What is the distribution of $(1+X^2)^{-1}$ when $X$ is standard Cauchy?

Question

I have this problem:

Let $X \sim \text{Cauchy} (0,1)$. Obtain the distribution of the next random variable $Y=(1+X^2)^{-1}$.

My attempt was the following:

$$f(x)= \frac {1}{\pi (1+x^2) }$$

$Y=g(x)$

$$g^{-1}(y)= \sqrt{\dfrac{1}{y}+1}$$

Then:

$$f_{y} (y)= \left|\frac{1}{2y^2 \sqrt{\frac{1}{y}+1}}\right|\cdot\frac{1}{\pi \left(1+\left(\sqrt{\frac{1}{y}+1}\right)^2\right)}$$

But I don't know if I'm correct

Thank you for your help

Answer 1

Your computation of $g^{-1}$ is wrong. If $\frac 1 {1+x^{2}}=y$ then $x^{2}=\frac 1 y -1$ and $x =\pm \sqrt {\frac 1 y -1}$. Once you use this you will be able to a neater form for $f_Y$. I will let you do the calculation.

The corect answer is $f_Y(y)=\frac 1 {\pi \sqrt {y(1-y)}}$ for $0<y<1$.