# Show that $\sum_{k=1}^{\infty} \sigma^2_k <\infty$ implies $|\sum_{k=1}^{\infty} X_k|<\infty$ almost surely.

Suppose that $$X_1, X_2, X_3,\ldots$$ are sequence of independent random variables such that $$\mu_k= 0$$ and $$\sigma^2_k =\operatorname{Var}(X_k)< \infty$$ for all $$k$$. Then show that $$\sum_{k=1}^{\infty} \sigma^2_k <\infty$$ implies $$|\sum_{k=1}^{\infty} X_k|<\infty$$ almost surely.

I was wondering how could I use the facts: 1) if a martingale $$M$$ is bounded in $$L^2$$ then $$\lim M_n$$ exists almost surely. 2) Orthogonality of increments of $$M$$ to prove the above statement. I would like to see the solution in explained way.

• If $P(|Z| \ge b) \ge a$ then $Var(Z) \ge a b^2$. Using that $Var(Y_n) < C$ and $Var(Y-Y_n) \to 0$ show that the sequence of random variables $Y_n = \sum_{k=1}^n X_k$ is uniformly bounded almost surely, and that it converges almost surely. Apr 7, 2017 at 0:15
• Then where are the orthogonality applied? Apr 7, 2017 at 11:56
• $Var(Y_n) = \sum_{k=1}^n Var(X_k)$ Apr 7, 2017 at 11:59
• Thank you, Do you have some idea on the following one: math.stackexchange.com/questions/2221753/… Apr 7, 2017 at 12:11
• Did you complete this question ? Write your own answer then. Apr 7, 2017 at 12:12

Let $$M_n:=\sum_{i=1}^nX_i$$. Defining $$\mathcal F_n$$ as the $$\sigma$$-algebra generated by $$X_i$$, $$1\leqslant i\leqslant n$$. Then $$(S_n,\mathcal F_n)$$ is a martingale and using orthogonality of increments, $$\mathbb E\left[M_n^2\right]=\sum_{k=1}^n\mathbb E\left[X_k^2\right]=\sum_{k=1}^n\sigma_k^2$$ hence $$\sup_n\mathbb E\left[M_n^2\right]\leqslant \sum_{k=1}^{+\infty}\sigma_k^2.$$