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Suppose $N\sim Poisson(\lambda)$ and $\lambda\sim Unif(0, 10)$ what is the expected value $E(N)$?

I feel like I should just plug in the expected value of Unif(0, 10) but that seems too easy. Anyone have any thoughts?

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1 Answer 1

up vote 3 down vote accepted

${\rm E}(N) = {\rm E}{\rm E}(N|\lambda ) = {\rm E}(\lambda ) = 5$.

Elaborating. By the law of total expectation, $$ {\rm E}(N) = {\rm E}{\rm E}(N|\lambda ). $$ To show that ${\rm E}(N|\lambda ) = \lambda$, note that, given $\lambda = s$, $N$ has mean $s$. Hence, ${\rm E}(N|\lambda =s) = s$, and in turn ${\rm E}(N|\lambda ) = \lambda$. Thus ${\rm E}(N) = {\rm E}(\lambda) = 5$.

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Awesome, thanks a lot. – Deddryk Jun 20 '11 at 7:52

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