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Let $X$ be a random variable with Cauchy distribution with its probability density function $$f(x)=\frac{1}{\pi(1+x^2)}.$$ How can I find the probability density function of $$Y=\frac{2X}{1-X^2}?$$

All the problems I came across so far didn't involve rational function where both numerator and denominator were functions of $X$.
We can find the cumulative distribution function of $Y$ as: $$F_Y(Y)=P(Y<y)=P\bigg(\frac{2X}{1-X^2}<y\bigg)$$ But I'm not sure how I can continue from there and express $F_Y(y)$ in terms of $F_X(x)$.

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Hint: Use $$\frac{2X}{1-X^2} = y \iff yX^2+2X-y=0\iff X =\frac{-1\pm\sqrt{1+y^2}}{y}$$ to find the regions where $$\frac{2X}{1-X^2} \le y.$$

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  • $\begingroup$ Did you try to follow your own lead to reach a solution? Doing so is most probably out of reach of the OP. $\endgroup$
    – Did
    Commented Jul 13, 2018 at 17:36

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