Connection between mean and CDF

Let's assume $\xi$ is an integrable random variable with density $f(x)$, $\xi > 0$ almost surely. In other words, $\int\limits_0^{\infty}xf(x)dx<\infty$, $f(x)=0$ $\forall x<0$. Let F(x) be a CDF of $\xi$. In case all further integrals exist, we can write: $$\int\limits_{0}^{\infty}(1-F(x))dx=\left.(1-F(x))x\right|_0^{\infty}+\int\limits_0^{\infty}xf(x)dx$$ So if $\lim\limits_{x\rightarrow\infty}(1-F(x))\cdot x=0$, we can safely assume that $\int\limits_{0}^{\infty}(1-F(x))dx=E(\xi)$

The question is: do we need any additional requirements (beside abovementioned integrability and non-negativity) to be sure that $\lim\limits_{x\rightarrow\infty}(1-F(x))\cdot x=0$? If we do, are there some more "natural" requirements than simply stating the value of this limit?

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A natural requirement would be that $\xi$ has a finite mean. –  Sasha Nov 16 '11 at 19:44
Actually, $\int\limits_{0}^{\infty}f(x)dx<\infty$ for nonnegative random variables requires exactly that ("integrability" is quite the same as "having a finite mean"). I just have doubts it is enough. –  Dmitriy Korolevich Nov 16 '11 at 19:50
Is there some typo in your displayed equation? Shouldn't $\int_0^\infty f(x) dx$ equal $1$ since $f(x)$ is a density? –  Dilip Sarwate Nov 16 '11 at 20:16
That $\mathbb E X = \int_0^\infty (1-F(x)) \,\mathrm{d}x$ and more is proven very simply using Fubini's theorem and a simple change of variables. In fact, you can derive a similar expression for $\mathbb E X^n$. Neither of these depend on the existence of a density nor the finiteness of the Lebesgue integral. A similar expression exists for $F$ having support on the whole real line, with only a minor additional constraint that disallows the simultaneous divergence of both the positive and negative parts. –  cardinal Nov 16 '11 at 20:20
Why-why, my bad. You're right, corrected. Anyway, it's a typo, the very essence of question remains. –  Dmitriy Korolevich Nov 16 '11 at 20:20

At the risk of distracting you from the reading of Bogachov's fundamental tome, let me recall why the fact that $x\cdot(1-F(x))\to0$ when $x\to+\infty$ is a simple consequence of the integrability of $X$.

The idea is simple and powerful: first write the quantities you are interested in as integrals of functions (expectations of random variables, in the probabilistic jargon), then invoke Lebesgue's revered name in one way or another.

Here, $x\cdot(1-F(x))=x\cdot\mathrm P(\xi\gt x)=\mathrm E(\xi_x)$ where $\xi_x=x\cdot[\xi\gt x]$. Phase one completed. Now, to phase two: $\xi_x\to0$ pointwise when $x\to+\infty$ hence we only need a tool to exchange a limit and an integral, since this exchange would yield $\lim\limits_{x\to\infty}\mathrm E(\xi_x)=\mathrm E\left(\lim\limits_{x\to\infty}\xi_x\right)=0$.

Well, the most classical is the better: by Lebesgue dominated convergence theorem, our task is over if $\xi_x\leqslant \zeta$ for every $x$, with $\zeta$ integrable. Let me leave you the pleasure to guess such an integrable $\zeta$.

Coming back to the problem which interests you, the relation $\mathrm E(\xi)=\int\limits_0^{+\infty}(1-F(x))\cdot\mathrm dx$ holds without any restriction as soon as $\xi\geqslant0$ almost surely. Simply, if $\xi$ is integrable the RHS is $\mathrm E(\xi)$ and if $\xi$ is not integrable the RHS is infinite. Let us use our favorite tool once again, that is, let us express every probability involved as the expectation of a random variable. Taking stock of our previous computations, we see that $1-F(x)=\mathrm E(\eta_x)$ with $\eta_x=[\xi\gt x]$ hence $$\int\limits_0^{+\infty}(1-F(x))\cdot\mathrm dx=\int\limits_0^{+\infty}\mathrm E(\eta_x)\cdot\mathrm dx=\mathrm E(\eta),\qquad \eta=\int\limits_0^{+\infty}\eta_x\cdot\mathrm dx,$$ where the last equality between integrals is a consequence of Fubini's theorem (always true for nonnegative functions/random variables) for the product measure of the Lebesgue measure and $\mathrm P$ on the measurable space $(0,+\infty)\times\Omega$. The proof is over since $$\eta=\int\limits_0^{+\infty}[\xi\gt x]\cdot\mathrm dx=\int\limits_0^{\xi}\mathrm dx=\xi.$$

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My dear, too much programming, too little math REALLY makes stupid. Thank you, Master, you've returned me a little bit of Zen :) –  Dmitriy Korolevich Nov 16 '11 at 21:30
The ten thousand questions are one question. If you cut through the one question, then the ten thousand questions disappear... :-) –  Did Nov 16 '11 at 21:50