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$$dX = \kappa_x (\theta_x - X)dt + \sigma_x \sqrt{X} \,dW_x$$ $$dY = \kappa_y (\theta_y - Y)dt + \sigma_y \sqrt{Y} \,dW_y$$ $$dW_x dW_y = \rho\, dt$$

we know that $X$ and $Y$ are marginally distributed as non-central $\chi^2$. What is the joint distribution of $X$ and $Y$ ?

flag	asked Jan 24 2011 at 13:18 startover 1

What if to write it with respect to independent Brownian motions? – Ilya Jan 25 2011 at 10:26

The Bridge · Answer 1 · 2011-01-26 10:20:13Z UTC

1

Hi,

So what you really want to know can be put into this form :

Does there exists a multivariate extension to $\chi^2$ laws, and if so, does it matches the joint law of the bivariate process $(X,Y)$. Is that right ?

I think that you should have a look at Whishart processes and distributions

Regards

answered Jan 26 2011 at 10:20

The Bridge
634●4●8

	Whishart process is a matrix extension of non-central $\chi^2$ where only the diagonal elements have non-central $\chi^2$ marginals. In a 2*2 matrix case there is no way to set the parameters such that the two diagonal elements have the above dynamics. In fact I am looking more for the joint characteristic function of $X$ and $Y$. – startover Jan 27 2011 at 23:24
	You are right sorry for this. Anyway, using the suggestion of Gortaur you can reduce the problem to the calculation of the characteristic function of the couple $(X,Y')$ (where the sde followed by $Y'$ is similar to $Y$'s one) with both sde driven by the same Brownian Motion. But I don't know if the calculation is tractable. – The Bridge Jan 28 2011 at 8:26
	Can you explain more? I thought that Gortaur is recommending something like replacing $dW_y$ by $\rho dW_x + \sqrt{1-\rho^2}dW_z$ where $dW_xdW_z=0$. Can you explain more dynamics of $Y'$ in your comment? – startover Jan 31 2011 at 0:27
	Hi startover, I tried to achieve the calculations but I wasn't able to find a tractable form sorry Regards – The Bridge Jan 31 2011 at 12:01

square root processes with correlated deriving Brownian motion

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