[proofplan] The statistic is distributionally represented as a weighted sum of independent fair Bernoulli variables: the absolute ranks contribute the deterministic weights $1,\dots,n$, and the independent random signs decide which weights are selected. The expectation and variance then follow by additivity, because the Bernoulli variables are independent. For the normal approximation, we compute the characteristic function of the standardized weighted Bernoulli sum and show that its logarithm converges to $-t^2/2$ for every fixed $t\in\mathbb R$. [/proofplan]

[step:Represent the signed-rank statistic by independent Bernoulli selectors]Let $(\Omega,\mathcal F,\mathbb P)$ be the probability space on which the random variables $Y_1,\dots,Y_n$ are defined. Let $S_n$ denote the symmetric group, that is, the set of all permutations of $\{1,\dots,n\}$. For each $i\in\{1,\dots,n\}$, define the sign selector $S_i:\Omega\to\{0,1\}$ by \begin{align*} S_i(\omega)=\mathbb{1}_{(0,\infty)}(Y_i(\omega)). \end{align*} Since the distribution of $Y_i$ is continuous and symmetric about $0$, we have $\mathbb P(Y_i=0)=0$ and \begin{align*} \mathbb P(S_i=1)=\mathbb P(Y_i>0)=\frac12. \end{align*} Independence of $Y_1,\dots,Y_n$ gives independence of $S_1,\dots,S_n$. For each Borel set $A\subset[0,\infty)$, symmetry of the law of $Y_i$ and continuity at $0$ give \begin{align*} \mathbb P(Y_i>0, |Y_i|\in A)=\mathbb P(Y_i\in A\cap(0,\infty))=\frac12\mathbb P(|Y_i|\in A). \end{align*} Thus $S_i$ is independent of $|Y_i|$. Since the random variables $Y_1,\dots,Y_n$ are independent, the pairs $(S_i,|Y_i|)$ are independent. To pass from coordinatewise independence to vector independence, take arbitrary values $s_i\in\{0,1\}$ and Borel sets $A_i\subset[0,\infty)$. Independence of the pairs and the preceding coordinatewise independence give \begin{align*} \mathbb P(S_i=s_i, |Y_i|\in A_i \text{ for all } i)=\prod_{i=1}^n\mathbb P(S_i=s_i, |Y_i|\in A_i)=\prod_{i=1}^n\mathbb P(S_i=s_i)\prod_{i=1}^n\mathbb P(|Y_i|\in A_i). \end{align*} Since such rectangles generate the product Borel $\sigma$-algebra, the vector $(S_1,\dots,S_n)$ is independent of the vector $(|Y_1|,\dots,|Y_n|)$. Let $\pi:\Omega\to S_n$ be the random permutation determined almost surely by the increasing order of the absolute values: \begin{align*} |Y_{\pi(1)}|<|Y_{\pi(2)}|<\dots<|Y_{\pi(n)}|. \end{align*} This map is well-defined almost surely. Indeed, for $i\neq j$, the [random variable](/page/Random%20Variable) $|Y_i|$ is non-atomic because $Y_i$ has a continuous distribution, and independence gives \begin{align*} \mathbb P(|Y_i|=|Y_j|)=\mathbb E\big[\mathbb P(|Y_i|=|Y_j|\mid |Y_j|)\big]=0. \end{align*} The finite union over pairs $i\neq j$ therefore has probability $0$. Since $\pi$ is a measurable function of $(|Y_1|,\dots,|Y_n|)$, the vector $(S_1,\dots,S_n)$ is independent of $\pi$. Therefore \begin{align*} W_{n,+}(Y_1,\dots,Y_n)=\sum_{k=1}^n k\,S_{\pi(k)} \end{align*} almost surely. Since $(S_{\pi(1)},\dots,S_{\pi(n)})$ has the same distribution as $(S_1,\dots,S_n)$, if $B_1,\dots,B_n$ are independent random variables with $B_k\sim\operatorname{Ber}(1/2)$, then \begin{align*} W_{n,+}(Y_1,\dots,Y_n)\stackrel{d}{=}\sum_{k=1}^n kB_k. \end{align*}[/step]

