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

The backward generator of the drift--diffusion is $\mathcal{L}=\alpha {\partial }_y+D{\partial }_y^2$. The survival probability satisfies the Kolmogorov backward equationwithThe target is absorbing, while reflection gives the Neumann boundary condition at $L$.

The mean first-passage time obeys the backward equationFor $\alpha \ne 0$, direct integration givesIn particular,This increases monotonically with drift away from the target, so the constrained optimum isExpanding the exponential at zero drift yields

The telegraph process has backward survival equationsInitially $Q^{\pm }(y,0)=1$. Only a left-moving trajectory reaches the target, while reflection reverses a right-moving velocity at the outer wall, so the hyperbolic boundary conditions are

The backward equations for the two mean hitting times arewith ${\tau }^{-}(0)=0$ and ${\tau }^{+}(L)={\tau }^{-}(L)$. Their difference obeysand the reflecting condition fixes ${\tau }^{+}-{\tau }^{-}=2(L-y)/s$. Integration then givesand

At the reflecting endpoint the two values coincide:Under the diffusion limit of the telegraph process $s,\lambda \to \infty$ with $s^2/(2\lambda )=D$,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.

Choose $\Delta t$ with $\lambda \Delta t\ll 1$ and $s\Delta t$ small relative to $L$ and the distance to the target. For each of many independent trajectories:

1. Set $x=y$, $v=s$, and $t=0$.
2. Propose $x_{\mathrm{new}}=x+v\Delta t$.
3. If $x_{\mathrm{new}}\le 0$, record the linearly interpolated target-crossing time and stop. If $x_{\mathrm{new}}\ge L$, reflect to $x_{\mathrm{new}}=2L-x_{\mathrm{new}}$ and set $v=-|v|$.
4. Otherwise reverse $v$ with probability $1-e^{-\lambda \Delta t}$, set $x=x_{\mathrm{new}}$ and $t=t+\Delta t$, and repeat.

The sample mean of the recorded times estimates ${\tau }^{+}(y)$. Sampling a Poisson number of reversals and their ordered times inside each step removes the at-most-one-reversal approximation, but the stated Bernoulli scheme converges as $\Delta t\to 0$.