Setup
Given the iid random sample $X_1,...,X_m$ with ranks $Q_1,...,Q_m$ and the iid random sample, $Y_1,...,Y_n$ with ranks $R_1,...,R_n$. Then the sum of ranks is, \begin{align} W = \sum_{i=1}^{n}R_i \end{align} Given the indicator function $\psi$ which is 1 when the parameter is positive, and zero otherwise. The Mann Whitney test is, \begin{align} U = \sum_{i=1}^m\sum_{j=1}^n \psi(Y_j-X_i) \end{align}
Question
Show that $$W = U + \frac{n(n+1)}{2}$$
Solution
Well I recognize that, \begin{align} \sum_{i=1}^n i = \frac{n(n+1)}{2} \end{align} Qualitatively, $U$ is the sum of the number of times that each $Y_i$ is greater than $X_i$.
I tried to make a summation like, \begin{align} W &= \sum_{i=1}^m\sum_{j=1}^n \psi(Y_j-X_i) + \sum_{j=1}^n j\\ &= \frac{1}{m}\sum_{i=1}^m \sum_{j=1}^n\left[\psi(Y_j-X_i) + j \right] \end{align} leading nowhere. What I really want is some sort of summation to rewrite, \begin{align} W = \sum_{i=1}^n R_i = \sum_{i=1}^{n+m} i (\text{if } Z_i \text{ is } X_i) ... \end{align}
Seems like a simple problem, any help is much appreciated.