Show that $\sum \pm X_n<\infty$ a.s. implies $\sum X_n^2<\infty$ a.s.

Question

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?

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

Answer 1

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 $\sum\limits_nX_n^2<\infty$ $a.s.$