Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 21 / 6 / Solution

For the exit law, the two identified processes have the same argument. Starting at zero, neither can skip $-a$ or $b$: Brownian motion is continuous, and the symmetric Poisson difference process has unit jumps with integer boundary levels. Therefore $|X_{t\wedge T}|\le K=\max (a,b)$, and if $T$ is finite, $X_T\in \{-a,b\}$.

To justify finiteness rather than presume it, apply the optional stopping theorem to the martingale $X_t^2-t$ at the bounded stopping time $t\wedge T$:The first boundary hitting time is a stopping time in both cases: in the continuous case use the continuous-path hitting criterion, and in the jump case inspect the successive jump times. Monotone convergence now gives $\mathbb{E}T\le K^2$, so $T<\infty$ almost surely. The stopped position tends to $X_T$ as $t\to \infty$ and is bounded, making dominated convergence applicable. Stopping $X$ at $t\wedge T$ and taking this limit yields $\mathbb{E}{X}_T=0$.

If $\pi =\mathbb{P}(X_T=-a)$, then $0=-a\pi +b(1-\pi )$. Thus in both cases the entire distribution isThis is the same balanced gambler's ruin law despite the different one-time distributions. As a further consistency check, passing to the limit in the stopped square-martingale identity givesAll uses of an unbounded exit time were obtained by limits of bounded-time sampling with an explicit bound on the stopped position.