Aggregate Loss Model

Study guides: Exam ASTAM, Exam MAS-I, Exam MAS-II, Exam 8

An Aggregate Loss Model models the total (aggregate) losses $S$ from a portfolio as the sum of a random number of individual claim amounts:

$S = X_1 + X_2 + \cdots + X_N$

$E[S] = E[N] \cdot E[X]$

$\text{Var}(S) = E[N]\cdot\text{Var}(X) + \text{Var}(N)\cdot(E[X])^2$

$N$ is the claim count (frequency) random variable; $X_i$ are i.i.d. claim severity random variables independent of $N$
This is called the collective risk model
When $N \sim \text{Poi}(\lambda)$ : $\text{Var}(S) = \lambda \cdot E[X^2]$ (simplification since $\text{Var}(N) = E[N] = \lambda$ )
When $N \sim \text{Poi}(\lambda)$ : $S$ has a compound Poisson distribution; the MGF is $M_S(t) = \exp\!\left(\lambda(M_X(t) - 1)\right)$
The Normal approximation via the Central Limit Theorem (CLT) applies when the portfolio is large

Mean and Variance of Aggregate Losses A portfolio has $N \sim \text{Poi}(100)$ claims per year. Individual claim sizes $X_i$ have $E[X] = 500$ and $\text{Var}(X) = 90{,}000$ . Find $E[S]$ and $\text{Var}(S)$ .

Answer $E[S] = E[N]\cdot E[X] = 100 \times 500 = 50{,}000$ $\text{Var}(S) = E[N]\cdot\text{Var}(X) + \text{Var}(N)\cdot(E[X])^2 = 100(90{,}000) + 100(500)^2 = 9{,}000{,}000 + 25{,}000{,}000 = 34{,}000{,}000$

Normal Approximation for Aggregate Losses Using the values above, approximate $P(S > 55{,}000)$ .

Answer $\sigma_S = \sqrt{34{,}000{,}000} \approx 5{,}831$ . $Z = \frac{55{,}000 - 50{,}000}{5{,}831} \approx 0.858$ $P(S > 55{,}000) = 1 - \Phi(0.858) \approx 1 - 0.8045 = 0.1955$