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Math Help - Poisson problem with unconditional probability

  1. #1
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    Poisson problem with unconditional probability

    Hi all,

    lets say that, the number of defects X per metre in a piece of fabric has a Poisson distribution with parameter \lambda.
    This condition on \lambda being known, that is, we could say that:
    P(X=k|\lambda)=f_{X|\lambda}(X|\lambda)=\frac{\lam  bda^{k}}{k!}e^{-k}.
    And we suppose that \lambda is also a random variable with probability density function given by
    f(\lambda)=e^{-\lambda}, \lambda > 0.
    how would you show that the unconditional marginal probability of X is
    P(X=k)=(\frac{1}{2})^{k+1}                   .......k=0,1,2..
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  2. #2
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    Re: Poisson problem with unconditional probability

    Hey kingcobra.

    Recall Conditional Probability which says P(A|B) = P(A and B)/P(B) so P(X = k, lambda = whatever) = P(X = k|lambda)*P(lambda)

    Now you have the joint distribution for X and lambda, so now given this can you sum out (or integrate out in continuous distribution) lambda to get P(X = k)?

    Hint:

    Marginal distribution - Wikipedia, the free encyclopedia
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  3. #3
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    Re: Poisson problem with unconditional probability

    P[X=k|λ].f(λ) = (e-2λ)(λk)/(k!)

    P[X=k] = ∫(e-2λ)(λk)/(k!) dλ

    = (1/k!) * ∫(λke-2λ)dλ

    Use Gamma function to find ∫(λ^k)(e^-2λ)dλ
    or integration by parts

    = (1/k!) * ∫(λk)(e-2λ)dλ

    = (1/k!) * k!/2k+1

    = (1/2)^k+1
    Last edited by sher84; October 17th 2012 at 01:39 AM.
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