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Year 2 of the mathematics course of the University of Oxford

Probability course of the University of Oxford

Past exam of the probability course of the University of Oxford

2015

Using the <probability mass function> of the <Poisson distribution> and the power series of the <exponential function>,
$$
M_X(t)=\sum_{j=0}^{\infty}e^{tj}e^{-\mu}\frac{\mu^j}{j!}
=e^{-\mu}\sum_{j=0}^{\infty}\frac{(\mu e^t)^j}{j!}
=\boxed{\exp\{\mu(e^t-1)\}.}
$$
This <moment-generating function of a Poisson distribution> is finite for every real $t$. It also covers $\mu=0$, when $X=0$ with <probability> one and the <moment-generating function> is identically one.

Solution

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

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

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

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