Normal-Normal Credibility for Claim Costs

Why Normal-Normal Credibility?

An insurer may have several years of claim-cost observations for a group, but those observations need not be enough to estimate the group’s underlying average cost precisely. Bayesian credibility combines the observed experience with a prior distribution for that unknown mean. In the normal-normal model, this combination has a particularly useful form: the posterior mean is a weighted average of the prior mean and the sample mean.

This tutorial develops that result for known individual claim-cost variance. The variance of individual costs describes their scatter around the group mean; the prior variance describes uncertainty about what the group mean is. Keeping those two sources of variation separate is essential to obtaining the right credibility estimate.

Model assumptions: Let \(\theta\) be the unknown mean claim cost for a group. Before observing data, assume \(\theta\sim N(\mu_0,\tau^2)\), where \(\mu_0\) and \(\tau^2>0\) are known. Conditional on \(\theta\), suppose \(X_1,\ldots,X_n\) are independent and identically distributed as \(N(\theta,\sigma^2)\), where the individual-cost variance \(\sigma^2>0\) is known. The observed sample is \(x_1,\ldots,x_n\). These assumptions give a normal likelihood and a normal prior for the unknown mean.

Write \(\bar{x}=\frac{1}{n}\sum_{j=1}^n x_j\) for the observed sample mean. Conditional on \(\theta\), the sample mean has distribution \(N(\theta,\sigma^2/n)\). The division by \(n\) is important: averaging independent observations reduces the variance of the sample mean to \(\sigma^2/n\). It is this variance—not \(\sigma^2\) alone—that determines how informative the observed average is about \(\theta\).

Deriving the Posterior Distribution

Bayes’ rule says that the posterior density of \(\theta\), given the observations, is proportional to the prior density times the likelihood. Ignoring factors that do not depend on \(\theta\), their product is

\pi(\theta\mid x_1,\ldots,x_n) \ \propto\ \exp\!\left[-\frac{1}{2}\left\{ \frac{(\theta-\mu_0)^2}{\tau^2} +\frac{\sum_{j=1}^n(x_j-\theta)^2}{\sigma^2} \right\}\right].

To identify the posterior distribution, collect the terms involving \(\theta\). In the likelihood, \(\sum_{j=1}^n(x_j-\theta)^2\) contributes \(n\theta^2-2n\bar{x}\theta\) as its \(\theta\)-dependent part. Together with the prior term, the coefficient of \(\theta^2\) is \(1/\tau^2+n/\sigma^2\), and the coefficient of \(-2\theta\) is \(\mu_0/\tau^2+n\bar{x}/\sigma^2\). Completing the square therefore gives a normal posterior.

\theta\mid x_1,\ldots,x_n\sim N(\mu_n,\tau_n^2), \qquad \tau_n^2=\left(\frac{1}{\tau^2}+\frac{n}{\sigma^2}\right)^{-1}, \qquad \mu_n=\tau_n^2\left(\frac{\mu_0}{\tau^2}+\frac{n\bar{x}}{\sigma^2}\right).

The posterior precision, \(1/\tau_n^2\), is the sum of the prior precision and the data precision. The data precision is \(n/\sigma^2\) because the sample mean has variance \(\sigma^2/n\). This explains both why more observations increase the information about \(\theta\) and why more variable observations contribute less information.

Theorem: Under the stated normal-normal assumptions, the posterior mean can be written as \(\mu_n=(1-Z)\mu_0+Z\bar{x}\), where \(Z=\frac{n\tau^2}{\sigma^2+n\tau^2}\). The posterior variance is \(\tau_n^2=\frac{\tau^2\sigma^2}{\sigma^2+n\tau^2}\). For \(n>0\) and positive variances, \(0<Z<1\). The posterior mean lies between the prior mean and the sample mean, and the posterior variance is smaller than the prior variance.

The weight \(Z\) is the credibility assigned to the observed sample mean in this model. A larger sample size or a larger prior variance increases \(Z\): more data, or less prior certainty about \(\theta\), shifts the estimate toward the data. A larger individual-cost variance \(\sigma^2\) decreases \(Z\), since noisy observations provide less information about the mean. This weight is specific to the assumptions and known variances in this model.

