Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 29 / 3

The Brownian first-passage time $T_a$ is finite almost surely. Indeed the Brownian reflection principle gives $\mathbb{P}(T_a\le t)=2\mathbb{P}(B_t\ge a)\to 1$ as $t\to \infty$. Continuity of the paths means the first passage to $x+y$ occurs after the first passage to $x$. By the Strong Markov property, the process $B_{T_x+s}-x$ is an independent fresh Brownian motion. Thus $T_{x+y}-T_x$ has the same probability distribution as $T_y$ and is independent of $T_x$. Taking the Laplace transforms of nonnegative random variables givesFor fixed $\lambda \ge 0$, these transforms lie in $(0,1]$. Put $g(a)=-\log {\phi }_a(\lambda )$. It is additive on positive arguments and nondecreasing, because the hitting times increase with the level. Writing $c(\lambda )=g(1)$, additivity gives $g(r)=rc(\lambda )$ for positive rational $r$, and rational upper and lower approximations then give $g(a)=ac(\lambda )$ for every $a>0$. Hence ${\phi }_a(\lambda )=e^{-ac(\lambda )}$, with $c(\lambda )\ge 0$.

The Brownian scaling identity $T_a\stackrel{d}{=}{a}^2{T}_1$ givesFor $\lambda >0$, take $a=\sqrt{\lambda }$ in this relation with the transform parameter one. It follows that $c(\lambda )=C\sqrt{\lambda }$, where $C=c(1)\ge 0$. At zero, finiteness of the hitting time gives $c(0)=0$. Thus $\boxed{{\phi }_a(\lambda )=e^{-aC\sqrt{\lambda }}}$, with the constant determined in part (b).

Fix $\lambda >0$ and choose the positive rootUse the Exponential martingale for Brownian motion $M_t=\exp (\theta B_t-{\theta }^{2}t/2)$. Before the linear-boundary stopping time $\tau$, path continuity gives $B_s<a+bs$. At the boundary there is equality. ConsequentlyThe optional stopping theorem at the bounded stopping time $t\wedge \tau$ yields $\mathbb{E}{M}_{t\wedge \tau }=1$. Separate the two events:The last term is at most $e^{\theta a-\lambda t}$ and tends to zero. The first term converges by the monotone convergence theorem. Taking $e^{-\lambda \infty }=0$ therefore provesThis bounded-stopping argument does not assume in advance that $\tau$ is finite. For $b\le 0$, letting $\lambda \downarrow 0$ makes the right side tend to one, so it also proves $\mathbb{P}(\tau <\infty )=1$. Setting $b=0$ gives the Brownian first-passage Laplace transformThe zero-parameter value follows from the just-proved almost-sure finiteness.