The bivariate poisson distribution.

Suppose that $\displaystyle X_i, i=1,2,3$ are independent Poisson random variables with respective means $\displaystyle \lambda_i $. Let $\displaystyle X=X_1+X_2$ and [TEXY=X_2+X_3[/TEX]. The random vector $\displaystyle X,Y$ is said to have a bivariate Poisson distribution. Find its joint probability mass function. That is, find $\displaystyle P{X=n, Y=m}$

I tried:

We know

$\displaystyle P{X=x}=\frac{e^{-(\lambda_1 +\lambda_2)}(\lambda_1+\lambda_2)^x}{x!}$

$\displaystyle P{Y=y}=\frac{e^{-(\lambda_2 +\lambda_3)}(\lambda_2+\lambda_3)^y}{y!}$

However I get stuck as these two are not independent and cannot therefore just be multiplied

I also tried to create a new variable Z and use the equation for functions of random variables, but the Jacobian is zero due to the zero row.