[guided]Let $(\Omega,\mathcal F,\mathbb P)$ be the probability space carrying the random variables $Y_1,\dots,Y_n$. Let $S_n$ denote the symmetric group, meaning the set of all permutations of $\{1,\dots,n\}$. The statistic $W_{n,+}$ first orders the observations by absolute value and then adds the ranks whose original observations are positive. Thus the ranks are the deterministic list $1,\dots,n$ after ordering, while the remaining randomness is whether the observation carrying rank $k$ has positive sign. For each $i\in\{1,\dots,n\}$, define the sign selector $S_i:\Omega\to\{0,1\}$ by \begin{align*} S_i(\omega)=\mathbb{1}_{(0,\infty)}(Y_i(\omega)). \end{align*} Because $Y_i$ has a continuous distribution, $\mathbb P(Y_i=0)=0$. Because the distribution is symmetric about $0$, the two events $\{Y_i>0\}$ and $\{Y_i<0\}$ have equal probability. Hence \begin{align*} \mathbb P(S_i=1)=\mathbb P(Y_i>0)=\frac12. \end{align*} The random variables $Y_1,\dots,Y_n$ are independent, so the sign selectors $S_1,\dots,S_n$ are independent. We also need the signs to be independent of the absolute ranks. Fix $i\in\{1,\dots,n\}$ and let $A\subset[0,\infty)$ be a Borel set. Symmetry of the law of $Y_i$ gives the same probability to $A\cap(0,\infty)$ and to $-A\cap(-\infty,0)$, while continuity gives no atom at $0$. Therefore \begin{align*} \mathbb P(Y_i>0, |Y_i|\in A)=\mathbb P(Y_i\in A\cap(0,\infty))=\frac12\mathbb P(|Y_i|\in A). \end{align*} This proves that $S_i$ is independent of $|Y_i|$. Since independence across $i$ makes the pairs determined by different $Y_i$ independent, we verify the resulting vector independence on rectangles. Choose arbitrary values $s_i\in\{0,1\}$ and Borel sets $A_i\subset[0,\infty)$. Independence of the pairs $(S_i,|Y_i|)$ gives \begin{align*} \mathbb P(S_i=s_i, |Y_i|\in A_i \text{ for all } i)=\prod_{i=1}^n\mathbb P(S_i=s_i, |Y_i|\in A_i). \end{align*} For each factor, the coordinatewise independence already proved gives \begin{align*} \mathbb P(S_i=s_i, |Y_i|\in A_i)=\mathbb P(S_i=s_i)\mathbb P(|Y_i|\in A_i). \end{align*} Multiplying these identities yields the product of the probability of the sign rectangle and the probability of the absolute-value rectangle. Since rectangles generate the product Borel $\sigma$-algebras, the full sign vector $(S_1,\dots,S_n)$ is independent of the absolute-value vector $(|Y_1|,\dots,|Y_n|)$. Now define $\pi:\Omega\to S_n$ to be the random permutation that arranges the absolute values in increasing order: \begin{align*} |Y_{\pi(1)}|<|Y_{\pi(2)}|<\dots<|Y_{\pi(n)}|. \end{align*} The ordering is defined almost surely because pairwise ties have probability $0$. To see this, fix $i\neq j$. The random variable $|Y_i|$ is non-atomic: for any $a\geq0$, continuity of the law of $Y_i$ gives $\mathbb P(|Y_i|=a)=0$. Since $|Y_i|$ and $|Y_j|$ are independent, \begin{align*} \mathbb P(|Y_i|=|Y_j|)=\mathbb E\big[\mathbb P(|Y_i|=|Y_j|\mid |Y_j|)\big]=0. \end{align*} There are only finitely many pairs $(i,j)$ with $i\neq j$, so the probability of at least one absolute-value tie is $0$. Since $\pi$ is determined by $(|Y_1|,\dots,|Y_n|)$, the independence just proved implies that $(S_1,\dots,S_n)$ is independent of $\pi$. With this notation, the observation carrying absolute rank $k$ is $Y_{\pi(k)}$, and it contributes $k$ precisely when $Y_{\pi(k)}>0$. Hence \begin{align*} W_{n,+}(Y_1,\dots,Y_n)=\sum_{k=1}^n k\,S_{\pi(k)} \end{align*} almost surely. Since $\pi$ is independent of the signs and the signs are independent fair Bernoulli variables, the permuted vector $(S_{\pi(1)},\dots,S_{\pi(n)})$ has the same distribution as a vector $(B_1,\dots,B_n)$ of independent $\operatorname{Ber}(1/2)$ variables. Consequently \begin{align*} W_{n,+}(Y_1,\dots,Y_n)\stackrel{d}{=}\sum_{k=1}^n kB_k. \end{align*}[/guided]

[step:Compute the expectation from the Bernoulli representation] Let $(\Omega_B,\mathcal F_B,\mathbb P_B)$ be an auxiliary probability space carrying independent random variables $B_1,\dots,B_n$ with $B_k\sim\operatorname{Ber}(1/2)$. Define the map $T_n:\Omega_B\to\mathbb R$ by \begin{align*} T_n(\omega)=\sum_{k=1}^n kB_k(\omega). \end{align*} By the distributional identity from the previous step, \begin{align*} \mathbb E[W_{n,+}(Y_1,\dots,Y_n)] = \mathbb E_B[T_n], \end{align*} where $\mathbb E_B$ denotes expectation with respect to $\mathbb P_B$. Using linearity of expectation and $\mathbb E_B[B_k]=1/2$, \begin{align*} \mathbb E_B[T_n] = \sum_{k=1}^n k\,\mathbb E_B[B_k] = \frac12\sum_{k=1}^n k = \frac12\cdot \frac{n(n+1)}{2} = \frac{n(n+1)}{4}. \end{align*} [/step]

