Let $B$ be <planar Brownian motion> started at $z\in D$, let $\tau_D$ be its exit time, and let $\phi:D\to D'$ be conformal. Define
$$
C_s=\int_0^{s\wedge\tau_D}|\phi'(B_r)|^2\,dr,
\qquad
\sigma_t=\inf\{s:C_s>t\}.
$$
The <conformal invariance of planar Brownian motion> states that
$$
\widetilde B_t=\phi(B_{\sigma_t}),
\qquad 0\leq t<C_{\tau_D},
$$
is Brownian motion started at $\phi(z)$ and stopped when it exits $D'$. Thus conformal maps preserve Brownian paths after this random time-change.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 203 / 2 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-203/2/a/solution