# Random signs - a remark of david williams:probability with martingales

This can be found in David Williams p.113-114.

Suppose that $(a_n)$ is a sequence of real numbers and that $(\epsilon_n)$ is a sequence of IID random variables with $P(\epsilon_n=\pm1)=\frac{1}{2}$. The result of 12.2 shows that

$(1) \sum \epsilon_na_n \text{ converges a.s. if and only if} \sum a_n^2<\infty$

and

$(2)\sum\epsilon_na_n \text{ oscillates infinitely if and only if} \sum a_n^2=\infty$

It's very clear to me why the first equivalence is true. Just the usage of 12.2 in David Williams which states that:

(12.2) $\sum X_k \text{ converges a.s. if and only if} \sum Var(X_k)<\infty$ for a sequence of independent zero mean random variables $(X_k)$.

I want to understand now why the second equivalence is true.

For $"\Rightarrow"$:

If $\liminf_{n\rightarrow \infty}\sum \epsilon_na_n\neq \limsup_{n\rightarrow \infty}\sum \epsilon_na_n$ then $\sum \epsilon_na_n$ does not converge and theorem 12.2 yields that $\sum Var(\epsilon_na_n)=\sum a_n^2=\infty$ as desired.

But now I don't how to show the other direction. Can someone help?

• You may be missing $=\infty$ at the end of (2) – Henry Jan 6 '15 at 13:29
• What's the definition of "oscillates infinitely"? – saz Jan 6 '15 at 17:05
• It's $\limsup_{N\rightarrow \infty} \sum_{n=1}^N \epsilon_na_n=-\liminf_{N\rightarrow \infty} \sum_{n=1}^{\infty}\epsilon_na_n=\infty$ – Epsilondelta Jan 6 '15 at 17:12
• Though you need an "almost surely" associated with the "oscillates infinitely" – Henry Jan 6 '15 at 21:20