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

Finite second moment implies $\mathbb{E}X<\infty$. For $\lambda$ near any positive ${\lambda }_0$, the derivative of $e^{-\lambda X}$ is bounded in absolute value by $X$, provided the neighbourhood stays in $(0,\infty )$. The dominated convergence theorem therefore permits differentiation under the expectation:For the expansion, put $r(z)=e^{-z}-1+z-z^2/2$. Taylor expansion gives $r(z)/z^2\to 0$ as $z\downarrow 0$. Also, for $z\ge 0$,The inequalities follow by integrating $0\le 1-e^{-u}\le u$ from zero to $z$. Consequently $r(\lambda X)/{\lambda }^2\to 0$ pointwise and its absolute value is bounded by $X^2/2$. Another application of the dominated convergence theorem gives the second-moment expansion of a Laplace transform:No third moment is needed for the remainder estimate.

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