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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!

Kyoto
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  • Remember that $\lim_{x \to \infty} F(x) = 1$ for any CDF $F(X)$, so $\lim_{x \to \infty} \overline{F}(x) = 0$. – Zain Patel Jul 21 '16 at 12:23
  • @ZainPatel I know that. What I don't know if $\overline{F}(x)$ is converging faster than $x$ to 0 as $x\to\infty$. – Kyoto Jul 21 '16 at 12:25
  • perhaps L'Hopital might be useful here? – Zain Patel Jul 21 '16 at 12:27
  • is this useful: http://math.stackexchange.com/questions/1201577/show-mathbbex-int-0-infty-1-f-xx-dx-for-a-continuous-random?rq=1 – Zain Patel Jul 21 '16 at 12:39
  • @ZainPatel I saw that answer. But I was wondering if the proof can be done as written above. – Kyoto Jul 21 '16 at 12:43
  • Kyoto, what about this? http://math.stackexchange.com/questions/354942/proof-that-ex-infty-entails-lim-n-to-inftyn-prx-ge-n-0?rq=1 (for the discrete version, perhaps you can adapt this to the continuous case) – Zain Patel Jul 21 '16 at 14:01

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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.

saz
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