$E[X^2] = \sum_{x=0}^{\infty}(2x+1)P[X>x]$

Question

How to prove $E[X^2] = \Sigma_{x=0}^{\infty}(2x+1)P[X>x]$? It is easy to use generating function to prove: $E[X]=\sum_{x=0}^{\infty}P[X>x]$. Given that $X$ is discrete random variable and with countable elements.

Answer 1

Also, $X$ need to be positive.

Hint

Briefly, \begin{align*} \mathbb E[X^2]&=\int_\Omega X^2\,\mathrm d \mathbb P\\ &=\int_\Omega \int_0^X2x\,\mathrm d x\,\mathrm d \mathbb P\\&=\int_\Omega \int_0^\infty 2x\boldsymbol 1_{\{X>x\}}\,\mathrm d x\,\mathrm d \mathbb P\\ &\underset{\text{Fubini}}{=}\int_0^\infty 2x\mathbb P\{X>x\}\,\mathrm d x. \end{align*} Using this previous formula yields : $$\mathbb E [X^2]=\sum_{k=0}^\infty \mathbb P\{X>k\}\int_k^{k+1}2x\,\mathrm d x=...$$