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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.2011-11-16
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    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.2011-11-16
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    Is there some typo in your displayed equation? Shouldn't $\int_0^\infty f(x) dx$ equal $1$ since $f(x)$ is a density?2011-11-16
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    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.2011-11-16
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    Why-why, my bad. You're right, corrected. Anyway, it's a typo, the very essence of question remains.2011-11-16
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    Yes, the question still remains. For some of the details of the use of Fubini's theorem as cardinal has suggested, see [this answer](http://math.stackexchange.com/questions/64186/intuition-behind-using-complementary-cdf-to-compute-expectation-for-nonnegative/64199#64199) by Didier Piau.2011-11-16
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    2 cardinal: extending support to negative numbers is not a problem really, but as for "simple change of variables", unfortunately, I can't see one. Anyway, thank you very much, it seems I've found an appropriate piece in a fundamental tome by Bogachov, so I'll try to understand this one.2011-11-16

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