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A planar Brownian motion exits a conformal image of a bounded planar domain in finite time almost surely. If the image lifetime were infinite, the inverse derivative is bounded away from zero on a small interior disc. Recurrence of planar Brownian motion and the Strong Markov property give infinite Brownian occupation time there, forcing the inverse conformal Brownian clock to diverge. It cannot exceed the finite original exit time. The expected image exit time can nevertheless be infinite.

The density of expected Brownian occupation time before exit from a planar domain. For planar Brownian motion with infinitesimal generator $\frac{1}{2}\Delta$, it obeys $-\frac{1}{2}{\Delta }_w{G}_D(z,w)={\delta }_z(w)$ and has singularity $-{\pi }^{-1}\log |w-z|$. A conformal bijection preserves this Dirichlet Green function: the squared derivative in the conformal Brownian clock cancels the area Jacobian determinant.

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.

Take the convention that the coordinates of planar Brownian motion are independent standard real Brownian motions, so its infinitesimal generator is $\frac{1}{2}\Delta$. Define the forward conformal Brownian clockSince a conformal bijection has nonzero derivative, $A$ is a strictly increasing continuous map from $[0,T)$ onto $[0,\tilde{T})$. The time used inside the original path is its inverse:Thus the interval direction for $\tau$ is the inverse-clock direction.

Write $\varphi =u+iv$. The Cauchy-Riemann equations and the Itô formula make $u(B_s),v(B_s)$ continuous local martingales, withIndeed both components are harmonic functions, and their gradients are orthogonal with the same squared norm. After the time change of a continuous process by $\tau$, their quadratic variations are $t$ and their quadratic covariation is zero. The Lévy characterization of multidimensional Brownian motion therefore identifies $\tilde{B}$ as planar Brownian motion started at $z^{\prime }$, up to its lifetime.

It remains to identify that lifetime as the exit time, rather than merely produce a local Brownian path. Boundedness of $D$ gives $T<\infty$ almost surely. For every compact subset $C\subset D^{\prime }$, its inverse image under $\varphi$ is compactly contained in $D$. As $s\uparrow T$, continuity gives $B_s\to B_T\notin D$, so $\varphi (B_s)$ eventually leaves $C$. Consequently the transformed path leaves every compact subset of $D^{\prime }$ at its lifetime. If $\tilde{T}<\infty$, its Brownian extension has a finite limit, and that limit is outside $D^{\prime }$; if $\tilde{T}=\infty$, the path never exits. In both cases its maximal lifetime is precisely the exit time from $D^{\prime }$.

The killed paths, together with their lifetimes, have the same law:

A finite original exit time alone does not imply that the integral defining the conformal Brownian clock is finite. To establish this, let $\psi ={\varphi }^{-1}$. On the event $\{\tilde{T}=\infty \}$, the Brownian path of part (a) lives forever in $D^{\prime }$, while its inverse clock satisfiesChoose a closed disc $C$ compactly contained in $D^{\prime }$. Its radius may be decreased so that $|{\psi }^{\prime }{|}^2\ge c>0$ on $C$. Recurrence of planar Brownian motion, together with the Strong Markov property, gives infinite total Brownian occupation time in $C$: return repeatedly to a smaller concentric disc, and use the fixed positive probability of remaining in $C$ for a fixed positive duration. The successive trials imply infinitely many such durations. This is the same mechanism as the divergence of a positive planar Brownian occupation integral.

Hence, on any infinite-lifetime Brownian path,This contradicts $\tau (t)<T<\infty$. Thus $\tilde{T}$ is finite almost surely, and equality in law from part (a) gives finite exit from a conformal image of a bounded planar domain:This asserts almost-sure finiteness, not finiteness of the expected exit time.

Use the Green function of killed planar Brownian motion, normalized as the density of expected occupation with respect to area:for nonnegative measurable $f$. Equivalently, $G_D(z,w)={\int }_0^{\infty }{p}_D(t,z,w)dt$, where $p_D$ is the killed Brownian transition density. With generator $\frac{1}{2}\Delta$, the distributional normalization is $-\frac{1}{2}{\Delta }_w{G}_D(z,w)={\delta }_z(w)$, and the singularity is $-{\pi }^{-1}\log |w-z|$ plus a locally harmonic function. This fixes the normalization of the Dirichlet Green function explicitly.

By conformal invariance of planar Brownian motion and its conformal Brownian clock,The Jacobian determinant of a conformal map is $|{\varphi }^{\prime }(w){|}^2$. Changing the area variable to $w^{\prime }=\varphi (w)$ givesUniqueness of the occupation density proves the desired equality almost everywhere. Both functions are continuous and harmonic away from their pole, so it holds at every $w\ne z$. The Green function is conformally invariant:If the Dirichlet Green function is instead normalized for $-\Delta$, both kernels are divided by two and the invariance statement is unchanged.