= Solution
For an <aggregate claims model>, let $X$ denote one claim size, with <expected value> $\mu$ and <variance> $\sigma^2$. The total $S$ is zero when the claim count is zero. Given $N=n$, <independence> gives
$$
\mathbb E(S\mid N)=N\mu,\qquad \operatorname{Var}(S\mid N)=N\sigma^2.
$$
The <law of total expectation> and <law of total variance>, together with $\mathbb EN=\operatorname{Var}(N)=\lambda$ for a <Poisson distribution>, imply
$$
\boxed{\mathbb ES=\lambda\mu,\qquad
\operatorname{Var}(S)=\lambda(\sigma^2+\mu^2)=\lambda\mathbb EX^2.}
$$
The raw second moment appears because the random count itself contributes $\lambda\mu^2$ to the <variance>.
Conditional on $N$, the <moment-generating function> of the sum is $M_X(t)^N$. Averaging with the <Poisson distribution> therefore gives the <compound Poisson distribution> transform
$$
\boxed{M_S(t)=\exp\{\lambda(M_X(t)-1)\}.}
$$
This is valid wherever $M_X(t)$ is finite. Finite first and second moments alone do not ensure positive exponential moments; for positive claims the corresponding <Laplace transform> always exists.
For the independent portfolios put $\Lambda=\lambda_1+\lambda_2>0$ and $w_i=\lambda_i/\Lambda$. Their <Poisson distribution> counts add to a <Poisson distribution> with parameter $\Lambda$. By <Poisson-multinomial conditioning>, conditional on the total count $n$ the first risk count has a <binomial distribution> with parameters $n,w_1$, and the second is the remaining count. Thus one can generate the same total loss by drawing $n$ independent risk labels with these weights, then drawing each claim from its label's law. The merged severity has <mixture distribution>
$$
f(x)=w_1f_1(x)+w_2f_2(x).
$$
It follows that $T$ has a <compound Poisson distribution> with count parameter $\Lambda$ and this severity law. This is the fixed-year version of <Poisson superposition of insurance portfolios>. Its <expected value> and <variance> are
$$
\boxed{\mathbb ET=\lambda_1\mu_1+\lambda_2\mu_2,\qquad
\operatorname{Var}(T)=\sum_{i=1}^2\lambda_i(\sigma_i^2+\mu_i^2).}
$$
Alternatively, multiplying the two independent aggregate <Laplace transforms> yields $\exp\{\Lambda(w_1M_{X_1}(t)+w_2M_{X_2}(t)-1)\}$ wherever finite, confirming the same <compound Poisson distribution>.
For the <retained compound Poisson aggregate> under per-claim <reinsurance>, replace each claim $X$ by its retained payment $g(X)$. Assume the retention is measurable, with $0\le g(x)\le x$ as usual. The count parameter remains $\Lambda$, and the severity law is the <pushforward measure> of the mixture severity under $g$. Thus the <retained compound Poisson aggregate> $T_I$ has a <compound Poisson distribution> with that transformed severity and
$$
\boxed{\mathbb ET_I=\sum_{i=1}^2\lambda_i\mathbb E g(X_i),\qquad
\operatorname{Var}(T_I)=\sum_{i=1}^2\lambda_i\mathbb E[g(X_i)^2].}
$$
Its <moment-generating function> is $\exp\{\sum_i\lambda_i(M_{g(X_i)}(t)-1)\}$ on its finite domain. For a general retention $g$, zero retained payments can occur and are allowed as compound-Poisson marks; they may equivalently be removed by <Poisson thinning>. The two specified contracts retain strictly positive payments for strictly positive claims.
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