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I think I'm missing something glaringly obvious here that's causing problems for me in the entire subject.

I have two independent standard normal random variables, X and Y ~N(0,1), and I need to find the density of U=Y/X. I start with f(x,y)=$\frac1{2\pi}e^{-x^2/2}e^{-y^2/2}$, then set Y=ux, and take a double integral of that. I'm leaving out my limits of integration, because my problem is that I know that I have to integrate with $\iint$y f(x,y), but I have no idea where that y (which is needed to integrate $e^{-y^2/2}$) comes from. Is this a Jacobian? None of my notes or our textbook mentions any use of a Jacobian in the CDF method. Is this just impossible with the CDF method?

user143719
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1 Answers1

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I'm not 100% sure what approach you are taking. I would approach this problem by considering the CDF that you want, which is $F(t) = P(X/Y<t)$ for $t>0$ (for now, in order to avoid some issues). Equivalently, consider $P(X<tY)$ for positive $Y$, and $P(X>tY)$ for negative $Y$. With $f(x,y)=\frac{1}{2\pi}\exp\left(\frac{1}{2}(x^2+y^2)\right)$, $$P(X<tY) = \int_{0}^{\infty} \int_{-\infty}^{tY} f(x,y) dx dy$$ This can be switched to polar coordinates using the same trick found on the page http://en.wikipedia.org/wiki/Gaussian_integral: $$P(X<tY) = \int_{\arctan(1/t)}^{\pi} \int_{0}^{\infty} \frac{1}{2\pi}r\mathrm{e}^{-r^2/2} dr d\theta = \frac{1}{2}-\frac{1}{2\pi}\arctan(1/t) $$ The $\arctan$ arrives by noticing that you are integrating $\theta$ from the line $x=ty$ to $\pi$. For $t>0$, $\arctan(1/t) = \pi/2 - \arctan(t)$. By symmetry, $$ F(t) = P(X<tY|Y>0) + P(X>tY|Y<0) \\ = 1 - \frac{1}{\pi} \arctan(1/t) \\ = \frac{1}{2}+\frac{1}{\pi}\arctan(t) $$ which is the CDF of the Cauchy distribution. You can continue this to $t=0$ by integrating over half of the plane and getting $1/2$. You can handle $t<0$ by itegrating $\theta$ from $\arctan(1/t)$ to $0$. The issue is that the range of $\arctan$ is $(-\pi/2,\pi/2)$.

Finally, differentiate to get the density function $F'(t)=\frac{1}{\pi}\frac{1}{1+t^2}$.

Adam B Kashlak
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