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Suppose $\{X_n\}$ are arbitrary random variables such that $\sum \pm X_n$ converges almost surely for all choices $\pm1$. Show that $\sum X_n^2$ converges almost surely.

Denote $\{B_n\}$ Bernoulli random variables, the statement above is to say $\sum B_n X_n$ converges a.s. implies $\sum X_n^2$ converges a.s.

Can anybody give a solution to this problem?

This problem comes from Kai Lai Chung's A Course in Probability Theory, Third Edition, pp.129

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    $\begingroup$ FYI, "the statement above" is not saying that $\sum B_n X_n$ converges a.s. implies $\sum X_n^2$ converges a.s. $\endgroup$
    – Did
    Commented Dec 8, 2016 at 9:34

1 Answer 1

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Since $\sum\limits_nB_nX_n<\infty$ $a.s.$, Thus $\exists K$ $s.t.$ $|B_nX_n|\le K$ $a.s.$

Notice that $E(B_nX_n)=EB_n\cdot EX_n=0$.

By Theorem 12.2 in Probability with Martingales (Williams)(Refer to The convergence of $\sum \pm a_n$ with random signs) we obtain $\sum\limits_nVar(B_nX_n)<\infty$. (This may have some problems without the independence of $\{X_n\}$ as mentioned by the comments)

Thus $\sum\limits_nEX_n^2=\sum\limits_nE(B_nX_n)^2<\infty$.

Using Fubini's Theorem we have $E(\sum\limits_nX_n^2)<\infty$, therefore $\sum\limits_nX_n^2<\infty$ $a.s.$

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  • $\begingroup$ The first line is $\sum a_n < \infty \to \lim a_n = 0 \to a_n \ \text{is convergent} \to a_n \ \text{is bounded}$ ? $\endgroup$
    – BCLC
    Commented Dec 2, 2016 at 5:21
  • $\begingroup$ How do you know 1. $\sum Var(B_n X_n) < \infty$? Do we have $Var(B_n X_n) < \infty$? 2. $E[B_n X_n] = E[B_n]E[X_n]$ ? $\endgroup$
    – BCLC
    Commented Dec 2, 2016 at 5:27
  • $\begingroup$ Dumb question: $\sum (a_n - b_n) < \infty \to \sum a_n < \infty$ and $ \sum b_n < \infty $? $\endgroup$
    – BCLC
    Commented Dec 2, 2016 at 5:29
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    $\begingroup$ No to the last, try for example $a_n = 1$ and $b_n = 1-1/2^n$ $\endgroup$
    – cats
    Commented Dec 2, 2016 at 5:49
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    $\begingroup$ There are some problems with this answer. The first sentence is completely wrong. Plus, you can only use Theorem 12.2 if the random variables $(X_n)$ are independent, but is not assumed in the original post. $\endgroup$
    – user940
    Commented Dec 2, 2016 at 17:05

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