Why does $\mathbb{E}[X] = \sum^\infty_{r=0}\mathbb{P}(X > r)$?

Question

Why does $\mathbb{E}[X] = \sum^\infty_{r=0}\mathbb{P}(X > r)$?

I understand if it was $\mathbb{E}[X] = \sum^\infty_{r=0}r\cdot\mathbb{P}(X = r)$, but for this I don't understand why.

Answer 1

Because, $$P\{X>r\}=\sum_{k=r+1}^\infty P\{X=k\}.$$ under the assumption that $P\{X\in \mathbb N\}=1$.

Added

Therefore, $$\sum_{r=0}^\infty P\{X>r\}=\sum_{r=0}^\infty \sum_{k=r+1}^\infty p\{X=k\}=\sum_{k=1}^\infty \sum_{r=0}^{k-1}p\{X=k\}=\sum_{k=1}kp\{X=k\}=\mathbb E[X].$$

Answer 2

Let me answer the question explicitly:

This only holds if $P(X \in \mathbb{N})=1$, a fact we use on the 3rd equality:

$\sum^{\infty}_{n=0}P(X > n)=\sum^{\infty}_{n=0}E[1_{X>n}]=E[\sum^{\infty}_{n=0}1_{X>n}]= E[\sum^{X-1}_{n=0}1_{X>n}]=E[\sum^{X-1}_{n=0}1]=E[X]$

The second equation is due to Fubini, as $\sum$ is an integral with respect to the counting measure.