Cauchy random variable question If X is a Cauchy random variable it has the density function as follows:
$f_X(x):= \frac{1}{\pi}\cdot\frac{1}{1+(x-\Theta)^2 }$ con $x \in R$ e $- \Theta<x<\Theta$
I have to demostrate that the random variable $Y=\frac{1}{X}$ is a Cauchy random variable too.
$$F_Y (a)=P(Y<a)=P(\frac{1}{X}<a)=P(X>\frac{1}{a})=1-F_x(\frac{1}{a})=1- \int_{- \infty}^{\frac{1}{a}} \frac{1}{\pi}\cdot\frac{1}{1+(x-\Theta)^2 }\, dx=1-\frac{1}{\pi} *[arctg(x- \Theta)]_{- \infty}^{\frac{1}{a}} =1-\frac{1}{\pi} *[arctg(\frac{1}{a} - \Theta)-(-\frac{\pi}{2}) ]=1-\frac{1}{\pi} *(\frac{\pi}{2}+arctg(\frac{1}{a} - \Theta) )=1-\frac{1}{2} -\frac{1}{\pi}*arctg(\frac{1}{a} - \Theta )=\frac{1}{2} -\frac{1}{\pi}*arctg(\frac{1}{a} - \Theta )$$
Throught the identity $\frac{\pi}{2}=\arctan (x)+\arctan (\frac{1}{x})$.
I can write the previous identity in
$$F_Y (a)=
\frac{1}{\pi}\cdot\arctan(\frac{x}{1-x*\theta})$$
Then I can derivate $F_Y (a)$ to find $f_Y (a)$
$$f_Y (a)=\frac{1}{\pi}\cdot\frac{1}{1+\frac{x}{(1- \Theta*x)^2}}$$p
Now I don't know how to continue
can someone help me if I have done some mistakes?
 A: Just the final steps: all is perfect up to
$$F_Y (a)=
\frac{1}{\pi}\cdot\arctan(\frac{x}{1-x*\theta})$$
"Then I can derivate $F_Y (a)$ to find $f_Y (a)$" also ok, only you have to set $x\to 1/a$ before you differentiate with respect to $a$.
Then you`ll get
$$f_{Y}(a) = \frac {1}{\pi} \frac{1}{a^2 \left(\theta ^2+1\right)-2 a \theta +1}\tag{1}$$
In order to identify this as a Cauchy distribution we write the pdf of a Cauchy distribution in $t$ with parameters $u$ and $v$ in the form
$$f_{C}(u,v,t)=\frac{1}{\pi  \left(\frac{(t-u)^2}{v}+v\right)}$$
the original parameters were 
$$\left\{u\to \theta, v\to 1, t\to x\right\}$$
and after the transformation we have found in $(1)$ a Cauchy distribution with the parameters
$$\left\{u\to \frac{\theta }{\theta ^2+1},v\to \frac{1}{\theta ^2+1}, t\to a\right\}$$
A: Let $h(x)=x^{-1}$. Then
\begin{align}
f_Y(y)&=f_X(x)(h^{-1}(y))\times\left\lvert \frac{dh^{-1}(y)}{dy}\right\rvert \\
&=\frac{1}{\pi}\times\frac{1}{1+(1/y-\theta)^2}\times\frac{1}{y^2} \\
&=\frac{1}{\pi}\times\frac{1}{\gamma\left(1+[\gamma^{-1}(y-\tau)]^2\right)},
\end{align}
where $\gamma\equiv (1+\theta^2)^{-1}$ and $\tau\equiv\gamma\cdot\theta$.

Alternatively (assume for simplicity that $\theta=0$), for $y>0$,
$$
\mathsf{P}(Y\le y)=\mathsf{P}(X\le 0)+\mathsf{P}(X\ge y^{-1})=1-\frac{\arctan(y^{-1})}{\pi}
$$
and for $y\le 0$,
$$
\mathsf{P}(Y\le y)=\mathsf{P}(y^{-1}\le X\le 0)=-\frac{\arctan(y^{-1})}{\pi}.
$$
Thus,
$$
f_Y(y)=F_Y'(y)=\frac{1}{\pi}\times \frac{1}{1+y^2}
$$
