Past exam of the stochastic differential equations and Brownian motion in complex analysis course of the University of Oxford / 2010 / B / 4

There is a printed domain error: $\mathrm{Re}z>0$ defines the right half-plane. Taken literally, continuity forces $\mathrm{Re}{Z}_{T_H}=0$ almost surely, so its law is a point mass at zero. The displayed density belongs to the intended upper half-plane $\mathrm{Im}z>0$, with $b>0$. We prove that intended assertion explicitly.

Write $Z_t=a+X_t+i(b+Y_t)$ for independent standard Brownian motions. The exit time $T$ is the first time $Y$ hits $-b$, and depends only on $Y$. Its density, by the supplied Brownian first-passage time law, is $b(2\pi t^3{)}^{-1/2}{e}^{-b^2/(2t)}$. Conditional on $T=t$, the real exit coordinate is normally distributed with mean $a$ and variance $t$. Independence therefore givesWith $u=[b^2+(x-a{)}^2]/(2t)$, the integral is $2/[b^2+(x-a{)}^2]$. Thus the Cauchy exit law from a Brownian half-plane isFor the literal right half-plane, the corresponding nondegenerate coordinate is $\mathrm{Im}{Z}_{T_H}$, whose density is $a/\{\pi [a^2+(y-b{)}^2]\}$ for $a>0$.

If $f:D\to D^{\prime }$ is a conformal map, the image of planar Brownian motion in $D$ is Brownian motion in $D^{\prime }$ after the increasing clock changeMore precisely, $f(Z_{A^{-1}(u)})$ is planar Brownian motion, starting at $f(Z_0)$ and run up to the relevant exit time. The Itô formula and the Cauchy-Riemann equations give image-coordinate local martingales with common bracket $A$ and zero cross bracket; the Lévy characterization of multidimensional Brownian motion identifies the time-changed process. In particular, conformal invariance of planar Brownian motion transports exit locations and harmonic measure, while changing traversal speed.

For $z=x+iy$ in the strip,so $\mathrm{Im}f(z)>0$. This is a bijective conformal map onto the upper half-plane. One way to verify bijectivity is to factor it as $w=e^{\pi z/a}$, followed by $(w+w^{-1})/2$. The first map gives $\mathrm{Im}w>0$, $|w|>1$. The two inverse roots of the second map have product one; for any point in the upper half-plane exactly one root has modulus greater than one, and its imaginary part is positive because $\mathrm{Im}[(w+w^{-1})/2]=(1-|w{|}^{-2})\mathrm{Im}w/2$.

The three boundary pieces map as follows: the lower edge goes to $(1,\infty )$, the upper edge to $(-\infty ,-1)$, and the vertical edge to $(-1,1)$. Thus the desired exit event becomes exit from the upper half-plane to the left of $-1$. WriteBy conformal invariance of harmonic measure and the Cauchy exit law from a Brownian half-plane,The strip exit is finite almost surely, since its imaginary coordinate exits $(0,a)$ in finite time; corners have probability zero because individual points are polar points for planar Brownian motion. This is the Top-edge exit probability of a Brownian half-strip.

Let $\alpha =\pi q/a\in (0,\pi )$. With $q$ and $a$ fixed,The principal arctangent satisfies $\arctan (\cot \alpha )=\pi /2-\alpha$ throughout this interval. Substitution givesFar from the vertical boundary, the limit agrees with the one-dimensional Brownian motion probability of reaching height $a$ before zero.