Let $X_n$ defined on a probability space for each $n\geq1$
The question is that $(X_n)_{n=1}^{\infty}$ is iid and $$
P\left(\limsup _{n \rightarrow \infty}\left|X_n\right| / n^{1 / p} \leq \varepsilon\right)>0
$$for some $p>0$ and some constant $\varepsilon>0$. Prove that $E|X_1|^p<\infty$.
My thinking is that we suppose $E|X_1|^p=\infty$, then we have $E|X_1|^p\leq 1+\sum_{n=1}^{\infty}P(|X_1|^p\geq n\varepsilon^p) = 1 + \sum_{n=1}^{\infty}P(\dfrac{|X_1|}{n^{1/p}}\geq \varepsilon) = \infty$
By iid and second Borel-Cantelli, and let $A_n = \{\omega:\dfrac{|X_1|}{n^{1/p}}\geq \varepsilon\}$, we have
$P(\limsup_nA_n) = 1$, by which we have $P(\liminf_n(A_n)^C)=0$
From this question Is it correct to say that ($\color{red}{(} \limsup |W_k|/k\color{red}{)} \le 1) \supseteq \limsup \color{red}{(}|W_k|/k \le 1\color{red}{)}$?
We have $\liminf_n(A_n)^C=\liminf_n\{\omega:\dfrac{|X_1|}{n^{1/p}}< \varepsilon\}\subseteq \{\omega:\limsup _{n \rightarrow \infty}\dfrac{|X_1|}{n^{1 / p}} < \varepsilon\}=:B_n$
But $P(B_n)\geq0$ tells us nothing, I guess there is some way to prove $P(B_n)\leq0$ and then we have $P(B_n)=0$, which yields a contradiction.
I don't know my thinking is correct, can anyone can help me to solve this?
Best Answer
$P\left(\limsup _{n \rightarrow \infty}\left|X_n\right| / n^{1 / p} \leq \varepsilon\right)>0$ implies that $P\left(\limsup _{n \rightarrow \infty}\left|X_n\right| / n^{1 / p} \leq \varepsilon\right)=1$ by Kolomogorov's 0-1 law (since the event on the left is measurable w.r.t the tail $\sigma-$ field). Hence, $P\left(\limsup _{n \rightarrow \infty}\left|X_n\right| / n^{1 / p} > \varepsilon\right)=0$ and $P\left(\limsup _{n \rightarrow \infty}(\left|X_n\right| / n^{1 / p} \geq \varepsilon\right))=0$. So Borel-Cantelli Lemma shows that $\sum_n P(|X_1| / n^{1 / p} > \varepsilon')<\infty$ if $0<\epsilon'<\epsilon$. In other words, $\sum_n P(|X_1|^{p}\varepsilon'^{-p} >n)<\infty$. This implies (by a standard result ) that $E(|X_1|^{p}\varepsilon'^{-p})<\infty$.