= Nested Poisson flood count
{title2=$G_N(z)=\exp\{\nu(e^{\lambda(z-1)}-1)\}$}
If a <Poisson distribution> count $M$ of clusters has mean $\nu$, and each cluster independently contains a Poisson number of claims of mean $\lambda$, then the total count is a <compound Poisson distribution> with the displayed <probability generating function>. Conditioning gives $\mathbb EN=\nu\lambda$ and $\operatorname{Var}N=\nu\lambda(1+\lambda)$, so it is not Poisson for positive parameters. With independent severities of mean $\mu$ and variance $\sigma^2$, the aggregate payment has mean $\nu\lambda\mu$ and variance $\nu[\lambda(\sigma^2+\mu^2)+\lambda^2\mu^2]$. Independent validity thinning replaces the per-cluster count mean by $\lambda(1-p)$ and multiplies expected payment by $1-p$.
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