Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 29 / 4

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.

Put $L_r={\theta }_{T_r}$. The hitting times $T_r$ are finite almost surely by recurrence of planar Brownian motion, and increase with $r$. At $T_r$, the position is $e^{-r}{e}^{i{L}_r}$. By the Strong Markov property, Brownian scaling, and rotational invariance of planar Brownian motion, the future processis a standard planar Brownian motion started at $1$, independent of ${\mathcal{F}}_{T_r}$. The random rotation does not change its conditional law, since that law is rotationally invariant.

The first time this rescaled process reaches radius $e^{-h}$ corresponds to the original time $T_{r+h}$. Its continuous argument starts at zero and equals the increment of the original winding angle. HenceFor any ordered finite collection of levels, apply this assertion successively. Earlier winding values are measurable at the current hitting time, so all increments are independent and their laws depend only on the level differences. This proves the Lévy process property of winding at logarithmic radial passage levels.

One can also verify stochastic continuity and identify this process. The logarithmic-radius and argument martingales have the common clockThe same Cauchy-Riemann equations calculation as in part (a), applied locally to the logarithm and joined along the continuous argument, gives independent Brownian motions $U,V$ such that $\log R_t=U_{H_t}$ and ${\theta }_t=V_{H_t}$. This clock diverges: if it had a finite limit, convergence of martingales with finite total quadratic variation would make $\log R_t$ converge to a finite value, forcing the clock integral itself to diverge.

Thus, with ${\eta }_r=\inf \{u:U_u=-r\}$, we have $L_r=V_{{\eta }_r}$. The Brownian first-passage Laplace transform is $\mathbb{E}{e}^{-q{\eta }_r}=e^{-r\sqrt{2q}}$. It follows either from the Exponential martingale for Brownian motion stopped at ${\eta }_r\wedge n$, or from the one-dimensional exit equation. Conditioning on the independent process $U$ givesSo this is a symmetric Cauchy process, with level parameter $r$. The formula implies stochastic continuity at zero and, by stationary increments, everywhere. Under the usual definition requiring càdlàg paths, use the right-continuous passage clock ${\eta }_r^{+}=\inf \{u:U_u<-r\}$. It agrees with ${\eta }_r$ for each fixed level almost surely, and $V_{{\eta }_r^{+}}$ is the càdlàg modification. This distinction concerns simultaneous passage levels and does not alter any increment law.