# Cauchy distribution

1. Dec 12, 2008

### dirk_mec1

1. The problem statement, all variables and given/known data

3. The attempt at a solution
I tried dividing the two pdf's but that isn't right. If you have X normal and Y normal distributed how can you derive the distribution function of X/Y?

2. Dec 12, 2008

### Dick

Of course you don't divide the pdf's. To get the quotient pdf as a function of Z, you take the dirac delta function $\delta(Z-X/Y)$ and integrate it times the pdf's for X and Y, dXdY.

3. Dec 12, 2008

Another method (based techniques typically presented early in mathematical stats) is this:

Define these two variables (one you already have)

$$W = \frac{Z_1}{Z_2}, \quad V = Z_2$$

Then

$$Z_1 = V \cdot W, \quad Z_2 = V$$

By their definitions both new random variables range over $$(-\infty, \infty)$$.

Use the basic ideas for transformation of a joint distribution to get the distribution of $$V$$ and $$W$$, then integrate out $$V$$.

4. Aug 1, 2010

### dmancevo

Hi,

I'm actually going over some probability problems and I got a bit stuck in this one too.

If you let:

W=Z1/Z2 and V=Z2

Then truly Z1=V*W and Z2=V

And if you calculate the Jacobian determinant of such transformations you get:

Jacobian determinant = V (here we take the absolute value when sticking into formula below)

Therefore:

f(w,v) =

and so all you need to get the probability density function of W is to integrate the joint probability with respect to v as follows:

First note that: d/dv (e-v2(1+w2)/2) = -v(1+w2)*e-v2(1+w2)/2

=>

and here is where I seem to be overlooking something, in order to get f(w) you must evaluate the integral from minus infinity to plus infinity and so I believe you get:

Which is just plainly equal to zero, so I must've done something wrong, can anyone spot what was it? I would appreciate if someone did. Thanks.

Last edited: Aug 1, 2010