How to show $E[X^k]=\sum_{i=0}^{\infty}((i+1)^k-i^k)P(X>i)$

Question

where $X$ is a nonnegative integer valued random variable

I can show the case when k=1, but I am completely stuck on the general case. Does anyone have any suggestions?

Answer 1

$\Bbb P(X=i+1)=\Bbb P(X>i)−\Bbb P(X>i+1)$, and $0^k\Bbb P(X>0)=0$.

Therefore:- $$\begin{align}\Bbb E(X^k)&=\sum_{j=1}^\infty j^k\Bbb P(X=j)\\&=\sum_{i=0}^\infty (i+1)^k\Bbb P(X=i+1)\\&=\sum_{i=0}^\infty (i+1)^k(\Bbb P(X> i)-\Bbb P(X> i+1))\\&=\sum_{i=0}^\infty (i+1)^k\Bbb P(X> i)-\sum_{i=0}^\infty (i+1)^k\Bbb P(X> i+1)\end{align}$$ However:- $$\begin{align}\sum_{i=0}^\infty (i+1)^k\Bbb P(X> i+1)&=\sum_{i=1}^\infty i^k\Bbb P(X> i)\\&=\sum_{i=0}^\infty i^k\Bbb P(X>i)\end{align}$$ ... so:- $$\Bbb E(X^k)=\sum_{i=0}^\infty ((i+1)^k-i^k)\,\Bbb P(X> i)$$

Answer 2

Hint: Use that $P(X > i) = \sum_{j > i}P(X = j)$ and then change the order of summation.

Another way: Summation by parts.