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$ X $ is a non-negative continuous random variable with density function $f$ and distribution function $F$.

Use integration by parts to show that

$ \int_0^{\infty} ( 1- F(x)) dx = \int_0^{\infty}xf(x)dx $

I'm quite puzzled on how to even integrate $F(X)$ to get $f(x)$ :S

neverloggedin
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1 Answers1

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Hint: In the usual calculus notation for integration by parts, let $u=1-F(x)$ and $dv=dx$. Then $du=-f(x)\,dx$ and we can take $v=x$.

Added: As pointed out by беркай. the fact that $\lim_{x\to\infty}x(1-F(x))=0$ requires justification. This takes a few lines. For details, please see this part of the Wikipedia article on cumulative distribution functions.

André Nicolas
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  • How does it follow that $x(1-F(x))=0$ as $x \to \infty$? It's not so clear to me. L'Hospital rule may help but it also needs some justifications. What's your argument? – Berkheimer Jul 21 '13 at 23:00