What is the law of the ratio of two independent gaussian randon variables? [duplicate]

Consider two independent randon variables $N , N' \sim \mathcal N (0,1)$.What is the law of $N/N'$ ?

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marked as duplicate by Did, Fabian, Erick Wong, Dilip Sarwate, cardinalJan 1 '13 at 20:14

Here is the first hit on Google for "ratio of Gaussian random variables": en.wikipedia.org/wiki/Ratio_distribution. Your answer is in the second paragraph, it is a Cauchy distribution. – Rahul Jan 1 '13 at 15:29
Just to amplify on Rahul's comment - the Cauchy distribution is a power law. If I remember it doesn't even have a convergent mean, certainly no variance. This long-tail behavior is intuitive if you think of the denominator often being close to zero ($\sim \mathcal N (0,1)$). – alancalvitti Jan 1 '13 at 15:50
An unrelated way to obtain a Cauchy distribution is via Gull's "Lighthouse" problem from 1988: gps.caltech.edu/classes/ge193/practicals/practical3/… – alancalvitti Jan 1 '13 at 15:55
Answers on this website can be found here as well as in many other questions (see for example, Didier's comment on the question). – Dilip Sarwate Jan 1 '13 at 17:26
@alancalvitti That, or the first hitting place of a given hyperplane by a Brownian particle. – Did Jan 1 '13 at 19:57

Write $(N,N')$ in polar coordinates as $(R,\Theta)$. Their joint distribution becomes $$\text{constant}\cdot e^{-r^2/2} r\,dr\,d\theta.\tag{1}$$ From $(1)$ we see that $R$, $\Theta$ are independent and $\Theta$ is uniformly distributed in $\mathbb R\bmod 2\pi$. The ratio $N/N'$ is $\tan\Theta$. Since $\tan$ has period one-half of a circle, nothing is lost by taking $\Theta$ to be in $(-\pi/2,\pi/2)$.
$$\frac{d}{d\theta}\Pr(\tan\Theta\le\theta) = \frac{d}{d\theta}\Pr(\Theta\le \arctan\theta) = \frac{d}{d\theta} \frac{\arctan\theta}{\pi} = \frac{1/\pi}{1+\theta^2}.$$