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


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.

  • $\begingroup$ 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. $\endgroup$ – 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.

  • $\begingroup$ 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! $\endgroup$ – Paruru Mar 22 at 16:58
  • $\begingroup$ I'm glad my answer was helpful :) Since this is your first question here, I just want to let you know that if you're happy my answer resolves the question, if you'd like you can accept it so that the question is marked as resolved on the site. $\endgroup$ – Rhys Steele Mar 22 at 17:04

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