Posterior Distribution

Study guide: Exam MAS-II

The posterior distribution $\pi(\theta \mid \mathbf{x})$ is the distribution of an unknown risk parameter $\theta$ after a risk's experience $\mathbf{x} = (x_1, \ldots, x_n)$ has been taken into account — the prior $\pi(\theta)$ updated by the likelihood of the data through Bayes Theorem. In Bayesian Credibility it is the distribution every forecast for that risk is averaged over.

$\pi(\theta \mid \mathbf{x}) = \frac{f(\mathbf{x} \mid \theta)\,\pi(\theta)}{\int f(\mathbf{x} \mid \theta)\,\pi(\theta)\,d\theta}$

$\pi(\theta \mid \mathbf{x}) \propto \pi(\theta)\prod_{i=1}^{n} f(x_i \mid \theta)$

The pieces. The prior $\pi(\theta)$ describes how $\theta$ varies across the portfolio before this risk's own experience is seen. The likelihood $f(\mathbf{x} \mid \theta) = \prod f(x_i \mid \theta)$ treats the $x_i$ as independent given $\theta$ . The denominator is the marginal probability of the data; it only normalises. With a discrete prior (risk classes) the integral is a sum.
Work with the kernel. Drop every factor that does not involve $\theta$ , multiply prior by likelihood, and recognise the family from what is left. For a conjugate prior this is the whole technique: Poisson–gamma, binomial–beta, normal–normal and exponential–inverse gamma each return a posterior in the prior's family with updated parameters.
What it is used for. The Bayesian premium is the posterior mean of the hypothetical mean, $E[\mu(\Theta) \mid \mathbf{x}]$ , the estimator that minimises expected squared error. The predictive distribution of the next observation integrates $f(x_{n+1} \mid \theta)$ against the posterior.
Prior versus data. With little data the posterior stays near the prior. As $n$ grows, the likelihood dominates and the posterior concentrates near the MLE. That movement is credibility. When the posterior mean is linear in $\bar{x}$ it equals the Bühlmann estimate exactly — see Exact Credibility.
Updating is sequential: this year's posterior is next year's prior, and processing the years one at a time gives the same result as processing them together.

Updating a Two-Class Driver Prior A portfolio is $75\%$ good drivers, whose annual claim counts are Poisson with mean $0.2$ , and $25\%$ bad drivers, Poisson with mean $0.6$ . A driver chosen at random has $1$ claim in a year. Find the posterior class probabilities and the expected number of claims next year.
Answer
 $\begin{align*} P(N = 1 \mid \text{good}) &= 0.2e^{-0.2} \\ &= 0.16375 \\ P(N = 1 \mid \text{bad}) &= 0.6e^{-0.6} \\ &= 0.32929 \\ P(N = 1) &= 0.75(0.16375) + 0.25(0.32929) \\ &= 0.12281 + 0.08232 \\ &= 0.20513 \end{align*}$ 
The posterior probabilities are $0.12281/0.20513 = 0.5987$ (good) and $0.08232/0.20513 = 0.4013$ (bad).
 $\begin{align*} E[N_2 \mid N_1 = 1] &= 0.5987(0.2) + 0.4013(0.6) \\ &= 0.1197 + 0.2408 \\ &= 0.3605 \end{align*}$ 
One claim moves the chance that this is a bad driver from $25\%$ to $40\%$ , and the forecast from the prior mean $0.30$ to $0.3605$ .

Beta Posterior for a Group Claim Probability A group disability policy covers $50$ employees. Each independently has probability $q$ of a claim in the year, and the prior for $q$ is Beta $(a = 2, b = 18)$ , with mean $0.10$ . This year $9$ employees claim. Find the posterior of $q$ , its mean and its standard deviation.
Answer Keep only the factors that involve $q$ :
 $\begin{align*} \pi(q \mid x) &\propto q^{2-1}(1-q)^{18-1} \cdot q^{9}(1-q)^{41} \\ &= q^{10}(1-q)^{58} \end{align*}$ 
This is the kernel of a Beta $(11, 59)$ distribution.
 $\begin{align*} E[q \mid x] &= \frac{11}{11 + 59} \\ &= 0.1571 \\ \mathrm{Var}(q \mid x) &= \frac{(11)(59)}{(70)^{2}(71)} \\ &= 0.001865 \end{align*}$ 
The posterior SD is $\sqrt{0.001865} = 0.0432$ , against a prior SD of $0.0655$ . The observed $18\%$ pulls the estimate from $10\%$ to $15.7\%$ , and the year's data cut the uncertainty about $q$ by about a third.