Past exam of the stochastic differential equations and Brownian motion in complex analysis course of the University of Oxford / 2010 / B / 6

Use first that planar Brownian motion avoids a fixed point, so the path starting at $1$ never reaches zero. Its continuous logarithmic lift, initialized at zero, has real part $U_t=\log |Z_t|$ and imaginary part the continuous unwrapped angle ${\Theta }_t$. Local branches of the complex logarithm, patched along the path, and the Itô formula show that both coordinates are continuous local martingales withThe clock is finite on every compact time interval, since a continuous nonvanishing path stays a positive distance from zero there. It is continuous and strictly increasing. Moreover $\tau (\infty )=\infty$: by recurrence, planar Brownian motion spends infinite total time in a fixed disk, and inside the unit disk the clock integrand is at least one. This is the divergence of a positive planar Brownian occupation integral.

The common-clock time change of orthogonal local martingales, using the Lévy characterization of multidimensional Brownian motion, supplies independent Brownian coordinates ${\zeta }^1,{\zeta }^2$ such that $U_t+i{\Theta }_t={\zeta }_{\tau (t)}$. Exponentiating givesTo identify the inverse clock, write $A={\tau }^{-1}$. The derivative of an inverse function and the representation just obtained giveConsequentlyThe strict inequality in this generalized inverse has the same infimum as the unique inverse value, since $A$ is continuous and strictly increasing onto the half-line.

Assume $0<r<1<R$, as required for the annulus and logarithms. The logarithmic radius of planar Brownian motion is a local martingale. Stopping at $T=T_{r,R}$ keeps it between the constants $\log r$ and $\log R$, so the stopped process is a bounded martingale. The exit time is finite almost surely because the annulus is bounded and Brownian motion eventually exits its containing disk.

The optional stopping theorem followed by dominated convergence gives $\mathbb{E}\log |Z_T|=\log |Z_0|=0$. Continuity puts the exit radius at either $r$ or $R$. If $p$ is the inner-exit probability, then $p\log r+(1-p)\log R=0$. HenceStopping before taking expectations matters: the unstopped logarithmic-radius process is a strict local martingale, so one cannot assume its unrestricted expectation is constant.

The second coordinate ${\zeta }^2$ is one-dimensional Brownian motion, whose zero set is unbounded almost surely by recurrence. The clock $\tau$ and its inverse are continuous, increasing, and unbounded, so the unwrapped angle ${\Theta }_t={\zeta }_{\tau (t)}^2$ has zeros at arbitrarily large original times.

To ensure the times are stopping times for the original Brownian filtration, define directlyThe logarithmic lift makes $\Theta$ a continuous adapted process. Hitting a closed set after a stopping time is a stopping time under the usual filtration conditions, and the unbounded zero set proves inductively that each $T_n$ is finite almost surely. Since $T_n\ge T_{n-1}+1$, the sequence tends to infinity, and continuity givesGeometrically, the Brownian path returns to the positive real ray at arbitrarily late times with zero net unwrapped winding angle about the origin. The assertion is stronger than principal argument zero, which also allows any integer number of completed turns.

The long-time Spitzer winding theorem states that, for the continuous unwrapped angle ${\Theta }_t$ of planar Brownian motion started at $1$,Here $C$ has the standard symmetric Cauchy distribution. The notation means convergence in distribution: for every real $x$,Equivalently, expectations converge for every bounded continuous test function. This is a statement about the laws at large deterministic times, without asserting almost-sure convergence of sample paths. The angle is unwrapped; a principal argument is bounded and would instead give zero after division by $\log t$.