[step:Compute the variance from independence] Since $B_k\sim\operatorname{Ber}(1/2)$, we have \begin{align*} \operatorname{Var}(B_k)=\frac12\left(1-\frac12\right)=\frac14. \end{align*} The random variables $B_1,\dots,B_n$ are independent, so the variables $kB_k$ are independent. Therefore additivity of variance for independent random variables gives \begin{align*} \operatorname{Var}(W_{n,+}(Y_1,\dots,Y_n)) = \operatorname{Var}_B(T_n) = \sum_{k=1}^n \operatorname{Var}_B(kB_k) = \sum_{k=1}^n k^2\operatorname{Var}_B(B_k). \end{align*} Thus \begin{align*} \operatorname{Var}(W_{n,+}(Y_1,\dots,Y_n)) = \frac14\sum_{k=1}^n k^2 = \frac14\cdot \frac{n(n+1)(2n+1)}{6} = \frac{n(n+1)(2n+1)}{24}. \end{align*} [/step]

[step:Normalize the weighted Bernoulli sum] For each $n\in\mathbb N$, define the variance scale \begin{align*} \sigma_n^2:=\frac{n(n+1)(2n+1)}{24}. \end{align*} Define the standardized random variable $Z_n:\Omega_B\to\mathbb R$ by \begin{align*} Z_n(\omega)=\frac{\sum_{k=1}^n k(B_k(\omega)-1/2)}{\sigma_n}. \end{align*} It is enough to prove $Z_n\xrightarrow{d}\mathcal N(0,1)$, because $W_{n,+}(Y_1,\dots,Y_n)$ has the same distribution as $\sum_{k=1}^n kB_k$. For fixed $t\in\mathbb R$, define the characteristic function $\varphi_n:\mathbb R\to\mathbb C$ by $\varphi_n(t):=\mathbb E_B[e^{itZ_n}]$. Independence of $B_1,\dots,B_n$ gives \begin{align*} \varphi_n(t)=\prod_{k=1}^n \mathbb E\left[\exp\left(it\frac{k(B_k-1/2)}{\sigma_n}\right)\right]. \end{align*} Since $B_k$ takes the values $0$ and $1$ with probability $1/2$ each, this becomes \begin{align*} \varphi_n(t)=\prod_{k=1}^n \frac12\left(e^{itk/(2\sigma_n)}+e^{-itk/(2\sigma_n)}\right)=\prod_{k=1}^n \cos\left(\frac{tk}{2\sigma_n}\right). \end{align*} [/step]

[step:Show the characteristic functions converge to the standard normal characteristic function] Fix $t\in\mathbb R$. Define \begin{align*} a_{n,k}:=\frac{tk}{2\sigma_n} \end{align*} for $k\in\{1,\dots,n\}$. Since $\sigma_n^2=n(n+1)(2n+1)/24$, we have \begin{align*} \max_{1\leq k\leq n}|a_{n,k}|\leq \frac{|t|n}{2\sigma_n}\to 0. \end{align*} Thus, for this fixed $t$, there exists $N_t\in\mathbb N$ such that $|a_{n,k}|\leq 1/2$ for every $n\geq N_t$ and every $k\in\{1,\dots,n\}$. For such $n$ and $k$, $\cos(a_{n,k})>0$, so the real logarithm is defined and $\log\varphi_n(t)=\sum_{k=1}^n\log(\cos a_{n,k})$. Using the Taylor expansion \begin{align*} \log(\cos u)=-\frac{u^2}{2}+O(u^4) \end{align*} as $u\to 0$, with the implicit constant uniform for $|u|\leq 1/2$, we obtain for $n\geq N_t$ \begin{align*} \log \varphi_n(t)=\sum_{k=1}^n \log(\cos a_{n,k})=-\frac12\sum_{k=1}^n a_{n,k}^2+O\left(\sum_{k=1}^n a_{n,k}^4\right). \end{align*} The quadratic term is \begin{align*} \sum_{k=1}^n a_{n,k}^2=\frac{t^2}{4\sigma_n^2}\sum_{k=1}^n k^2=\frac{t^2}{4\sigma_n^2}\cdot \frac{n(n+1)(2n+1)}{6}=t^2, \end{align*} because $\sigma_n^2=n(n+1)(2n+1)/24$. For the error term, \begin{align*} \sum_{k=1}^n a_{n,k}^4 \leq \left(\max_{1\leq k\leq n}a_{n,k}^2\right)\sum_{k=1}^n a_{n,k}^2 \to 0. \end{align*} Therefore \begin{align*} \log \varphi_n(t)\to -\frac{t^2}{2}, \end{align*} and hence \begin{align*} \varphi_n(t)\to e^{-t^2/2}. \end{align*} The function $t\mapsto e^{-t^2/2}$ is the characteristic function of $\mathcal N(0,1)$, so the [continuity theorem](/theorems/1145) for characteristic functions gives \begin{align*} Z_n\xrightarrow{d}\mathcal N(0,1). \end{align*} Substituting the definition of $Z_n$ and using the distributional representation of $W_{n,+}(Y_1,\dots,Y_n)$ gives \begin{align*} \frac{W_{n,+}(Y_1,\dots,Y_n)-n(n+1)/4}{\sqrt{n(n+1)(2n+1)/24}} \xrightarrow{d} \mathcal N(0,1). \end{align*} This is the desired normal approximation. [/step]