Find the Mean for Non-Negative Integer-Valued Random Variable Let $X$ be a non-negative integer-valued random variable with finite mean.
Show that
$$E(X)=\sum^\infty_{n=0}P(X>n)$$
This is the hint from my lecturer.
"Start with the definition $E(X)=\sum^\infty_{x=1}xP(X=x)$. Rewrite the series as double sum."
For my opinion. I think the double sum have the form of $\sum\sum f(x)$, but how to get this form? And how to continue?
 A: $\sum_{n=0}^{\infty}P(X>n)=\sum_{n=1}^{\infty}\sum_{x=n+1}^{\infty}P(X=x)=\sum_{x=0}^{\infty}\sum_{n=0}^{x-1}P(X=x)=\sum_{x=0}^{\infty}xP(X=x)=EX$
The interchange of the infinite sums is justified since $X$ has finite mean.
A: you could also proof this using telescoping series:
$\begin{align*}
\sum_{x=0}^\infty xP(X=x)&=\sum_{x=0}^\infty x(P(X>x-1)-P(X>x))\\
&=\sum_{x=0}^\infty xP(X>x-1)-\sum_{x=0}^\infty xP(X>x)\\
&= \sum_{x=1}^\infty  xP(X>x-1)-\sum_{x=1}^\infty  (x-1)P(X>x-1)\\
&=\sum_{x=1}^\infty  (x-(x-1))P(X>x-1)\\
&=\sum_{x=1}^\infty  P(X>x-1)\\
&=\sum_{x=0}^\infty  P(X>x)
\end{align*}$
A: \begin{array}
& & 0P(X=0) & + & 1P(X=1) & + & 2 P(X=2) & + & 3P(X=3) & + & \cdots \\[18pt]
= &  &  & P(X=1) & + & P(X=2)  & + & P(X=3) & + & \cdots \\
&  &   &        &  + & P(X=2) & + & P(X=3) & + & \cdots \\
&  &   &        &    &        & + & P(X=3) & + & \cdots\\
& & & & & & &  & + & \cdots
\end{array}
The sum in the first row is $P(X>0)$; that in the second row is $P(X>1)$; that in the third row is $P(X>2)$, and so on.
