Let $X$ be a random variable with Cauchy distribution, compute the density function of $Y=\frac{1}{1+X^2}$

Let $X$ be a random variable with Cauchy distribution, if $$Y=\frac{1}{1+X^2}$$ then compute the density function of $Y$

I have tried to use some random variable transformation theorem using the functions $\frac{1}{1+X}$ and $X^2$ but I have not been able to get very far, any suggestion or help would be greatly appreciated.

Solution 1:

$$P(Y \leq y)=P(X^{2} \geq \frac 1y -1)$$ $$=2P(X \geq \sqrt {\frac 1 y -1})$$ $$=2\int_{\sqrt {\frac 1 y -1}}^{\infty} \frac 1 {\pi} \frac 1 {1+t^{2}} dt.$$ Now just differentiate w.r.t. $y$.

Solution 2:

The problem can also be solved via the change of the variable in PDF. $$X:\,f(x)=\frac{1}{\pi}\frac{1}{1+x^2}; \,\,Y=\frac{1}{1+X^2}; \,\,y=f(x);\,\,g(y)-?$$ $$1+x^2=\frac{1}{f(x)}\,\Rightarrow x=\sqrt{\frac{1-f(x)}{f(x)}}$$ $$ dx=-\frac{1}{2}\bigg(\frac{1}{\sqrt{1-f(x)}\sqrt f(x)}+\frac{\sqrt{1-f(x)}}{(\sqrt f(x))^3}\bigg)df=-\frac{1}{2}\frac{df}{\sqrt{f(x)}\sqrt{1-f(x)}f(x)}$$ Therefore, after changing the variable, the PDF looks ($y=f(x)$) $$\frac{dx}{\pi}\frac{1}{1+x^2}=-\frac{1}{2\pi}\frac{f(x)\,df}{\sqrt{f(x)}\sqrt{1-f(x)}f(x)}=-\frac{1}{2\pi}\frac{dy}{\sqrt{y}\sqrt{1-y}}$$ We also have to note that $y>0$ at any $x$, and $y\in(1;0)$ for $x\in(0;\infty)$; taking the usual order of integration (from $0$ to$1$) and multiplying PDF by $2$ (taking into consideration the part of $x\in(-\infty;0)\,$), we get $$g(y)=\frac{1}{\pi}\frac{1}{\sqrt{y}\sqrt{1-y}}$$ $$\frac{1}{\pi}\int_0^1\frac{dy}{\sqrt{y}\sqrt{1-y}}=\frac{1}{\pi}B\Big(\frac{1}{2};\frac{1}{2}\Big)=1$$