Calculating Bühlmann-Straub Class Estimates

Why Unequal Weights Matter

In ratemaking, experience for one class may cover observations with different exposure volumes. A period with 10,000 exposures usually provides more information about the class’s expected loss cost than a period with 1,000 exposures. The Bühlmann-Straub model accounts for that difference by weighting each observation according to its exposure.

The calculation has three central parts: the weighted mean of the class’s observations, the sum of their weights, and the credibility assigned to that experience. The final estimate blends the class’s weighted mean with a collective mean. The greater the total exposure, the more weight the class’s own experience receives.

Definition: For class $i$, let $X_{ij}$ be the observed loss cost or other rate for observation period $j$, and let $w_{ij}>0$ be its exposure weight. The class’s total weight and weighted mean are $$ W_i=\sum_j w_{ij}, \qquad \bar X_i=\frac{\sum_j w_{ij}X_{ij}}{W_i}. $$ Under the Bühlmann-Straub model, conditional on the class parameter $\theta_i$, observations are independent, have conditional mean $\mu(\theta_i)$, and have conditional variance $\sigma^2(\theta_i)/w_{ij}$. Define the expected process variance $\phi=E[\sigma^2(\theta_i)]$, the variance of hypothetical means $a=\operatorname{Var}(\mu(\theta_i))$, and $K=\phi/a$, with $a>0$. Given a collective mean $\mu=E[\mu(\theta_i)]$, the credibility factor and class estimate are $$ Z_i=\frac{W_i}{W_i+K}, \qquad \widehat{\mu}_i=Z_i\bar X_i+(1-Z_i)\mu. $$

The weights must be on a consistent exposure basis, and $X_{ij}$ must represent a rate per unit of that exposure. For example, if one observation is total loss rather than loss per exposure, it must first be put on the appropriate rate basis before this formula is used. A zero-weight observation contributes neither to the weighted mean nor to total weight.

From the Class Data to an Estimate

Use the calculation in a fixed order. First multiply each observation by its weight, then sum those weighted observations and the weights separately. Divide the first total by the second to obtain the weighted class mean. Finally, calculate credibility using total weight and blend the class mean with the collective mean.

Check the data basis.
Confirm that each $X_{ij}$ is a rate per unit of exposure and each $w_{ij}$ is the corresponding exposure volume.

Calculate the class summaries.
Compute $\sum_j w_{ij}X_{ij}$ and $W_i=\sum_jw_{ij}$, then divide to get $\bar X_i$.

Calculate credibility.
Use $K=\phi/a$ and $Z_i=W_i/(W_i+K)$. The same $K$ applies across classes when the structural parameters are common.

Blend and interpret.
Calculate $Z_i\bar X_i+(1-Z_i)\mu$. The class’s own weighted mean receives fraction $Z_i$; the collective mean receives the remainder.

Worked Example: One Class With Three Unequal Weights

A hypothetical coverage class has three observations of loss cost per exposure: $80 with weight 2, $100 with weight 3, and $70 with weight 1. The collective mean is $72. Suppose $\phi=900$ and $a=100$. Calculate the weighted class mean, total weight, credibility factor, and class estimate.

Method. Apply the Bühlmann-Straub weighted-mean and credibility formulas. The weights measure exposure, so each rate is multiplied by its corresponding weight.

1. Total weight. $$ W_i=2+3+1=6. $$

2. Weighted loss-cost total and mean. $$ \sum_jw_{ij}X_{ij} =(2)(80)+(3)(100)+(1)(70) =160+300+70=530, $$ $$ \bar X_i=\frac{530}{6}=\frac{265}{3}=88.3333\ldots. $$

3. Structural ratio and credibility. $$ K=\frac{\phi}{a}=\frac{900}{100}=9, \qquad Z_i=\frac{6}{6+9}=\frac{6}{15}=0.4. $$

4. Credibility-weighted estimate. $$ \widehat{\mu}_i =(0.4)\left(\frac{265}{3}\right)+(0.6)(72) =\frac{106}{3}+\frac{216}{5} =\frac{530+648}{15} =\frac{1178}{15} =78.5333\ldots. $$ Rounded to the nearest cent, the estimate is $78.53 per exposure. As a check, the estimate is between the class mean of about $88.33 and the collective mean of $72, as it must be for a credibility blend with $0\leq Z_i\leq1$.

Common mistake avoided: Treating the three observations as equally weighted gives $(80+100+70)/3=83.33$, which is not the class weighted mean. Using that incorrect mean would give an estimate of about $76.53. The correct calculation gives more influence to the observation with weight 3.

Why the Credibility Formula Has This Form

The class estimate can be understood as a linear projection of the class’s hypothetical mean onto its weighted experience. This provides a short justification for why total weight enters the credibility factor in the stated way.

