I'm looking for some help with this probability problem.

Here's the question:

Suppose that $X$ and $Y$ are independent standard normal random variables. Show that the probability density function of $Z = X / |Y|$ is given by $$ f(t) = \frac{1}{\pi(1+t^2)}, \quad (-\infty < t < \infty). $$

Thanks for looking through my question.

-updated- Hi guys, I've looked through the solution and am still thinking, i'm also thinking if its possible to log the expressions and apply convolution theorem... thanks for the answers once again.

-updated- tried the solution using first method but couldn't figured out how to complete the last step.

Heres how i did it:

$f(x,y) = f_{x}(x)f_{y}(y) = \frac{1}{2\pi}e^{-\frac{1}{2}(x^2 + y^2)}$

let Z = X/|Y|, V = X then x = v, y = v/z

the Jacobian J = $x_{z}y_{v}-x_{v}y_{z} = \frac{v}{z^2}$

by Transformation theorem,

$w(z,v)=f(v,\frac{v}{z})|J| = ???$

Will someone please point out to me how do i proceed from here to obtain f(t)? thanks a lot!

-update 3- this is actually a Cauchy density!! thanks guys, i think i got it figured out.

  • 1
    $\begingroup$ Take $\varphi$ a continuous function with compact support, and compute $E\left[\varphi\left(\frac X{|Y|}\right)\right]$ using the fact that you know a probability density of $(X,Y)$. $\endgroup$ Commented Nov 6, 2011 at 19:16
  • $\begingroup$ Please see responses to this question which addresses your particular question as an example. $\endgroup$
    – Sasha
    Commented Nov 6, 2011 at 20:00
  • $\begingroup$ Like @Davide said. Try to emulate this. $\endgroup$
    – Did
    Commented Nov 6, 2011 at 20:41
  • $\begingroup$ @Sasha, about your answer to the other question, note that a crucial hypothesis missing there is available here. $\endgroup$
    – Did
    Commented Nov 6, 2011 at 20:45

3 Answers 3


I imagine you could use the following theorem:

Let $X_1$ and $X_2$ be jointly continuous r.v.'s with joint density $f_{X_1,X_2}$. Let $Y_1=g(X_1,X_2)$ and $Y_2=g_2(X_1,X_2)$.


1) $y_1=g(x_1,x_2)$ and $y_2=g_2(x_1,x_2)$ can be uniquely solved for $x_1$ and $x_2$ by, say, $x_1=h_1(y_1,y_2)$ and $x_2=h_2(y_1,y_2)$


2) $g_1$ and $g_2$ have continuous partials with $$ J(x_1,x_2) ={\partial g_1\over\partial x_1}{\partial g_2\over\partial x_2} - {\partial g_1\over\partial x_2}{\partial g_2\over\partial x_1} \ne0, $$ then

$$ f_{Y_1,Y_2}(y_1,y_2) = f_{X_1,X_2} (x_1,x_2) |J(x_1,x_2)|^{-1}, $$ where

and $x_1=h_1(y_1,y_2)$ and $x_2=h_2(y_1,y_2)$.

Use the above to find the joint distribution of $Y_1=X$ and $Y_2=X/Y$ (with $X_1=X$ and $X_2=Y$); then integrate to obtain the distribution of $Y_2$.

This, by the way, is a problem in Sheldon Ross' ``A first Course in Probability'' (Chapter 6, Theoretical Exercise 33.)

Edit: This may be a ``sledgehammer''. Refer to the post suggested by Sasha...


One way is polar coordinates: $(X,Y)= R(\cos \Theta,\sin\Theta)$. Observe that $X/Y=\cos\Theta/\sin\Theta$. Find the distribution of $(\cos \Theta,\sin\Theta)$ by thinking about the density of $(X,Y)$. You'll see that that's easy and the answer you get is simple. Then think about tangents and cotangents.

  • 1
    $\begingroup$ Also, this works whenever the joint density of $X$ and $Y$ is rotationally symmetric around the origin. (If $X$ and $Y$ are independent, though, then that forces $X$ and $Y$ to be mean-zero normal.) $\endgroup$ Commented Nov 6, 2011 at 20:33

This is just a variation on the theme of my earlier answer. Maybe some will find this way of saying it to be clearer.

Write $(X,Y)$ 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 $(-\pi/2,\pi/2)$.

$$ \frac{d}{d\theta}\Pr(\tan\Theta\le\theta) = \frac{d}{d\theta}\Pr(\Theta\le \arctan\theta) = \frac{d}{d\theta} \frac{\arctan\theta}{\pi} = \frac{1/\pi}{1+\theta^2}. $$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .