Cauchy Distribution

• Parameter
  • $a$
• CDF
  • $\frac{1}{\pi }\left(\arctan \frac{x}{a}+\frac{\pi }{2}\right)$
• PDF
  • $\frac{1}{\pi }\frac{a}{x^2+a^2}$
• Mean
  • Undefined
• Variance
  • Undefined
• MGF
  • Undefined
• Characteristic Function
  • $e^{-a|t|}$

Characteristic Function

The Characteristic Function of a Cauchy Random Variable is

$$\varphi (t)={\int }_{\mathbb{R}}{e}^{ixt}\frac{a}{\pi }\frac{1}{x^2+a^2}dx.$$

From Residue Theorem, we know that

$$\varphi (t)=2\pi i\frac{a}{\pi }\frac{e^{ix|t|}}{\frac{\mathrm{d}}{\mathrm{d}x}{x}^2+a^2}{|}_{x=ai}=e^{-a|t|}.$$

Note that we need different semi-circle contour for $t\lessgtr 0$.

Problem

Consider $P(0,a)\in {\mathbb{R}}^2$. A line passing $P$ has the angle with the y-axis $\theta \sim \mathcal{U}\left(-\frac{\pi }{2},\frac{\pi }{2}\right)$. Let $X$ be the intersection of the line with x-axis. Then the distribution of $X$ is Cauchy distribution with parameter $a$.

Reproducibility and Sample Mean

Just like MGF, a CF uniquely determine a distribution. Let $\{X_i{\}}_{i=1}^n$ be a set of Cauchy random variables with parameter $\{a_i\}$. Then we have

$$\mathbb{E}\left[e^{it{\sum }_{i=1}^n{X}_i}\right]=\prod _{i=1}^n\mathbb{E}[e^{it{X}_i}]=\prod _{i=1}^n{e}^{-a_i|t|}=e^{-\left({\sum }_{i=1}^n{a}_i\right)|t|}.$$

Thus, ${\sum }_{i=1}^n{X}_i$ is also a Cauchy random variable, with parameter ${\sum }_{i=1}^n{a}_i$.

The sample mean r.v. of $\{X_i\}$ is $\overline{X}=\frac{{\sum }_{i=1}^n{X}_i}{n}$. And we have

$$\mathbb{E}[e^{it\overline{X}}]=\prod _{i=1}^n\mathbb{E}[e^{it{X}_i/n}]=\prod _{i=1}^n{e}^{-a_i|t/n|}=e^{-\left({\sum }_{i=1}^n{a}_i\right)/n|t|}.$$

Thus, $\overline{X}$ is also a Cauchy random variable, with parameter ${\sum }_{i=1}^n{a}_i/n$. Specifically, when $a_i$ are the same, then $\overline{X}$‘s parameter is also $a$.

PDF Approach

Another way to show that the sample mean of Cauchy r.v.s is still Cauchy is to calculate its PDF. We start with $(X_1+X_2)/2$. We know the PDF of $X_1/2$ is $2f_1(2x):={\hat{f}}_1(x)$, where $f_1$ is the PDF of $X_1$. Therefore, the PDF of $(X_1+X_2)/2$ is

$$\begin{array}{rl}f(x)& ={\hat{f}}_1\ast {\hat{f}}_2(x)\\ & =\frac{4a_1{a}_2}{{\pi }^2}{\int }_{\mathbb{R}}\frac{1}{((2t{)}^2+a_1^2)((2(x-t){)}^2+a_2^2)}dt\\ & =\frac{4a_1{a}_2}{{\pi }^2}\frac{2x{a}_1{a}_2(\log (4t^2+a_1^2)-\log (4(x-t{)}^2+a_2^2))+a_1(4x^2+a_1^2-a_2^2){\tan }^{-1}\left(\frac{2(t-x)}{a_2}\right)+a_2(4x^2+a_2-a_1){\tan }^{-1}\left(\frac{2x}{a_1}\right)}{2a_1{a}_2(16x^4+8x^2(a_1^2+a_2^2)+(a_1^2-a_2^2{)}^2)}{|}_{t=-\infty }^{\infty }\\ & =\frac{2}{{\pi }^2(16x^4+8x^2(a_1^2+a_2^2)+(a_1^2-a_2^2{)}^2)}\pi (a_1(4x^2+a_1^2-a_2^2)+a_2(4x^2+a_2^2-a_1^2))\\ & =\frac{2}{\pi }\frac{a_1+a_2}{4x^2+(a_1+a_2{)}^2}\\ & =\frac{\frac{a_1+a_2}{2}}{\pi }\frac{1}{x^2+{\left(\frac{a_1+a_1}{2}\right)}^2}\end{array}$$

That is, $(X_1+X_2)/2$ is a Cauchy r.v. with parameter $(a_1+a_2)/2$. By deduction, we know that the sample mean of Cauchy r.v.s is still Cauchy.