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?
Best Answer
you could also proof this using telescoping series: $\begin{align*} \sum_{x=0}^\infty xP(X=x)&=\sum_{x=0}^\infty x(P(X>x-1)-P(X>x))\\ &=\sum_{x=0}^\infty xP(X>x-1)-\sum_{x=0}^\infty xP(X>x)\\ &= \sum_{x=1}^\infty xP(X>x-1)-\sum_{x=1}^\infty (x-1)P(X>x-1)\\ &=\sum_{x=1}^\infty (x-(x-1))P(X>x-1)\\ &=\sum_{x=1}^\infty P(X>x-1)\\ &=\sum_{x=0}^\infty P(X>x) \end{align*}$