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Let $X,Y\in \mathcal{N}(0,1)$ with $X$ and $Y$ independent. I am interested in the pdf of $Z=X^2/Y^2$ which has a $\mathcal{F}(1,1)$ Fisher distribution. I know that his can be done by first finding the pdfs of $U=X^2$ and $V=Y^2$ and find the pdf of $U/V$. However, I am interested in the following approach: let $h$ be any Borel measurable function then $$E[h(Z)]=E[h(X^2/Y^2)]=\int_0^\infty\int_0^\infty h\left(\frac{x^2}{y^2}\right)e^{-x^2/2-y^2/2}\frac{1}{2\pi}dxdy.$$ I'm not sure but my intuition tells me that there is change of variable one can do to obtain $$E[h(Z)]=\int_0^\infty h(z)\cdot f(z)dz,$$ and hence $f(z)$ would be the pdf of $Z$. Do you think one cna find this change of variable?

UserA
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    Why not begin with the well-known and easily established result that $X/Y$ has a Cauchy distribution? As far as your particular approach goes, a change to polar coordinates $(r,\theta)$ is the evident choice, where $x^2/y^2$ simplifies to $\cot^2\theta$ and everything else is a function of $r.$ – whuber Dec 14 '23 at 21:04
  • Sure, after the change of variable we get $$\int_0^{\frac{\pi}{2}}\int_0^\infty h(\cot^2\theta)e^{-r^2/2}\frac{r}{2\pi}drd\theta $$ which after computing the first integral yields $$\int_0^{\frac{\pi}{2}} h(\cot^2\theta)\frac{1}{\pi}d\theta.$$ But then how to proceed from here? We do another change of variable $z=\cot^2\theta$? – UserA Dec 15 '23 at 07:05
  • That appears to be what you are asking for ;-). – whuber Dec 15 '23 at 18:28

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