Mixing Distribution

Study guide: Exam MAS-I

A Mixing Distribution is the distribution of a model parameter that varies randomly from risk to risk. In a Mixed Poisson Process it is the distribution $g(\lambda)$ of the claim rate $\Lambda$ across the portfolio: each insured has a fixed but unknown rate, and a randomly chosen insured's count distribution is the Poisson averaged over $g$ .

$P(N(t) = n) = \int_0^\infty \frac{e^{-\lambda t}(\lambda t)^n}{n!}\, g(\lambda)\, d\lambda$

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

With a discrete mixing distribution (a few risk classes with weights $w_j$ and rates $\lambda_j$ ) the integral becomes a sum, $\sum_j w_j\, e^{-\lambda_j t}(\lambda_j t)^n/n!$ . The mean is always $t\,E[\Lambda]$ ; the extra $t^2\,\text{Var}(\Lambda)$ in the variance is the between-risk heterogeneity.
Gamma mixing gives the negative binomial. If $\Lambda \sim$ Gamma $(\alpha, \theta)$ (shape $\alpha$ , scale $\theta$ ), then $N(t)$ is negative binomial in its failures-count form with $r = \alpha$ and $p = 1/(1 + \theta t)$ :

$P(N(t) = n) = \frac{\Gamma(n + \alpha)}{n!\,\Gamma(\alpha)}\, p^{\alpha}(1-p)^{n}$

The mixing distribution is the prior. After $n$ claims in $t$ years, the gamma rate's posterior is Gamma with shape $\alpha + n$ and scale $\theta/(1 + \theta t)$ ; its mean equals $Z(n/t) + (1-Z)\alpha\theta$ with $Z = t/(t + 1/\theta)$ . This is the gamma–Poisson case of Bayesian Credibility, where Bayes and Bühlmann agree (exact credibility).
Mixing is why a policyholder's past predicts its future — the basis of Experience Rating and a Bonus-Malus System. It also costs the process its independent increments (it keeps stationary ones).
The same device mixes any parameter: an exponential severity whose rate is gamma-distributed is Pareto unconditionally.

Gamma Mixing Across an Auto Portfolio Annual claim rates across a personal auto portfolio are Gamma with $\alpha = 2$ and $\theta = 0.1$ (mean $0.2$ ). For a randomly selected insured, find the probabilities of $0$ and $1$ claims next year, and compare with a $\text{Poi}(0.2)$ model.
Answer Here $p = 1/(1 + 0.1) = 10/11$ :
 $\begin{align*} P(N = 0) &= \left(\tfrac{10}{11}\right)^2 \\ &= 0.8264 \\ P(N = 1) &= \frac{\Gamma(3)}{1!\,\Gamma(2)}\left(\tfrac{10}{11}\right)^2\left(\tfrac{1}{11}\right) \\ &= 0.1503 \end{align*}$ 
The Poisson gives $e^{-0.2} = 0.8187$ and $0.2e^{-0.2} = 0.1637$ . Same mean, but mixing moves probability to zero claims and to the tail: the variance is $2(0.1)(1.1) = 0.22$ against $0.20$ .

Updating One Insured's Rate In the same portfolio, an insured has $3$ claims in $2$ years. Find the posterior mean of its claim rate.
Answer The posterior is Gamma with shape $2 + 3 = 5$ and scale $0.1/(1 + 0.2) = 1/12$ :
 $\begin{align*} E[\Lambda \mid N(2) = 3] &= \frac{5}{12} \\ &= 0.4167 \end{align*}$ 
As a credibility weighting, $Z = 2/(2 + 10) = 1/6$ on the observed $1.5$ per year and $5/6$ on the prior $0.2$ : $0.25 + 0.1667 = 0.4167$ . The rate estimate more than doubles on two years of data, yet most of the weight stays on the portfolio mean.

Two Risk Classes and a Claim-Free Year $70\%$ of drivers have rate $0.1$ and $30\%$ have rate $0.5$ . Find the probability that a randomly chosen driver is claim-free for a year, and the probability that a claim-free driver is high-risk.
Answer
 $\begin{align*} P(N = 0) &= 0.7e^{-0.1} + 0.3e^{-0.5} \\ &= 0.63339 + 0.18196 \\ &= 0.81535 \end{align*}$ 
 $\begin{align*} P(\text{high} \mid N = 0) &= \frac{0.18196}{0.81535} \\ &= 0.2232 \end{align*}$ 
One clean year lowers the chance the driver is high-risk from $30\%$ to $22\%$ — the logic a claim-free discount rests on.