# Cauchy Distribution

## Cauchy Distribution

A continuous random variable $X$ is said to follow Cauchy distribution with parameters $\mu$ and $\lambda$ if its probability density function is given by

 $$\begin{eqnarray}\label{eqch1} f(x; \mu, \lambda) &=& \begin{cases} \frac{\lambda}{\pi}\cdot \frac{1}{\lambda^2+(x-\mu)^2}, & -\infty < x < \infty; \\ & -\infty < \mu < \infty, \lambda>0; \\ 0, & Otherwise. \end{cases} \end{eqnarray}$$

In notation it can be written as $X\sim C(\mu, \lambda)$. The parameter $\mu$ and $\lambda$ are location and scale parameters respectively.

Here $f(x;\mu, \lambda)>0$ for all $x$. Now we check that $\int_{-\infty}^\infty f(x;\mu, \lambda)\; dx =1$.

 $$\begin{eqnarray*} \int_{-\infty}^\infty f(x;\mu, \lambda)\; dx &=& \int_{-\infty}^\infty\frac{\lambda}{\pi}\cdot \frac{1}{\lambda^2+(x-\mu)^2} \; dx \\ &=& \frac{\lambda}{\pi}\int_{-\infty}^\infty \frac{1}{\lambda^2+(x-\mu)^2} \; dx \\ &=& \frac{\lambda}{\pi}\bigg[\frac{1}{\lambda}\tan^{-1}\bigg(\frac{x-\mu}{\lambda}\bigg)\bigg]_{-\infty}^\infty \\ &=& \frac{1}{\pi}\bigg[\tan^{-1}(\infty)-tan^{-1}(-\infty)\bigg]\\ &=& \frac{1}{\pi}\bigg[\frac{\pi}{2}-\bigg(-\frac{\pi}{2}\bigg)\bigg]\\ &=& \frac{1}{\pi}(\pi)=1. \end{eqnarray*}$$

Hence, function defined in \eqref{eqch1} is p.d.f.

## Graphs of Cauchy Distribution

Following is the graph of probability density function of Cauchy distribution for different values of $\mu$ by keeping $\lambda$ constant.

It is clear that as we change the value of $\mu$, the location of the graph changes.

## Graph of Cauchy distribution

Following is the graph of probability density function of Cauchy distribution for different values of $\lambda$ by keeping $\mu$ constant.

It is clear that as we change the value of $\lambda$, the scale of the graph changes.

## Standard Cauchy Distribution

If $\mu=0$ and $\lambda =1$, then the distribution is called Standard Cauchy Distribution. The p.d.f. of standard Cauchy distribution is

 \begin{align*} f(x) &= \begin{cases} \frac{1}{\pi}\cdot \frac{1}{1+x^2}, & -\infty < x < \infty; \\ 0, & Otherwise. \end{cases} \end{align*}

## Mean and Variance of Cauchy Distribution

Cauchy distribution does not possesses finite moments of order greater than or equal to 1. Hence, mean and variance does not exists for Cauchy distribution.

## Median of Cauchy Distribution

If $M$ is the median of the distribution, then then median of Cauchy distribution is $\mu$.

#### Proof

Let $M$ be the median of Cauchy distribution. Then

 $$\begin{equation*} \int_{-\infty}^M f(x)\; dx =\frac{1}{2}. \end{equation*}$$

Let $\frac{x-\mu}{\lambda} = z$ $\Rightarrow$ $dx = \lambda\; dz$.

$x=-\infty$ $\Rightarrow z = -\infty$ and $x = M \Rightarrow z=\frac{M-\mu}{\lambda}$. Hence, we have

