Past exam of the stochastic differential equations and Brownian motion in complex analysis course of the University of Oxford / 2010 / B / 5

The disk exit time is finite almost surely; for example, either coordinate of the planar Brownian motion eventually leaves $(-1,1)$. By path continuity, the exit lies on the unit circle. Starting at the origin, rotational invariance of planar Brownian motion makes the exit law invariant under every circle rotation. The unique rotation-invariant probability measure is normalized arc length. Consequently $Z_{T_D}$ is uniform on the unit circle, or equivalently its angle has density $1/(2\pi )$ on $[0,2\pi )$.

By conformal invariance of planar Brownian motion, $w(Z_{T_D})$ has the uniform boundary law from part (a), because the image process starts at zero. A disk automorphism extends to an orientation-preserving boundary diffeomorphism. Choose a continuous angular lift $\varphi :\mathbb{R}\to \mathbb{R}$ satisfyingThe principal-angle version may have a branch jump, which is harmless for the density. The change of variables for probability densities yields the exit-angle density ${\varphi }^{\prime }(\theta )/(2\pi )$. Differentiating the boundary relation givesTaking absolute values proves ${\varphi }^{\prime }(\theta )=|w^{\prime }(e^{i\theta })|$. ThereforeThe angular lift explains rigorously the derivative appearing in the printed hint despite the principal-angle branch cut.

By Brownian scaling, $r^{-1}{Z}_{r^{2}t}$ is planar Brownian motion started at $w_0/r$. Its unit-disk exit angle is exactly the original radius-$r$ exit angle. Substitute $z_0=w_0/r$ into the supplied Poisson kernel for the unit disk:This density is with respect to angle $d\theta$. Its density with respect to arc length would be smaller by a factor $r$.

Put $\rho =|z|$. The triangle inequality gives $1-\rho \le |z-e^{i\theta }|\le 1+\rho$, so the Poisson kernel for the unit disk obeysIf $\rho \le r$, replace the lower bound by $(1-r)/(1+r)$ and the upper bound by $(1+r)/(1-r)$. Integrating against the positive measure $h(e^{i\theta })d\theta /(2\pi )$ givesHere $u(0)$ is exactly the uniform boundary average from part (a). Positivity permits multiplication of the pointwise kernel bounds; continuity on the compact circle ensures all integrals are finite. This is the Harnack inequality on the unit disk.