If for any $\alpha > 1$, $\frac{P(|X| > \alpha n)}{P(|X| > n)}\to 0$, then any moment of $X$ exists.

Question

For any $\alpha > 1$, $\frac{P(|X| > \alpha n)}{P(|X| > n)}\to 0$. Prove that any moment of $X$ exists.

I know $E|X|^r=r\int_{0}^{\infty}x^{r-1}P(|X|>x)dx$. How can I use it to prove above theorm?

Answer 1

A rough argument is that $P(|X|>a^k) < \varepsilon^k P(|X|>1)$, for some $\varepsilon\in[0,1)$ (prove by induction), so the tail probability decays exponentially and all moments exist.

Answer 2

In view of the mentioned formula for the $r$-th moment of $\left\lvert X\right\rvert$, it suffices to prove that for each positive $r$, the series $$\tag{*} \sum_{n=1}^{+\infty}2^{nr}\Pr\left\{\left\lvert X\right\rvert>2^n\right\} $$ converges. To this aim, use the assumption with $\alpha=2$. Letting $a_n:=\Pr\left\{\left\lvert X\right\rvert>2^n\right\}$, we can write $a_{n+1}\leqslant a_n\varepsilon_n$, where $\varepsilon_n\to 0$ as $n$ goes to infinity hence the convergence of (*) follows from the ratio test.