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