Perform calculations for sums of compound Poisson models.

Two independent books of business each follow compound Poisson laws. Their aggregate is also compound Poisson, with one frequency parameter and one weighted severity. That single fact powers most exam problems in this LO.

The sum theorem. Suppose $S_1, S_2, \ldots, S_n$ are independent compound Poisson random variables. The i-th has Poisson frequency $\lambda_i$ and severity CDF $F_i$ with severity random variable $X_i$. Then the aggregate is itself compound Poisson.

$$S = \sum_{i=1}^{n} S_i \sim \text{CP}\!\left(\lambda = \sum_{i=1}^{n}\lambda_i,\; F(x) = \frac{1}{\lambda}\sum_{i=1}^{n}\lambda_i F_i(x)\right)$$

The mechanics: each individual Poisson stream of claims arrives independently. Superposing independent Poisson processes gives a Poisson process whose rate is the sum of the rates. Conditional on a claim arriving in the merged stream, the probability it came from stream $i$ is $\lambda_i/\lambda$. So the severity of a generic merged claim is a mixture across the $F_i$ with those weights.

KEY: The merged severity is a mixture, not a sum. You weight individual CDFs by $\lambda_i/\lambda$. You do NOT convolve them.

Common mistakes

• Convolving the severity CDFs instead of mixing them. The merged severity is $F = \sum (\lambda_i/\lambda) F_i$, never a convolution.
• Using mixture weights of $1/n$. Weights are $\lambda_i/\lambda$, proportional to frequency, not equal across components.
• Writing $\text{Var}(S) = \lambda \text{Var}(X)$. The correct formula uses the second raw moment: $\text{Var}(S) = \lambda E[X^{2}]$.

Bottom line

• If $S_i \sim \text{CP}(\lambda_i, F_i)$ are independent, then $S = \sum S_i$ is $\text{CP}(\lambda, F)$ with $\lambda = \sum \lambda_i$ and $F = \sum (\lambda_i/\lambda) F_i$.
• The combined severity is a mixture with mixing weights $\lambda_i/\lambda$ (proportional to frequency), not a convolution.
• $E[S] = \sum \lambda_i E[X_i]$ and $\text{Var}(S) = \sum \lambda_i E[X_i^2]$ use the second raw moment and bypass the mixture entirely.
• MGF factorizes: $M_S(t) = \prod \exp(\lambda_i(M_{X_i}(t)-1))$, so exponents add and total rate is $\lambda$.

Exam shortcut

For aggregate mean and variance on a sum, skip the mixture: compute $\sum \lambda_i E[X_i]$ and $\sum \lambda_i E[X_i^{2}]$ directly. For deductible or excess-of-loss problems, multiply each stream's lambda by $\Pr(X_i > d)$ to get the ceded frequency, then use the conditional excess as the ceded severity. Check closure via the MGF: if $M_S(t) = \exp(\lambda(M_X(t) - 1))$ holds for some valid $\lambda$ and severity $X$, the sum is compound Poisson.

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Learning objectives

• 2c

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