 $$\begin{eqnarray*} & & \int_{-\infty}^M f(x)\; dx =\frac{1}{2} \\ &\Rightarrow& \frac{1}{\pi}\int_{-\infty}^{\frac{M-\mu}{\lambda}}\frac{dz}{1+z^2} =\frac{1}{2} \\ &\Rightarrow&\frac{1}{\pi} \bigg[\tan^{-1}z\bigg]_{-\infty}^{\frac{M-\mu}{\lambda}}=\frac{1}{2}\\ &\Rightarrow&\frac{1}{\pi} \bigg[\tan^{-1}\frac{M-\mu}{\lambda}-\tan^{-1}(-\infty)\bigg]=\frac{1}{2}\\ &\Rightarrow&\frac{1}{\pi} \bigg[\tan^{-1}\frac{M-\mu}{\lambda}-\bigg(-\frac{\pi}{2}\bigg)\bigg]=\frac{1}{2}\\ &\Rightarrow&\frac{1}{\pi} \tan^{-1}\frac{M-\mu}{\lambda}=0\\ &\Rightarrow& \frac{M-\mu}{\lambda}=0\\ & \Rightarrow & M =\mu. \end{eqnarray*}$$

Hence, median of the Cauchy distribution is $\mu$.

## Mode of Cauchy Distribution

The mode of Cauchy distribution is $\mu$.

#### Proof

Mode is the value at which density function becomes maximum.

Taking $\log_e$ on both the sides of \eqref{eqch1}, we have

 $$$$\label{eqch2} \log_e f(x) = \log (\lambda/\pi) - \log (\lambda^2+(x-\mu)^2)$$$$

By the principle of maxima and minima, differentiating \eqref{eqch2} w.r.t. $x$, and equating to zero, we get

 $$\begin{eqnarray*} \frac{\partial \log_e f(x)}{\partial x} &=& 0\\ \Rightarrow & & \frac{-2(x-\mu)}{\lambda^2+(x-\mu)^2} =0 \\ \Rightarrow & & (x-\mu)=0\\ \Rightarrow & & x =\mu. \end{eqnarray*}$$

And

 $$\begin{eqnarray*} \frac{\partial^2 \log_e f(x)}{\partial x^2} &=& -2\bigg[\frac{[\lambda^2+(x-\mu)^2] (1)-(x-\mu)[2(x-\mu)]}{[\lambda^2+(x-\mu)^2]^2}\bigg]_{x=\mu} \\ &=& -2\bigg[\frac{[\lambda^2+(x-\mu)^2] (1)-2(x-\mu)^2}{[\lambda^2+(x-\mu)^2]^2}\bigg]_{x=\mu} \\ &=& -2\bigg[\frac{\lambda^2}{\lambda^4}\bigg] = \frac{-2}{\lambda^2} < 0. \end{eqnarray*}$$

Hence, $f(x)$ is maximum at $x=\mu$. Therefore, mode of Cauchy distribution is $\mu$.

## Quartiles of Cauchy Distribution

First Quartile :

If $Q_1$ is the first quartile of the distribution, then

 $$\begin{equation*} P(X\leq Q_1)=\int_{-\infty}^{Q_1} f(x)\; dx =\frac{1}{4}. \end{equation*}$$

Let $\frac{x-\mu}{\lambda} = z$ $\Rightarrow$ $dx = \lambda\; dz$.

$x=-\infty$ $\Rightarrow z = -\infty$ and $x = Q_1 \Rightarrow z=\frac{Q_1-\mu}{\lambda}$. Hence, we have

 $$\begin{eqnarray*} \int_{-\infty}^{Q_1} f(x)\; dx =\frac{1}{4} &\Rightarrow& \frac{1}{\pi}\int_{-\infty}^{\frac{Q_1-\mu}{\lambda}}\frac{dz}{1+z^2} =\frac{1}{4} \\ &\Rightarrow&\frac{1}{\pi} \bigg[\tan^{-1}z\bigg]_{-\infty}^{\frac{Q_1-\mu}{\lambda}}=\frac{1}{4}\\ &\Rightarrow&\frac{1}{\pi} \bigg[\tan^{-1}\frac{Q_1-\mu}{\lambda}-\tan^{-1}(-\infty)\bigg]=\frac{1}{4}\\ &\Rightarrow&\frac{1}{\pi} \bigg[\tan^{-1}\frac{Q_1-\mu}{\lambda}-\bigg(-\frac{\pi}{2}\bigg)\bigg]=\frac{1}{4} \end{eqnarray*}$$

 $$\begin{eqnarray*} &\Rightarrow& \tan^{-1}\frac{Q_1-\mu}{\lambda}=-\frac{\pi}{4}\\ &\Rightarrow& \frac{Q_1-\mu}{\lambda}=\tan\bigg(-\frac{\pi}{4}\bigg)=-1 \; \Rightarrow Q_1 =\mu-\lambda. \end{eqnarray*}$$

Therefore for the Cauchy distribution, first quartile $Q_1=\mu-\lambda$.

