Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 101 / 4 / a / Solution

Let $\tau =T_{-a}\wedge T_b$ for the simple symmetric random walk. From any interior point, $a+b$ successive $+1$ increments force exit through $b$. The conditional probability of this block is $2^{-(a+b)}$, so a blockwise geometric tail bound from a uniform escape probability gives $\tau <\infty$ almost surely.

Apply the bounded optional stopping theorem to $\tau \wedge n$. Since $S_n$ is a martingale, ${\mathbb{E}}_0{S}_{\tau \wedge n}=0$. The stopped positions lie in $[-a,b]$, so the bounded convergence theorem gives ${\mathbb{E}}_0{S}_{\tau }=0$. If $h={\mathbb{P}}_0(T_{-a}<T_b)$, nearest-neighbor jumps ensure $S_{\tau }=-a$ on that event and $S_{\tau }=b$ otherwise. ThusNext apply the bounded optional stopping theorem to the square-minus-time martingale of a simple symmetric random walk:Again the right side converges by the bounded convergence theorem, while the left side increases to ${\mathbb{E}}_0\tau$ by the monotone convergence theorem. ThereforeThis argument justifies passage to the unbounded stopping time rather than applying optional stopping without an integrability check.