Suppose X$\sim$ Cauchy(0,1). Then what will be the distribution of $\frac{1-X}{1+X}$? In order to find distribution of $\frac{1-X}{1+X}$ below approach I followed,
Let,
\begin{align}
    Y = \frac{1-X}{1+X}
\end{align}
Then, cdf of Y is
\begin{align}
    F_{Y}(y) = P(Y \leq y)
\end{align}
\begin{align}
    = P\left(\frac{1-X}{1+X} \leq y\right)
\end{align}
\begin{align}
    = 1 - P\left(X < \frac{1-y}{1+y}\right)
\end{align}
\begin{align}
    = 1 -  \int_{-\infty}^{\frac{1-y}{1+y}} f(x) \,dx
\end{align}
\begin{align}
    = 1 -  \int_{-\infty}^{\frac{1-y}{1+y}} \frac{1}{\pi}\cdot \frac{1}{1+x^2} \,dx
\end{align}
\begin{align}
    = 1 -  \frac{1}{\pi}\cdot \left[tan^{-1}x\right]_{-\infty}^{\frac{1-y}{1+y}}
\end{align}
\begin{align}
    F_{Y}(y) = \frac{1}{\pi}\cdot \left[-\frac{\pi}{2}+tan^{-1}\left({\frac{1-y}{1+y}}\right)\right]
      = \frac{1}{2} -\frac{1}{\pi}.tan^{-1}\left({\frac{1-y}{1+y}}\right)
\end{align}
and then
\begin{align}
    f_{Y}(y) = \frac{d F_{Y}(y)}{dy} = \frac{1}{\pi}\cdot \frac{1}{1+y^2}
\end{align}
But I have a little confusion here how to find range of Y from X? And CDF of Y is doesn't looks like cdf of a cauchy distribution.

Remark (Theorem of transformation for one random variable): Let be $X$ a continuous r.v. with pdf given by $f_X$ and let be Y=g(X) with g a diffeomorphism, then the pdf $f_Y$ is given by:

$f_Y(y)=f_X(g^{-1}(y))\cdot|{{d\over{dy}}g^{-1}(y)}|$.

If $Y=g(X)={{1+X}\over{1-X}}$ then $X=g^{-1}(Y)={{1-Y}\over{1+Y}}$. We find that ${dx\over{dy}}=-{{2\over{1+y^2}}}$. Now:

$f_Y(y)=f_X({{1-y}\over{1+y}})\cdot{2\over{1+y^2}}$.

I'll let you do the math, then you get:

$f_Y(y)={1\over{\pi{(1+y^2)}}}$.

So $Y\sim{Cauchy(0,1)}$.
