Collective Risk Model

Let annual claim count be $N$ and severities be $X_1,X_2,\ldots$. Aggregate loss is

with $S=0$ when $N=0$.

1. Baseline assumptions

The classical compound model assumes:

1. $X_i$ are independent and identically distributed;
2. $N$ is independent of all $X_i$;
3. losses use one period, currency, and contract basis.

These assumptions make the derivation possible. They are not universal insurance facts.

2. Mean by conditioning

Given $N=n$,

Using the law of total expectation,

If expected count is 40 and expected severity £5,000, expected annual aggregate is £200,000. Realised loss is not fixed at that amount.

3. Variance by conditioning

Given $N=n$,

The law of total variance gives

The first term is severity variation for a given count; the second is count variation acting through mean severity.

Compound Poisson simplification

If $N\sim\operatorname{Poisson}(\lambda)$, then $E[N]=\operatorname{Var}(N)=\lambda$:

For lognormal severity with mean £5,000 and log-scale SD $\sigma=1$, $E[X^2]=£5{,}000^2e^1$. At $\lambda=40$:

4. Probability-generating and moment-generating functions

If $P_N(z)=E[z^N]$ is the count PGF and $M_X(t)=E[e^{tX}]$ exists, then

For Poisson count,

This identity supports moment derivation and analytic approximations. For heavy-tailed severities such as lognormal, the MGF is infinite for $t>0$; use Laplace transforms, characteristic functions, recursion, or simulation instead of forcing an MGF argument.

5. Distribution methods

Panjer recursion is not a new risk model; it is an algorithm for computing the compound distribution under eligible count laws and discretised severity.

6. Reinsurance enters at the right level

For a per-loss treaty $g(X)$ retained by the insurer,

Replace severity moments by moments of $g(X)$. For an annual aggregate treaty $h(S)$, first construct gross $S$, then apply $h$:

Applying an aggregate treaty claim by claim answers a different contract.

7. Dependence breaks the simple product

If count and severity share a catastrophe state $Z$,

which need not equal $E[N]E[X]$. If high-count states also have high severity, the independence formula understates expected loss and usually understates tail risk.

Practice

1. $E[N]=10$, $\operatorname{Var}(N)=15$, $E[X]=3$, and $\operatorname{Var}(X)=20$. Find $E[S]$ and $\operatorname{Var}(S)$.
2. Why can a Normal approximation produce impossible results for a low-frequency portfolio?
3. Where should a per-occurrence catastrophe treaty be applied?

Next, retain risk-specific structure in the Individual Risk Model.