A Formula of $E(X^n)$

Question

The problem 1.1 on p.46 of Stochastic Processes, Sheldon M. Ross, The Second Edition said:

If $X$ is nonnegative with distribution $F$, then $$ E[X^n]=\int_0^\infty{nx^{n-1}\bar{F}(x)\mathrm{d}x}. $$

Here $\bar{F}(x) := 1-F(x)$.

Here's my solution:

\begin{equation} \begin{split} E(X^n) &= \int_0^\infty{x^n\mathrm{d}F(x)}\\ &= -\int_0^\infty{x^n\mathrm{d}\bar{F}(x)}\quad(F(x)+\bar{F}(x)=1)\\ &= -\biggl.x^n\bar{F}(x)\biggr|_0^\infty+\int_0^\infty{\bar{F}(x)\mathrm{d}x^n}\\ &= -\biggl.x^n\bar{F}(x)\biggr|_0^\infty+\int_0^\infty{nx^{n-1}\bar{F}(x)\mathrm{d}x}\\. \end{split} \end{equation} I'm almost there, but it suffices, if all above are right, to show $$ \lim_{x\to\infty}{x^n\bar{F}(x)}=0,\tag{1} $$ which I'm afraid is wrong.

Can someone figure out my problem, or just show that (1) is actually right?

Answer 1

Your equation $(1)$ is correct, assuming that $E|X^n|<\infty$. For any $x>0$, if you start with the inequality between random varaibles $$ x^n{\bf 1}(X>x)\le |X|^n,\qquad (x>0) $$ and take expectations, you get $$ x^n\overline{F}(x)\le E|X^n|\qquad (x>0) $$ Furthermore, you have $\lim_{x\to\infty} x^n {\bf 1}(X>x)=0$, almost everywhere in the probability space. Assuming $E|X^n|$ is integrable, the dominated convergence theorem lets you upgrade the almost-everywhere limit to a limit of expectations, so $\lim_{x\to\infty}x^n\overline{F}(x)=0$ as well.