Is $\lim_{x\to\infty} x\overline{F}(x)=0$?

Question

With partial integration I wanted to prove that for non-negative random variable with CDF F(x) holds $$ \int_0^{\infty}\overline{F}(x)dx=E[X]. $$ Here is $\overline{F}(x)= 1-F(x)$. I got this far $$ \int_0^{\infty}\overline{F}(x)dx=\lim_{x\to{\infty}} x\overline{F}(x)-0+\underbrace{\int_{0}^\infty xf(x)dx}_{E[X]}. $$

But now I don't know hot to calcute the upper limit.

Does anyone have a clue how to prove this?

Have a nice day!

Answer 1

Since $\bar{F}(x) = 1- \mathbb{P}(X \leq x) = \mathbb{P}(X>x)$, we have

$$x \bar{F}(x) = \int_{\{X>x\}} x \, d\mathbb{P} \leq \int_{\{X>x\}} X \, d\mathbb{P}$$

for any $x>0$. If $\mathbb{E}(X)<\infty$, then we can let $x \to \infty$ using the dominated convergence theorem to conclude

$$\lim_{x \to \infty} x \bar{F}(x)=0.$$

If $\mathbb{E}(X)=\infty$ then $x \bar{F}(x)$ does not necessarily converge to $0$ as $x \to \infty$ (consider for example a random variable with Cauchy distribution).

A remark concerning your proof: Mind that you have to assume the existence of a density, i.e. your proof works only for random variables which have a density with respect to Lebesgue measure.