Third Quartile :

If $Q_3$ is the second quartile of the distribution, then

 $$\begin{equation*} P(X\leq Q_3)=\int_{-\infty}^{Q_3} f(x)\; dx =\frac{3}{4}. \end{equation*}$$

Let $\frac{x-\mu}{\lambda} = z$ $\Rightarrow$ $dx = \lambda\; dz$.

$x=-\infty$ $\Rightarrow z = -\infty$ and $x = Q_3 \Rightarrow z=\frac{Q_3-\mu}{\lambda}$. Hence, we have

 $$\begin{eqnarray*} \int_{-\infty}^{Q_3} f(x)\; dx =\frac{3}{4} &\Rightarrow& \frac{1}{\pi}\int_{-\infty}^{\frac{Q_3-\mu}{\lambda}}\frac{dz}{1+z^2} =\frac{3}{4} \\ &\Rightarrow&\frac{1}{\pi} \bigg[\tan^{-1}z\bigg]_{-\infty}^{\frac{Q_3-\mu}{\lambda}}=\frac{3}{4}\\ &\Rightarrow&\frac{1}{\pi} \bigg[\tan^{-1}\frac{Q_3-\mu}{\lambda}-\tan^{-1}(-\infty)\bigg]=\frac{3}{4}\\ &\Rightarrow&\frac{1}{\pi} \bigg[\tan^{-1}\frac{Q_3-\mu}{\lambda}-\bigg(-\frac{\pi}{2}\bigg)\bigg]=\frac{3}{4} \end{eqnarray*}$$

 $$\begin{eqnarray*} &\Rightarrow& \tan^{-1}\frac{Q_3-\mu}{\lambda}=\frac{\pi}{4}\\ &\Rightarrow& \frac{Q_3-\mu}{\lambda}=\tan\bigg(\frac{\pi}{4}\bigg)=1 \; \Rightarrow Q_3 =\mu+\lambda. \end{eqnarray*}$$

Therefore for the Cauchy distribution, third quartile $Q_3=\mu+\lambda$.

Hence the Quartile Deviation for Cauchy distribution is

 $$\begin{equation*} QD = \frac{Q_3-Q_1}{2} =\frac{\mu+\lambda-(\mu-\lambda)}{2}=\lambda. \end{equation*}$$

## Distribution Function of Cauchy Distribution

The distribution function of Cauchy distribution is

 $$\begin{equation*} F(x) =\frac{1}{\pi}\tan^{-1}\bigg(\frac{x-\mu}{\lambda}\bigg) + \frac{1}{2}. \end{equation*}$$

#### Proof

The distribution function of Cauchy distribution is

 $$\begin{equation*} F(x)=P(X\leq x) = \int_{-\infty}^x f(x)\; dx. \end{equation*}$$

Let $\frac{x-\mu}{\lambda} = z$ $\Rightarrow$ $dx = \lambda\; dz$.

$x=-\infty$ $\Rightarrow z = -\infty$ and $x = x \Rightarrow z=\frac{x-\mu}{\lambda}$. Hence, we have

 $$\begin{eqnarray*} F(x) & = & \frac{1}{\pi}\int_{-\infty}^{\frac{x-\mu}{\lambda}}\frac{dz}{1+z^2}\\ & = &\frac{1}{\pi} \bigg[\tan^{-1}z\bigg]_{-\infty}^{\frac{x-\mu}{\lambda}}\\ & = &\frac{1}{\pi} \bigg[\tan^{-1}\bigg(\frac{x-\mu}{\lambda}\bigg)-\tan^{-1}(-\infty)\bigg]\\ & = &\frac{1}{\pi} \bigg[\tan^{-1}\bigg(\frac{x-\mu}{\lambda}\bigg)-\bigg(-\frac{\pi}{2}\bigg)\bigg]\\ & = & \frac{1}{\pi}\tan^{-1}\bigg(\frac{x-\mu}{\lambda}\bigg) + \frac{1}{2}. \end{eqnarray*}$$

