# probability with martingales 12.2 sum of zero-mean independent variables in L^2

I am struggling with the following theorem from David Williams, Probability with Martingales:

THEOREM

Suppose that $$(X_{k}:k\in\mathbb{N})$$ is a sequence of independent random variables such that, for every $$k$$, $$E(X_{k})=0, \sigma_{k}^2:=Var(X_{k})<\infty$$.

(a) Then $$\sum\sigma_{k}^2<\infty\Rightarrow\sum X_{k}\text{ converges a.s. .}$$

(b) If the variables $$(X_{k})$$ satisfies $$\exists K \in [0,\infty),\forall k, \omega,\\ |X_{k}(\omega)|\leq K,$$ then $$\sum X_{k}\text{ converges a.s.}\Rightarrow\sum\sigma_{k}^2<\infty.$$

The proof for the statement (a) is easy to understand, but I cannot get the other one. According to the proof, "since $$\sum X_{n}$$ converges a.s., the partial sums of $$\sum X_{k}$$ are a.s. bounded, and it must be the case that for some $$c$$, $$P(T=\infty)>0$$." Here $$T$$ is the stopping time

$$T = \inf\{r: |\sum_{k=1}^r X_k| > c\}.$$

The problem is that I cannot find this $$c$$. I know that it's trivial if its boundedness is uniform in $$\Omega$$, but it's not the case, is it? Can anyone figure out which $$c$$ meets this condition? Thanks in advance.

• Could you edit to include the definition of $T$ for those who don't have the book available? Questions on this site should be self-contained. – Rhys Steele Mar 22 at 16:20

Using the notation from Williams, $$M_n = X_1 + \dots + X_n$$, we know that $$M_n$$ is a bounded sequence almost surely since it converges almost surely.
I will write $$T_n = \inf\{r : |M_r| > n\}$$. Notice that we have $$\{\sup_{r \geq 1} |M_r| < \infty\} = \bigcup_{n \geq 1} \{T_n = \infty\}.$$ So if $$\mathbb{P}(T_n = \infty) = 0$$ for each $$n$$ then $$1 = \mathbb{P}(\sup_{r \geq 1} |M_r| < \infty) = \mathbb{P}\big(\bigcup_{n \geq 1} \{T_n = \infty\} \big) \leq \sum_n \mathbb{P}(T_n = \infty) = 0$$ which is a contradiction and hence there is an $$n$$ such that $$\mathbb{P}(T_n = \infty) > 0$$ as desired.
• Thank you very much for your quick response and clarification of my question! I can fully understand the proof. It is not necessary to bound $|M_{r}|$ uniformly. Thank you again! – Paruru Mar 22 at 16:58