a problem about boundedness implying integrability

Let $X_k,k\in N$ be independent identically distributed random variables, and suppose that for some $p\in (0;2)$ the sequence $\{\frac{1}{n^{1/p}}\sum_{k\leq n} X_k\}_{n\ge 1}$ is a.s. bounded. Prove that $X_1$ is in $L^p$.

The question cannot be that hard, but I got lost at the moment :(

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1 Answer

Hints: Let $S_n=\sum\limits_{k\leqslant n}X_k$.

(1.) The event $[n^{-1/p}S_n\ \text{bounded}]$ is included in the event $[n^{-1/p}X_n\ \text{bounded}]$.

(2.) For each $x\gt0$, the series $\sum\limits_n\mathrm P(|X_1|\gt xn^{1/p})$ converges if and only if $X_1$ is in $L^p$.

(3.) By Borel-Cantelli, if the series $\sum\limits_n\mathrm P(|X_n|\gt xn^{1/p})$ diverges, then $\limsup\limits_nn^{-1/p}|X_n|\geqslant x$ almost surely.

(4.) Assume that $X_1$ is not in $L^p$, then use the contrapositive of (2.), then (3.), and finally (1.) to conclude that $n^{-1/p}S_n$ is not almost surely bounded.

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+1 for nice argument! For (1), $$\frac{X_n}{n^{1/p}} = \frac{S_n}{n^{1/p}} - \left(1-\frac{1}{n}\right)^{1/p}\frac{S_{n-1}}{(n-1)^{1/p}},$$ which shows that boundedness of $S_n/n^{1/p}$ indeed implies the boundedness of $X_n / n^{1/p}$. – Sangchul Lee Aug 24 '12 at 15:49