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Math Help - Poisson distribution moment generating function

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    Poisson distribution moment generating function

    Let Y1, Y2,...,Yn be independent Poisson random variables with means ƛ1, ƛ2, ...,ƛn, respectively.
    Find the: Probability function of Yi using the method of moment generating functions.

    I am unsure of how to do this question so any help would be great.
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  2. #2
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    Quote Originally Posted by redwings6 View Post
    Let Y1, Y2,...,Yn be independent Poisson random variables with means ƛ1, ƛ2, ...,ƛn, respectively.
    Find the: Probability function of Yi using the method of moment generating functions.

    I am unsure of how to do this question so any help would be great.
    Do you mean find the pdf of U = Y1 + Y2 + .... + Yn using the method of moment generating functions? What have you tried and where do you get stuck?
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    Yeah that is exactly what i mean. The sum of Yi (ie. from Y=1 to Y=n). I am unsure of how to set it up and then do the first line of calculations.
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    Quote Originally Posted by redwings6 View Post
    Yeah that is exactly what i mean. The sum of Yi (ie. from Y=1 to Y=n). I am unsure of how to set it up and then do the first line of calculations.
    Well, do you know what the mgf is for a Poisson distribution? And do you realise that for a sum of independent rv's you multiply the mgf's? Please show what you've done.
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    Quote Originally Posted by mr fantastic View Post
    Well, do you know what the mgf is for a Poisson distribution? And do you realise that for a sum of independent rv's you multiply the mgf's? Please show what you've done.
    Well I know that the mgf of Poisson is: exp[ƛ(e^t - 1)]
    I also realize that since they are indepedent you can multiply them together. ie. My1(t) = exp[ƛ1(e^t - 1)], My2(t) = exp[ƛ2(e^t - 1)],....,Myn(t) = exp[ƛn(e^t - 1)] ------> My1(t)*My2(t)*...*Myn(t)

    I just don't know how that when multiplying these together you obtain the Poisson probability function: p(n) = (ƛ^n*e^-ƛ)/n!
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    Quote Originally Posted by redwings6 View Post
    Well I know that the mgf of Poisson is: exp[ƛ(e^t - 1)]
    I also realize that since they are indepedent you can multiply them together. ie. My1(t) = exp[ƛ1(e^t - 1)], My2(t) = exp[ƛ2(e^t - 1)],....,Myn(t) = exp[ƛn(e^t - 1)] ------> My1(t)*My2(t)*...*Myn(t)

    I just don't know how that when multiplying these together you obtain the Poisson probability function: p(n) = (ƛ^n*e^-ƛ)/n!
    Use a basic rule for multiplying exponentials to get M(t) = \exp[(e^t - 1) (\lambda_1 + \lambda_2 + .... + \lambda_n)]. The conclusion is crystal clear.
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    Quote Originally Posted by mr fantastic View Post
    Use a basic rule for multiplying exponentials to get M(t) = \exp[(e^t - 1) (\lambda_1 + \lambda_2 + .... + \lambda_n)]. The conclusion is crystal clear.
    Thank you very much!
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