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Let $X$ be a Cauchy random variable with parameter $1$ i.e. with density $\dfrac{1}{\pi(1+x^2)}$. What is the density function of $Z:=\dfrac{1}{1+X^2}$?

My attempt:

Say $\phi(x) = \frac{1}{1+x^2}$ so then $\phi^{-1}(z) = \sqrt{\frac{1}{z}-1}$ and $\phi^{-1}\prime(z) = \frac{1}{2}\left(\frac{1}{z}-1\right)^{-\frac{1}{2}} .\frac{-1}{z^2}$

Finally we can say $$f_Z(z)=f_X(\phi^{-1}(z)) \left|\frac{d}{dz}\phi^{-1}(z)\right|= \left|\frac{-1}{2z^2\sqrt{\frac{1}{z}-1}}\right| \sqrt{\frac{1}{z}-1} = \frac{1}{2z^2}$$

Firstly is this right? If so do I need to exclude $z=0$ or adjust the range? I think the Cauchy distribution works over all $\mathbb{R}$ but am not sure I am ok if $z=0$.

Did
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Luskentyrian
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1 Answers1

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$Z$ is supported on $(0,1)$ and for every $a\in(0,1)$: $$ \mathbb{P}\left[Z\geq a\right]=\mathbb{P}\left[\left|X\right|\leq \sqrt{\frac{1}{a}-1}\right]=\frac{2}{\pi}\,\arctan\sqrt{\frac{1}{a}-1}$$ hence by differentiating with respect to $a$ we have that the pdf of $Z$ is: $$ f_Z(a) = \frac{1}{\pi\sqrt{a(1-a)}}\cdot\mathbb{1}_{(0,1)}(a).$$

Jack D'Aurizio
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  • Am I right in thinking the $\frac{2}{\pi}\arctan$ is twice, because of symmetry, the integral from $0$ to $\sqrt{\frac{1}{a}-1}$ of $\frac{1}{\pi(1+x^2)}$? Also how do you obtain the support for $Z$? – Luskentyrian Jun 15 '15 at 14:14
  • @Luskentyrian: you are right about the symmetry. About the support: since the support of $X$ is $\mathbb{R}$, the support of $\frac{1}{1+X^2}$ is $(0,1]$. – Jack D'Aurizio Jun 15 '15 at 14:49