Wilcoxon Signed-Rank Statistic: Asymptotic Normality

Proposition

Under \(H_0\),

\[W_- \xrightarrow{d} \mathcal{N}\!\left(\frac{n(n+1)}{4},\, \frac{n(n+1)(2n+1)}{24}\right).\]

Proof

Write \(W_- = \sum_{i=1}^n R_i\,\mathbf{1}\{D_i < 0\}\). Under \(H_0\) each sign is symmetric and independent of the ranks, so

\[W_- \stackrel{d}{=} \sum_{i=1}^n i\,\varepsilon_i, \qquad \varepsilon_i \overset{\text{iid}}{\sim} \mathrm{Ber}(1/2).\]

Expectation.

\[\mathbb{E}[W_-] = \sum_{i=1}^n i\,\mathbb{E}[\varepsilon_i] = \frac{1}{2}\sum_{i=1}^n i = \frac{n(n+1)}{4}.\]

Variance. Since the \(\varepsilon_i\) are independent and \(\mathrm{Var}(\varepsilon_i) = 1/4\),

\[\mathrm{Var}(W_-) = \frac{1}{4}\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{24}.\]

CLT. The summands \(X_i = i\,\varepsilon_i\) are independent. Check Lyapunov’s condition with \(\delta=2\):

\[\frac{\sum_{i=1}^n \mathbb{E}[|X_i - \mathbb{E}X_i|^4]}{\mathrm{Var}(W_-)^2} \leq \frac{\sum_{i=1}^n i^4}{\left(\sum_{i=1}^n i^2\right)^2} = O\!\left(\frac{n^5}{n^6}\right) = O(n^{-1}) \to 0.\]

Hence \(\dfrac{W_- - \mathbb{E}[W_-]}{\sqrt{\mathrm{Var}(W_-)}} \xrightarrow{d} \mathcal{N}(0,1)\). \(\blacksquare\)

Appendix: Lyapunov’s CLT

Lyapunov’s Central Limit Theorem

Let \(X_1, X_2, \dots\) be independent random variables (not necessarily identically distributed) with \(\mathbb E[X_k] = \mu_k\) and finite variance \(\mathrm{Var}(X_k) = \sigma_k^2\). Set \[ S_n = \sum_{k=1}^n X_k, \qquad s_n^2 = \mathrm{Var}(S_n) = \sum_{k=1}^n \sigma_k^2. \]

If there exists \(\delta > 0\) such that \[ \lim_{n \to \infty}\; \frac{1}{s_n^{2 + \delta}}\,\sum_{k=1}^n \mathbb E\!\left[\,|X_k - \mu_k|^{2 + \delta}\,\right] \;=\; 0, \tag{Lyapunov} \] then \[ \frac{S_n - \mathbb E[S_n]}{\sqrt{\mathrm{Var}(S_n)}} \;\xrightarrow[n \to \infty]{d}\; \mathcal N(0, 1). \]

Intuition. The condition says that no single term \(X_k\) dominates the sum: the contribution of each term to the higher-order moment \(\sum_k \mathbb E|X_k - \mu_k|^{2+\delta}\) is asymptotically negligible compared to the total variance raised to the power \((2+\delta)/2\). In the i.i.d. case (\(\sigma_k^2 = \sigma^2\)), Lyapunov’s condition is automatically satisfied and reduces to the classical CLT.

Common choice. Taking \(\delta = 1\) requires controlling third absolute moments: \[ \frac{1}{s_n^3}\sum_{k=1}^n \mathbb E\!\left[|X_k - \mu_k|^3\right] \to 0. \]

In the Wilcoxon proof above we use \(\delta = 2\) (fourth moments) because the bounded Bernoulli terms make fourth moments especially easy to compute.