Showing $\mathbb{E}(X^p)=p\int_{0}^{\infty} (1-F_X(u))u^{p-1} \ du$

Question

Let $X$ be a positive continuous random variable with density function $f_X$ and cumulative distribution function $F_X$. By interchanging the double integral, show that for $p\in\mathbb{N}^+$ $$\mathbb{E}(X^p)=\int_{0}^{\infty} x^pf_X(x) \ dx=p\int_{0}^{\infty} (1-F_X(u))u^{p-1} \ du$$ where $x^p=\int_{0}^{x} pu^{p-1} \ du$.

My solution:

\begin{align} \mathbb{E}(X^p)&=\int_{0}^{\infty} x^pf_X(x) \ dx \\ &=\int_{0}^{\infty} \left(\int_{0}^{x} pu^{p-1} \ du\right)f_X(x) \ dx \\ &=\int_{0}^{\infty} f_X(x) \ dx \left(\int_{0}^{x} pu^{p-1}\ du\right) \ \\ &=\int_{0}^{\infty} pu^{p-1}\ du \left(\int_{u}^{\infty} f_X(x) \ dx\right) \ \ \ \ \ \text{(changing from u-simple to x-simple)} \ \\ &=p\int_{0}^{\infty} \left(1-F_X(u)\right)u^{p-1}\ du \end{align}

I am not so confident that the logic in my answer is correct, especially when I switch the order of integration. Is my working from the third to the forth line correct?

Your argument is correct. In fact there is a similar formula for any random variable, with or without density. — Kavi Rama Murthy, Jul 12 '18 at 05:31
Yes, the derivation is correct, In general $$\mathbb{E}(g(X))=\int_{-\infty}^0 g'(u)F_X(u)du+\int_0^{\infty}g'(u)(1-F_X(u))du$$ The derivation for $-\infty$ to $0$ is similar to what you have done with a small change in limits giving the different result. — Abhishek Bakshi, Jul 12 '18 at 06:30
@KaviRamaMurthy Any example where i can use this result ? I want to use it . — Daman, Dec 15 '18 at 05:14
Just adding linking this to the node of this network of related posts. See the meta post and the measure-theoretic treatment. — Lee David Chung Lin, Jan 31 '20 at 11:54

Showing $\mathbb{E}(X^p)=p\int_{0}^{\infty} (1-F_X(u))u^{p-1} \ du$

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