Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 :(

share|cite|improve this question
up vote 2 down vote accepted

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.

share|cite|improve this answer
+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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.