cauchy distribution

[dirk_mec1]

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

http://img132.imageshack.us/img132/1/48572399ly5.png (http://imageshack.us)



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?

[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.

[statdad]

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 .

[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) =http://www3.wolframalpha.com/Calculate/MSP/MSP91219bfg00e7b70caif00005aa636h4egb540i1?MSPStor eType=image/gif&s=39&w=148&h=44

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

=>http://www3.wolframalpha.com/Calculate/MSP/MSP243119bff8ch835e1b7i00001639e2c5d96b97gg?MSPSto reType=image/gif&s=39&w=366&h=54

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:

http://www3.wolframalpha.com/Calculate/MSP/MSP237019bff8ch83i380ib0000641a5786ggg043f3?MSPSto reType=image/gif&s=39&w=124&h=43

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.