$\lim_{x\rightarrow+\infty} x^{\alpha+\beta}P(|X|>x)=0\Leftrightarrow\lim_{x\rightarrow+\infty}x^{\alpha}E||X|^{\beta}\cdot1_{\{|X|>x\}}|=0$

Question

I have met with a probability problem which I have no idea to deal with. It says:

Let $\alpha>0$, $\beta\geq0$, prove:

$\lim_{x\rightarrow+\infty} x^{\alpha+\beta}\mathbb{P}(|X|>x)=0\Leftrightarrow\lim_{x\rightarrow+\infty}x^{\alpha}\mathbb{E}||X|^{\beta}\cdot1_{\{|X|>x\}}|=0$.

The sufficiency is easy to prove. However, I have no ideas how to prove the necessity. Can you give me some advice? Any hints would be appreciated.

Thanks for your time and patience.

Answer 1

Suppose $\lim_{x\to \infty} x^{\alpha+\beta}P(|X|>x) = 0$. Note that \begin{align} E[|X|^\beta \cdot 1_{|X|>x}] &= P(|X|>x)E[|X|^\beta | |X| > x] \\ &= x^\beta P(|X|>x) E[|X/x|^\beta | |X/x|>1] \\ &\le x^\beta P(|X|>x) E[|X|^\beta], \quad \text{for large $x$}. \end{align} Note that for large $x$, $|X/x| < |X|$, and thus $|X/x|^\beta < |X|^\beta$.

Assuming $E[|X|^\beta] < \infty$, $$ \lim_{x \to \infty} x^\alpha E[|X|^\beta \cdot 1_{|X|>x}] = 0. $$

Answer 2

You can simplify the statement by writing $Y$ for $|X|^{\beta}$ and repalcing $x^{\beta} $ by $t$. The statement then becomes the following: $t^{1+s}P(Y>t) \to 0$ iff $t^{s}EYI_{Y>t} \to 0$ as $t \to \infty$.[Here $s=\frac {\alpha} {\beta}$. I will leave the case $\beta=0$ to you]. As noted by you the íf'part follows easily by Chebychev's Inequality. For the ónly if'part write $t^{s}EYI_{Y>t} \to 0$ as $t^{s}\int_t^{\infty} P(Y>u)du$ which is less than or equal to $t^{s}\int_t^{\infty} \epsilon u^{-1-s}du=\epsilon$ for $t$ sufficiently large.