Let \(X_{1}, \ldots, X_{n}\) be a set of independent and identically distributed random variables from a continuous distribution \(F\) that is symmetric about a point \(\theta\). Let \(\mathbf{R}\) be the vector of ranks of \(\left|X_{1}-\thetaight|, \ldots,\left|X_{n}-\thetaight|\) and let \(\mathbf{C}\) be an \(n \times 1\) vector with \(i^{\text {th }}\) element \(c_{i}=\delta\left\{X_{i}-\theta ;(0, \infty)ight\}\) for \(i=1, \ldots, n\). The signed rank statistic was seen in Example 11.3 to have the form \(W=\mathbf{C}^{\prime} \mathbf{R}\). In Example 11.12 is was further shown that

\[n^{1 / 2}\left(\begin{array}{l}n \\2\end{array}ight)^{-1}[W-E(W)] \xrightarrow{d} Z_{2}\]

as \(n ightarrow \infty\), where \(Z_{2}\) has a \(\mathrm{N}\left(0, \frac{1}{3}ight)\) distribution.

a. Using direct calculations, prove that under the null hypothesis that \(\theta=0\) it follows that \(E(W)=\frac{1}{4} n(n+1)\).

b. Using direct calculations, prove that under the null hypothesis that \(\theta=0\) it follows that \(V(W)=\frac{1}{24} n(n+1)(2 n+1)\).

c. Prove that

\[\frac{W-\frac{1}{4} n(n+1)}{\left[\frac{1}{24} n(n+1)(2 n+1)ight]^{1 / 2}} \xrightarrow{d} Z,\]

as \(n ightarrow \infty\) where \(Z\) has a \(\mathrm{N}(0,1)\) distribution.

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Example 11.3. Let (X1,Y),..., (X,Y) be a set of independent and identi- cally distributed paired random variables from a continuous bivariate distribu- tion F. Let G and H be the marginal distributions of X and Yn, respectively. Assume that H(x) = G(x0) for some shift parameter and that G is a symmetric distribution about zero. Let Z = X;-Y; and let R,..., R denote the ranks of the absolute differences |Z1,..., Zn. Define C = {Zi; (0,)} for i = 1,...,n, with C' = (C,..., Cn). The Wilcoxon signed rank statistic is then given by W = C'R, which is the sum of the ranks of the absolute differ- ences that correspond to positive differences. When testing the null hypothesis Ho: 0=00 the value of 0 in W, is replaced by 60. Under this null hypothesis it follows from Theorem 11.2 that W, is distribution free. The distribution of W under the null hypothesis can be found by enumerating the value of Wn over all possible equally likely permutations of the elements of C and R. For further details see Section 3.1 of Hollander and Wolfe (1999).