Explain why $E(|X|^p)=p\int_{0}^{\infty}y^{p-1}P(|X|>y) dy$

Question

I'm trying to understand a passage from the book: A Basic Course in Probability Theory, Rabi Bhattacharya Edward C. Waymire, in the page 21. The calculation is the following:

If $X$ is a random variable on $(\Omega,\mathcal{ F}, \mathbb{P} )$, then for any $p > 0, x ≥ 0$, writing $x^p=p\int_0^x y^{p−1} dy$ in the formula $\mathbb{E}|X|^p = \int_Ω |X(\omega)|^p \mathbb{P} (d\omega)$ we obtains:

$$ \mathbb{E}|X|^p=\int_{\Omega}\left(p\int_0^{|X(\omega)|}y^{p-1}dy \right)d \mathbb{P} ~~ {\color{red}=^{??}}~ p\int_{0}^{\infty}y^{p-1}\mathbb{P}(|X|>y) dy $$ I do not understand how the left side of the above equation became the right hand side. I know that the Fubini's theorem should be used but do not know the complete details.

Answer 1

It's Fubini's theorem . The domain of the double integral is: $(\omega,y)\in\Omega\times [0, |X(\omega)|]$, however the outer integral is with respect to the measure of the outcomes. $\mathrm d\mathbb P\equiv \mathbb P(\mathrm d \omega)$. So when we apply Fubini's theorem the new inner integral is simply the cumulative measure.

Via:

$$\begin{align}\mathbb{E}|X|^p & =\int_{\Omega}p\left(\int_0^{|X(\omega)|}y^{p-1}\,\mathrm d y \right)\,\mathbb{P}(\mathrm d\omega) \\ & = p\int_0^\infty y^{p-1}\left(\int_{\Omega:y<|X(\omega)|} \mathbb{P}(\mathrm d\omega)\right)\,\mathrm d y \\ & = p\int_0^\infty y^{p-1}\mathbb P(y<|X|)\,\mathrm d y \end{align}$$

Answer 2

Because $|X|^p=\displaystyle p\int_{0}^{\infty}y^{p-1}\mathbf 1_{|X|>y}\,\mathrm dy$.