pr.probability - connection between the Gaussian and the Cauchy distribution

Robin, a simple explanation for why the 2-dim Brownian motion stopped when hitting the real line is that Brownian motion is conformally invariant. Let $f:Omega rightarrow Omega'$ be a conformal mapping and $B_{z,Omega}(t)$ be a Brownian motion started at $zin Omega$ and stopped at the first time $T$ when it hits the boundary of $Omega$. The conformal invariance of Brownian motion is the fact that $f(B_{z,Omega}(t))$ for $tin[0,T]$ has the same distribution as a Brownian motion in $Omega'$ started at $f(z)$ and stopped when reaching the boundary of $Omega'$ for the first time.

To connect this with the problem above of a Brownian motion started at $(0,1)$ and stopped when hitting the real line, just map the upper half plane onto the unit circle in such a way that $(1,0)$ is mapped to the origin. A Brownian motion started from the center of the circle obviously hits the boundary of the circle and a uniformly distributed point $P'$ on the boundary of the circle. Thus, the angle of the line from the center of the circle to $P'$ with another fixed line through the center of the circle is uniformly distributed between $-pi$ and $pi$. Since the conformal map from the upper half-plane to the circle maps lines through $(0,1)$ to lines through the origin, then conformal invariance of Brownian motion implies that the angle between the $y$-axis and the line from $(0,1)$ to the point $P$ where the Brownian motion hits the $x$-axis is also uniform between $-pi$ and $pi$.

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