At the reflecting endpoint the two values coincide:
$$
\tau^+(L)=\tau^-(L)=\frac Ls+\frac{\lambda L^2}{s^2}.
$$
Under the <diffusion limit of the telegraph process> $s,\lambda\to\infty$ with $s^2/(2\lambda)=D$,
$$
\boxed{\tau^\pm(L)\longrightarrow\frac{L^2}{2D}=\tau_0.}
$$
Rapid velocity reversals erase directional persistence. Their integrated velocity converges to Brownian motion with diffusivity $D$, so its first-passage statistic converges to the zero-drift result.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 356 / 1 / e / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-356/1/e/solution