I am trying to understand a particular proof of the relation between Walsh averages and the Wilcoxon signed rank test statistic, as given in the book Statistical Inference Based on Ranks by T.P. Hettmansperger.
Suppose $X_1,X_2,\ldots,X_n$ are i.i.d with cdf $F(x-\theta)$ where $F$ is continuous and symmetric about $0$. Then the signed rank test statistic for testing $H_0:\theta=0$ is $$T=\sum\limits_{i=1}^n I(X_i>0)R_i^+\,,$$ where $R_i^{+}$ is the rank of $|X_i|$ among $\{|X_1|,|X_2|,\ldots,|X_n|\}$.
The problem is to show that $T$ can be expressed as the sum of positive Walsh averages:
$$T=\sum_{1\le i\le j\le n}I\left(\frac{X_i+X_j}{2}>0\right)$$
The following extract from the book seems to give a neat proof:
But I am having difficulty following this argument. I know that rank of $X_j$ can be written as $\sum\limits_{i=1}^n I(X_i\le X_j)$. And based on the actual result, I should have something like $$I(X_j>0)R_j^{+}=\sum\limits_{i=1}^j I(X_i+X_j>0)$$
I can understand an algebraic proof but I am interested to know how they justify this in words.
Previously discussed at Prove the relationship between Walsh averages and Wilcoxon signed rank test.
