connection between the Gaussian and the Cauchy distribution I have always been surprised by the fact that the quotient of two independent Gaussian random variables is a Cauchy Random variable - as this is often the case, coincidence in mathematics are not accidental: is there any deep explanations behind this connection between the Gaussian and the Cauchy distribution ?
other examples:


*

*if a $2$-dimensional Brownian motion $(X_t, Y_t)$ is started at $(0,1)$ and stopped the first time $T$ that it hits the real axis, then $X_T$ is also distributed as a Cauchy distribution.

*the Cauchy distribution also shows up when studying how a complex brownian motion winds around the origin.

 A: 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 $z\in \Omega$ 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 $t\in[0,T]$ 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 $\pi$. 
A: The bivariate distribution formed by two independent
normalized Gaussians is rotationally symmetric (think about the
usual argument for evaluating the probability integral). The
quotient of two random variables $X$ and $Y$ is the tangent of the angle
between $(0,0)$ and $(X,Y)$ with the $x$-axis. If one has a rotationally
symmetric distribution for $X$ and $Y$ (with no point mass at the origin)
then $Y/X$ is a tangent of a uniformly distributed angle. This is the Cauchy
distribution.
Added
Your example with the Brownian motion states in effect that if
$P$ is the first point that the motion hits the $x$-axis then the
angle between the line from $P$ to the starting point and
the $y$-axis is uniformly distributed between $-\pi$ and $\pi$.
I can't see any reason why this should be so, but perhaps someone (unlike me)
who actually knows something about Brownian motion might know why.
A: $$ F_Z(z)=\mathbb P(Z\leq z)=\mathbb P(Y/X\leq z)=\mathbb P(Y\leq zX)\\ =\mathbb P(Y\leq zX,\,X> 0)+ \mathbb P(Y\geq zX,\,X< 0),\,\, \mbox{that implies}\ f_Z(z)= \frac{dF_Z(z)}{dz}=\int_{-\infty}^{+\infty}|x|f_Y(zx)f_X(x)\, dx\ =\frac{1}{2\pi}\int_{-\infty}^{+\infty}|x|e^{-(z^2+1)x^2/2}\, dx=\frac{1}{\pi(x^2+1)}. $$
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