The distribution function of Cauchy distribution is

 $$\begin{equation*} F(x) =\frac{1}{\pi}\tan^{-1}\bigg(\frac{x-\mu}{\lambda}\bigg) + \frac{1}{2}. \end{equation*}$$

## MGF of Cauchy Distribution

Moment generating function of Cauchy distribution does not exists

## Characteristics Function of Cauchy Distribution

If $X$ is standard Cauchy variate then its characteristics function is

 $$$$\label{cfc1} \phi_X(t) = \int_{-\infty}^\infty e^{itx} f(x)\; dx=\frac{1}{\pi}\int_{-\infty}^\infty \frac{e^{itx}}{1+x^2}\; dx.$$$$

Consider the p.d.f. of standard Laplace distribution

 $$\begin{equation*} f_1(z) = \frac{1}{2} e^{-|z|}, \; \; -\infty < z < \infty. \end{equation*}$$

The characteristics function of $Z$ is

 $$\begin{equation*} \phi_1(t) = E(e^{itz}) = \frac{1}{1+t^2}. \end{equation*}$$

Here, $\phi_1(t)$ is absolutely integrable in $(-\infty, \infty)$, we have by Inversion Theorem,

 $$\begin{eqnarray*} f_1(z) &=& \frac{1}{2\pi} \int_{-\infty}^\infty e^{-itz} \phi_1(t)\;dt \\ &=& \frac{1}{2\pi} \int_{-\infty}^\infty \frac{e^{-itz}}{1+t^2}\; dt\\ \therefore \frac{1}{2}e^{-|z|} & = &\frac{1}{2\pi} \int_{-\infty}^\infty \frac{e^{-itz}}{1+t^2}\; dt\\ e^{-|z|} & = &\frac{1}{\pi} \int_{-\infty}^\infty \frac{e^{itz}}{1+t^2}\; dt\qquad \text{ changing t to -t} \end{eqnarray*}$$

Interchanging $t$ and $z$, we get

 $$$$\label{cfc2} e^{-|t|} = \frac{1}{\pi} \int_{-\infty}^\infty \frac{e^{itz}}{1+z^2}\; dz$$$$

From \eqref{cfc1} and \eqref{cfc2}, we get
 $$\begin{equation*} \phi_X(t) = e^{-|t|}. \end{equation*}$$

Let $Y\sim C(\mu, \lambda)$, then $X=\frac{Y-\mu}{\lambda}\sim C(0,1)$. Hence, $Y=\mu + \lambda X$ has a characteristics function

 $$\begin{eqnarray*} \phi_Y(t)& = & E(e^{itY})\\ & = & e^{i\mu t}E(e^{it\lambda X}) \\ & = & e^{i\mu t}\phi_X(t\lambda)\\ & =& e^{i\mu t}e^{-\lambda|t|}\\ & =& e^{i\mu t -\lambda|t|}. \end{eqnarray*}$$

If $X_1$ and $X_2$ are two independent Cauchy random variables with parameters $(\mu_1, \lambda_1)$ and $(\mu_2, \lambda_2)$ respectively, then $X_1+X_2$ is a Cauchy random variable with parameter $(\mu_1+\mu_2, \lambda_1+\lambda_2)$.

#### Proof

The characteristics function of $X_1$ and $X_2$ are $\phi_{X_1}(t) =e^{i\mu_1 t -\lambda_1|t|}$ and $\phi_{X_1}(t) = e^{i\mu_2 t-\lambda_2|t|}$ respectively.

Hence the characteristics function of $X_1+X_2$ is

 $$\begin{eqnarray*} \phi_{X_1+X_2} (t) &=& \phi_{X_1}(t)\cdot \phi_{X_2}(t), \quad \text{ X_1 and X_2 are independent} \\ &=& e^{i\mu_1 t -\lambda_1|t|}\cdot e^{i\mu_2 t-\lambda_2|t|}\\ & = & e^{i(\mu_1+\mu_2) t -(\lambda_1+\lambda_2)|t|}. \end{eqnarray*}$$

Hence by uniqueness theorem of characteristics function, $X_1+X_2$ follows Cauchy distribution with parameters $\mu_1+\mu_2$ and $\lambda_1+\lambda_2$.

