Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 201 / 4 / b

The strong law for Brownian motion gives $B_t/t\to 0$ almost surely. Thereforealmost surely, and hence ${\tilde{B}}_t\to -\infty$ almost surely.

Let ${\sigma }_x=\inf \{t\ge 0:{\tilde{B}}_t=x\}$. Continuity gives $\{S\ge x\}=\{{\sigma }_x<\infty \}$. On this event, the Strong Markov property says thatis an independent Brownian motion with drift $-\mu$. It reaches level $y$ with probability $\mathbb{P}(S\ge y)$. Therefore

Under the Cameron-Martin theorem for a linear drift, the probability that Brownian motion with drift $-\mu$ reaches $x$ isUsing the supplied Laplace transform with $\lambda ={\mu }^2/2$ givesand henceThis is the survival function of the exponential distribution with rate $2\mu$.