# Thread: The bivariate poisson distribution.

1. ## 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.

2. 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.

3. If we use the formula

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

what gives us the justification to ignore the second term on the RHS?

4. 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:

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

All terms on the RHS are easy to compute. i also assume the sum will simplify conveniently but i haven't checked.

5. Yes, that's what I had in mind. Sorry I forgot to mention it!

6. 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.

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### Suppose that Xi, i = 1, 2, 3 are independent Poisson random variables with respective means λi, i = 1, 2, 3. Let X = X1 X2 and Y = X2 X3. The random vector X, Y is said to have a bivariate Poisson distribution. Find its joint probability mass functio

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