Timeline for connection between the Gaussian and the Cauchy distribution
Current License: CC BY-SA 2.5
4 events
| when toggle format | what | by | license | comment | |
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| Aug 21, 2010 at 13:57 | comment | added | Attar Reda | 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 ? | |
| Aug 9, 2010 at 16:53 | comment | added | Jon Peterson | 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. | |
| Aug 6, 2010 at 18:17 | comment | added | George Lowther | 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 $[-\pi/2,\pi/2]$. | |
| Aug 6, 2010 at 13:31 | history | answered | Jon Peterson | CC BY-SA 2.5 |