Position of 2D Brownian motion exiting quarter plane

Question

Let $X_t = (X_t^1,X_t^2)$ a planar brownian motion without drift with independent components startet at $X_0 = (1,1)$ and $\tau := \inf \lbrace t\ge 0: X_t \notin (0,\infty)^2 \rbrace$ the first time the process leaves the positive quadrant. So one component of $X_\tau$ needs to be $0$. What is the distribution of the other one? Does this distribution have finite expectation?

Answer 1

Starting at any $(x_0,y_0)$, the exit distribution is the Cauchy distribution proportional to $1/((x-x_0)^2/y_0^2+1)$. You'll recognize this as the fundamental solution for $(x,0)\in \partial\mathbb{H}$ of the Laplace equation of the upper half plane $\mathbb{H}$, evaluated at $(x_0, y_0)$. Read any textbook on stochastic analysis to learn about this connection between Brownian motion and the Laplace equation.

EDIT: For some reason I misread the question as asking about the half plane. For the quarter plane, the solution is still found by finding the Green function for the corresponding Laplace equation, which can easily be googled