Past exam of the stochastic modelling of biological processes course of the University of Oxford / 2023 / 2

The native PDF repeats $d{W}_y$ in its list of independent noise terms, whereas its third stochastic differential equation uses $d{W}_z$. We use the intended three independent coordinate Brownian motions $W_x,W_y,W_z$.

Put $\mathbf{X}=(X,Y,Z)$, $r=|\mathbf{X}|$ and $S=r^2$. Before the origin is reached, Itô's formula givesThe $6$ is the sum of three quadratic variation terms, each $2dt$. The drift contribution is $-2\alpha dt$ because the numerator is $|\mathbf{X}{|}^2$. The radial stochastic integral can be written using a scalar Brownian motion $B$:Thus $S/2$ is a squared Bessel process of dimension $\delta =3-\alpha$ until it reaches zero. If this equation is taken globally with the usual nonnegative squared Bessel process continuation, taking expected values givesThe stochastic integral has zero expected value: stopping first at large levels bounds the integrands, yields a finite first moment for $S$, and then makes $\mathbb{E}{\int }_0^{t}S(u)du$ finite, so the unrestricted integral is a square-integrable martingale.

Boundary qualification. The punctured domain in the question does not itself specify that continuation. The boundary classification of a Bessel process says that the origin is inaccessible from $r_0>0$ when $\delta \ge 2$, equivalently $\alpha \le 1$. In this parameter range the original stochastic differential equation is global in the stated domain and the boxed result holds as written.

For $1<\alpha <3$, we have $0<\delta <2$: the origin is reached in finite time with probability one. One can also see accessibility directly from the radial generator of an SDE diffusionFor $r<b$, the harmonic function $1-(r/b{)}^{\alpha -1}$ is the probability of reaching zero before $b$. Finite-time access follows by using the nonnegative functionin the stopped Itô formula; it bounds the expected value of the time to reach zero or $b$. Letting $b\to \infty$ gives almost sure finite hitting of zero. The coefficients of the original vector stochastic differential equation are undefined there, so its stated punctured-domain version has no prescribed long-time evolution in this range.

The natural continuing radial squared Bessel process gives $6-2\alpha$, but stopping and holding the particle at the origin would give a different answer. Indeed, with $T_0$ the first hit of zero, Itô's formula for the stopped process givesSince $T_0<\infty$ almost surely, dominated convergence applied to $(t\wedge T_0)/t\le 1$ givesTherefore $6-2\alpha$ is the usual continued-radial answer; for $1<\alpha <3$ a boundary convention is necessary. It is not a uniquely defined limit for the punctured-domain model alone.

At $\alpha =0$ the transverse coordinates are independent Brownian motions with diffusion coefficient $D=1$. The cylinder is infinitely long, so the axial coordinate cannot affect its first contact time. By rotational invariance of Brownian motion, the mean first-passage time $u$ depends only on the transverse radius $\rho =\sqrt{x^2+y^2}$.

The Kolmogorov backward equation for a mean first-passage time is $Lu=-1$. Here $L={\partial }_x^2+{\partial }_y^2$, so the radial boundary value problem isThe absorbing boundary condition for a diffusion is imposed on the circular cross-section. At $\rho =0$, the solution must extend smoothly over the axis. Multiplying by $\rho$ and integrating givesSmoothness at the axis excludes the logarithm term, and $u(R)=0$ fixes $C_0=R^2/4$. ConsequentlyThis also verifies the finiteness assumption behind the Kolmogorov backward equation. Let $T$ be the first contact time and apply the stopped Itô formula to the bounded nonnegative function $u$ just obtained. Its bounded derivatives on the disk make the stopped stochastic integral a martingale, givingBy monotone convergence, $T$ has finite expected value; continuity and dominated convergence then make the final term vanish as $t\to \infty$. Hence $u({\hat{r}}_0)=\mathbb{E}T$. In particular, the mean first-passage time is $R^2/4$ on the axis and tends to zero at the adsorbing wall.

For a radial function $f(r)$ in three dimensions, the Laplacian is $f^{\prime \prime }+2f^{\prime }/r$, while the imposed radial drift contributes $-\alpha f^{\prime }/r$. Thus the radial generator of an SDE diffusion isThe hitting probability for the inner sphere is harmonic for this diffusion generator in the shell, with value one on the inner sphere and zero on the outer sphere. At $\alpha =1$, the boundary value problem is thereforeIntegrating $(r{p}^{\prime }{)}^{\prime }=0$ gives $p(r)=A\log r+B$. The two boundary conditions determineBoth arguments of the logarithms are dimensionless. The result decreases from one to zero across the shell and lies strictly between zero and one for $R_1<r_0<R_2$.

For a direct probability interpretation, stop at the first contact $T$ with either sphere. The stopped process $p(r(t\wedge T))$ is a bounded martingale by Itô's formula. The shell has positive inner radius, so its drift is bounded and its diffusion is nondegenerate; exit occurs almost surely. Optional stopping and dominated convergence therefore giveThe logarithm appears because the inward drift cancels one of the two geometric $f^{\prime }/r$ terms in the three-dimensional radial Laplacian, leaving the same radial diffusion generator as a planar Brownian motion with diffusion coefficient one.