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I'm working on a following problem:

Let $X$ be a positive random variable with CDF $F(x)$ and $\textbf{E}X^{\alpha} < \infty$ for some $\alpha > 0$. Prove that $$\textbf{E}X^{\alpha}=\alpha \int_{0}^{\infty}x^{\alpha-1}(1-F(x))dx$$

So far I got to the point where I want to use integration by parts formula: $$(\phi(x)F(x))'=\phi'(x)F(x)+\phi(x)F'(x)=\phi'(x)F(x)+\phi(x)f(x)$$ where $\phi(x)=x^{\alpha}$ and $f$ is a PDF of $X$. Now by rearranging and integrating I get: $$\phi(x)f(x)=(\phi(x)F(x))'-\phi'(x)F(x)$$ and $$\int_{0}^{\infty}\phi(x)f(x)dx=\int_{0}^{\infty}(\phi(x)F(x))'dx-\int_{0}^{\infty}\phi'(x)F(x)dx$$ so $$\textbf{E}X^{\alpha}=\int_{0}^{\infty}(\phi(x)F(x))'dx-\int_{0}^{\infty}\phi'(x)F(x)dx=\int_{0}^{\infty}(\phi(x)F(x))'dx-\alpha\int_{0}^{\infty} x^{\alpha-1}F(x)dx$$ and finally: $$\textbf{E}X^{\alpha}=[\ \phi(x)F(x) ]\_0^\infty -\alpha\int_{0}^{\infty} x^{\alpha-1}F(x)dx$$ Using integration by parts is even hinted at in the solutions part but without actual derivation and apparently it would imply that $[\ \phi(x)F(x) ]\_0^\infty = \int_{0}^{\infty}\phi'(x)dx$ which I struggle to see.

I'm looking for an extra hint to either see this equality or how this problem can be approached in another way. Thanks!

borg
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  • I would suggest you to use the same line of arguments but by starting with the fact that, when the integral is well defined for a function $\Psi$, we have that $$ \int_{0}^\infty \Psi(x) dx = \lim_{M \to \infty} \int_{0}^M \Psi(x) dx.$$ Try to do the computations with a finite $M$ and then take the limit (which must exist by your assumptions). – Zoms96 Aug 21 '23 at 18:05
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    This has been answered before, by the way. I recommend checking this answer: https://math.stackexchange.com/a/172857/291100 – Nap D. Lover Aug 21 '23 at 20:55

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Instead of using $F$ in the integration by parts, use $1-F$. Then, $$ (\phi(x)(1-F(x)))'=\phi'(x)(1-F(x))-\phi(x)f(x). $$ Therefore, $$ \textbf{E}X^{\alpha}=[ -x^\alpha(1-F(x)) ]_0^\infty + \alpha\int_{0}^{\infty} x^{\alpha-1}(1-F(x))dx. $$ The term $x^\alpha(1-F(x))$ vanishes for $x=0$ and also vanishes for $x\to\infty$ because $\textbf{E}X^{\alpha} < \infty$.

P.S. Dester
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