Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 34 / 3

For $h(x,y)=\frac{1}{2}\log (x^2+y^2)$ away from the origin,The Itô formula for planar Brownian motion therefore has no drift term before the radius reaches $r$. Since the initial logarithm is zero,The integrand's squared norm is $|B_s{|}^{-2}\le r^{-2}$ on this stopped interval, so this is a square-integrable martingale on every finite horizon, in particular a local martingale. One may first stop in an outer annulus to apply the formula on a compact smooth domain and then remove that localization.

Let $\tau =S_r\wedge T_R$. The stopped logarithm lies between $\log r$ and $\log R$, so the bounded local martingale criterion makes it a true bounded martingale. Also $\tau$ is finite almost surely. Indeed $|B_t{|}^2-2t$ is a martingale, and applying its stopped identity through $t\wedge T_R$ givesmonotone convergence theorem implies $\mathbb{E}{T}_R\le (R^2-1)/2<\infty$, hence $\tau \le T_R<\infty$ almost surely. Boundedness permits dominated convergence theorem in the stopped logarithm as $t\to \infty$. ConsequentlyThe standard results used are Itô's formula, the bounded-local-martingale criterion, the stopped martingale identity at bounded stopping times, and dominated/monotone convergence theorem. At exit the radius is either $r$ or $R$ by continuity. Solving $p\log r+(1-p)\log R=0$ also gives the planar Brownian annulus hitting probability

For $k\ge 2$, substitute $r_k=k^{-k}$ and $R=k$ into the exit formula:Let ${\tau }_0$ be the first hit of the origin. On any path with ${\tau }_0<\infty$, continuity makes the radius bounded on $[0,{\tau }_0]$. For all sufficiently large $k$, that path reaches radius $r_k$ before ${\tau }_0$ and has not yet reached radius $k$, so it belongs to $\{S_{r_k}\le T_k\}$. Thus $\{{\tau }_0<\infty \}$ is contained in the liminf of these events. Fatou's inequality for their indicators gives probability at most ${\mathrm{liminf}}_{k}1/(k+1)=0$ for that liminf. Hence the polar point for planar Brownian motion conclusion isThe argument controls one event for all times, rather than merely proving zero hitting probability at each fixed time.

On a compact time interval a continuous path avoiding the origin has a strictly positive minimum radius. Thus the times $S_{1/n}$ tend to infinity almost surely. Part (a) localizes $M_t=\log |B_t|$, so $M$ is a continuous local martingale.

To show that it is not a martingale, compute its expectation at a positive deterministic time. In distribution $B_t=(1,0)+W_t$, where $W_t$ is a centered isotropic Gaussian vector. Conditional on $\rho =|W_t|$, its angle is uniform. The angular average of a logarithmic potential isFor $\rho <1$ this follows by averaging the real part of the power series for $\log (1+\rho e^{i\theta })$; for $\rho >1$ factor out $\rho$ and apply the same argument to ${\rho }^{-1}$. The value at $1$ follows by the integrable logarithmic singularity, or is irrelevant for the continuously distributed radius.

The radius has density $(\rho /t)e^{-{\rho }^2/(2t)}$. The logarithm of $|B_t|$ is absolutely integrable: near the origin its Gaussian density is bounded and ${\int }_0^{1}r|\log r|dr<\infty$; away from the origin its positive part is bounded by a constant plus $|B_t|$. Therefore conditioning is legitimate, and integration by parts yieldsSince $M_0=0$, this contradicts preservation of expectation by a martingale. Thus the logarithmic radius of planar Brownian motion is a strict local martingale. It is not a nonnegative local martingale; its negative values allow its expectation to increase.