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Thread: Distribution of a + X

  1. #1
    Super Member Anonymous1's Avatar
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    Distribution of a + X

    Suppose that Θ is a random variable that follows a gamma distribution with parameters λ and α, where α is an integer, and suppose that, conditional on Θ, Χ follows a poisson distribution with parameter Θ. Find the unconditional distribution of α + Χ. (Hint: Find the mgf by using iterated conditional expectations)
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  2. #2
    MHF Contributor matheagle's Avatar
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    Use the hint.

    $\displaystyle E(e^{(a+X)t}) =E(E(e^{(a+X)t})|\theta)$

    $\displaystyle =e^{at}E(E(e^{tX})|\theta) =e^{at}E(e^{\theta(e^t-1)})$

    $\displaystyle =e^{at}M_{\theta}(e^t-1)=e^{at}M_{\theta}(s)$

    SO, plug in $\displaystyle s=e^t-1$ into the MGF of your gamma distribution.
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  3. #3
    MHF Contributor matheagle's Avatar
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    Did you get an interesting result?
    I didn't finish this, because I think that's your job.
    But I would like to see what your final result is.
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  4. #4
    Super Member Anonymous1's Avatar
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    I'll post the result when I'm finished.

    One interesting thing I've found:

    $\displaystyle X \sim Negative Binomial(\alpha, \frac{1}{1+\lambda})$
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  5. #5
    Super Member Anonymous1's Avatar
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    I just did this in the past fifteen minutes so please let me know if things go awry.

    $\displaystyle M_{a+X} = E[e^{(a+X)t}] = e^{at}M_{\theta} (s) = e^{at} (\frac{\lambda}{\lambda -(e^t - 1)})^a$

    Now,

    $\displaystyle M'_{a+X} (0) = a(\frac{\lambda + a}{\lambda}) = E[a+X]$

    Also,

    $\displaystyle E[a+X] = \int_{-\infty}^{\infty} (a+x)f_{a+X} (x) dx$

    Therefore,

    $\displaystyle \int_{-\infty}^{\infty} (a+x)f_{a+X} (x) dx = a(\frac{\lambda + 1}{\lambda})$

    And I'm stuck here. Can I take the derivative of both sides or something?

    $\displaystyle f_{a+x} (x) = \frac{1}{(a+x)}$
    Last edited by Anonymous1; Feb 7th 2010 at 06:20 PM.
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  6. #6
    Super Member Anonymous1's Avatar
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    $\displaystyle \alpha + X \sim Negative Binomial$
    Last edited by Anonymous1; Feb 8th 2010 at 08:41 PM.
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