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Given two independent normal variables, $X\sim N(0, \sigma^2_X)$ and $Y\sim N (0, \sigma_Y^2)$, find the probability that $\frac{X}{Y} < 1$.

I seem to prove: $Z = \frac X Y \sim $ Cauchy distribution with pdf $f_Z(z) = \frac{1}{\pi} \frac{\frac{\sigma_X}{\sigma_Y}}{z^2 + \frac{\sigma^2_X}{\sigma^2_Y}}$ (I'm not sure though).

Anyway, help would be appreciated.

EDIT Note that the ratio follows Cauchy distribution is something I stumbled on while trying find the probability. As aleady noted on comments, this is not mandatory, and can be done by other technique. The linked question only wants to prove the distribution is Cauchy necessarily.

hola
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    $X/Y$ will only be Cauchy if $\mu_X=\mu_Y=0$. Otherwise, the resulting distribution is [quite complicated](https://en.wikipedia.org/wiki/Ratio_distribution#Uncorrelated_noncentral_normal_ratio). – COOLSerdash Nov 28 '19 at 14:44
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    What is your _question?_ The result you have obtained is not quite right in all the details but is plausible nonetheless: the ratio of two independent zero-mean equal-variance normal random variables is indeed a standard Cauchy random variable and it seems reasonable to suspect that with unequal variances you would get the result you got. The nonzero means, though, change the problem considerably and what you _should_ have obtained is more complicated, as the comment by COOLSerdash says. – Dilip Sarwate Nov 28 '19 at 14:48
  • @COOLSerdash Yep, correct. Fixed my question. – hola Nov 28 '19 at 18:17
  • @DilipSarwate Yes, the means are zero. Fixed. Is the expression correct now? – hola Nov 28 '19 at 18:18
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    Thanks for the clarification. Now you can use the [CDF](https://en.wikipedia.org/wiki/Cauchy_distribution#Cumulative_distribution_function) of a Cauchy distribution to get the general formula for $X/Y <1$. – COOLSerdash Nov 28 '19 at 18:23
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    While you can do this with a Cauchy it's not necessary to do it that way. Noting that $P(Y=0)=0$ you can multiply through by $Y$ and the inequality splits into two cases ($Y<0$, $Y>0$) for a joint normal, so you can also write it purely in terms of the joint distribution of a standardized X and Y, exploit the rotational symmetry and it in effect boils down to twice the area of a segment of a unit circle (which size depends on the ratio of the $σ$s). Mathematically it's the same thing either way, (of course) but if your mind is more geometrical than algebraic you may find it a bit easier – Glen_b Nov 29 '19 at 01:13
  • See [Wikipedia](https://en.wikipedia.org/wiki/Cauchy_distribution#Related_distributions), third bullet. – BruceET Nov 29 '19 at 05:21

1 Answers1

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Let's generalize the problem in order to draw out the concepts on which it rests.

To this end, let $(U,V)$ be any bivariate random variable whose distribution is invariant under all rotations around the origin. Let $p$ be the probability of the punctured line $U\ne0\, V=0.$ The event $(U,V)\ne 0$ is the disjoint union of infinitely many such punctured lines, each with probability $p$ by the rotational invariance. Since the chance of this union is finite, it must be that $p=0,$ implying the distribution of these lines is continuous. The angle made by such a line is a value in $[0,\pi)$ given by the angle between the positive $u$ axis and any point on the line in the upper half plane). The rotational invariance implies this angle has a uniform distribution. Because the chance $(U,V)$ lies on one of these lines is $1-\Pr(0,0),$ the density of this uniform distribution must be $(1-\Pr(0,0))/\pi.$

Let $\sigma_X$ and $\sigma_Y$ be positive numbers and $(X,Y) = (\sigma_X U, \sigma_Y V).$ We seek a formula for

$$\Pr\left(\frac{X}{Y} \le a\right)$$

with $a=1$ (generalizing to any $a\gt 0$).

Rewriting this event as

$$a \ge \frac{X}{Y} = \frac{\sigma_X U}{\sigma_Y V} = \frac{\sigma_X}{\sigma_Y} \frac{U}{V}$$

reduces the problem to finding

$$\Pr\left(\frac{U}{V} \le \frac{a\,\sigma_Y}{\sigma_X}\right).$$

This event is the union of (a) the origin $(0,0)$ and (b) all punctured lines making angles between $\operatorname{arccot}(a\sigma_Y/\sigma_X)$ and $\pi.$ Computing the probability in (b) according to the uniform distribution previously found and adding in the chance of (a) gives

$$\Pr\left(\frac{X}{Y} \le a\right) = \Pr(0,0) + \frac{1-\Pr(0,0)}{\pi}\left(\pi - \operatorname{arccot}(a\,\sigma_Y/\sigma_X) \right).$$

In the question $(X,Y)$ has a continuous distribution, whence $\Pr(0,0)=0.$ The result simplifies to

$$\Pr\left(\frac{X}{Y} \le a\right) = 1 - \frac{\arctan\left(\sigma_X/\left(a\,\sigma_Y\right)\right)}{\pi}.$$

whuber
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  • Just a quick question: What is the probability of $|\frac{X}{Y}|<1$? $$1-\frac{\color{red}{2} \tan^{-1}({\sigma_X}/{\sigma_Y})}{\pi},$$ right? – hola Dec 01 '19 at 06:49
  • The factor of $2$ cannot be correct. To see why not, consider that the answer must be a number between $1/2$ and $1,$ with the minimum value approached as $\sigma_X/\sigma_y$ grows large. The value of $1/2$ reflects the possibility that $X/Y$ is negative. – whuber Dec 01 '19 at 15:40
  • OK, so could it be $$\color{red}{2} -\frac{2 \tan^{-1}{(\sigma_X/\sigma_Y)}}{\pi}$$? – hola Dec 02 '19 at 11:34
  • I invite you to identify any error you might find in my derivation. Know, however, that before posting this answer I checked it through simulations--so unless some typographical error was introduced in posting it, the formula likely is correct. – whuber Dec 02 '19 at 14:31
  • Oops, probably I offended you. Apologies for that. I don't have any problem with your solution. I was trying to figure out the probability that $\color{blue}{|} \frac{X}{Y} \color{blue}{|} < 1$, using your method (notice the modulous sign). – hola Dec 03 '19 at 04:43
  • I'm not offended, just puzzled that you keep proposing different variations of the answer. Finding the chance that $|X/Y|\lt 1$ uses exactly the same argument. It helps to draw a picture of this event in the $(x,y)$ plane: you'll be able to read the answer right off the picture without any calculation. – whuber Dec 03 '19 at 14:41
  • Is the general answer for a number a instead of 1, be Pr(XY≤a)=a−arctan(σX/σY)π ? – gciriani Jan 08 '22 at 20:58
  • @gciriani Start at the beginning, replacing "1" by "a." You will discover that the answer depends only on $a\sigma_Y/\sigma_X.$ – whuber Jan 08 '22 at 22:27
  • So you obtain Pr( X / Y ≤ a ) = 1 − arctan( σX / (a σY) ) / π, right? – gciriani Jan 09 '22 at 13:54
  • @gciriani Right. I have made that generalization explicit in this answer. – whuber Jan 09 '22 at 18:37