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.
  • $\begingroup$ The Gaussian and Cauchy families are both stable; were you looking for another reason? Considering both distributions as limits of the distribution of sums of random variables seems to make the fact you refer to a lot less surprising. en.wikipedia.org/wiki/Stable_distribution $\endgroup$ Aug 6, 2010 at 14:28

3 Answers 3


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.

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    $\begingroup$ In fact, the Gaussian distribution is, up to scaling, the unique distribution such that an independent product is rotationally invariant, and can be derived from that characterization. This is equivalent to the mysterious fact that a random walk on a rectangular grid renormalizes to something anisotropic; somehow the grid axes disappear. $\endgroup$
    – Tracy Hall
    Aug 6, 2010 at 10:27

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 $\pi$.

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    $\begingroup$ Why should the conformal map take lines to lines ( instead of circles). I don't think it does, and it doesn't preserve the angle. Instead it halves the angle, so that it is uniformly distributed on $[-\pi/2,\pi/2]$. $\endgroup$ Aug 6, 2010 at 18:17
  • $\begingroup$ George, you are correct about the lines and circles. It's been a while since I've reviewed my complex analysis I guess. I guess it's easiest just to note that a conformal map from the unit disc to the upper half-plane maps the uniform distribution on the circle to the Cauchy distribution. For instance, the complex mapping $z \mapsto i \frac{1-z}{1+z}$ maps $e^{i\theta}$ to $\tan(\theta/2)$. If $\theta$ is uniformly distributed on $[-\pi,\pi]$ then $\tan(\theta/2)$ has a Cauchy distribution. $\endgroup$ Aug 9, 2010 at 16:53
  • $\begingroup$ Jon Peterson, you are correct. But why does it directly the calcul of cdf and pdf of $Y/X$, ie \begin{eqnarray*} 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)}. \end{eqnarray*} The difficulty I encountered is how to prove that the characteristic function of the variable $ Y / X $ is the same as the Cauchy distribution ? $\endgroup$
    – Attar Reda
    Aug 21, 2010 at 13:57

$$ F_Z(z)=\mathbb P(Z\leq z)=\mathbb P(Y/X\leq z)=\mathbb P(Y\leq zX)\\ =\mathbb P(Y\leq zX,\,X> 0)+ \mathbb P(Y\geq zX,\,X< 0),\,\, \mbox{that implies}\ f_Z(z)= \frac{dF_Z(z)}{dz}=\int_{-\infty}^{+\infty}|x|f_Y(zx)f_X(x)\, dx\ =\frac{1}{2\pi}\int_{-\infty}^{+\infty}|x|e^{-(z^2+1)x^2/2}\, dx=\frac{1}{\pi(x^2+1)}. $$

I believe that the jsmath is the same as LaTeX, but I think really have a problem with jsmath, I have no time to learn it. So if my source does not work you can take this source .tex and compile it in LaTeX, you have my answer .. Thank you


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