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Math Help - The bivariate poisson distribution.

  1. #1
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    The bivariate poisson distribution.

    Suppose that X_i, i=1,2,3 are independent Poisson random variables with respective means \lambda_i . Let X=X_1+X_2 and [TEXY=X_2+X_3[/TEX]. The random vector X,Y is said to have a bivariate Poisson distribution. Find its joint probability mass function. That is, find P{X=n, Y=m}

    I tried:

    We know
    P{X=x}=\frac{e^{-(\lambda_1 +\lambda_2)}(\lambda_1+\lambda_2)^x}{x!}
    P{Y=y}=\frac{e^{-(\lambda_2 +\lambda_3)}(\lambda_2+\lambda_3)^y}{y!}

    However I get stuck as these two are not independent and cannot therefore just be multiplied

    I also tried to create a new variable Z and use the equation for functions of random variables, but the Jacobian is zero due to the zero row.
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  2. #2
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    Here's my suggestion:

    Try writing the conditional joint pdf of (X,Y) conditioned on X_2.
    (Notice that, given X_2, X and Y are independent)
    Then you multiply that by the pdf of X_2.
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  3. #3
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    If we use the formula

    P(A) = P(A | B)P(B) + P(A | B^c)P(B^c)

    what gives us the justification to ignore the second term on the RHS?
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  4. #4
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    what gives us the justification to ignore the second term on the RHS?
    I dont think that's required. As ilanshom suggests, is the following result not true:

    P(X,Y) = \sum_{k=0}^{\infty} P(X2=k) P(X,Y|X2)

    All terms on the RHS are easy to compute. i also assume the sum will simplify conveniently but i haven't checked.
    Last edited by SpringFan25; May 19th 2011 at 05:23 AM.
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  5. #5
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    Yes, that's what I had in mind. Sorry I forgot to mention it!
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  6. #6
    MHF Contributor matheagle's Avatar
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    did it work?
    Let me know. I'm good with these calculations
    I haven't been around lately.
    Classes are finally over, but I just closed on a condo on the lake.
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