# Dependent Poissons

• Jul 18th 2013, 09:10 PM
casio415
Dependent Poissons
Hi everyone,

Just had a quick question regarding the covariance between two dependent poisson distributions with stochastic intensity.

Say we have P_A which is distributed as Poi(aG) where a is constant and G is Gamma(a,b)
We also have P_B which is distributed as Poi(bG) where b is a constant.

Then how do we work out Cov(P_A, P_B)?

I have tried the conditional covariance formula but still cannot obtain Cov(P_A, P_B | G).

Any help would be much appreciated.
• Jul 18th 2013, 10:53 PM
chiro
Re: Dependent Poissons
Hey casio415.

Can you show us specifically what you have tried (in terms of the details of your calculations and attempts)?

Also have you tried using a Bayesian approach for this problem?
• Jul 20th 2013, 04:21 AM
casio415
Re: Dependent Poissons
Well by conditioning, I have Cov(P_A, P_B) = E[Cov(P_A, P_B | G)] + Cov[E(P_A)E(P_B) | G) ]

The latter term Cov[E(P_A)E(P_B) | G) ] = ab Cov(G,G) = ab Var(G) = ab (ab^2) = a^2 b^3

but i cannot work out the E[Cov(P_A, P_B | G)] term.

I have also considered using the result that a Poisson-Gamma mixture is Negative binomially distributed so it boils down to covariance of two NB variables but that didnt get anywhere either.

Also can you elaborate on your suggested Baynesian approach?
• Jul 20th 2013, 06:22 PM
chiro
Re: Dependent Poissons
The Bayesian idea is that P(A|B) is proportional to P(B|A)*P(A). As it turns out, the Bayesian conjugate prior for a Poisson distribution is a gamma.

Basically if one mean parameter for one random variable is mu1 = aG then the other is c*mu2 = bG so c = b/a (which is a constant).

I recommend you represent one variable as a function of the other (i.e. if X1 is random variable 1, then X2 = f(X1) for some function f) and then use the covariance results to calculate Cov[X1,f(X1)]

Because the only difference will be a b/a term, it should (hopefully) be straight-forward.

If you are stuck with getting the probability functions, then look up conjugate priors for Poisson distributions.
• Jul 21st 2013, 04:29 PM
casio415
Re: Dependent Poissons
Ok this is what I tried:

P(mu | X1 ) proportional to P(X1 | mu) P (mu)

where

(X1 | mu) is poisson distributed
mu is gamma distributed
can be shown that (mu | X1) is gamma distributed.

Now, what I am stuck on is how the above procedure helps to get to Cov[X1,(b/a)X1]=(b/a)Var(X1)