# Bivariate Poisson

by jimmy1
Tags: bivariate, poisson
 P: 61 I have 2 dependent random Poisson distributed variables, $$X$$ and $$Y$$. I have that $$E[X] = mu$$ and $$E[Y] = c*mu$$ where $$c$$ is just a constant. Now I'm trying to get the joint distribution of $$XY$$. I've found the expression of the bivariate Poisson distribution but the problem is in order to use it I have to define $$X$$ and $$Y$$ as $$X = X' + Z$$ and $$Y = Y' + Z$$ where $$X', Y', Z'$$ are independent Poisson distributions with $$E[X'] = (mu - d)$$, $$E[Y'] = (c*mu - d)$$ and $$E[Z'] = d$$. So basically my question is how do I get the parameter $$d$$?? Is there any formal way to get it??
 Sci Advisor P: 6,080 You have not been given enough information. X and Y could be independent or else Y=cX or something in between.
 P: 61 Well, X and Y are definitley dependent, it is always $$E[Y] = cE[X]$$. Does that help?? If not, what more information is needed?? In the paper I have about these bivariate Poisson distribution it also states that $$P(X|Y) = d/(c*mu + d)$$ and also $$P(Y|X) = d/(mu + d)$$, if that's any help?
P: 6,080
Bivariate Poisson

 Well, X and Y are definitley dependent, it is always E[Y]=cE[X].
Not so, they can be independent and their means happen to obey the equation.

Your additional equation could be the key to the solution.
Emeritus
PF Gold
P: 16,091
 Quote by jimmy1 In the paper I have about these bivariate Poisson distribution it also states that $$P(X|Y) = d/(c*mu + d)$$ and also $$P(Y|X) = d/(mu + d)$$, if that's any help?
You sure you have that right? It doesn't make notational sense. (Incidentally, if you write \mu, LaTeX will convert that into a mu)
 P: 371 Ummm, if P(Y|X) is a function that doesn't depend on X, then Y and X are independent.
P: 61
 Quote by mathman Not so, they can be independent and their means happen to obey the equation.
If this is the case, then how to you formally define a dependent variable?
Emeritus