Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 29 / 2

Set $\beta =\pi /\alpha$ and use the branch $\varphi (z)=z^{\beta }$ with argument in $(-\alpha /2,\alpha /2)$. This is a conformal map from the wedge to the right half-plane, fixes the starting point $1$, and sends the outer circle of radius $r$ to that of radiusThe two wedge sides map to the imaginary axis.

By conformal invariance of planar Brownian motion, the image of the stopped path is planar Brownian motion after the increasing conformal Brownian clock $A(t)={\int }_0^t|{\varphi }^{\prime }(B_s){|}^{2}ds$. This clock does not change which boundary portion is reached first. Localization away from the vertex justifies the map even when its derivative is unbounded there; the vertex is a polar point for planar Brownian motion and has zero hitting probability from $1$.

Consequentlywhere the probability on the right is for the right half-plane. The power-map reduction for Brownian exit from a wedge also works at $\alpha =2\pi$, when the wedge is the plane slit along the negative real axis.

Let $E_{+}$ and $E_{-}$ denote positive and negative real part at the circular exit. On $\{S<T(r)\}$, reflect the portion of the planar Brownian motion after $S$ by $z\mapsto -\overline{z}$. This reflection fixes the imaginary axis and preserves distances from the origin. The Strong Markov property and reflection symmetry show that the resulting path has the same law, its circular exit time is unchanged, and $E_{+}$ is exchanged with $E_{-}$. ThereforeAn exit with negative real part must first cross the imaginary axis. An exit before $S$ has positive real part. The two points $ir,-ir$ have zero circular exit probability, since circular harmonic measure has no atoms. HenceSubtracting provesThis is the reflection identity for Brownian exit from a half-disc.

Scale the disk to the unit disc, so that the starting point is $a=1/r\in (0,1)$. Use the Möbius transformationwhich maps the disk onto itself, sends $a$ to $0$, and sends $1$ to $1$. By conformal invariance of planar Brownian motion, the exit image is the circular exit of a Brownian motion starting at $0$. Rotational invariance of planar Brownian motion makes that exit uniform in angle.

The endpoints of the right semicircle satisfyIts image is the arc through $1$ between these points, of angular length $2\phi$. Thus the Möbius calculation of circular Brownian exit givesSince $E_{+}$ and $E_{-}$ partition the exit almost surely, part (b) yieldsThe last equality uses $2\arctan (1/r)\in (0,\pi /2)$, so the double-angle tangent identity uses the stated principal branch. As a check, the boundary angle derivative of ${\psi }_a$ is $(1-a^2)/|e^{i\theta }-a{|}^2$; integrating this circular exit density over the right semicircle gives exactly the integral supplied in the question.

Substitute $R=r^{\pi /\alpha }$ from part (a) into part (c). The Brownian wedge-exit probability isIt tends to $1$ as $r\downarrow 1$, and is asymptotic to $(4/\pi )r^{-\pi /\alpha }$ as $r\to \infty$. The exponent reflects how the conformal map stretches wedge angles; wider wedges give slower decay.