= Brownian exit law from a quadrant
{c}
{title2=$q_a(r)=\frac{4a^2r}{\pi(r^4+4a^4)}$}
Planar <Brownian motion> started at $a+ia$, $a>0$, exits the positive quadrant through either positive coordinate axis with <probability> one-half. The density along each axis is $q_a(r)$ for $r>0$, measured with respect to distance along that axis. Squaring maps the quadrant to the upper half-plane and sends the starting point to $2ia^2$. The boundary square therefore has the centered <Cauchy distribution> of scale $2a^2$; pulling it back by $r\mapsto r^2$ gives the displayed density. The radius has distribution function $2\arctan(r^2/(2a^2))/\pi$, independently of which axis is chosen. There is no atom at the origin.
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