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Question; Given two independent standard normally distributed random variables $X,Y,$ find the cumulative density function of $\frac{X}{Y}.$

My attempt:

Note that \begin{align*} F(x) & = P(\frac{X}{Y}\leq x) \\ & = P(\frac{X}{Y}\leq x | Y >0)P(Y>0) + P(\frac{X}{Y}\leq x | Y<0) P(Y<0) \\ & = \frac{1}{2} \left[ P(X\leq xY | Y >0) + P(X\geq xY | Y<0) \right]. \end{align*} However, I have trouble evaluating $P(X\leq xY | Y >0).$

I know the final answer is that the distribution of $\frac{X}{Y}$ is Cauchy. I am trying to prove it by considering CDF and then differentiating CDF to obtain PDF.

Idonknow
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  • The region $y \leq xz,z>0$ in the $(y,z)$ plane can be nicely described as a polar region (involving the angle $\arctan(1/x)$) which enables you to calculate the associated probability by direct integration in a manner very similar to how the Gaussian integral is normally evaluated. – Ian Dec 12 '19 at 04:43
  • Would you mind to elaborate it? Actually I always have difficult evaluating probability involving at two random variables. – Idonknow Dec 12 '19 at 04:47
  • Do you know how the Gaussian integral is evaluated? It is pretty much exactly like that, only with different $\theta$ bounds. – Ian Dec 12 '19 at 04:48
  • Yes, one can use polar coordinate to evaluate the gaussian integral. – Idonknow Dec 12 '19 at 04:49
  • See Quotient of two random variables for finishing this calculation. When both means are zero the result is the Cauchy distribution, see also ratio distribution more generally. – Conifold Dec 12 '19 at 05:12
  • I meant this one: https://math.stackexchange.com/questions/77873/how-calculate-the-probability-density-function-of-z-x-1-x-2?noredirect=1&lq=1. – StubbornAtom Dec 13 '19 at 13:20

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