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Math Help - Moment Generating Functiom to determine Distribution

  1. #1
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    Moment Generating Functiom to determine Distribution

     \lambda_{1},...,\lambda_{n} > 0 and X_{1},...,X_{n} are independent random variables such that X_{k} ~ Pois(\lambda,r_{k}) for each k = 1,...,n. Use moment generating functions to determine the distribution of X = X_{1} +...+ X_{n}

    So far I have M_{X_1 + X_2 +...+X_n}(t)=M_{x_1}(t)M_{x_2}(t)...M_{x_n}(t)

    = e^{\lambda_{1}(e^t-1)} \star e^{\lambda_{2}(e^t-1)} \star ... \star e^{\lambda_{n}(e^t-1)}

    = e^{e^t-1 [\lambda_{1}+\lambda_{2}+...+\lambda_{n}}]

    I'm not sure what kind of distribution this looks like, if I'm even doing it right.
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  2. #2
    MHF Contributor matheagle's Avatar
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    you're missing a ( )
    I fixed it below, I don't know how to do the red highlight stuff here.


    Quote Originally Posted by Flipz4226 View Post
     \lambda_{1},...,\lambda_{n} > 0 and X_{1},...,X_{n} are independent random variables such that X_{k} ~ Pois(\lambda,r_{k}) for each k = 1,...,n. Use moment generating functions to determine the distribution of X = X_{1} +...+ X_{n}

    So far I have M_{X_1 + X_2 +...+X_n}(t)=M_{x_1}(t)M_{x_2}(t)...M_{x_n}(t)

    = e^{\lambda_{1}(e^t-1)} \star e^{\lambda_{2}(e^t-1)} \star ... \star e^{\lambda_{n}(e^t-1)}

    = e^{(e^t-1) [\lambda_{1}+\lambda_{2}+...+\lambda_{n}]}

    I'm not sure what kind of distribution this looks like, if I'm even doing it right.
    so the sum of indep poissons is a poisson (as I said in class today)
    clearly the mean of the sum is the sum of the means
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  3. #3
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    ok thanks!
    Last edited by Flipz4226; October 5th 2009 at 05:10 PM.
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