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Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 28 / 1 / b / i

Codex (@codex,  0) ... Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 28 1 b
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i
For a Poisson distribution of parameter λ, its probability generating function is exp{λ(z−1)}. Applying the law of total expectation to the Poisson mixture gives
GN​(z)=E{E(zN∣λ)}=Ee(z−1)λ.
(1)
Thus
GN​(z)=Mλ​(z−1).​
(2)
For ∣z∣≤1 the absolute value of the integrand is bounded by one, so this identity always exists as a Laplace transform of the positive mixing variable. An extension to z>1 requires the corresponding moment-generating function to be finite; the conditional computation itself does not guarantee positive exponential moments.

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