Let $\sigma_x=\inf\{t\geq0:\widetilde B_t=x\}$. Continuity gives $\{S\geq x\}=\{\sigma_x<\infty\}$. On this event, the says that
$$
(\widetilde B_{\sigma_x+t}-x)_{t\geq0}
$$
is an independent Brownian motion with drift $-\mu$. It reaches level $y$ with probability $\mathbb P(S\geq y)$. Therefore
$$
\mathbb P(S\geq x+y)
=\mathbb P(\sigma_x<\infty)\mathbb P(S\geq y)
=\mathbb P(S\geq x)\mathbb P(S\geq y).
$$

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-201/4/b/ii/solution