the entire block supplies exactly $r$ positive averages when $X_{(r)}>0$, and none otherwise. Summing the blocks provesThe two tests therefore have exactly the same statistic, critical values and p-values, not merely the same limiting distribution. The Walsh averages are dependent, so their positivity indicators must not instead be treated as an independent binomial sample. For a translation $X_i=Z_i+\theta$ of any null-symmetric sample, every Walsh average increases by $\theta$; their positive count is nondecreasing, which also verifies the direction of the one-sided test.