Past exam of the probability course of the University of Oxford / 2015 / 1 / b / ii

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- Solution ii

Solution

Using the probability mass function of the Poisson distribution and the power series of the exponential function,

M_{X} (t) = \sum_{j = 0}^{\infty} e^{t j} e^{- μ} \frac{μ ^{j}}{j !} = e^{- μ} \sum_{j = 0}^{\infty} \frac{( μ e ^{t} ) ^{j}}{j !} = exp {μ (e^{t} - 1)} .

This moment-generating function of a Poisson distribution is finite for every real

t

. It also covers

μ = 0

, when

X = 0

with probability one and the moment-generating function is identically one.

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