Introduction
A credibility-weighted rate estimate combines class experience with a collective estimate. In an integrated calculation, the challenge is not just applying a credibility formula: the exposure weights, estimated variance components, model assumptions, and normalization basis must all agree. This tutorial develops those steps for hypothetical loss-cost data.
The model treats each class as having its own underlying expected loss cost. Experience within a class varies around that expectation, while the underlying expectations themselves vary across classes. These are distinct sources of variation, and the credibility assigned to a class depends on their relative sizes.
This is a model of loss costs, not claim counts or individual claim sizes. The examples use no discounting, mortality assumption, or claim severity distribution. Exposures and loss costs are hypothetical; no external claim data or fitted distribution is being assumed.
Class Summaries and Variance Components
For each class, add its exposures to obtain \(W_i=\sum_j w_{ij}\). Its exposure-weighted average loss cost is \(\bar X_i=\sum_j w_{ij}X_{ij}/W_i\). Across all classes, let \(W=\sum_i W_i\) and \(\bar X=\sum_i W_i\bar X_i/W\).
A method-of-moments estimate of \(\phi\) uses the variation among periods within classes. If there are \(N\) observations across \(I\) classes, the estimator is the total weighted within-class sum of squares divided by its degrees of freedom, \(N-I\). To estimate \(\tau^2\), first measure the weighted variation among class averages. That variation includes both true differences between class means and process noise in the observed averages.
The denominator \(D\) is positive when there are at least two classes with positive exposure. The raw method-of-moments estimate of \(\tau^2\) can be negative in a small or noisy data set. A negative estimate is not a possible variance; an implementation may constrain it to zero, but that constraint changes the estimator and should be stated. The examples below have positive estimates.
Worked Example: Estimating Both Variance Components
A hypothetical portfolio has three classes, each observed in two periods. All exposures are one unit. Estimate the process and between-class variance components.
| Class | Period 1 loss cost | Period 2 loss cost | Class average |
|---|---|---|---|
| A | 79 | 81 | 80 |
| B | 99 | 101 | 100 |
| C | 119 | 121 | 120 |
Method. Use the within-class weighted residual sum of squares for \(\widehat{\phi}\), then subtract the expected process contribution from \(B\) before dividing by \(D\). Because every weight is one, all weighted sums here reduce to ordinary sums.
Process component. Each class average is the midpoint of its two observations. The within-class sums of squares are \(1^2+1^2=2\) for A, \(1^2+1^2=2\) for B, and \(1^2+1^2=2\) for C. Thus the total is \(6\). There are \(N=6\) observations and \(I=3\) classes, so
Between-class component. The total exposure is \(W=6\), each \(W_i=2\), and the grand mean is \(\bar X=(80+100+120)/3=100\). Therefore,
Check and common mistake. The expected process contribution to \(B\) is \((I-1)\widehat{\phi}=2(2)=4\), much smaller than the observed \(B=1600\), leaving a positive between-class estimate. Substituting back gives \(D\widehat{\tau^2}+(I-1)\widehat{\phi}=4(399)+4=1600=B\). Do not use \(N-I\) as the denominator for \(\widehat{\tau^2}\); it is the degrees of freedom for the within-class estimate of \(\phi\), not the between-class adjustment.
Why the Between-Class Adjustment Works
The subtraction in the estimator is justified by the expected value of \(B\). Write each observation as \(X_{ij}=\Theta_i+\varepsilon_{ij}\), where the conditional process errors have mean zero and variance \(\phi/w_{ij}\). The weighted average is \(\bar X_i=\Theta_i+\bar\varepsilon_i\), with \(\operatorname{Var}(\bar\varepsilon_i\mid\Theta_i)=\phi/W_i\). Under the stated uncorrelatedness assumptions, different class averages have zero covariance.
For a weighted sum of squared deviations from the weighted grand mean, the expected contribution from the class means is \(\tau^2 D\). The expected contribution from process error is \(\phi(I-1)\): each class average has process variance \(\phi/W_i\), and the grand-mean adjustment removes one degree of freedom. Hence \(E[B]=\tau^2D+(I-1)\phi\), which rearranges to the method-of-moments estimator above when the parameters are replaced by estimates.
The mutual uncorrelatedness of the class means is essential. Knowing only that each class mean has variance \(\tau^2\) does not establish the expected value just derived. For example, if distinct class means have a common covariance \(c\), while process errors remain uncorrelated across classes, then the between-class contribution is \((\tau^2-c)D\), not \(\tau^2D\). The covariance structure must be modeled if class means are dependent.
Credibility-Weighted Loss Costs
Given variance components, the Bühlmann-Straub credibility factor for class \(i\) is \(Z_i=W_i/(W_i+K)\), where \(K=\phi/\tau^2\). The credibility-weighted estimate blends the class average with the collective mean: \(\widehat{\Theta}_i=Z_i\bar X_i+(1-Z_i)\mu\). More exposure increases \(Z_i\); greater process variation increases \(K\) and reduces credibility, while greater between-class variation reduces \(K\) and makes class experience more informative.
When parameters are estimated from data, a practical calculation often substitutes \(\widehat{\phi}\), \(\widehat{\tau^2}\), and an estimated collective mean for their model values. The resulting credibility-weighted loss cost is a plug-in estimate. It does not automatically account for uncertainty in the estimated parameters themselves.
Worked Example: Credibility-Weighting Unequal Class Experience
A hypothetical insurer has two classes. The collective loss cost is \(\mu=100\), and the assumed model parameters are \(\phi=400\) and \(\tau^2=1600\), in consistent loss-cost units. Class A has period loss costs 120 and 80 with exposures 1 and 3. Class B has period loss costs 140 and 100 with exposures 0.5 and 1.5. Calculate the weighted loss-cost estimate for each class.
