Here's my suggestion:
Try writing the conditional joint pdf of (X,Y) conditioned on X_2.
(Notice that, given X_2, X and Y are independent)
Then you multiply that by the pdf of X_2.
Suppose that are independent Poisson random variables with respective means . Let and [TEXY=X_2+X_3[/TEX]. The random vector is said to have a bivariate Poisson distribution. Find its joint probability mass function. That is, find
I tried:
We know
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.
I dont think that's required. As ilanshom suggests, is the following result not true:what gives us the justification to ignore the second term on the RHS?
All terms on the RHS are easy to compute. i also assume the sum will simplify conveniently but i haven't checked.