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Math Help - Finding the marginal distribution of a random variable w/ a random variable parameter

  1. #1
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    Finding the marginal distribution of a random variable w/ a random variable parameter

    I am a little shaky on my probability, so bear with me if this is a dumb question...

    Anyway, these two random variables are given:

    X : Poisson ( \lambda)
    \lambda : exponential ( \theta)

    And I simply need the marginal distribution of X and the conditional density for \lambda given a value for X

    I have all the equations for dependent distributions, but do not know how to apply them to this ostensibly easy problem...

    Any help?
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  2. #2
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    That's called a mixture of distributions where

    f(x) = \int^{\infty}_0 f(X| \lambda) f(\lambda) d \lambda

    where f(X| \lambda) : Poisson (\lambda) and f(\lambda ) : exponential (\theta)

    http://www.mathhelpforum.com/math-he...ributions.html
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  3. #3
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    Thanks, that sounds good.

    But I have an even dumber question -- how do I get rid of the thetas if I am only integrating out the lambdas? Are the thetas constants? Am I even setting the integral up right?

    <br />
f_X(x) = \int^{\lambda=\infty}_{\lambda=0} \frac{\lambda^{x}}{x!} e^{-\lambda} \times \theta e^{-\theta x} d \lambda<br />
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  4. #4
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    Hello,

    The \theta are treated as constants.
    But you set your integral wrong : f(\lambda)=\theta e^{-\theta {\color{red}\lambda}}


    But I'm not sure it's really a mixture... See the wikipedia articles : http://en.wikipedia.org/wiki/Mixture_model and http://en.wikipedia.org/wiki/Mixture_density for the definitions...
    Also, I can't see the formula you've been given

    What d'you think ? I would be pleased if I'm wrong !
    Last edited by Moo; January 27th 2010 at 11:51 PM.
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  5. #5
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    Quote Originally Posted by Moo View Post
    But I'm not sure it's really a mixture... See the wikipedia articles : Mixture model - Wikipedia, the free encyclopedia and Mixture density - Wikipedia, the free encyclopedia for the definitions...
    Also, I can't see the formula you've been given
    Maybe it's not. In my textbook, there's a section called "Mixture Distributions" and the questions/examples are similiar in nature.
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  6. #6
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    That makes more sense.

    "Also, I can't see the formula you've been given"

    What formula? When I said, "I have all the equations for dependent distributions", I was referring to the fact that I have all the books and material required for this type of problem, but I just do not know which one to use.

    "What d'you think ? I would be pleased if I'm wrong !"

    I was thinking of using Bayes' theorem. Would that yield the same answer? (I guess I will find out... working on it now.)
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  7. #7
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    It is impossible to evaluate the integral above, so that is no bueno.
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  8. #8
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    In order to evaluate, you'll need to read up on the Gamma Function:



    You'll be able to see it easier if you take out the constants from the integral.

    http://en.wikipedia.org/wiki/Gamma_function
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  9. #9
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    OKAY, I take that back. With a little manipulation, it is doable.

    u = lambda(1+theta)
    du=(1+theta)dlambda
    lambda^x=u^x/(1+theta)^x

    So you are left with the integral,

    C*integral(u^k*e^(-u),u,0,inf)

    Where C = theta/(x!(1+theta)^(x+1))

    Which makes the answer theta/(1+theta)^(x+1) for x=0,1,...
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  10. #10
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    Yeah, I started typing that before I read your last post

    Thanks, man.
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