# r.v Law of the ratio

Earlier the following question was asked, I think by a classmate:

r.v. Law of the min

I posted my solution to it, but I am stumped on the second half of the problem:

If the law of $(X,Y)$ is $\displaystyle \frac{1}{2\pi}e^{-\frac{x^2+y^2}{2}}dxdy$, what is the law of $\displaystyle \frac{X}{Y}$?

I'm trying to figure out how to set up the integral, and how I'm supposed to approach this one. I'd appreciate any input.

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Let $Z=X/Y$. Then $$F_Z(z) = \Pr(Z \leq z)$$ $$= \Pr(X/Y \leq z)$$ $$= \Pr(X \leq Yz)\bf{1}_{Y>0}+\it \Pr(X > Yz)\bf{1}_{Y<0}$$ with

$$\Pr(X \leq Yz)\bf 1 \it _{ Y>0}=\int_{\rm 0}^{\infty}\Pr(X \leq yz)\frac{1}{\sqrt{2\pi}}e^{-\frac{y^2}{2}}\ \mathrm{d} y$$

and

$$\Pr(X \leq yz)=\int_{-\infty}^{yz}\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}\mathrm{d} x$$

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Hint: fix $\phi\colon\Bbb R^2\to\Bbb R$ a continuous bounded function. Then compute $E[\phi(XY^{—1})]$ using a substitution like $t=x/y$ for a $y$ fixed. Then you will be able to find a density of $\frac XY$.

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