Aggregate claims model ($S={\sum }_{j=1}^N{X}_j$)

An aggregate claims model separates the number $N$ of claims from their sizes $X_j$, with total $S={\sum }_{j=1}^N{X}_j$ and the empty sum equal to zero. Under independent and identically distributed random variables for claim sizes, independent of $N$, conditional expectation gives tractable moments and transforms.

Independent claim Poisson processes of rates ${\lambda }_i$ merge into a Poisson process of rate $\lambda ={\sum }_i{\lambda }_i$, by the Superposition theorem for Poisson point processes. The merged claim law is a mixture distribution of the individual claim laws with weights ${\lambda }_i/\lambda$. Equivalently, multiplication of the individual compound-Poisson transforms produces $\exp (\lambda [{\sum }_i({\lambda }_i/\lambda )M_i(r)-1])$.

For independent claim sizes with common expected value $m$ and variance $v$, independent of the nonnegative integer count $N$, the aggregate satisfies $\mathbb{E}S=m\mathbb{E}N$ and $\mathrm{Var}(S)=v\mathbb{E}N+m^2\mathrm{Var}(N)$. Its moment-generating function is $G_N(M_X(t))$ wherever finite. These identities follow from the law of total expectation and law of total variance.

The law of a sum of independent identically distributed claims with an independent Poisson distribution count of parameter $\lambda$. If the claim moment-generating function is $M_X$, the aggregate transform is $\exp (\lambda (M_X(r)-1))$ wherever finite. It has an atom at zero when the claims are positive. This is the fixed-time law of a Compound Poisson process.

For positive claim sizes, a random sum of independent claims is zero exactly when its count is zero. Its law is a mixture distribution of a zero atom of mass $p=\mathbb{P}(N=0)$ and, with weight $1-p$, a random sum whose count has the zero-truncated claim-count distribution. An independent Bernoulli random variable multiplying that positive component gives the same law. This is distinct from arbitrarily inserting additional zeros into an otherwise unchanged count law.

If $\Lambda$ has gamma distribution with shape $2$ and rate $p/q$, $N\mid \Lambda$ has Poisson distribution with intensity $\Lambda$, and the independent claim sizes have exponential distribution of mean $\mu$, where $p+q=1$, the aggregate law is $p^2{\delta }_0+2pq\mathrm{Exp}(p/\mu )+q^2\mathrm{Gamma}(2,\text{rate }p/\mu )$. This follows by expanding its moment-generating function as $[p+qp/(p-\mu t){]}^2$.