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Math Help - 3 part Hierarchical Probability and Stats Question

  1. #1
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    Question 3 part Hierarchical Probability and Stats Question

    Here's my problem:

    Suppose we have a three-stage hierarchical model

    Y|N~ Binomial(N,p), N|λ ~ Poisson(λ), λ ~ Gamma(a,b)

    Find a marginal distribution for Y.


    What I did was first find the pdf for f(N,λ) by multiplying f(N|λ) * f(λ).

    Next step was to find a marginal probability for N. I integrated f(N,λ) with respect to λ from 0 to infinity which worked out somewhat nicely giving me the marginal distribution for N.

    here's what I got: Γ(N+a) ((b/(b+1))^(N+a)/(N!b^aΓ(a)),
    or, in a more easily looked at format, here

    Now, I found f(N,Y) which is just f(N)*f(Y|N).

    The problem arises when I try to obtain the marginal for Y. I've been looking for a pdf to help make this easier, but I can't see any that do any good.

    Thanks for you help!
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  2. #2
    MHF Contributor matheagle's Avatar
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    I would multiply all three orginal dsitributions then integrate out lambda and finally sum out the N.
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  3. #3
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    Thanks for the advice-- I'll give that a try.
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  4. #4
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    I gave it a try, and I'm still winding up with the same issue...
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  5. #5
    MHF Contributor matheagle's Avatar
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    Maybe try MGFs.
    I'm stuck on the last part, but try this

    M_Y(t)=E(E(E(e^{Yt})|N)|\lambda)

    =E(E((q+pe^t)^N)|\lambda)

    let q+pe^t=e^s

    M_Y(t)=E(E(e^{sN}|\lambda))

     =E(e^{\lambda (e^s-1)}) =E(e^{\lambda (q+pe^t-1)})

     =E(e^{\lambda (pe^t-p)}) =E(e^{\lambda p(e^t-1)})

    Let  u=p(e^t-1)

     =E(e^{\lambda u})=M_{\lambda}(u)

    Plug in your gamma MGF and see what you have.
    I'm going to bed, it's after 2 AM.
    Last edited by matheagle; March 4th 2010 at 12:16 AM.
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  6. #6
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    hee hee- same here. For some reason I can't stop though! G'night.
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  7. #7
    MHF Contributor matheagle's Avatar
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    I was stuck with that same nasty sum.
    I wanted to claim it was a Poisson and the garbage in the sum was just a constant.
    Because you can pull out the

     \left({p\over 1-p}\right)^y {1\over y!}

    which is two-thirds of a Poisson.

    But this gamma seems odd.
    Are you sure of that distribution?
    The parameters a and b are annoying.
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  8. #8
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    I double checked and, yes, the parameters are a&b. I got something that looked a lot like what you have too... I feel like it should turn into the poisson given the parameter of N but I don't know how to manipulate it. I will be going to the professor tomorrow for some help.
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  9. #9
    MHF Contributor matheagle's Avatar
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    1 Y is discrete
    2 It support once N is gone is from 0 to infinity.
    3 The sum looks like the Taylor Series for e, but that a and b mess it up.
    That leads support to a Poisson, but I should recognize this MGF.
    Last edited by matheagle; March 4th 2010 at 07:25 AM.
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  10. #10
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    So apparently my professor said it's a negative binomial. I feel like that doesn't work though... the a's get in the way!
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  11. #11
    MHF Contributor matheagle's Avatar
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    well, I figured that gamma was wrong
    The neg binomial and gamma are quite different.
    I figured it was some kind of exponential
    now I need to know exactly how you write your neg binomial
    You need to know if you're counting the trial in which the r{th} success occurs...
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  12. #12
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    I honestly don't know if it includes rth trial. He said something to the effect that all my N's would get summed into 1 if I could set up the right negative binomial. So far, what I have is something to this effect:

    1/(b^a*y*(a-1)!) * (b/(b+1))^a * Summation from n=0 to infinity [ (N+a-1)!p^y(1-p)^(N-y)* b^N /((y-1)!(N-y)! (b+1)^N)

    This looks an awful lot like a negbin, but instead of (N+a-1)!, I need (N-1)! and I don't know what to do with (b/(b+1))^N.
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  13. #13
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    I'm just about there: It just needs a bit of fine tuning, but I think I can make it look like I want it to. I'll try and post an answer tonight.
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  14. #14
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    Blast! I'm stuck again. Is this a legal negbin?

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  15. #15
    MHF Contributor matheagle's Avatar
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    Oh, I thought that intead of a gamma the distribution of lambda was a neg binominal.
    You're saying that the final distribution is a neg bi, well thats possible and my MGF technique is close,
    but I think there needs to be a relationship between b and p.
    Here r is just a.
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