
Sum of two Poissons
Hi there people! Im doing a project based on predicting Football (soccer) results but have hit a slight roadblock when investigating the dependance between the goals scored by the home and away team.
Im modelling the goals scored using the poisson distribution as follows:
H = C + Eh
A = C + Ea
C is independent with H, A, Eh and Ea. H and A are also independent.
Where C is poisson distributed with mean Theta, Eh is poisson distributed with mean (LambdaTheta) and Ea is poisson distributed with mean (MuTheta).
Now, my question is, how woud I calculate the probability H=1,A=1.
Would appreciate any help I could get, my stats isnt too great to be honest. Thanks in advance.

To calculate P(H = 1, A = 1), note that
$\displaystyle
(H = 1, A = 1)
= (H = 1, A = 1, C = 1) + (H = 1, A = 1, C = 0)$
$\displaystyle
= (C = 1, E_h = 0, E_a = 0) + (C = 0, E_h = 1, E_a = 1)
$
Here, the + operator represents a disjoint union. To get other sorts of probabilities, I think this sort of decomposition is what you need to work with: break up the set of interest into a disjoint union of sets where you are varying over C.
I haven't looked at the problem beyond noticing this, but I'm not terribly optimistic that you can do a whole lot better than this. Maybe you can though.