# Thread: Joint PDF of Cauchy - Integrating by partial fractions

1. ## Joint PDF of Cauchy - Integrating by partial fractions

Hi MathHelpForum,

My question relates to the use of the "Jacobian" function to find the joint pdf of two cauchy random variables. I have searched previous posts and do not seem to be able to find any solutions. I have attempted this problem but seem to have problems simplifying (integrating by partial fractions). Can the international maths community kindly help with solutions?

Thanks, PK.

2. Hello,

Find the Jacobian of the transformation $\phi ~:~ (x,y) \mapsto (x/y,y)$
Then use this to find the joint pdf of U,V.
And then integrate with respect to the second variable to get the pdf of U.

Is there something you don't know how to do ?

3. Hi Moo,

I am stuck with whats in the red boxes.

Box 1: I am not sure how the transformation gave the Cauchy variable.

Box 2: I am not sure how the solution used partial fractions to split up the equation. (I understand that since pdf is normal we can evaluate half the integral and multiply by 2).

Thanks, PK.

4. Well, I think you worry too much

Box 1 : the joint pdf of X & Y is the product of their pdf since they're independent. So it's $\frac{1}{\pi(1+x^2)}\cdot\frac{1}{\pi(1+y^2)}$

So since $x=uv$ and $y=v$, you get $\frac{1}{\pi^2} \cdot\frac{1}{(1+u^2v^2)(1+v^2)}$, which you have to multiply by |det Jacobian| to get the joint pdf of U & V. Which is exactly what you did.
(However, I must admit that I don't understand your remark "I am not sure how the transformation gave the Cauchy variable.")

Box 2 : your partial fractions are correct. The partial fractions and the factor 2 are two different things (two steps in one).
We indeed have $\frac{1}{(1+u^2v^2)(1+v^2)}=\frac{1}{1-u^2}\cdot \left(\frac{1}{1+v^2}-\frac{u^2}{1+u^2v^2}\right)$

Note : if you want to do it in a "reverse" way, you may have a look at this : http://www.mathhelpforum.com/math-he...-integral.html