connection between the Gaussian and the Cauchy distribution - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T06:03:19Z http://mathoverflow.net/feeds/question/34745 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/34745/connection-between-the-gaussian-and-the-cauchy-distribution connection between the Gaussian and the Cauchy distribution Alekk 2010-08-06T08:31:30Z 2010-08-21T14:27:12Z <p>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 ?</p> <p><strong>other examples:</strong></p> <ul> <li>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.</li> <li>the Cauchy distribution also shows up when studying how a complex brownian motion winds around the origin.</li> </ul> http://mathoverflow.net/questions/34745/connection-between-the-gaussian-and-the-cauchy-distribution/34746#34746 Answer by Robin Chapman for connection between the Gaussian and the Cauchy distribution Robin Chapman 2010-08-06T08:37:21Z 2010-08-06T10:40:19Z <p>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.</p> <p><strong>Added</strong> 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.</p> http://mathoverflow.net/questions/34745/connection-between-the-gaussian-and-the-cauchy-distribution/34765#34765 Answer by Jon Peterson for connection between the Gaussian and the Cauchy distribution Jon Peterson 2010-08-06T13:31:15Z 2010-08-06T13:31:15Z <p>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. </p> <p>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 $\pi$. </p> http://mathoverflow.net/questions/34745/connection-between-the-gaussian-and-the-cauchy-distribution/36279#36279 Answer by Attar Reda for connection between the Gaussian and the Cauchy distribution Attar Reda 2010-08-21T13:57:33Z 2010-08-21T13:57:33Z <p>Jon Peterson, you are correct. But why does it directly the calcul of cdf and pdf of $Y/X$, ie</p> <p>\begin{eqnarray*} F_Z(z)&amp;=&amp;\BBp(Z\leq z)=\BBp(Y/X\leq z)=\BBp(Y\leq zX)\ &amp;=&amp;\BBp(Y\leq zX,\,X> 0)+ \BBp(Y\geq zX,\,X&lt; 0),\,\, \mbox{that implies}\ f_Z(z)&amp;=&amp; \frac{dF_Z(z)}{dz}=\int_{-\infty}^{+\infty}|x|f_Y(zx)f_X(x)\, dx\ &amp;=&amp;\frac{1}{2\pi}\int_{-\infty}^{+\infty}|x|e^{-(z^2+1)x^2/2}\, dx=\frac{1}{\pi(x^2+1)}. \end{eqnarray*}</p> <p>The difficulty I encountered is how to prove that the characteristic function of the variable $Y / X$ is the same as the Cauchy distribution ?</p> http://mathoverflow.net/questions/34745/connection-between-the-gaussian-and-the-cauchy-distribution/36282#36282 Answer by Attar Reda for connection between the Gaussian and the Cauchy distribution Attar Reda 2010-08-21T14:20:41Z 2010-08-21T14:20:41Z <p>$$F_Z(z)=\BBp(Z\leq z)=\BBp(Y/X\leq z)=\BBp(Y\leq zX)\ =\BBp(Y\leq zX,\,X> 0)+ \BBp(Y\geq zX,\,X&lt; 0),\,\, \mbox{that implies}\ f_Z(z)= \frac{dF_Z(z)}{dz}=\int_{-\infty}^{+\infty}|x|f_Y(zx)f_X(x)\, dx\ =\frac{1}{2\pi}\int_{-\infty}^{+\infty}|x|e^{-(z^2+1)x^2/2}\, dx=\frac{1}{\pi(x^2+1)}.$$</p> <p>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</p> http://mathoverflow.net/questions/34745/connection-between-the-gaussian-and-the-cauchy-distribution/36283#36283 Answer by Attar Reda for connection between the Gaussian and the Cauchy distribution Attar Reda 2010-08-21T14:27:12Z 2010-08-21T14:27:12Z <p>$$F_Z(z)=P(Z\leq z)=P(Y/X\leq z)=P(Y\leq zX)\ =P(Y\leq zX,\ X> 0)+ P(Y\geq zX,\ X&lt; 0)$$ that implies $$f_Z(z)= \frac{dF_Z(z)}{dz}=\int_{-\infty}^{+\infty}|x|f_Y(zx)f_X(x)\, dx\ =\frac{1}{2\pi}\int_{-\infty}^{+\infty}|x|e^{-(z^2+1)x^2/2}\ dx=\frac{1}{\pi(x^2+1)}.$$ Againt, The difficulty I encountered is how to prove that the characteristic function of the variable YX is the same as the Cauchy distribution ?</p>