Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 356 (Stochastic Processes in Biology)

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 $\Delta t\to 0$.

Let ${\mathbf{e}}_i$ be the $i$th coordinate vector and take $p$ to be zero whenever any argument is negative. Right jumps $i\to i+1$, left jumps $i\to i-1$, and absorption from compartment $1$ give the chemical master equation

Multiplying the master equation by $a_i$, summing over every state, and shifting indices in each gain term gives, for $2\le i\le m-1$,At the absorbing left edge and reflecting right edge the corresponding equations are

A centered nearest-neighbour discretization of drift--diffusion is obtained fromFor sufficiently small $h$ these rates are nonnegative. Taylor expansion of the mean equations givesAbsorption into the target to the left of the first compartment giveswhile the absence of outward jumps at the right endpoint gives the zero-flux boundary condition

The Fokker-Planck equation corresponds to the Itô diffusionabsorbed at $0$ and reflected at $L$.

For $v(x)=x$, an Euler--Maruyama proposal isIf $Y\le 0$, kill the path. If $X_n>0$ and $Y>0$, an endpoint-only test can miss a crossing. Conditional on the endpoints, the local Brownian bridge crossing probability isKill the path with this probability; otherwise impose reflection at $L$ by replacing an overshoot $Y>L$ with $2L-Y$ and set $X_{n+1}=Y$. Repeated reflection handles very rare multiple overshoots, and the approximation converges as $\Delta t\to 0$.

Under the independence closure, the mean rightward flux across the bond $(i,i+1)$ isThe mean equation is the discrete conservation law ${\dot{M}}_i=J_{i-1/2}-J_{i+1/2}$. Substitution of the rates from part c and Taylor expansion show that the exclusion factors cancel from the symmetric diffusive contribution but remain in the biased contribution:Therefore the mean-field asymmetric simple exclusion process limit isThe factor $1-c$ is the probability that the destination site is vacant.

Use stochastic mass-action propensitiesWriting $E_x^{r}f(x,y)=f(x+r,y)$ and similarly for $E_y$, the fast and slow forward operators areEach shift operator acts on everything to its right, including the propensity. Birth of $Y$ and consumption of $Y$ by $X+Y\to X$ are fast; $2Y\to X$ and $2X\to X$ are slow. The slow species is $X$.

At leading order, ${\mathcal{L}}_0^{\ast }[p_0(y\mid x)p_0(x,t)]=0$. For fixed $x$, the conditional stationary law therefore satisfiesThis is an immigration--death process, whose stationary distribution is Poisson with meanThe mean is finite only for $x>0$. The assumption $X(0)\ne 0$ and the pair-coalescence propensity $x(x-1)$ ensure that the slow process cannot remove its last $X$ molecule, so the reduced model remains in that domain.

Averaging the slow propensities over the conditional Poisson distribution uses $\mathbb{E}[Y(Y-1)\mid X=x]=q(x{)}^2$. Thus the effective birth and death rates of $X$ areConsequently

Applying the reduced Markov jump-process generator to $f(x)=x$ gives the exact moment equationUnder the stated moment closure this becomes the same expression evaluated at $X=m$. Set $m=V\bar{x}$ and let $V\to \infty$ to obtainIts positive stable equilibrium is

The stochastic quasi-steady-state simulation evolves only $X$:

1. At the current integer $x\ge 1$, compute $q(x)={\alpha }_3{V}^2/({\alpha }_{4}x)$ and the averaged rates ${\lambda }_1(x)={\alpha }_{1}q(x{)}^2/V$ and ${\lambda }_2(x)={\alpha }_{2}x(x-1)/V$.
2. Draw a waiting time $\Delta t\sim \mathrm{Exp}({\lambda }_1+{\lambda }_2)$.
3. Set $x\leftarrow x+1$ with probability ${\lambda }_1/({\lambda }_1+{\lambda }_2)$; otherwise set $x\leftarrow x-1$.
4. Advance time by $\Delta t$ and repeat.

This is a Gillespie algorithm for the averaged slow master equation. It samples the fast conditional equilibrium analytically through its factorial moment and never simulates individual fast $Y$ births or deaths.