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Where to from here...

Problem: If $U$ and $Y$ are continuous random variables, show that if ${U}$ has uniform $[0,1]$ distribution, then the random variable: $$Y = \tan(\pi U −\frac{\pi}{2})$$ has the Cauchy distribution.

Start of a Solution: Firstly, since $U$ is the uniform distribution on $[0,1]$, $u \in [0,1]$. Now, let $F_{Y}(y)$ be the probability distribution function where: $$F_{Y}(y) = \mathbb{P}(\{x \in [0,1] : \tan(\pi u - \pi/2 \leq y \})$$

This is equivalent to: $$F_{Y}(y) = \mathbb{P}(\{ x \in [0,1] : u \leq \frac{1}{\pi}(\arctan( y )) - \frac{1}{2}\}) $$

This is where I need further justification: intuitively, I'd say hey, that condition on the right looks an awful lot like the distribution function I want... but how do I get there from here?

Thanks!

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More clearly, \begin{align*} F_Y(y)&=\mathbb P(Y\leqslant y)=\mathbb P\Big(\tan(\pi U -\pi/2)\leqslant y\Big)=\mathbb P\Big(U\leqslant \frac{1}{2}+\frac{1}{\pi}\arctan y\Big)\\ &=F_U\Big(\frac{1}{2}+\frac{1}{\pi}\arctan y\Big)=\frac{1}{2}+\frac{1}{\pi}\arctan y. \end{align*} Then $Y$ forms the (standard) Cauchy distribution.

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  • $\begingroup$ ah, yes. Thank you. Very clear. And when you say it like that it looks so easy ;-) $\endgroup$ – Mathachist Jul 27 '17 at 18:36

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