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If we have a iid random variables $X_i$ with probability generating function $\xi(t) = E[t^{x_i}]$ and $N$ is Poisson with mean $\lambda$ with Probability generating function :
$$ \begin{aligned} \phi(t) = \sum_{i \geq 0} { e^{- \lambda} \lambda ^i \over i!} t^i = e^{-\lambda (1-t)} \end{aligned} $$

Take $Y = \sum_1^NX_i$ what would the probability generating function be in this case?

$$ \begin{aligned} E[t^Y] = E[t^{\sum_1^N X_i}] = E[t^{X_1} ... t^{X_N} ] = E[t^{X_1}] ... E[t^{X_N}] = \xi(t)^N \end{aligned} $$ But this would be assuming that N is not a random variable. I am confused as to how one would take them both into account?

EDIT

So the probability generating function would be: $$ \begin{aligned} E[E[t^{\sum_1^N X_i}|X_i]|N] = E[\xi(t)^N | N] = \sum_{i \geq 0} { e^{- \lambda} \lambda ^i \over i!} \xi(t)^i = e^{-\lambda (1- \xi (t))} \end{aligned} $$ correct?

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    $\begingroup$ You've gotten part way there; now take the expectation of your last term with respect to the probability distribution of $N$. $\endgroup$
    – jbowman
    Sep 21, 2017 at 17:00
  • $\begingroup$ @jbowman is the edit correct? $\endgroup$ Sep 21, 2017 at 17:13
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    $\begingroup$ Right. Might want to post your edit as an answer to your question, then accept it - that way everyone knows it was answered and accepted, plus you did figure it out with just one hint! $\endgroup$
    – jbowman
    Sep 21, 2017 at 17:17

1 Answer 1

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Based on comment: You've gotten part way there; now take the expectation of your last term with respect to the probability distribution of NN. – jbowman

$$ \begin{aligned} E[E[t^{\sum_1^N X_i}|X_i]|N] = E[\xi(t)^N | N] = \sum_{i \geq 0} { e^{- \lambda} \lambda ^i \over i!} \xi(t)^i = e^{-\lambda (1- \xi (t))} \end{aligned} $$

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