Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 28 / 2 / b / Solution

First the exit time has sufficiently many finite moments. Over any interval of length $a^2$, a Brownian motion increment has probability $q=\mathbb{P}(|B_{a^2}|>2a)>0$ of absolute size larger than $2a$. If the process is still inside $(-a,a)$, such an increment forces it to exit before the next interval endpoint. Independence of increments at deterministic times therefore givesThus $T<\infty$ almost surely and all its positive moments are finite. This justifies extracting moments from its transform rather than presupposing them.

For $\lambda >0$, put $\theta =\sqrt{2\lambda }$. The processis a martingale, being the average of two exponential martingales for Brownian motion. Apply the optional stopping theorem at the bounded time $T\wedge t$. Since $|B_{T\wedge t}|\le a$, the stopped values are bounded by $\cosh (\theta a)$. Continuity gives $B_T\in \{-a,a\}$. Letting $t\to \infty$ by the dominated convergence theorem yieldsHence the Laplace transform of symmetric Brownian interval-exit time isThe value at zero follows from almost-sure finiteness. Expanding the reciprocal givesCompare with part (a), now applicable because of the geometric tail bound. The Brownian symmetric interval-exit moments are