- #1
csprh
- 6
- 1
Hi all
I have a univariate distribution of the form
[itex]\frac{xγ}{((x^2+γ^2)^{1.5})}[/itex]
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
[itex]\frac{(γ/∏)}{(x^2+γ^2)}[/itex]
where x and y (in this case) are real numbers which can be positive and negative.
it's bivariate distribution can be defined as
[itex]\frac{(γ/(2∏))}{(x^2+y^2+γ^2)^{1.5}}[/itex]
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.
Thanks in advance
Paul
I have a univariate distribution of the form
[itex]\frac{xγ}{((x^2+γ^2)^{1.5})}[/itex]
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
[itex]\frac{(γ/∏)}{(x^2+γ^2)}[/itex]
where x and y (in this case) are real numbers which can be positive and negative.
it's bivariate distribution can be defined as
[itex]\frac{(γ/(2∏))}{(x^2+y^2+γ^2)^{1.5}}[/itex]
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.
Thanks in advance
Paul