Proving $E(X^p)=\int_0^\infty px^{p-1}P(X>x)dx$

Question

Let $X$ be a positive random variable on the $(\Omega,\mathscr{A},P)$. Show that if $X\in L_p$ for $1<p<\infty$. Prove $E(X^p)=\int_{0}^{\infty}px^{p-1}P(X>x)dx$

I have been thinking about this question but it does not come to my mind how should I go from $E(X^p)=\int |X|^p dP$ into a Riemann integral. I understand that $\frac{d(x^p)}{dx}=px^{p-1}$ however I cannot see how the Riemann integral appears. $\int\mathbb{1}_{X>x}dP=P(X>x)$.

Question:

How should I solve this question?

Thanks in advance!

Answer 1

We can write $$\mathbb{E}[X^p] = \int_\Omega X(\omega)^p \mathbb{P}(d \omega)$$ The trick is to then write $x^p = \int_0^x ps^{p-1} ds$ to get that \begin{align} \mathbb{E}[X^p] &= \int_\Omega \int_0^{X(\omega)} ps^{p-1} ds \mathbb{P}(d \omega) \\& = \int_0^\infty \int_\Omega 1_{X(\omega) > s} ps^{p-1} \mathbb{P}(d \omega) ds \\& = \int_0^\infty ps^{p-1} \mathbb{P}(X>s) ds \end{align} where I was able to swap the integrals around using Fubini's theorem since $X$ is non-negative.