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Math Help - Help with probability generating function

  1. #1
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    Help with probability generating function

    Let X1, X2, .... be a sequence of iid random variables with PMF:

    P(X=x)= [a(1-p)^x]/x, x=1,2,3...., with 1>1-p>0 and a=-1/log(p)

    If N is independent of the Xi and has a Poisson distribution with parameter u (au is an integer) then show that S, the sum from i=1 to N of the Xi has a Negative Binomial distribution with PGF:


    Gs(S)=[p/1-s(1-p)]^au

    Any hints how to do this?

    I know that Gs(S)=GN(Gx(s))

    I have tried this many times by trying first to calculate Gx(s), with the thought that I should get something like the PGF of a geometric but am getting nowhere.

    Can anyone hint of how to get started etc
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  2. #2
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    Re: Help with probability generating function

    Hey darren86.

    What did you get for the product of PGF's and the PGF itself?

    Take a look here for more details (you need to get the PGF for one variable first using E[z^x])

    Probability-generating function - Wikipedia, the free encyclopedia
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  3. #3
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    Re: Help with probability generating function

    Quote Originally Posted by chiro View Post
    Hey darren86.

    What did you get for the product of PGF's and the PGF itself?

    Take a look here for more details (you need to get the PGF for one variable first using E[z^x])

    Probability-generating function - Wikipedia, the free encyclopedia
    This is where I'm having trouble, Gx(s)=E[s^x]=???

    a*the sum of {[s(1-p)]^x}/x for x=1 to infinity.

    I presume this is a common power series or something, but I cant find what it is????
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  4. #4
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    Re: Help with probability generating function

    Have you considered transforming the random variable to a geometric random variable and then using the property that the sum of independent geometric has a negative binomial?

    (Hint: Transform it by transforming the parameter and not the distribution itself since mean of geometric is 1/p and variance is [1-p]/p^2)

    If you transform the parameter and show that it has the distribution form you can take it from there.
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  5. #5
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    Re: Help with probability generating function

    Quote Originally Posted by chiro View Post
    Have you considered transforming the random variable to a geometric random variable and then using the property that the sum of independent geometric has a negative binomial?

    (Hint: Transform it by transforming the parameter and not the distribution itself since mean of geometric is 1/p and variance is [1-p]/p^2)

    If you transform the parameter and show that it has the distribution form you can take it from there.
    I have thought about showing that the Xi are geometric random variables, but how can this be done when there is a x in the denominator? I have thought about different parameters for a geometric random variable but cannot see how we can have that x in the denominator, any hints?
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  6. #6
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    Re: Help with probability generating function

    I just took a look at the Wiki site and the sum of geometric random variables is definitely a Negative Binomial.

    Since distributions are unique it means that the only way a sum of IID random variables gives a Negative Binomial is that all random variables (the Xi's) are also geometric random variables or some other formulation which happens to be (from Wikipedia) the logarithmic distribution.

    Negative binomial distribution - Wikipedia, the free encyclopedia

    Logarithmic distribution - Wikipedia, the free encyclopedia

    I don't think in all honesty that I would have seen the connection myself, but the wiki page outlines exactly what you need and why it is the case.
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