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\title{Asymptotic Normality of the Wilcoxon Signed-Rank Statistic via Lindeberg CLT}
\author{chatgpt.com}
%\date{}

\begin{document}
\maketitle

\section{Model and notation}
Let $X_1,\dots,X_n$ be i.i.d. continuous random variables with a distribution symmetric about $0$.
Define
\[
R_i = \mathrm{rank}(|X_i|) \in \{1,\dots,n\}, \quad S_i = \mathbf{1}(X_i > 0).
\]
The Wilcoxon signed-rank statistic is
\[
T^+ = \sum_{i=1}^n R_i S_i.
\]
Under symmetry:
\begin{itemize}
\item $\mathbb{P}(S_i=1)=\mathbb{P}(S_i=0)=1/2$,
\item $(S_i)$ are independent of $(|X_i|)$ and hence of $(R_i)$,
\item $(R_1,\dots,R_n)$ is a uniform random permutation of $\{1,\dots,n\}$.
\end{itemize}

\section{Expectation and variance}
We use linearity and exchangeability.
\[
\mathbb{E}[T^+] = \sum_{i=1}^n \mathbb{E}[R_i]\mathbb{E}[S_i]
= n \cdot \frac{n+1}{2} \cdot \frac{1}{2}
= \frac{n(n+1)}{4}.
\]

The variance is known to be
\[
\mathrm{Var}(T^+) = \frac{n(n+1)(2n+1)}{24}.
\]

\section{Representation as a linear rank statistic}
Define centered signs:
\[
Z_i = S_i - \frac{1}{2}, \quad \mathbb{E}[Z_i]=0, \quad \mathrm{Var}(Z_i)=\frac{1}{4}.
\]
Then
\[
T^+ - \mathbb{E}[T^+] = \sum_{i=1}^n R_i Z_i.
\]
Conditionally on $R_1,\dots,R_n$, this is a sum of independent mean-zero random variables.

Define weights
\[
a_{ni} = R_i.
\]
Then
\[
T^+ - \mathbb{E}[T^+] = \sum_{i=1}^n a_{ni} Z_i.
\]

This is a triangular array structure suitable for the Lindeberg-Feller CLT.

\section{Variance normalization}
Let
\[
V_n = \mathrm{Var}(T^+) = \frac{n(n+1)(2n+1)}{24} \sim \frac{n^3}{12}.
\]
Define normalized sum
\[
U_n = \frac{T^+ - \mathbb{E}[T^+]}{\sqrt{V_n}}.
\]

Then
\[
U_n = \sum_{i=1}^n b_{ni} Z_i, \quad b_{ni} = \frac{R_i}{\sqrt{V_n}}.
\]

\section{Lindeberg condition}
We verify the Lindeberg condition for the triangular array $\{b_{ni} Z_i\}$.

Since $|Z_i| \le 1/2$, we have for any $\varepsilon>0$:
\[
|b_{ni} Z_i| \le \frac{R_i}{2\sqrt{V_n}} \le \frac{n}{2\sqrt{V_n}} \to 0.
\]
Hence for large $n$, no single term contributes a large amount, and Lindeberg's condition is trivially satisfied:
\[
\sum_{i=1}^n \mathbb{E}[b_{ni}^2 Z_i^2 \mathbf{1}(|b_{ni} Z_i|>\varepsilon)] = 0 \quad \text{for large } n.
\]

\section{Asymptotic variance normalization}
We compute:
\[
\sum_{i=1}^n \mathbb{E}[b_{ni}^2 Z_i^2]
= \frac{1}{4V_n} \sum_{i=1}^n \mathbb{E}[R_i^2].
\]
Since $R_i$ is uniform on $\{1,\dots,n\}$:
\[
\mathbb{E}[R_i^2] = \frac{(n+1)(2n+1)}{6}.
\]
Thus
\[
\sum_{i=1}^n \mathbb{E}[b_{ni}^2 Z_i^2]
= \frac{n(n+1)(2n+1)}{24V_n} = 1.
\]

\section{Application of Lindeberg-Feller CLT}
Since:
\begin{itemize}
\item $Z_i$ are independent, mean zero,
\item the variance normalization holds,
\item Lindeberg condition is satisfied,
\end{itemize}
we conclude:
\[
U_n \xrightarrow{d} \mathcal{N}(0,1).
\]

\section{Conclusion}
Therefore,
\[
\frac{T^+ - \frac{n(n+1)}{4}}{\sqrt{\frac{n(n+1)(2n+1)}{24}}}
\xrightarrow{d} \mathcal{N}(0,1).
\]

This establishes the asymptotic normality of the Wilcoxon signed-rank statistic.

\end{document}