Show that $\sum \pm X_n<\infty$ a.s. implies $\sum X_n^2<\infty$ a.s. 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
Since $\sum\limits_nB_nX_n&lt;\infty$ $a.s.$, 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 $\sum\limits_nVar(B_nX_n)&lt;\infty$. (This may have some problems without the independence of $\{X_n\}$ as mentioned by the comments)

Thus $\sum\limits_nEX_n^2=\sum\limits_nE(B_nX_n)^2&lt;\infty$.

Using Fubini's Theorem we have $E(\sum\limits_nX_n^2)&lt;\infty$, therefore $\sum\limits_nX_n^2&lt;\infty$ $a.s.$
Since $\sum\limits_nB_nX_n&lt;\infty$ $a.s.$, 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 $\sum\limits_nVar(B_nX_n)&lt;\infty$. (This may have some problems without the independence of $\{X_n\}$ as mentioned by the comments)

Thus $\sum\limits_nEX_n^2=\sum\limits_nE(B_nX_n)^2&lt;\infty$.

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