optional sampling theorem question

Question

Let $X_1,X_2,\ldots$ be iid random variables with $$\mathbb{P}\{X_i = 2\} = \frac{1}{3}, \mathbb{P}\{X_i = \frac{1}{2}\} = \frac{2}{3}.$$ Consider $M_n = X_1 \cdots X_n$. Then $M_n$ is a martingale. The question is to use the optional sampling theorem to find the probability that $M_n$ ever attains a value of at least $64$. It seems like you're meant to apply the theorem where the stopping time $T$ is defined to be the least $n$ such that $M_n = 64$. But doesn't one need to know that $\mathbb{P}\{T < \infty\} = 1$ to apply the theorem, which would mean the answer is $1$? This can't be the correct approach.

Answer 1

The common trick in this situation is to consider the stopped martingale $M^T$ instead of $M$, defined for all $n\in\mathbb N^*$ by $M^T_n=M_{n\wedge T}$.

Here $M^T$ is a nonnegative martingale, bounded by $64$, therefore we can apply the optional stopping theorem to it. Because it is a bounded martingale, it converges almost surely to a random variable $M^T_\infty$. Note that on $\{T<+\infty\}$ we have of course $M^T_\infty=M_T=64$, but on $\{T=+\infty\}$ it could be anything.

Actually here we can easily prove that $M^T_\infty=0$ on $\{T=+\infty\}$. Indeed, fix an outcome $\omega\in\{T=+\infty\}$ and set $x=M^T_\infty(\omega)$. If we had $x>0$, then for $n$ large enough we should have $\frac{3x}{4}<M_n(\omega)<\frac{5x}{4}$. But as $X_{n+1}(\omega)\in\{\frac12,2\}$, $M_{n+1}(\omega)$ would then necessarily leave the interval $(\frac{3x}{4}, \frac{5x}{4})$. Therefore $x=0$.

So $M^T_\infty=1_{\{T<+\infty\}}\times64$, and the optional stopping theorem yields $$ \mathbb E[M^T_\infty]=\mathbb P(T<+\infty)\times64=\mathbb E[M_1]=1, $$ hence $\mathbb P(T<+\infty)=\frac{1}{64}$.