Past exam of the stochastic differential equations and Brownian motion in complex analysis course of the University of Oxford / 2007 / B / 8

The printed upper-tail identity cannot be a probability formula: it vanishes at zero and increases in $x$. Its correction isEquivalently the cumulative distribution function is $1/2+{\pi }^{-1}\arctan x$. Differentiating that cumulative function gives the densityThe normalization and sign can also be checked directly by conformal invariance of planar Brownian motion: the Cayley transform $(z-i)/(z+i)$ sends the starting point to the center of the unit disc. Its exit angle is uniform by rotational symmetry. The inverse on the unit circle gives $X=-\cot (\Theta /2)$, $\Theta$ uniform on $(0,2\pi )$, whose change-of-variable density is exactly the one above.

For a start $z_0=x_0+i{y}_0$, $y_0>0$, translation and positive scaling send it to $i$. Thus the exit coordinate has law $x_0+y_{0}C$, where $C$ is standard Cauchy. The corresponding formulas areThis is the upper-half-plane Poisson kernel. The printed density is correct; it is the preceding asserted tail from which it is meant to follow that needs correction.

Using the Cauchy distribution from the preceding part gives one common probability space for all starting points:As $(x,y)\to (0,0)$ with $y>0$, $x+yC\to 0$ almost surely, since $C$ is a finite real random variable almost surely. Continuity of $f$ at zero gives $f(x+yC)\to f(0)$ almost surely, and boundedness supplies an integrable constant bound. Dominated convergence theorem therefore provesThe argument applies to every sequence approaching zero, including tangential approaches. Along the real boundary the extension is already $f$, which is continuous. Hence the extension is continuous at zero in the closed half-plane. No finite moment of the Cauchy distribution is required.