Worked Applications

Worked Example: Updating a Prior With Four Observed Costs

A hypothetical group has individual annual claim costs \(X_j\) measured in dollars. Before observing the group’s experience, an analyst assigns the unknown mean \(\theta\) the prior distribution \(N(100,25)\). Conditional on \(\theta\), each cost is independently \(N(\theta,400)\). The four observed costs are \(120\), \(90\), \(110\), and \(100\). Find the posterior distribution. Use the stated variances in dollars squared.

Method and sample mean. This is a normal-normal update with known individual variance. The observed sample mean is \(\bar{x}=(120+90+110+100)/4=420/4=105\). Here \(\mu_0=100\), \(\tau^2=25\), \(\sigma^2=400\), and \(n=4\).

Posterior variance. The prior precision is \(1/25=0.04\), and the data precision is \(4/400=0.01\). Thus \(\tau_n^2=1/(0.04+0.01)=1/0.05=20\).

Posterior mean. Using precision weights, \(\mu_n=(100(0.04)+105(0.01))/0.05=(4+1.05)/0.05=101\). Equivalently, \(Z=4(25)/(400+4(25))=100/500=0.2\), so \(\mu_n=(1-0.2)(100)+0.2(105)=80+21=101\).

Result and check. The posterior distribution is \(N(101,20)\). The posterior mean lies between 100 and 105, closer to the prior mean because the four observations provide relatively little precision compared with the prior. The posterior variance, 20, is less than the prior variance, 25, as expected.

Common mistake avoided: Do not use \(4/25\) as prior precision or \(4/400\) as prior precision. The prior contributes \(1/\tau^2=1/25\); the four observations contribute \(n/\sigma^2=4/400\).

Worked Example: A More Variable Claim-Cost Sample

For a hypothetical portfolio, the prior distribution for the unknown mean annual claim cost is \(N(500,400)\), in dollars and dollars squared, respectively. Conditional on that mean, individual annual costs are independent \(N(\theta,1600)\). Four observations are \(480\), \(520\), \(560\), and \(520\) dollars. Calculate the posterior mean and variance.

Method and sample mean. The sample mean is \(\bar{x}=(480+520+560+520)/4=2080/4=520\). The prior parameters are \(\mu_0=500\), \(\tau^2=400\); the known individual variance is \(\sigma^2=1600\), and \(n=4\).

Posterior variance. The prior precision is \(1/400=0.0025\), while the sample precision is \(4/1600=0.0025\). Hence \(\tau_n^2=1/(0.0025+0.0025)=200\).

Posterior mean. The two precisions are equal, so the posterior mean is the midpoint: \(\mu_n=(500(0.0025)+520(0.0025))/0.005=(1.25+1.30)/0.005=510\). The credibility-weight calculation gives \(Z=4(400)/(1600+4(400))=1600/3200=0.5\), and \(500+0.5(520-500)=510\).

Result and check. The posterior distribution is \(N(510,200)\). Its mean is halfway between the prior and sample means, consistent with their equal precisions, and its variance is half the prior variance.

Common mistake avoided: Treating \(\bar{x}\) as one observation would give it variance \(1600\) and incorrectly yield \(Z=400/(1600+400)=0.2\), hence a posterior mean of \(504\). The sample mean of four independent observations has variance \(1600/4=400\), so its correct credibility weight is \(0.5\).

Worked Example: Updating Sequentially

A hypothetical program has prior mean claim cost \(\mu_0=40\) dollars and prior variance \(\tau^2=100\) dollars squared. Individual costs, conditional on the unknown mean, are independent normal observations with known variance \(\sigma^2=400\) dollars squared. The first two observed costs are \(30\) and \(50\) dollars; a third cost of \(60\) dollars is then recorded. Find the posterior distribution after all three observations, and verify that sequential updating agrees with a single update.

Single update. The three-cost sample mean is \(\bar{x}=(30+50+60)/3=140/3\). Prior precision is \(1/100=0.01\), and data precision is \(3/400=0.0075\). Therefore \(\tau_n^2=1/(0.01+0.0075)=1/0.0175=400/7\approx57.1429\).

The posterior mean is \(\mu_n=(40(0.01)+(140/3)(0.0075))/0.0175=(0.4+0.35)/0.0175=300/7\approx42.8571\) dollars.

