Let $X$ be a discrete random variable whose range is $\{0,1,2,3,\cdots,N\}$. I want to prove that $ \sum\limits_{n=0}^{N-1} \sum\limits_{k=n+1}^N P(X=k)= \sum\limits_{k=1}^N \sum\limits_{n=0}^{k-1} P(X=k)$
In lecture we proove this with a picture. One time we summing vertically and the other time horizontally. But my teacher said that you can also prove it in analytically way with characteristic functions, change the the order of the two summations, represent the charactersitic function as function of the other variable..Can anyone help what proof he mean?
As Did commented, just observe that $\mathsf P(X=k)$ is some function of $k$, and invariant wrt $n$.
Let $p(k):=\mathsf P(X=k)$, then::
$$\begin{align}\sum\limits_{n=0}^{N-1}\;\sum\limits_{k=n+1}^N p(k) ~=~& \sum_{n\in\Bbb N\,:\,0\leq n < N}\;\sum_{k\in\Bbb N\,:\,n< k\leq N} p(k)\\[1ex] ~=~& \mathop{\sum\!\!\sum}\limits_{(n,k)\in\Bbb N^2\,:\,0\leq n< k\leq N}p(k) \\[1ex] ~=~& \sum_{k\in\Bbb N\,:\,0<k\leq N}\;\sum_{n\in\Bbb N\,:\,0\leq n<k} p(k) \\[1ex] ~=~& \sum\limits_{k=1}^N\; \sum\limits_{n=0}^{k-1} p(k) & \blacksquare\\[2ex] ~=~&\sum\limits_{k=1}^N\, k\, p(k) \end{align}$$