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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.

Let $X_i=Y_i-{\theta }_0$. Under the symmetric continuous null hypothesis, positive and negative signs have probability $1/2$ and zeros have probability zero. For the sign test, take $N_{+}={\sum }_i{1}_{\{X_i>0\}}$ and reject for small values: $N_{+}\sim \mathrm{Bin}(n,1/2)$ under the null. Choose $k$ with ${\mathbb{P}}_0(N_{+}\le k)\le \alpha$; randomization at the next value attains exact size if desired. The lower tail is appropriate for $\theta <{\theta }_0$.

For the Wilcoxon signed-rank test, rank $|X_i|$ increasingly and put $W_{+}={\sum }_i{R}_i{1}_{\{X_i>0\}}$. Continuous observations have no absolute-value ties almost surely. Conditional on the absolute values, the signs are independent fair signs because symmetry makes the sign independent of magnitude. Hence $W_{+}$ has the distribution of ${\sum }_{r=1}^{n}r{B}_r$ for independent Bernoulli random variables $B_r$ with parameter $1/2$. Again reject in the lower tail, using this exact subset-sum distribution or a normal approximation. Independence gives the requested null moments:For a symmetric kernel $h$ of fixed order $m$, a one-sample U-statistic is $U_n={\left(\genfrac{}{}{0ex}{}{n}{m}\right)}^{-1}{\sum }_{i_1<\cdots <i_m}h(X_{i_1},\dots ,X_{i_m})$. State the U-statistic central limit theorem: if $\mathbb{E}{h}^2<\infty$, $\eta =\mathbb{E}h$, and ${\zeta }_1=\mathrm{Var}(\mathbb{E}[h\mid X_1])>0$, then $\sqrt{n}(U_n-\eta )\Rightarrow N(0,m^2{\zeta }_1)$. The nondegeneracy condition is important.

For $h(x,y)=1_{\{x+y>0\}}$, its order-two U-statistic satisfies the exact Wilcoxon signed-rank statistic identityTo prove it, in an unordered pair the sign of the sum is the sign of whichever observation has larger absolute value. Counting the pairs for which a positive observation is larger contributes its rank minus one; its own positive-sign indicator supplies the remaining one.

Write $F_0$ for the distribution function of $X$ under the null. Symmetry gives $\eta =1/2$ andThe probability integral transform makes $F_0(X_1)$ uniform on $(0,1)$, so ${\zeta }_1=1/12$. Therefore $\sqrt{n}(U_n-1/2)\Rightarrow N(0,1/3)$. The centred $N_{+}$ is $O_p(\sqrt{n})$ and is negligible on the scale $n^{3/2}$ of $W_{+}$. Combining the exact decomposition and its mean givesThe lower-tail approximate level-$\alpha$ rejection rule compares this standardized statistic with ${\Phi }^{-1}(\alpha )$; a half-unit continuity correction can refine discrete-tail calculations.

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.