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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.

Start the planar Brownian motion $X_t+i{Y}_t$ at $-x$ and stop at $\tau =\inf \{t:X_t=0\}$. The complex exponential is holomorphic with nonzero derivative. By conformal invariance of planar Brownian motion, the process $e^{X_t+i{Y}_t}$, run on its conformal Brownian clockis planar Brownian motion started at $\epsilon =e^{-x}$, until its first exit from the unit disk. The clock is strictly increasing before $\tau$, and $A_{\tau }\le \tau <\infty$ almost surely. It is legitimate to use the exponential although it is not globally injective: the Brownian coordinate brackets depend locally on the nonzero derivative, and the clock construction gives the stopped planar Brownian law. Before $\tau$, $|e^{X_t+i{Y}_t}|<1$; at $\tau$ it equals one. Hence the disk exit time in the new clock is exactly $A_{\tau }$.

The continuous lifted argument of this path, starting from angle zero, is $Y_t$. Its total argument at exit is therefore $Y_{\tau }$, which has distribution $xC$ by part (b). If the winding count records only completed integer turns, it differs from $Y_{\tau }/(2\pi )$ by a remainder $R_{\epsilon }$ with $|R_{\epsilon }|\le 1$. If fractional turns are retained, that remainder is zero. In either convention, since $\log \epsilon =-x$,The first term has distribution $-C/(2\pi )$ for every $x$, and the second tends to zero deterministically in absolute value as $x\to \infty$. Symmetry of the Cauchy distribution removes the minus sign, so Slutsky theorem gives the small-radius Brownian winding law:

The conformal invariance of planar Brownian motion is a time-change statement. If $\varphi :D\to D^{\prime }$ is a conformal bijection of planar domains and $B$ is planar Brownian motion started in $D$, let ${\tau }_D$ be its exit time and defineBefore exit this clock is strictly increasing. Its inverse $C_u$ makes $\varphi (B_{C_u})$ a planar Brownian motion started at $\varphi (B_0)$, up to its transformed lifetime $A_{{\tau }_D}$. The image of the stopped trajectory is therefore invariant in law after this change of time. A holomorphic map with nonzero derivative gives the corresponding local statement even without global injectivity.

Here is the proof. Write $\varphi =u+iv$ and $B=(B^1,B^2)$. The real and imaginary parts are harmonic. The Itô formula therefore gives the two continuous local martingalesstopped on compact subsets of $D$. The Cauchy-Riemann equations implyThus both coordinate quadratic variations equal $A_t$ and their quadratic covariation is zero. After inverse time change, the covariance matrix of the two continuous local martingales is $u{I}_2$. The multidimensional Lévy characterization of multidimensional Brownian motion states that a continuous vector local martingale starting at zero with this covariance matrix is standard vector Brownian motion. Apply it after subtracting $\varphi (B_0)$, and exhaust $D$ by compact subsets. This proves the theorem, with killing at the transformed exit lifetime. The clock is the conformal Brownian clock; omitting it would generally give an incorrect speed.

The exponential map is a conformal bijection from the stripto the wedge containing the starting point $1$. The restriction $\alpha +\beta <2\pi$ makes it injective. A planar Brownian motion $U+iV$ started at zero in this strip maps, after the conformal Brownian clock, to the required Brownian motion in the wedge.

Let $\sigma$ be the first time $V$ exits $(-\alpha ,\beta )$. It is finite almost surely, and $V_{t\wedge \sigma }$ is a bounded martingale. The optional stopping theorem givesThe strip exit maps to the corresponding bounding ray. The exponential clock is finite at $\sigma$, because its integrand is continuous on the compact time interval $[0,\sigma ]$. Thus it does not remove or reverse the exit event. We obtainThis is the Brownian wedge-exit probability obtained directly from the strip, using conformal invariance of planar Brownian motion.

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.

For a Brownian path in the unit disc from $0$ conditioned to exit at $1$, let ${\Phi }_U:U\to D$ fix $0$ and $1$, and suppose $U$ agrees with the disc near $1$. The probability of avoiding $D\setminus U$ is the ratio of the boundary Poisson-kernel densities, hence $|{\Phi }_U^{\prime }(1)|$. Given avoidance, the conformally mapped path has the original law up to the conformal Brownian clock. There is no conformal-radius factor at the interior starting point.

For planar Brownian motion started at radius $\epsilon$ and stopped on first hitting radius one, the net winding count divided by $\log \epsilon$ converges to a standard Cauchy random variable divided by $2\pi$. The complex exponential maps Brownian coordinates in the left half-plane to the punctured unit disk after the conformal Brownian clock. The unwrapped argument at exit is the vertical half-plane exit coordinate, with the Cauchy exit law from a Brownian half-plane. Counting integer turns introduces only a uniformly bounded error in winding units.