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Thread: Bernoulli distribution and PGF

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    Bernoulli distribution and PGF




    let
    Yi,j be independent Bernoulli random variables with E[Yi,j ] = 1/j. Let Zj = sumYi,j for i=1 to j . Let MZj(s) be the probability generating function of Zj. Calculate
    M∞(s) = limj→∞ MZj(s). Do you recognize this function?

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    Re: Bernoulli distribution and PGF

    What have you tried so far?

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    Re: Bernoulli distribution and PGF

    Yij~Bernoulli
    My(s)= ((1-P)+pe^s)
    Mzj(s)= ((1-P)+pe^s)^j
    Limit Mzj(s)= limit((1-P)+pe^s)^j, when j goes infinity
    = (1-(1/j)+(e^s)/(i/j))^j j goes infinity
    =(1+(e^s-1)/(1/j))^j j goes infinity
    =e^((e^s)-1) ~ Poisson (mean=1, Var=1)

    This is my solution but not sure if it is correct

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    Re: Bernoulli distribution and PGF

    The arithmetic looks good to me but it seems that you are working on the mgf instead of the pgf.

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    Re: Bernoulli distribution and PGF

    As I underestand the difference between mgf and pgf is the fisrt one is for continouse distribution and the second is for discrete. Is there any other diferences?

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    Re: Bernoulli distribution and PGF


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    Re: Bernoulli distribution and PGF


    You are right. I uede the pgf and the result was the same.

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