Past exam of the mathematics course of the University of Cambridge 2008 iii Paper 37 4 c Solution Created 2026-10-03 Updated 2026-10-14
Start the planar Brownian motion at and stop at . The complex exponential is holomorphic with nonzero derivative. By conformal invariance of planar Brownian motion, the process , run on its conformal Brownian clockis planar Brownian motion started at , until its first exit from the unit disk. The clock is strictly increasing before , and 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 , ; at it equals one. Hence the disk exit time in the new clock is exactly .
The continuous lifted argument of this path, starting from angle zero, is . Its total argument at exit is therefore , which has distribution by part (b). If the winding count records only completed integer turns, it differs from by a remainder with . If fractional turns are retained, that remainder is zero. In either convention, since ,The first term has distribution for every , and the second tends to zero deterministically in absolute value as . Symmetry of the Cauchy distribution removes the minus sign, so Slutsky theorem gives the small-radius Brownian winding law: