Conjugate Prior

Study guide: Exam MAS-II

A conjugate prior is a prior distribution for a risk parameter that, combined with a given likelihood, produces a posterior in the same family. Updating then reduces to arithmetic on the parameters — the reason Bayesian Credibility is computable by hand on an exam.

$\pi(\theta) \in \mathcal{F} \;\text{and}\; f(x \mid \theta) \;\Longrightarrow\; \pi(\theta \mid \mathbf{x}) \in \mathcal{F}$

The conjugate pairs on the syllabus:

Likelihood	Prior	Posterior	Posterior mean
Poisson $(\lambda)$	Gamma $(\alpha, \theta)$	Gamma $\left(\alpha + \sum x_i,\; \frac{\theta}{1 + n\theta}\right)$	$Z\bar{x} + (1-Z)\alpha\theta$ , $Z = \frac{n}{n + 1/\theta}$
Bernoulli / Binomial	Beta $(a, b)$	Beta $\left(a + \sum x_i,\; b + n - \sum x_i\right)$	$Z\bar{x} + (1-Z)\frac{a}{a+b}$ , $Z = \frac{n}{n + a + b}$
Normal $(\theta, \sigma^2)$	Normal $(\mu, \tau^2)$	Normal, precision-weighted	$Z\bar{x} + (1-Z)\mu$ , $Z = \frac{n}{n + \sigma^2/\tau^2}$
Exponential $(\theta)$	Inverse Gamma	Inverse Gamma	credibility-weighted

Every posterior mean above is linear in $\bar{x}$ , which is exactly the form Bühlmann Credibility assumes — so for a conjugate pair the Bühlmann and Bayesian estimates are identical, not merely close
The prior parameters act like pseudo-observations: a Gamma $(\alpha, \theta)$ prior behaves like having already seen $1/\theta$ exposures with $\alpha$ claims
Outside a conjugate pair the posterior generally has no closed form and Bühlmann is a linear approximation to it

Gamma-Poisson Update A driver's annual claim count is Poisson $(\Lambda)$ with prior $\Lambda \sim$ Gamma $(\alpha = 2, \theta = 0.10)$ . Over 3 years the driver reports 1 claim. Find the posterior and the Bayesian estimate of next year's frequency.

Answer Posterior: Gamma with $\alpha^* = \alpha + \sum x_i = 2 + 1 = 3$ and $\theta^* = \dfrac{\theta}{1 + n\theta} = \dfrac{0.10}{1 + 0.3} = 0.0769$ . $E[\Lambda \mid \mathbf{x}] = \alpha^*\theta^* = 3(0.0769) = 0.2308$ Check against Bühlmann: $k = 1/\theta = 10$ , $Z = 3/13 = 0.2308$ , $\bar{x} = 1/3$ , $\mu = \alpha\theta = 0.20$ : $0.2308\left(\tfrac{1}{3}\right) + 0.7692(0.20) = 0.0769 + 0.1538 = 0.2308 \;\checkmark$