Sequential check. After the first two costs, their mean is \((30+50)/2=40\). The intermediate posterior precision is \(0.01+2/400=0.015\), so the intermediate posterior variance is \(1/0.015=200/3\), and its mean is \(40\). Adding the third observation gives precision \(1/(200/3)+1/400=0.015+0.0025=0.0175\). The updated mean is \((40(0.015)+60(0.0025))/0.0175=(0.6+0.15)/0.0175=300/7\).

Result and check. Both approaches give posterior distribution \(N(300/7,400/7)\), approximately \(N(42.8571,57.1429)\). Sequential updating agrees because each observation contributes its likelihood to the same prior-data product.

Common mistake avoided: After two observations, the intermediate posterior is a distribution for the unknown mean, not a new individual-cost distribution. When incorporating the third cost, combine the intermediate posterior precision with the observation precision \(1/400\), not with the original prior precision again.

Interpretation, Scope, and Exam Traps

The posterior variance \(\tau_n^2\) measures uncertainty about the unknown mean after the observed data are taken into account. It is not the variance of a future individual claim cost, which under the model is \(\sigma^2\) conditional on \(\theta\). Nor is the posterior mean automatically a predicted future cost distribution; here the task is to update beliefs about \(\theta\).

The normal model is mathematically convenient, but a normal distribution assigns positive probability to negative values. If claim costs are strictly nonnegative and a substantial negative-cost probability would be implausible, the model may be unsuitable without a defensible approximation or a different model. The normal-normal calculations also require a known \(\sigma^2\); substituting an estimated variance changes the assumptions and is not the update derived here.

A reliable exam procedure is to identify which variance belongs to the prior and which belongs to an individual observation, calculate the sample mean, then use either precision formulas or the credibility-weight form. Check that the posterior mean lies between the two means and that the posterior variance is below the prior variance. Those checks can reveal a mistaken variance, missing factor of \(n\), or reversed weight. This material is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Write down the model.
Record \(\mu_0,\tau^2,\sigma^2,n\), and confirm that \(\sigma^2\) is the known variance of an individual observation.

Summarize the observations.
Compute \(\bar{x}\); its variance conditional on \(\theta\) is \(\sigma^2/n\).

Update using precision.
Calculate \(\tau_n^2=(1/\tau^2+n/\sigma^2)^{-1}\) and \(\mu_n=\tau_n^2(\mu_0/\tau^2+n\bar{x}/\sigma^2)\).

Check the answer.
Verify that the posterior mean is between \(\mu_0\) and \(\bar{x}\), and that \(\tau_n^2<\tau^2\).

Check Your Understanding

For each problem, assume the stated normal-normal model and use dollars and dollars squared as indicated. Check the posterior mean and variance before comparing with the answer.

A prior for a group’s unknown mean cost is \(N(200,100)\). Individual costs conditional on the mean are independent \(N(\theta,900)\), and three costs have sample mean \(230\). Find the posterior mean and variance. (Answer: posterior distribution \(N(207.50,75)\).)
The prior is \(N(80,64)\), individual-cost variance is \(256\), and the two observed costs are \(70\) and \(90\). Find the posterior distribution. (Answer: posterior distribution \(N(80,42.6667)\), with variance rounded to four decimal places.)
A prior is \(N(50,25)\), individual-cost variance is \(100\), and three observed costs are \(45\), \(55\), and \(65\). Find the credibility weight, posterior mean, and posterior variance. (Answer: \(Z=3/7\), posterior mean \(365/7\approx52.1429\), and posterior variance \(100/7\approx14.2857\).)
A prior is \(N(120,36)\); the known individual-cost variance is \(144\). The observed costs are \(110\), \(130\), and \(160\). Find the posterior mean and variance. (Answer: posterior distribution \(N(880/7,144/7)\), approximately \(N(125.7143,20.5714)\).)
Suppose the prior is \(N(\mu_0,\tau^2)\), the individual-cost variance is \(\sigma^2\), and a sample of \(n\) independent costs has mean \(\bar{x}\). State the posterior variance and explain whether it increases or decreases if \(n\) increases while the other quantities remain fixed. (Answer: \(\tau_n^2=(1/\tau^2+n/\sigma^2)^{-1}\); it decreases as \(n\) increases.)

What You'll Learn