Method. First calculate each exposure-weighted class average and total exposure. Then calculate \(K\), each class's \(Z_i\), and the credibility blend. These are supplied model parameters, not variance estimates from these four observations.
Class A. Its total exposure is \(W_A=1+3=4\), and its average is \(\bar X_A=(1(120)+3(80))/4=(120+240)/4=90\). The common credibility constant is \(K=400/1600=0.25\). Thus \(Z_A=4/(4+0.25)=16/17\), and
Class B. Its total exposure is \(W_B=0.5+1.5=2\), and its average is \(\bar X_B=(0.5(140)+1.5(100))/2=(70+150)/2=110\). Its credibility is \(Z_B=2/(2+0.25)=8/9\), so
Check and common mistake. Each estimate lies between its class average and the collective mean: A is between 90 and 100, while B is between 100 and 110. The calculations retain exposures when forming class averages; using the simple unweighted average of the two period loss costs would incorrectly give 100 for A and 120 for B. Values are rounded to four decimal places only at the final step.
Converting Estimates to Normalized Relativities
A credibility-weighted loss cost is an estimated level for a class. To express class differences as relativities, divide each estimate by a specified weighted average of estimates. If \(q_i\) is the exposure or portfolio-mix weight selected for normalization, define the average loss-cost estimate as \(\bar L_q=\sum_i q_i\widehat{\Theta}_i/\sum_i q_i\). The normalized relativity is \(r_i=\widehat{\Theta}_i/\bar L_q\). This makes the weighted average relativity equal to one for that selected mix.
The normalization weights \(q_i\) need not equal the experience exposures \(W_i\) used to calculate credibility. They answer a different question: which portfolio mix should have average relativity one? A changed mix can change normalized relativities even when the class loss-cost estimates stay fixed.
Worked Example: Normalizing Relativities for a Target Mix
Use the loss-cost estimates from the preceding example: \(\widehat{\Theta}_A=1540/17\) and \(\widehat{\Theta}_B=980/9\). A hypothetical target mix has normalization weights \(q_A=2\) and \(q_B=3\). Calculate normalized relativities and check their weighted average.
Method. Compute the target-mix weighted average loss cost, divide each class estimate by that average, and verify the weighted-average constraint.
Target-mix average. The weighted total is \(2(1540/17)+3(980/9)=3080/17+980/3\). Using denominator 51 gives \(9240/51+16660/51=25900/51\). The total weight is \(2+3=5\), so
Relativities. For A, \(r_A=(1540/17)/(5180/51)=4620/5180=0.8919\). For B, \(r_B=(980/9)/(5180/51)=8330/7770=1.0721\). (The exact fractions reduce to \(693/777\) and \(833/777\), respectively.)
Check and common mistake. With unrounded values, \(2r_A+3r_B=2(693/777)+3(833/777)=1386/777+2499/777=5\); dividing by the total weight 5 gives a weighted average of exactly one. Rounding the displayed relativities to four decimals can produce a tiny discrepancy in the check. Do not normalize each estimate by an arbitrary class's loss cost or by an unweighted mean when the stated target mix requires weighted normalization.
Assumption Checks and Exam Traps
A numerical result is meaningful only to the extent that the model assumptions are plausible. The setup requires a common collective mean and common process and between-class variance parameters, with process variance inversely proportional to exposure. Exposure weights should represent the amount of comparable risk underlying each loss-cost observation. If periods are correlated, if their loss-cost definitions or risk mix change, or if process variance does not scale as \(\phi/w_{ij}\), the stated estimators and credibility factor may not be appropriate without adjustment.
The cross-class condition deserves a separate check. A common variance for each \(\Theta_i\) does not imply mutually uncorrelated class means. For a concrete diagnostic, suppose \(\tau^2=399\), \(\phi=2\), \(I=3\), and \(D=4\), but distinct class means have common covariance \(c=100\). The independent-class calculation would predict \(E[B]=399(4)+2(2)=1600\). Under the stated common covariance, the correct expectation is \((399-100)(4)+2(2)=1200\). Using the independent-class adjustment in that dependent setting would therefore overstate the expected between-class variation by 400. This is not a small technical detail: the variance-component estimator must match the assumed dependence structure.
Check Your Understanding
For each problem, use the stated Bühlmann-Straub assumptions and keep full precision until the final answer.
- A class has two observations with exposures 1 and 2 and loss costs 70 and 100. Given \(\phi=180\) and \(\tau^2=720\), calculate its credibility factor. (Answer: \(Z=12/13=0.9231\))
- A class has exposure 2 at loss cost 90 and exposure 1 at loss cost 60. The collective mean is 75, with \(\phi=300\) and \(\tau^2=1200\). Calculate its exposure-weighted average and credibility-weighted estimate. (Answer: \(\bar X=80\), \(\widehat{\Theta}=79.6154\))
- There are four classes and eight observations. The total weighted within-class sum of squares is 84. Each class has total exposure 2; the observed between-class statistic is \(B=420\). Estimate \(\phi\) and \(\tau^2\). (Answer: \(\widehat{\phi}=21\), \(\widehat{\tau^2}=59.5\))
- Two credibility-weighted loss-cost estimates are 72 and 96. Normalize them using target-mix weights 1 and 2, respectively. (Answer: target average \(=88\); relativities \(=0.8182\) and \(1.0909\))
- Suppose \(I=3\), \(D=5\), \(\tau^2=50\), \(\phi=4\), and distinct class means have common covariance \(c=10\); process errors are uncorrelated across classes. What is the expected value of \(B\)? (Answer: \(E[B]=(50-10)(5)+(3-1)(4)=208\))