Expectation of a random variable that takes points in the natural numbers Let $X : \Omega \to \mathbb{N} $ be a r.v. Then $$ \mathbb{E} \{ X \} = \sum_{n=0}^{\infty} P(X > n) $$
Attempt:
$$ \mathbb{E} \{ X \} = \sum_j j P(X = j ) = \sum_j j(1 - P(X > j) - P(X<j ) )$$
But here it gets complicated. Am I on the right track?

\begin{equation}
E(X) = \sum_{j=0}^{\infty}jP(X=j) = \sum_{j=0}^{\infty}\sum_{k=0}^{j-1}P(X=j) = \sum_{k=0}^{\infty}\sum_{j=k+1}^{\infty}P(X=j) = \sum_{k=0}^{\infty}P(X>k)
\end{equation}
The crucial step is to swap the two sums.
