Exact Credibility

Study guide: Exam MAS-II

Exact credibility is the case in which the Bühlmann credibility estimate equals the Bayesian premium $E[\mu(\Theta) \mid \mathbf{X}]$ exactly, not approximately. It happens precisely when the Bayesian premium is itself linear in the observations, which is guaranteed when the model distribution belongs to the linear exponential family and the prior is its natural conjugate prior.

$E[\mu(\Theta) \mid \mathbf{X}] = Z\bar{X} + (1 - Z)\,\mu$

$Z = \frac{n}{n + v/a}$

$\mu = E[\mu(\Theta)]$ is the collective mean, $v$ the EPV and $a$ the VHM, so $v/a$ is the Bühlmann $k$ .
Why linearity is the whole story. The Bühlmann estimate minimises $E\big[(\mu(\Theta) - \alpha_0 - \sum_j \alpha_j X_j)^2\big]$ over linear functions of the data. The Bayesian premium minimises the same expected squared error over all functions. If the Bayesian premium happens to be linear, it is in the smaller class and must be the Bühlmann optimum. Otherwise Bühlmann is the best linear approximation to it.
The sufficient condition. Take a model of the form $f(x \mid \theta) = p(x)e^{r(\theta)x}/q(\theta)$ . This covers the Poisson, binomial, negative binomial, normal with known variance, exponential, and gamma with known shape. Pair it with its natural conjugate prior (subject to a mild condition on the prior's tails). The posterior is then in the prior's family, its mean is linear in $\bar{X}$ , and a prior parameter plays the role of $k$ .
The pairs to know, with k = v/a:
- Poisson $(\Lambda)$ with $\Lambda \sim$ gamma( $\alpha$ , scale $\beta$ ): $k = 1/\beta$
- Bernoulli $(Q)$ with $Q \sim$ beta $(a, b)$ : $k = a + b$
- Normal $(\Theta, \sigma^2)$ with $\Theta \sim$ normal $(\mu, \tau^2)$ : $k = \sigma^2/\tau^2$
- Exponential with mean $\Theta$ , $\Theta \sim$ inverse gamma $(\alpha, \beta)$ : $k = \alpha - 1$
How it is examined. Recognise a conjugate pair and take the shortcut: read $k$ from the prior and skip the posterior algebra. Or, given a discrete prior or a uniform prior on a Poisson mean, compute the two estimates separately, because they will differ.
The normal–normal case reappears in mixed models: with known variance components, the BLUP of a normal random effect is its posterior mean, which is credibility-weighted for exactly this reason.

Severity Credibility with an Inverse Gamma Prior A risk's claim sizes are exponential with mean $\Theta$ . Across the portfolio, $\Theta$ is inverse gamma with $\alpha = 3$ and $\beta = 2{,}000$ , so $E[\Theta] = \beta/(\alpha - 1) = 1{,}000$ . The risk has $4$ claims totalling $\$6{,}000$ . Compute the Bayesian premium and the Bühlmann estimate.
Answer Bayes. The posterior is inverse gamma $(\alpha + n, \beta + \sum x_i) = (7,\ 8{,}000)$ :
 $\begin{align*} E[\Theta \mid \mathbf{x}] &= \frac{8{,}000}{7 - 1} \\ &= 1{,}333.33 \end{align*}$ 
Bühlmann. Here $\mu(\Theta) = \Theta$ and $\sigma^2(\Theta) = \Theta^2$ . For the inverse gamma, $E[\Theta^2] = \beta^2/[(\alpha-1)(\alpha-2)] = 2{,}000{,}000$ :
 $\begin{align*} v &= E[\Theta^2] \\ &= 2{,}000{,}000 \\ a &= E[\Theta^2] - E[\Theta]^2 \\ &= 1{,}000{,}000 \\ Z &= \frac{4}{4 + 2} \\ &= \frac{2}{3} \\ \hat{\mu} &= \tfrac{2}{3}(1{,}500) + \tfrac{1}{3}(1{,}000) \\ &= 1{,}333.33 \end{align*}$ 
The two agree exactly, and $k = \alpha - 1 = 2$ could have been read straight off the prior.

When They Do Not Coincide: A Two-Point Prior Drivers are $75\%$ good (Poisson mean $0.2$ ) and $25\%$ bad (Poisson mean $0.6$ ). The Bayesian premiums after one year with $0$ , $1$ and $2$ claims are $0.2731$ , $0.3605$ and $0.4672$ . Compute the Bühlmann estimates and explain the difference.
Answer
 $\begin{align*} \mu &= 0.75(0.2) + 0.25(0.6) \\ &= 0.30 \\ v &= E[\Lambda] \\ &= 0.30 \\ a &= 0.75(0.04) + 0.25(0.36) - 0.30^2 \\ &= 0.03 \\ Z &= \frac{1}{1 + 0.30/0.03} \\ &= \frac{1}{11} \end{align*}$ 
The Bühlmann estimate is $\frac{1}{11}N + \frac{10}{11}(0.30)$ , giving $0.2727$ , $0.3636$ and $0.4545$ for $N = 0, 1, 2$ .

Bühlmann rises by the same $0.0909$ per claim. The Bayesian premium rises by about $0.087$ , then by about $0.107$ . It is not linear in $N$ , because a two-point prior is not conjugate to the Poisson, so no straight line can match it. Bühlmann is the best straight-line fit to it, close but not exact.