# Almost surely convergence for a martingale

Let $X_1, \dots$ be a sequence of independent random variables with $P(X_i = \frac{1}{2})=\frac{1}{2}$ and $P(X_i = \frac{3}{2})=\frac{1}{2}$. Let $S_n := \prod_{i=1}^n X_i$.

I've proven that $S_n$ is a martingale and using the convergence theorem for martingales there exists an $S_\infty$ such that $S_n \to S_\infty$ a.s. Now I'd like to compute $S_\infty$ and this is my attempt so far:

Using the strong law of large numbers it is clear that:

$\frac{1}{n}\log(S_n) = \frac{1}{n}\sum_{i=1}^n \log(X_i) \to E(\log(X_1)) = \frac{1}{2}\log(\frac{3}{2})-\frac{\log(2)}{2} < 0$ a.s.

Therefore $\log(S_n) \to -\infty$ a.s. and so $S_n \to 0$ a.s which means $S_\infty = 0$.

Is the proof correct or is there something lacking?

• @saz My question is whether or not the proof is fine or wrong. – PeterMcCoy Dec 18 '16 at 18:21
• Looks correct for me. – zhoraster Dec 18 '16 at 18:25
• @zhoraster Thank you, I was a little bit worried if the argument is sufficient enough and thought for some reason that I did a mistake. – PeterMcCoy Dec 18 '16 at 18:34