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Fun distribution problem

  1. Feb 16, 2012 #1
    1. The problem statement, all variables and given/known data

    I need to find the distribution of [itex]A[/itex]

    2. Relevant equations

    [itex]A|B\sim \text{Poisson}(B)[/itex]

    [itex]B\sim \text{Exponential}(\mu)[/itex]

    3. The attempt at a solution
    [itex]\displaystyle P(A=k)=E[P(A=k|B)]=E\left[e^{-B}\cdot\frac{B^k}{k!}\right]=\ldots[/itex]

    Not sure how to calculate this...
     
  2. jcsd
  3. Feb 16, 2012 #2
    Let me guess the A|B is Poisson with parameter B, where B is exponential with parameter μ, so A is discrete and B is continuous, the problem is finding the marginal distribution of the discrete A, the joint pdf of A and B is f(A,B)=Poisson(A;B)*Exp(B;μ), so that the marginal pdf of A is ∫dB f(A,B)=∫dB Poisson(A;B)*Exp(B;μ);
    If it ends up becoming evaluating [itex]\int dB e^{-B} B^k[/itex] for some k, start with k=0, and try integration by parts to come up with a recursive relation from k to k+1, and evaluate in closed form if possible.
     
    Last edited: Feb 16, 2012
  4. Feb 16, 2012 #3
    [tex]=\displaystyle\int_{0}^{\infty}e^{-b}\frac{b^k}{k!}\mu e^{-\mu b}db[/tex]

    [tex]=\frac{\mu}{k!}\displaystyle\int_{0}^{\infty}b^ke^{-b(1+\mu)} db[/tex]

    This is where I am at... not sure what to do now.
     
  5. Feb 16, 2012 #4
    Integration by parts to go from k to k+1
     
  6. Feb 16, 2012 #5

    Ray Vickson

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    Change variables in the integral; you ought to be able to get a Gamma function; seehttp://en.wikipedia.org/wiki/Gamma_function .

    RGV
     
  7. Feb 20, 2012 #6
    [itex]= \frac{\mu}{k!}\frac{\Gamma(k+1)}{(1+\mu)^{k+1}}= \frac{\mu}{(1+\mu)^{k+1}}[/itex]

    [itex]= \frac{\mu}{1+\mu}\left(1-\frac{\mu}{1+\mu}\right)^k[/itex]

    Would this be geometric then? What is the parameter?
     
    Last edited: Feb 20, 2012
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