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Math Help - Sum of two Poissons

  1. #1
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    Feb 2011
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    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 (Lambda-Theta) and Ea is poisson distributed with mean (Mu-Theta).

    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.
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  2. #2
    Senior Member
    Joined
    Oct 2009
    Posts
    340
    To calculate P(H = 1, A = 1), note that

    <br />
(H = 1, A = 1) <br />
= (H = 1, A = 1, C = 1) + (H = 1, A = 1, C = 0)
    <br />
= (C = 1, E_h = 0, E_a = 0) + (C = 0, E_h = 1, E_a = 1)<br />

    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.
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