$\mathbb{E}(|X|^p)=\int_0^{\infty}pt^{p-1}\mathbb{P}(|X|\ge t)dt$

Question

I found the following formula in a proof of Kinchin's inequality :
$\mathbb{E}(|X|^p)=\int_0^{\infty}pt^{p-1}\mathbb{P}(|X|\ge t)dt\ \ p\ge 1$
where X is a finite sum of random independant variables $X_i$ satisfying $\mathbb{P}(X_i=1)=\mathbb{P}(X_i=-1)=\frac{1}{2}$. I couldn't however prove this formula.
I thought about using $\mathbb{E}(|X|^p)=\int_{\mathbb{R}}|t|^pf(t)dt\ (1)$ where $f$ is the derivative of $t\mapsto \mathbb{P}(|X|\le t)$ and then integrating by parts but I'm not even sure $(1)$ is true (I know it is true if $X$ is a continuous random variable with density $f$ but is it true for $X$ discrete ?)
Thank you in advance.

Answer 1

Note that $$ \begin{align} \int_0^\infty pt^{p-1}P(|X|\ge t) \,dt &=\int_0^\infty pt^{p-1}\left(\int I(t\leq |X|)\,dP\right)\,dt\\ &=\int\int_0^\infty pt^{p-1}I(t\leq |X|)\, dt\, dP\\ &=\int\int_0^{|X|}pt^{p-1}dt\, dP\\ &=\int |X|^p dP\\ &=E|X|^p \end{align} $$ where the interchanging of order of integration is justified by Tonelli's theorem.

Answer 2

Hint

Let $f_X$ then density function of $|X|$. Then, $$\mathbb P\{|X|>t\}=\int_t^\infty f_X(x)\,\mathrm d x,$$

and use Fubini.