Möbius calculation of circular Brownian exit
= Möbius calculation of circular Brownian exit
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{title2=$\psi_a(w)=(w-a)/(1-aw)$}
A <planar Brownian motion> beginning at $a\in(0,1)$ in the <unit disc> has circular exit law obtained by pulling back uniform angular measure through $\psi_a(w)=(w-a)/(1-aw)$. This <Möbius transformation> sends the starting point to zero. The right semicircle maps to an arc with endpoints at angles $\pm(\pi/2+2\arctan a)$, giving positive-half-plane exit probability $1/2+(2/\pi)\arctan a$.