# Prove $\lim_{x\to\infty} x . [1 - F(x)] = 0$ [duplicate]

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I have no idea how to proceed with proving this. If $$X$$ is a continuous random variable, $$P(X > 0) = 1$$, $$E(X)$$ is defined and $$F(x)$$ is the CDF, then prove $$\lim_{x\to\infty} x . [1 - F(x)] = 0$$

## marked as duplicate by Mike Earnest, StubbornAtom, Community♦Apr 19 at 7:12

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• Assuming you don't have tools of dominated convergence theorems from the Foobaz answer, you can write, for all real numbers $x>0$: $$\infty>E[X] = \int_0^{\infty} t f_X(t)dt = \int_0^x tf_X(t)dt+\underbrace{\int_x^{\infty} tf_X(t)dt}_{\mbox{operate on this}}$$ Can you operate to produce $x[1-F(x)]$ ? – Michael Apr 17 at 22:41
• Thanks for the comment, @Michael. I'm unable to produce $x[1-F(x)]$ directly from the integral, but with help, I could see that the integral is always greater than or equal to $x[1-F(x)]$, and is 0. So $x[1-F(x)]$ must be 0 as $x \to \infty$. Right? – Neo M Hacker Apr 18 at 0:12
• @NeoMHacker : Yes, I intended you to notice that the integral identified by the underbrace is $\geq x[1-F(x)]$. I don't know what you mean by "and is 0." That integral is not necessarily zero, but you need to show the limit is zero. This is indeed the same technique that Mike Earnest points out in his link. – Michael Apr 18 at 3:22
• @Michael: thanks, I had typed “and is 0” by mistake. – Neo M Hacker Apr 19 at 7:11

## 1 Answer

Note that for $$x>0$$ $$xI(X> x)\leq XI(X>x)$$ where $$I$$ is the indicator function. By taking expectations we find that $$0\leq xP(X>x)\leq EXI(X>x)\to 0$$ as $$x\to \infty$$ by the dominated convergence theorem.