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 \(\delta=2\):

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

where \(s_n^2 = \mathrm{Var}(W_-)\). Hence \(\dfrac{W_- - \mathbb{E}[W_-]}{\sqrt{\mathrm{Var}(W_-)}} \xrightarrow{d} \mathcal{N}(0,1)\). \(\blacksquare\)