## Generalization of additive property

Let $X_1, X_2, \cdots, X_n$ be a sample of $n$ independent observations from Cauchy distribution such that $X_j \sim C(\lambda_j, \mu_j)$, $j=1,2\cdots, n$. Then $\overline{X}$, sample mean follows a Cauchy distribution with parameters $\overline{\lambda}$ and $\overline{\mu}$, where $\overline{\lambda} = \frac{1}{n}\sum_{j=1}^n \lambda_j$ and $\overline{\mu} = \frac{1}{n}\sum_{j=1}^n \mu_j$.

## Distribution of Reciprocal of Standard Cauchy Distribution

Let $X$ be a random variable having Cauchy distribution with parameter $\mu=0$ and $\lambda=1$. Find the distribution of $\frac{1}{X}$. Identify the distribution.

#### Solution

$X\sim C(0, 1)$. Therefore, the p.d.f. of $X$ is $f(x) =\frac{1}{\pi}\cdot \frac{1}{1+x^2}$, $-\infty < x < \infty$. Let $y=\frac{1}{x}$. Therefore $dx = -\frac{1}{y^2}dy$. Hence, p.d.f. of $Y$ is

 $$\begin{eqnarray*} f(y) &=& f(x)\cdot\bigg|\frac{dx}{dy} \bigg|\\ &=& \frac{1}{\pi}\cdot \frac{1}{1+(1/y^2)}\cdot \frac{1}{y^2}\\ &=& \frac{1}{\pi}\cdot \frac{1}{1+y^2},\quad -\infty < y < \infty, \end{eqnarray*}$$

which is the p.d.f. of standard Cauchy distribution. Hence $Y\sim C(0,1)$ distribution.

## Example of Cauchy Distribution

Let $X\sim C(\mu, \lambda)$. Let $X_1$ and $X_2$ are two independent observations on $X$. Find

(a) $P[X_1+X_2 > 2(\mu+\lambda)]$
(b) $P[X_1+X_2 < 2(\mu-\lambda)]$
(c) $P[2(\mu-\lambda) < X_1+X_2 < 2(\mu+\lambda)]$.

#### Solution

Let $X_1\sim C(\mu,\lambda)$ and $X_2\sim C(\mu, \lambda)$. Moreover $X_1$ and $X_2$ are independent. Hence $\frac{X_1+X_2}{2}=\overline{X} \sim C(\mu, \lambda)$. Hence

 $$\begin{eqnarray*} P[X_1+X_2> 2(\mu+\lambda)] &=& P[\overline{X}> (\mu+\lambda)] \\ &=& P[\overline{X}> Q_3],\quad \text{ since Q_3=\mu+\lambda} \\ &=& 1-P[\overline{X}\leq Q_3]\\ &=& 1-\frac{3}{4}=\frac{1}{4}. \end{eqnarray*}$$

## Distribution of Square of standard Cauchy distribution

A random variable $X$ has a standard Cauchy distribution. Find the p.d.f. of $X^2$ and identify the distribution.

#### Solution

Let $X\sim C(0,1)$. The p.d.f. of $X$ is $f(x) =\frac{1}{\pi}\cdot \frac{1}{1+x^2}$, $-\infty < x< \infty$. Let $Y=X^2$.

Distribution function of $Y$ is given by

 $$\begin{eqnarray*} G(y) &=& P(Y\leq y) \\ &=& P(X^2\leq y)\\ &=& P(-\sqrt(y)\leq X\leq \sqrt{y})\\ &=& \int_{-\sqrt{y}}^{\sqrt{y}}f(x) \; dx\\ &=& \frac{1}{\pi}\int_{-\sqrt{y}}^{\sqrt{y}}\frac{1}{1+x^2} \; dx\\ &=& \frac{2}{\pi}\int_{0}^{\sqrt{y}}\frac{1}{1+x^2} \; dx\\ &=& \frac{2}{\pi}\big[\tan^{-1}x\big]_{0}^{\sqrt{y}}\\ &=& \frac{2}{\pi}\cdot\tan^{-1}\sqrt{y}. \end{eqnarray*}$$

