= Moment-generating function of a Poisson distribution
{title2=$M_N(t)=\exp(\lambda(e^t-1))$}
= Poisson moment-generating function
{c}
{synonym}
For $N\sim\operatorname{Pois}(\lambda)$ with $\lambda\geq0$, insert its <probability mass function> into the <expected value>:
$$
\mathbb E e^{tN}=e^{-\lambda}\sum_{j\geq0}\frac{(\lambda e^t)^j}{j!}
=\exp(\lambda(e^t-1)),\qquad t\in\mathbb R.
$$
The <exponential series> shows finiteness for every real $t$. Differentiating this <moment-generating function> gives <mean> and <variance> both $\lambda$. Taking the <logarithm> gives the <cumulant-generating function> $\lambda(e^t-1)$, so every positive-order <cumulant> is also $\lambda$.
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