# Trying to generate a bivariate distribution from a univariate distribution

1. May 10, 2012

### csprh

Hi all

I have a univariate distribution of the form

$\frac{xγ}{((x^2+γ^2)^{1.5})}$

where both parameters are real and non-negative.

How do I go about finding the bivariate form, where x and y (the new bivariate variables) are still both real and positive?

To explain further what I mean, the univariate cauchy distribution can be similarily is defined as

$\frac{(γ/∏)}{(x^2+γ^2)}$

where x and y (in this case) are real numbers which can be positive and negative.

it's bivariate distribution can be defined as

$\frac{(γ/(2∏))}{(x^2+y^2+γ^2)^{1.5}}$

How was this calculation done? One guess I had was to transform the first equation above into the Fourier domain and therefore obtain it's characteristic function. Transform this somehow in the CF domain and invert. Hmmmm. I seem to be a long way off this at the moment.

Any help would be gratefully received. I have spent at least three weeks mucking around with Matlab, Maple and Mathematica, but not to too much avail.

Paul

2. May 10, 2012

### viraltux

OK, there goes my 2 cents on this... as far as I remember that's up on how you define your multivariate distribution, let's imaging $(X_1...X_n)$ follows a multivariate Normal distribution or the Cauchy distribution, then every linear combination of its components is also a Normal or a Cauchy distribution.

But the Normal and the Cauchy are both symmetric distributions and yours is not, so there's no way you can get a bivariate distribution where the linear combination of its components will behave as your distribution.

Now, if you forget about that condition and X,Y are independent you only need to multiply their PDF. $f(x,y)=f(x) f(y)$ and if they are not then $f(x,y)=f(x|y) f(y|x)$

I hope this helps.

3. May 10, 2012

### chiro

Hey csprh and welcome to the forums.

If you have a form of a PDF which needs to be normalized to make it a 'true' PDF, then the best way to handle this is to first decide a domain for your PDF (i.e. x is between blah and blah and y is between blah and blah) and then calculate the volume under this function using a double integral, which will give you a number. Also before you do this, make sure your PDF is > 0 at all points in your domain.

When you have done this you will get integral = V which is the volume under the surface. To make this a PDF simply multiply your PDF by 1/V and this will create a proper PDF that you can use for doing distribution calculations.

If you are trying however to treat your random variables as 'particular' distributions and want to find the corresponding distribution of something in relation to this, then you will need to use a transformation theorem to find your PDF in terms of these other variables. So for example lets say you have X and Y which are distributions and want to find the distribution of f(X,Y): you need to use the properties of these (as well as independence properties and conditional properties if they exist) to get your final PDF.