Let $p>0$ be a constant. Prove that if $E{|X|}^p<\infty$, then $x^pP(|X| > x)\to 0(x\to \infty)$.

Question

My idea is $$x^pP(|X|>x)=x^pEI(|X|>x)\le x^p E\frac{|X|^p}{x^p}I(|X|>x)\le E|X|^p<\infty$$.Then I have no idea. Is there anyone can help me?

Answer 1

$$x^p P(|X|>x) = x^p \int_\Omega 1(|X| >x) dP \leq \int_\Omega |X|^p 1(|X|>x) dP $$

If anything in the above inequality does not make sense, feel free to ask.

Anyway, since $E|X|^p < \infty$ and $E|X|^p = \int_\Omega |X|^p dP$, it must be that

$$\int_\Omega |X|^p1(|X|>x) dP \to 0$$

as $x \to \infty$, since if it did not, then $E|X|^p$ could not be finite. But we have shown

$x^p P(|X|>x) \leq \int_\Omega |X|^p 1(|X|>x) dP $.

Hence $x^p P(|X|>p) \to 0$ as $x \to \infty$ also.

Answer 2

Your estimate of $x^pP(|X|>x)$ by $E\left\lvert X\right\rvert^p$ is a little bit too crude. As zhoraster suggests, integrating the pointwise inequality $$ x^p\mathbf 1\{\left\lvert X\right\rvert \gt x\}\leqslant \left\lvert X\right\rvert^p\mathbf 1\{\left\lvert X\right\rvert \gt x\} $$ and then using the monotone convergence theorem gives the result.