Applying Doob's Optional Stopping Theorem for a random walk

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Let $X_1,X_2,\dots$ be i.i.d. random variables equalling $1$ with probability $p\in(0,1)$ and $-1$ with probability $q=1-p$. Consider natural numbers $0<a<b$.

For $n\geq 0$, put $$S_n=a+X_1+\cdots X_n.$$

Let $(\mathcal F_n)_{n\geq 0}$ be the natural filtration corresponding to the $X_n$.

Lastly, let $T=\inf\{n\geq 0\mid S_n=0 \text{ or } S_n=b\}$.

The goal is to compute $\mathbb ES_T$.


This is part of a bigger exercise. I have shown that

  • $T$ is a stopping time with finite expectation;
  • If we define $M_n=\left(\frac qp\right)^{S_n}$, then $(M_n)_{n\geq 0}$ is a martingale (wrt. the filtration $(\mathcal F_n)_{n\geq 0}$).

One can notice that to determine $\mathbb E S_T$, it suffices to determine $\mathbb P\{S_T=b\}$, since $\mathbb ES_T=b\mathbb P\{S_T=b\}$.

My idea was to apply Doob's Optinal Stopping Theorem for the martingale $(M_n)_n$ and the stopping time $T$, yielding $$\mathbb P\{S_T=0\}+\left(\frac qp\right)^b\mathbb P\{S_T=b\}=\left(\frac qp\right)^a.$$ Replacing $\mathbb P\{S_T=0\}$ by $1-\mathbb P\{S_T=b\}$, then isolation $\mathbb P\{S_T=b\}$ allows to conclude.

However, am I allowed to even apply Doob here? In fact, I believe that

  • $T$ isn't bounded,
  • the $M_n$ aren't uniformly bounded,
  • the the differences $|M_{n+1}-M_n|$ aren't uniformly bounded.

Am I overseeing something? Is there a different path to take?