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Bivariate poisson - probability

  1. Apr 30, 2008 #1
    1. The problem statement, all variables and given/known data
    Let {Mi(t), t >= 0 }, i=1, 2 be independent Poisson processes with respective rates λi, i=1, 2, and set

    N1(t) = M1(t) + M2(t), N2(t) = M2(t) + M3(t)

    The stochastic process {(N1(t), N2(t)), t >= 0} is called a bivariate Poisson process.
    (a) Find P{N1(t) = n, N2(t) = m}
    (b) Find Cov (N1(t), N2(t))



    2. Relevant equations



    3. The attempt at a solution
    I am trying to solve this problem as follows:
    (a)
    P{M1(t) +M2(t) = n, M2(t) +M3(t) = m} = P{M1(t) +M2(t) = n | M2(t) +M3(t) = m} / P{ M2(t) +M3(t) = m}
    which then equals by independence of M1(t) +M2(t), M2(t) + M3(t)
    = P{M1(t) +M2(t) = n}P{M2(t) +M3(t) = m} / P{ M2(t) +M3(t) = m}

    Now, are they independent? Or is my assumption wrong here? I'm starting to think I should condition on the value of M2(t) here, but conditioning on one of 2 variables is starting to confuse me!

    AND

    (b) By using the fact that:

    Cov (X, Y) = E[XY] - E[X]E[Y]

    (substituting in for N1(t) and N2(t))

    Cov (N1(t), N2(t)) = E[N1(t)N2(t)] - E[N1(t)]E[N2(t)]

    and I can get:
    E[N1(t)] = E[M1(t) + M2(t)] = E[M1(t)] + E[M2(t)] = (λ1t) + (λ2t)
    E[N2(t)] = E[M2(t) + M3(t)] = E[M2(t)] + E[M3(t)] = (λ2t) + (λ3t)

    Is my following assumption correct?
    E[N1(t)] N2(t)] = E[{M1(t) + M2(t)}{M2(t) + M3(t)}]
    = E[M1(t)M2(t) + M2(t)M2(t) + M1(t)M3(t) + M2(t)M3(t)]
    = E[M1(t)]E[M2(t)] + E[M2(t)]E[M2(t)] + E[M1(t)]E[M3(t)] + E[M2(t)]E[M3(t)]
    = (λ1t)(λ2t) + (λ2t)(λ2t) + (λ1t)(λ3t) + (λ2t)(λ3t)

    which would give me the Covariance? Or should I solve this by using the Conditional Covariance formula? I am assumming that M1(t), M2(t), and M3(t) are all independent.
     
  2. jcsd
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