Past exam of the probability, measure and martingales course of the University of Oxford / 2021 / 3 / b / iii

The optional stopping theorem at a bounded stopping time states that an integrable martingale $Z_n$ satisfies $\mathbb{E}{Z}_T=\mathbb{E}{Z}_0$ whenever $T\le n$ for a fixed finite $n$. Apply this first to $S$ at $T=n\wedge \tau$. The simple symmetric random walk cannot cross an integer endpoint without hitting it, soThe preceding part gives $\tau <\infty$ almost surely, hence $S_{n\wedge \tau }\to S_{\tau }$ almost surely. The constant bound $C=\max \{-a,b\}$ permits the dominated convergence theorem, yielding $\mathbb{E}{S}_{\tau }=0$. Since $S_{\tau }$ is either $a$ or $b$, solving $ap+b(1-p)=0$ gives its full probability distribution:Now letThis counts the visit at time $0$, and excludes the exit time. Apply the same bounded-time optional stopping theorem to $M_n=|S_n|-L_n$. For each $n$ it givesThe visit counts increase to $L_{\tau }$, so monotone convergence applies to the left. Dominated convergence applies to the right using $|S_{n\wedge \tau }|\le C$. Consequently $L_{\tau }$ is an integrable random variable andThis is the origin case of expected occupation before interval exit for a simple symmetric random walk. The argument obtains integrability of the total occupation count before using any unbounded-time optional stopping for $M$, so it does not presuppose the uniform integrability asked about next. The endpoint probabilities also agree with the standard gambler's ruin calculation.^[ref]