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Thread: y follows a Poisson distribution with rate lambda*x

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    y follows a Poisson distribution with rate lambda*x




    Assume that the number of faults in the roll, say y, follows a Poisson distribution with rate lambda*x, where x is the length of the roll.
    Does this mean the pmf is
    f(y; lambda*x) = exp(-lambda*x)(lambda*x)^y / y! ?

    Thanks

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    Re: y follows a Poisson distribution with rate lambda*x

    Yes, you are correct

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    Re: y follows a Poisson distribution with rate lambda*x

    Thanks!!

    So if lambda follows a Gamma distribution with shape and scale parameters alpha and beta, how do I go about finding the distribution of y (which follows the Poisson distribution with rate lamba*x) ?

    Thanks again!

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    Re: y follows a Poisson distribution with rate lambda*x

    Given

    Y|\{\Lambda=\lambda\} \sim \mbox{Poisson}(\lambda x)

    \Lambda \sim \Gamma(\alpha, \beta)

    Then

    \Pr\{Y = y\}
= \int_0^{+\infty} \Pr\{Y = y|\Lambda = \lambda\} f_{\Lambda}(\lambda)
d\lambda

    The conditional p.m.f. and the density is already known.
    And you just need to evaluate this integral.

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    Re: y follows a Poisson distribution with rate lambda*x


    So I have
    y_i \sim Poisson(\lambda x_i) and \hat{\lambda}_{MLE}={\bar{y}/\bar{x}}

    Now I need to find var(\lambda)
    Should I have
    var(\lambda)=var({\bar{y}/\bar{x}})=(E(\bar{y}))^2 var(1/\bar{x})+(E(1/\bar{x}))^2 var(\bar{y})+var(1/\bar{x})var(\bar{y}

    or should I have
    \lambda = E(y_i)/x_i
    var(\lambda)=var({y_i/x_i})=(E(y_i))^2 var(1/x_i)+(E(1/x_i))^2 var(y_i)+var(1/x_i)var(y_i)

    Which one makes sense??
    Last edited by zzzc; 12-05-2010 at 06:16 PM.

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