# The bivariate poisson distribution.

• May 18th 2011, 08:13 AM
FGT12
The bivariate poisson distribution.
Suppose that $X_i, i=1,2,3$ are independent Poisson random variables with respective means $\lambda_i$. Let $X=X_1+X_2$ and [TEXY=X_2+X_3[/TEX]. The random vector $X,Y$ is said to have a bivariate Poisson distribution. Find its joint probability mass function. That is, find $P{X=n, Y=m}$

I tried:

We know
$P{X=x}=\frac{e^{-(\lambda_1 +\lambda_2)}(\lambda_1+\lambda_2)^x}{x!}$
$P{Y=y}=\frac{e^{-(\lambda_2 +\lambda_3)}(\lambda_2+\lambda_3)^y}{y!}$

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.
• May 18th 2011, 10:42 PM
ilanshom
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.
• May 19th 2011, 02:29 AM
FGT12
If we use the formula

$P(A) = P(A | B)P(B) + P(A | B^c)P(B^c)$

what gives us the justification to ignore the second term on the RHS?
• May 19th 2011, 04:55 AM
SpringFan25
Quote:

what gives us the justification to ignore the second term on the RHS?
I dont think that's required. As ilanshom suggests, is the following result not true:

$P(X,Y) = \sum_{k=0}^{\infty} P(X2=k) P(X,Y|X2)$

All terms on the RHS are easy to compute. i also assume the sum will simplify conveniently but i haven't checked.
• May 19th 2011, 06:17 AM
ilanshom
Yes, that's what I had in mind. Sorry I forgot to mention it!
• May 23rd 2011, 06:40 PM
matheagle
did it work?
Let me know. I'm good with these calculations
I haven't been around lately.
Classes are finally over, but I just closed on a condo on the lake.