I ask this because I'm trying to understand a proof of the expected value of a non negative random variable is equal to $\int_{0}^{\infty}(1-F(x))dx$ ($F$ is the distribution function of $X$) when $E[X]$ exists, and in some step I have $\lim\limits_{x\to\infty}({x·P[X>x])}=0$ and I don't know why it verifies. The problem I see is the indetermination in the limit $\infty·0$.
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This is a consequence of Chebyshev's inequality. – Aug 20 '18 at 16:54
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1@YvesDaoust Chebychev's only gives you $xP(X>x)\le E[X;X>x]$. You need something like dominated convergence to conclude that $\lim_{x\to\infty} E[X;X>x]=0$. – Mike Earnest Aug 20 '18 at 16:58
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A similar question in Cross Validated limit of $x \left[1-F(x) \right]$ as $x \to \infty$. – Amir May 10 '24 at 22:56
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Here is a more general question, where $x$ is replaced with any function $g(x)$: https://math.stackexchange.com/questions/4419069/the-conditions-for-lim-x-to-inftygx-barfx-0 – Mike Earnest Nov 04 '24 at 23:11
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If $EX<\infty$, then you can conclude $\lim_{x\to\infty}xP(X>x)=0$. In order for this proof to work in full generality, you need to be comfortable with the Lebesgue-Stieltjes integral. This simultaneously generalizes the formulas $EX=\sum_{x\in \mathcal X}xp(x)$ and $EX=\int_{0}^\infty xp(x)\,dx$ for the expectation of discrete and continuous random variables.
Assume$^1$ that $X\ge 0$ always. Then $$ EX=\int_0^\infty t\,dF(t) = \int_0^x t\,dF(t)+\int_x^\infty t\,dF(t)\tag{see 2} $$ Also, you have that $$ EX=\lim_{x\to\infty} \int_0^x t\,dF(t) $$ Therefore, it follows that $$ \lim_{x\to\infty} \int_x^\infty t\,dF(t)=0 $$ You can then conclude by noting $$ \int_x^\infty t\,dF(t)\ge \int_x^\infty x\,dF(t)=x\int_x^\infty dF(t)=x[(\lim_{N\to\infty}F(N))-F(x)]=xP(X>x)\ge 0 $$ and using the squeeze theorem.
If you want to generalize to allow $X\le 0$, apply the result to $Y=\max(X,0)$.
If you are only interested in the case when $X$ has a continuous distribution, then you can replace $\int t\,dF(t)$ with $\int t f(t)\,dt$ throughout, where $f(t)$ is the pdf.
It is also possible to give a proof which works when $X$ is discrete, by replacing the integrals in question with sums.
Mike Earnest
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