Past exam of the continuous martingales and stochastic calculus course of the University of Oxford / 2023 / 3 / b

Let $\rho =|z_0|$ and $f(x,y)=\log \sqrt{x^2+y^2}$. On the annulus $r<\sqrt{x^2+y^2}<R$, its gradient and Laplace operator areThus $f$ is a harmonic function there. The stopped Itô formula givesThe sum of the squared coefficients is $|B_s{|}^{-2}\le r^{-2}$ before $\tau$. The Itô isometry therefore makes the stochastic term a true square-integrable martingale on every finite interval. Continuity prevents overshoot at exit, so $\log r\le \log |B_{t\wedge \tau }|\le \log R$. This proves the required bounded martingale property.

Before taking the terminal expectation, prove that $\tau$ is finite. Set ${\theta }_R=\inf \{t\ge 0:|B_t|=R\}$. The Itô formula for $|B{|}^2$ shows thatis a martingale, since its stopped stochastic integrands are bounded. Taking expectations yieldsThe monotone convergence theorem gives $\mathbb{E}{\theta }_R\le (R^2-{\rho }^2)/2<\infty$. Hence $\tau \le {\theta }_R<\infty$ almost surely.

The bounded convergence theorem, equivalently the optional stopping theorem applied to this bounded martingale, gives $\mathbb{E}\log |B_{\tau }|=\log \rho$. The two exit events partition the sample space up to a null set. Writing $p=\mathbb{P}({\tau }_r<{\tau }_R)$,Solving gives the planar Brownian annulus hitting probability

Fix an outer radius $R>\rho =|z_0|$. By continuity, reaching zero before radius $R$ requires first reaching every radius $\epsilon \in (0,\rho )$. The planar Brownian annulus hitting probability from (i) therefore impliesThus $\mathbb{P}({\tau }_0<{\tau }_R)=0$ for every fixed $R>\rho$.

On any continuous sample path that reaches zero in finite time, the radius has a finite maximum before that time. There is therefore some integer $m>\rho$ for which zero is reached before radius $m$. ConsequentlyEvery event on the right has probability zero, soThis proves that planar Brownian motion avoids a fixed point even though it can reach arbitrarily small positive radii.

First suppose $0<r<\rho =|z_0|$. For every $R>\rho$, the planar Brownian annulus hitting probability givesHence ${\tau }_r<\infty$ almost surely. This uses increasing outer radii; letting the inner radius tend to zero instead proves the different claim in part (ii).

For $r>\rho$, the disk-exit estimate in (i), with outer radius $r$, directly yields $\mathbb{E}{\tau }_r\le (r^2-{\rho }^2)/2<\infty$.

For $r=\rho$, retain the definition ${\tau }_r=\inf \{t>0:|B_t|=r\}$. The first case shows that radius $\rho /2$ is reached in finite time. Starting there, the Strong Markov property of Brownian motion and the disk-exit argument show that radius $\rho$ is subsequently reached in finite time. Thus the infimum over positive hitting times is finite in this case as well. Altogether,The conclusion can also be made simultaneous in all $r>0$: intersect the probability-one events for positive rational radii, then use continuity to cross each intermediate radius.