Mixed Poisson Process

Study guide: Exam MAS-I

A Mixed Poisson Process treats the arrival rate itself as random: each risk has its own rate $\Lambda$ drawn from a mixing distribution, and conditional on $\Lambda = \lambda$ the counts follow an ordinary Poisson Process. It is the standard way to model a heterogeneous portfolio in which some insureds are simply more claim-prone than others.

$N(t) \mid \Lambda = \lambda \ \sim\ \text{Poi}(\lambda t)$

$E[N(t)] = t\,E[\Lambda], \qquad \text{Var}(N(t)) = t\,E[\Lambda] + t^2\,\text{Var}(\Lambda)$

The variance always exceeds the mean (for $t$ where $\text{Var}(\Lambda) > 0$ ): mixing produces overdispersion, the signature that a plain Poisson model is too tight for the data
The moments follow from conditioning: $E[N] = E[E[N \mid \Lambda]]$ and $\text{Var}(N) = E[\text{Var}(N\mid\Lambda)] + \text{Var}(E[N\mid\Lambda])$
With a Gamma mixing distribution the unconditional counts are exactly negative binomial — the single most useful fact in this section
Increments are not independent: observing many claims early raises the posterior for $\Lambda$ and so raises the expectation for later intervals. This is the probabilistic basis of experience rating and Bayesian Credibility
Contrast with the Nonhomogeneous Poisson Process, where the rate varies over time but is known; here it varies across risks and is unobserved
Detecting it in data is a job for Exploratory Data Analysis: if the sample variance of counts far exceeds the sample mean, fit a negative binomial rather than a Poisson Regression

Gamma Mixing Gives a Negative Binomial Annual claim counts for a randomly chosen insured are Poisson with rate $\Lambda$ , where $\Lambda \sim \text{Gamma}$ with mean $0.20$ and variance $0.04$ . Find the mean and variance of the number of claims in one year.

Answer $E[N(1)] = E[\Lambda] = 0.20$ $\text{Var}(N(1)) = E[\Lambda] + \text{Var}(\Lambda) = 0.20 + 0.04 = 0.24$ The variance-to-mean ratio is $0.24/0.20 = 1.2 > 1$ , so the portfolio is overdispersed. Because the mixing distribution is Gamma, $N(1)$ is exactly negative binomial.

Why Overdispersion Appears A book of business is half "good" drivers with $\lambda = 0.10$ and half "poor" drivers with $\lambda = 0.30$ . Compare the mean and variance of the annual claim count for a randomly selected driver against a single Poisson model with the same mean.

Answer $E[\Lambda] = 0.5(0.10) + 0.5(0.30) = 0.20$ $E[\Lambda^2] = 0.5(0.01) + 0.5(0.09) = 0.05 \ \Rightarrow\ \text{Var}(\Lambda) = 0.05 - 0.04 = 0.01$ $E[N] = 0.20, \qquad \text{Var}(N) = 0.20 + 0.01 = 0.21$ A single $\text{Poi}(0.20)$ model would claim a variance of $0.20$ . The extra $0.01$ is the between-driver heterogeneity — ignoring it understates the spread of portfolio results.