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Let $X_{i}$ be independent and identically distributed random variables such that $S_n/n\to 0$ almost surely where $S_n=X_1+\dots+X_n$. How to prove that $E|X_1|<\infty$ and therefore $EX_1=0.$

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up vote 9 down vote accepted

For every positive $n$ call $U_n=S_n/n$. There are two simple facts at the basis of the characterization of the behaviour of the sequence $(U_n)$ when the increments $(X_n)$ are i.i.d. but not integrable.

(1) Stochastic fact: if $(X_n)$ is i.i.d. and $X_1$ is not integrable then, for every finite $x$, infinitely many events $A_n=[|X_n|\ge xn]$ are realized, almost surely.

(2) Deterministic fact: consider a deterministic sequence $(x_n)$ and for every positive $n$ define $u_n=(x_1+\cdots+x_n)/n$. If there exists a positive $x$ such that $|x_n|\ge xn$ for infinitely many indices $n$, then the sequence $(u_n)$ is divergent.

Putting facts (1) and (2) together yields that $(U_n)$ is almost surely divergent, in particular the event $[U_n\to0]$ has probability zero.


To prove fact (1), note that $(A_n)$ is an independent sequence of events and that the series $\sum P(A_n)$ is divergent hence Borel-Cantelli lemma yields the result.

To prove fact (2), assume that $(u_n)$ is convergent and that $u_n\to u$. Then, $$ x_n/n=u_n-u_{n-1}(n-1)/n\to u-u=0, $$ hence $|x_n|\le\frac12xn$ for every $n$ large enough, which is a contradiction.


Added later on Thought I might as well recall the proof that, in (1), the series $\sum P(A_n)$ in divergent. The idea is that $P(A_n)=P(|X_1|\ge xn)$ and that $$ \sum_{n\ge0}P(|X_1|\ge xn)=\sum_{n\ge0}(n+1)P(x(n+1)>|X_1|\ge xn)\ge x^{-1}E(|X_1|). $$ And finally, this upper bound loses almost nothing, since one also has $$ \sum_{n\ge1}P(|X_1|\ge xn)=\sum_{n\ge1}nP(x(n+1)>|X_1|\ge xn)\le x^{-1}E(|X_1|). $$

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Suppose for a contradiction that ${\rm E}|X_1| = \infty$. If ${\rm E}(X_1^ + )=\infty$ (respectively, ${\rm E}(X_1^ + ) < \infty$) and ${\rm E}(X_1^ - ) < \infty$ (respectively, ${\rm E}(X_1^ - ) = \infty$), then, a.s., $S_n / n \to \infty$ (respectively, $S_n / n \to -\infty$) as $n \to \infty$. If ${\rm E}(X_1^ + )=\infty$ and ${\rm E}(X_1^ - )=\infty$, then one of the following holds: 1) $S_n / n \to \infty$ a.s.; 2) $S_n / n \to -\infty$ a.s.; 3) $\lim \sup _{n \to \infty } n^{ - 1} S_n = \infty $ and $\lim \inf _{n \to \infty } n^{ - 1} S_n = -\infty $ a.s. Hence, $S_n/n \to 0$ a.s. implies ${\rm E}|X_1| < \infty$.

EDIT: For further details, including a reference, see the first three paragraphs of my answer to this question.

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Obviously, this is not the intended solution (as it relies on a "heavy" theorem, which you are not allowed to use), but it was relevant to post it here. (Note that it gives much more than what you actually asked for.) –  Shai Covo Jul 3 '11 at 12:30

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