## 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
```
$$
\begin{equation*}
f(x; \mu, \lambda) =\left\{
\begin{array}{ll}
\frac{\lambda}{\pi}\cdot \frac{1}{\lambda^2+(x-\mu)^2}, & \hbox{$-\infty < x< \infty$;} \\
& \hbox{$-\infty < \mu< \infty$, $\lambda>0$;} \\
0, & \hbox{Otherwise.}
\end{array}
\right.
\end{equation*}
$$
```

## Distribution Function of Cauchy Distribution

The distribution function of Cauchy distribution is

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

## Example

Let $X\sim C(2,4)$. Find the probability that

a. $X$ is less than 3,

b. $X$ is greater than 4,

c. $X$ is between 1 and 3.5

### Solution

The distribution function of Cauchy distribution is

```
$$
\begin{aligned}
F(x) &= \frac{1}{2}+ \frac{1}{\pi}tan^{-1}\big(\frac{x-\mu}{\lambda}\big)\\
&=0.5+\frac{1}{\pi} tan^{-1}\big(\frac{x-2}{4}\big)
\end{aligned}
$$
```

a. The probability that $X$ is less than $3$ is

```
$$
\begin{aligned}
P(X \leq 3) &=F(3)\\
&=0.5+\frac{1}{\pi} tan^{-1}\big(\frac{3-2}{4}\big)\\
&=0.5 + \frac{1}{3.1416}tan^{-1}\big(0.25\big)\\
&=0.5 + \frac{1}{3.1416}(0.245)\\
&= 0.578
\end{aligned}
$$
```

b. The probability that $X$ is greater than $4$ is

```
$$
\begin{aligned}
P(X >4) &=1- P(X<4)\\
&= 1- F(4)\\
&=1-\bigg(0.5+\frac{1}{\pi} tan^{-1}\big(\frac{4-2}{4}\big)\bigg)\\
&=0.5 - \frac{1}{3.1416}tan^{-1}\big(0.5\big)\\
&=0.5 - \frac{1}{3.1416}(0.4636)\\
&= 0.3524
\end{aligned}
$$
```

c. The probability that $X$ is between $1$ and $3$ is

```
$$
\begin{aligned}
P(1 \leq X \leq 3)&=P(X\leq 3)-P(X\leq 1)\\
&=F(3) -F(1)\\
&=\bigg[0.5+\frac{1}{\pi} tan^{-1}\big(\frac{3-2}{4}\big)\bigg]-\bigg[0.5+\frac{1}{\pi} tan^{-1}\big(\frac{1-2}{4}\big)\bigg]\\
&=\frac{1}{\pi} tan^{-1}\big(0.25\big)-\frac{1}{\pi} tan^{-1}\big(-0.25\big)\\
&=\frac{1}{3.1416}(0.245)-\frac{1}{3.1416}(-0.245)\\
&=0.156
\end{aligned}
$$
```