Compound Poisson Process

Study guide: Exam MAS-I

A Compound Poisson Process adds a random amount to each event of a Poisson Process: if claims arrive at rate $\lambda$ and the $i$ -th claim costs $X_i$ , the aggregate loss by time $t$ is the random sum $S(t) = \sum_{i=1}^{N(t)} X_i$ . It is the standard model for total claims on a portfolio and the process form of the Aggregate Loss Model.

$S(t) = \sum_{i=1}^{N(t)} X_i, \qquad N(t) \sim \text{Poi}(\lambda t)$

$E[S(t)] = \lambda t\, E[X], \qquad \text{Var}(S(t)) = \lambda t\, E[X^2]$

The severities $X_1, X_2, \ldots$ are i.i.d., independent of the counting process $N(t)$
The variance uses the second moment $E[X^2]$ , not the variance of $X$ — this falls out of the compound-variance formula $\text{Var}(S) = E[N]\text{Var}(X) + \text{Var}(N)E[X]^2$ with $E[N] = \text{Var}(N) = \lambda t$
The sum of independent compound Poisson processes is compound Poisson with rate $\lambda = \sum \lambda_j$ and severity distribution the $\lambda$ -weighted mixture of the $F_j$
Thinning carries over: claims of a given type (e.g. those exceeding a Deductible) form their own compound Poisson process, with rate $\lambda P(X > d)$ and severity distributed as the conditional excess
The moment generating function is $M_{S(t)}(r) = \exp\{\lambda t\,[M_X(r) - 1]\}$ , which is how higher moments and approximations are derived
$S(t)$ is not Poisson and is usually right-skewed; for large $\lambda t$ a Normal approximation is workable, but the skewness $\lambda t E[X^3] / (\lambda t E[X^2])^{3/2}$ vanishes only slowly

Mean and Variance of Annual Aggregate Losses Claims arrive at $\lambda = 50$ per year. Severity is exponential with mean $2{,}000$ . Find the mean and standard deviation of aggregate losses in one year.

Answer For an exponential with mean $\theta = 2{,}000$ : $E[X] = 2{,}000$ and $E[X^2] = 2\theta^2 = 8 \times 10^6$ . $E[S(1)] = 50 \times 2{,}000 = 100{,}000$ $\text{Var}(S(1)) = 50 \times 8 \times 10^6 = 4 \times 10^8$ $\text{SD}(S(1)) = \sqrt{4 \times 10^8} = 20{,}000$

Aggregate Losses Above a Deductible With $\lambda = 50$ and exponential severity of mean $2{,}000$ , a $1{,}000$ Deductible is applied to every claim. Find the mean aggregate payment.

Answer Claims that pierce the deductible are a thinned Poisson process with rate $\lambda' = 50 \times P(X > 1{,}000) = 50\,e^{-1000/2000} = 50\,e^{-0.5} \approx 30.33$ By the memoryless property, each such payment is again exponential with mean $2{,}000$ , so $E[S'] = 30.33 \times 2{,}000 \approx 60{,}650$