Consider a circle $\mathcal{C}$ in $\mathbf R^2$ of unit radius and a Brownian motion $\{X_t\}_{t\in \mathbf R^+} $ starting from its center. Let $\mu$ be the measure on $\mathcal{C}$ defined on arcs $I$ as $$ \mu(I) = Prob(X_\tau \in I), $$ where $\tau = \text{inf}_t (X_t \in \mathcal{C})$. Namely, $\mu(I)$ is the probability that $X$ leaves the circle through the arc $I$.
It is well known that $\mu$ is just Lebesgue measure on $\mathcal{C}$, simply by isotropy of Brownian motions.
My question is: what happens if we replace the circle with a square? Is there any closed formula for the probability density of the exit point of a Brownian motion from a square?
Guess: a friend of mine proposed this function $$ p(x) = \frac{\frac{1}{1+x^2} - \frac{1}{2}}{\frac{\pi}{2} - 1} $$ as probability density on a single side (perhaps a factor of $1/4$ is needed to normalize), where $x$ is distance from the middle point of the square side.
