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Probability course of the University of Oxford

Past exam of the probability course of the University of Oxford

2015

The <Poisson process> gives $N_k\sim\operatorname{Pois}(k)$, with <expected value> and <variance> both equal to $k$. Consequently $Y_k=(N_k-k)/\sqrt{k}$. Scaling and translating the argument in the <moment-generating function of a Poisson distribution> gives, for every real $t$,
$$
\begin{aligned}
M_{Y_k}(t)
&=e^{-t\sqrt{k}}M_{N_k}(t/\sqrt{k})\\
&=\boxed{\exp\!\left\{k\left(e^{t/\sqrt{k}}-1-\frac{t}{\sqrt{k}}\right)\right\}.}
\end{aligned}
$$
The centering term must remain inside the exponent: omitting it would leave a divergent first-order contribution.

Solution

Past exam of the probability course of the University of Oxford / 2015 / 1 / c / i / Solution

Past exam of the probability course of the University of Oxford / 2015 / 1 / c / i / Solution

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