$\sum \xi_n<\infty \;\text{a.e.}\; \rightarrow \sum \sigma_n^2 < \infty$ if $E(\xi_n)=0$ and $V(\xi_n)=\sigma_n^2<\infty$.

Question

Let $\xi_n$ be a random variable with zero mean and finite variances $\sigma_n^2$. If $|\xi_n|\le C$ for all $n\in \mathbb{N}$, show that we have
$$\sum \xi_n<\infty \;\text{a.e.}\; \rightarrow \sum \sigma_n^2 < \infty.$$

I'm looking at the following solution where we define $M_n:= (\xi_1+\cdots +\xi_n)^2 - (\sigma_1^2 + \cdots + \sigma_n^2) =:S_n^2 - A_n$.

However, at the bottom of the proof, I can't see why $P(\tau=\infty)=1$ for sufficiently large $\kappa$ since $\sum_j \xi_j$ converges almost surely. I would greatly appreciate any explanation on this.

Answer 1

I assume the random variables are independent (how else is Mn a Martingale right?)

Here's the proof in Williams' Probability with Martingales:

Convergence of the sum means that the partial sums are bounded. Here, I leave you to deduce a bound for |Mn|.

For a K not less than this bound, |Mn| will never be able to exceed K. Never being able to exceed is the same thing as it will take forever to exceed because $\inf(\emptyset)=\infty$.