Hence, p.d.f. of $Y$ is

 $$\begin{eqnarray*} g(y) &=& \frac{d}{dy} G(y) \\ &=& \frac{2}{\pi}\cdot\frac{1}{1+(\sqrt{y})^2}\cdot \frac{1}{2\sqrt{y}}\\ &=& \frac{2}{\pi}\cdot\frac{y^{-1/2}}{1+y}\\ &=& \frac{1}{\pi}\cdot\frac{y^{\frac{1}{2}-1}}{(1+y)^{\frac{1}{2}+\frac{1}{2}}}\\ &=& \frac{1}{B\big(\frac{1}{2},\frac{1}{2}\big)}\cdot\frac{y^{\frac{1}{2}-1}}{(1+y)^{\frac{1}{2}+\frac{1}{2}}} \end{eqnarray*}$$

which is the p.d.f. of $\beta$ distribution of second kind with $m=\frac{1}{2}, n=\frac{1}{2}$, i.e., $F$ distribution with (1,1) degrees of freedom. Hence, $Y=X^2 \sim \beta_2\big(\frac{1}{2},\frac{1}{2}\big)= F(1,1)$.

## Ratio of two independent standard normal is Cauchy

Let $X\sim N(0,1)$ and $Y\sim N(0,1)$ be independent random variable. Find the distribution of $X/Y$.

#### Solution

Let $X\sim N(0,1)$ and $Y\sim N(0,1)$. Moreover, $X$ and $Y$ are independently distributed. Therefore, the joint p.d.f. of $X$ and $Y$ is given by

 $$\begin{eqnarray*} f(x,y) &=& f(x)\cdot f(y) \\ &=& \frac{1}{2\pi} e^{-\frac{1}{2}(x^2+y^2)}. \end{eqnarray*}$$

Let $U=X/Y$ and $V=Y$. Hence $X=UV$ and $Y=V$. Hence, the Jacobian of the transformation is

 $$\begin{equation*} J=\bigg|\frac{\partial(x,y)}{\partial(u,v)}\bigg|=\bigg| \begin{array}{cc} v & u \\ 0 & 1 \\ \end{array} \bigg|=|v|. \end{equation*}$$

Hence the joint p.d.f. of $U$ and $V$ is

 $$\begin{eqnarray*} f(u,v) &=& f(x,y)\cdot |J| \\ &=& \frac{1}{2\pi} e^{-\frac{1}{2}(u^2v^2+v^2)}|v|. \end{eqnarray*}$$

Hence, the marginal p.d.f. of $U$ is

 $$\begin{eqnarray*} f(u) &=& \int_{-\infty}^\infty f(u,v) \; dv \\ &=& \frac{1}{2\pi} \int_{-\infty}^\infty e^{-\frac{1}{2}(u^2+1)v^2}|v|\; dv\\ &=& \frac{1}{\pi} \int_{0}^\infty e^{-\frac{1}{2}(u^2+1)v^2}v\; dv\\ & & \qquad \qquad \int_{0}^\infty e^{-ax^2}x^{2n-1}\; dx=\frac{\Gamma(n)}{2a^n}\\ &=& \frac{1}{\pi} \frac{\Gamma(2)}{(2(1+u^2)/2)^{2/2}}\\ &=& \frac{1}{\pi} \frac{1}{1+u^2}. \end{eqnarray*}$$

which is the p.d.f. of Cauchy distribution with parameter $\mu=0$ and $\lambda =1$. Hence $X/Y \sim C(0,1)$ distribution.

## Conclusion

In this tutorial, you learned about theory of Cauchy distribution like the probability mass function, median, mode, quartiles, characteristics function and other properties of Cauchy distribution.

To read more about the step by step examples and calculator for Cauchy distribution refer the link Cauchy Distribution Calculator with Examples. This tutorial will help you to understand how to calculate how to calculate probabilities and cumulative probabilities for Cauchy distribution with the help of step by step examples.