Thread: Ratio of uniforms is Cauchy

1. Ratio of uniforms is Cauchy

Hi all!

I have U is uniform(0,1) and V is uniform(-1,1) such that U^2 + V^2 <= 1. Furthermore, let X=V/U. Show that X is Cauchy.

I tried to solve this problem using transformation:
X = V/U => V = XY
Y = U => U = Y
Then the Jacobian is |J|=y and the joint pdf of X and Y is just f(x,y)=.5*y. This seems to be wrong since integrated out y won't give me a Cauchy.

Then I tried using the CDF:
P(X <= x) = P(V/U <= x) = P(V <= Ux) = x -.5 but this does not work either.

Obviously I have to incorporate the condition that U^2 + V^2 <= 1 but I have no idea how and I would really appreciate some help.

Thanks,
Jenny

2. Dont we need to know what is joint pdf of <X,Y> ?

Are they independant?

3. I mean joint pdf of <U, V>. Without this problem is not solvable.

4. Originally Posted by JennySton
Hi all!

I have U is uniform(0,1) and V is uniform(-1,1) such that U^2 + V^2 <= 1. Furthermore, let X=V/U. Show that X is Cauchy.

I tried to solve this problem using transformation:
X = V/U => V = XY
Y = U => U = Y
Then the Jacobian is |J|=y and the joint pdf of X and Y is just f(x,y)=.5*y. This seems to be wrong since integrated out y won't give me a Cauchy.

Then I tried using the CDF:
P(X <= x) = P(V/U <= x) = P(V <= Ux) = x -.5 but this does not work either.

Obviously I have to incorporate the condition that U^2 + V^2 <= 1 but I have no idea how and I would really appreciate some help.

Thanks,
Jenny
1) I'm betting the joint pdf is $\frac{2}{\pi}$; one over the area of the circle that U,V are defined on.

2) you then want

$U^2+V^2\leq 1\Rightarrow Y^2+(XY)^2\leq 1\Rightarrow Y^2(1+X^2)\leq 1\Rightarrow Y\leq\sqrt{ \frac{1}{1+X^2}}$

3) then compute

$\int\limits_{0}^{ \sqrt{\frac{1}{1+X^2}}} \frac{2}{\pi}ydy$

5. Thanks for your help! I see what I did wrong!

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