Write $\mu(\theta_i)$ for the class’s hypothetical mean and $\bar X_i$ for its weighted mean. Conditional independence and the model variances imply

\operatorname{Var}(\bar X_i\mid\theta_i) =\frac{1}{W_i^2}\sum_j w_{ij}^2 \frac{\sigma^2(\theta_i)}{w_{ij}} =\frac{\sigma^2(\theta_i)}{W_i}.

Taking the expected conditional variance gives $\phi/W_i$. The weighted mean has conditional expectation $\mu(\theta_i)$, so the covariance of $\mu(\theta_i)$ with $\bar X_i$ is $a$. By the law of total variance, $\operatorname{Var}(\bar X_i)=a+\phi/W_i$. Therefore, the coefficient on $\bar X_i$ in the best linear predictor of $\mu(\theta_i)$ is

\frac{\operatorname{Cov}(\mu(\theta_i),\bar X_i)} {\operatorname{Var}(\bar X_i)} = \frac{a}{a+\phi/W_i} = \frac{W_i}{W_i+\phi/a} = \frac{W_i}{W_i+K}.

The predictor is centered at the collective mean $\mu$, giving the linear blend $Z_i\bar X_i+(1-Z_i)\mu$. In particular, larger $W_i$ reduces the process-variance contribution $\phi/W_i$, so the class experience receives more credibility. This justification relies on the stated model structure; the formula should not be applied mechanically if observations are not on a consistent rate and exposure basis or the model assumptions do not reasonably describe them.

Comparing Classes and Avoiding Weight Errors

When two classes share the same structural parameters and collective mean, their credibility factors differ only through their total weights. Their weighted means can differ in either direction; credibility determines how strongly each mean moves its estimate away from the collective mean. A higher class mean does not, by itself, imply a higher estimate if that class has much less exposure.

Worked Example: Comparing Two Classes

A hypothetical insurer is estimating loss cost per exposure for two classes. The collective mean is $50, and common structural parameters are $\phi=1{,}200$ and $a=80$. Class M has observations $60, $54, and $66 with weights 2, 1, and 3, respectively. Class N has observations $48, $56, and $62 with weights 5, 2, and 1. Find each class’s weighted mean, total weight, credibility, and estimate.

1. Calculate the common $K$. $$ K=\frac{1{,}200}{80}=15. $$

2. Class M. Its total weighted amount and total weight are $$ (2)(60)+(1)(54)+(3)(66)=120+54+198=372, \qquad W_M=2+1+3=6. $$ Thus $\bar X_M=372/6=62$, and $$ Z_M=\frac{6}{6+15}=\frac{2}{7}. $$ The estimate is $$ \widehat{\mu}_M =\frac{2}{7}(62)+\frac{5}{7}(50) =\frac{124+250}{7} =\frac{374}{7} =53.428571\ldots, $$ or $53.43 per exposure to the nearest cent.

3. Class N. Its total weighted amount and total weight are $$ (5)(48)+(2)(56)+(1)(62)=240+112+62=414, \qquad W_N=5+2+1=8. $$ Thus $\bar X_N=414/8=51.75$, and $$ Z_N=\frac{8}{8+15}=\frac{8}{23}. $$ The estimate is $$ \widehat{\mu}_N =\frac{8}{23}(51.75)+\frac{15}{23}(50) =\frac{414+750}{23} =\frac{1{,}164}{23} =50.608695\ldots, $$ or $50.61 per exposure to the nearest cent.

Check. Class M has the higher weighted mean and the lower total weight, so its estimate moves toward its class mean by less than half the difference from the collective mean: $53.43$ is between $50$ and $62$. Class N’s estimate is also between its weighted mean and the collective mean. Using the unweighted average of each class’s observations would ignore the exposure differences and would not produce these estimates.

Adding Exposure to a Class

A class’s credibility depends on its accumulated weight, not simply on the number of observation periods. The formula also makes it possible to determine how much total weight corresponds to a target credibility. Solving $Z=W/(W+K)$ for $W$ gives $W=ZK/(1-Z)$, for $0\leq Z<1$. This is useful for checking exposure requirements, but an estimated class mean still depends on the observed rates and weights.

Worked Example: Updating an Estimate After New Experience

A hypothetical class has three observed loss costs: $100 with weight 3, $120 with weight 4, and $90 with weight 5. Its collective mean is $90, with $\phi=1{,}200$ and $a=100$. Calculate its estimate, then determine the updated estimate if a new block of experience has weight 24 and a weighted mean loss cost of $110.

1. Find $K$ and the original class summaries. $$ K=\frac{1{,}200}{100}=12, \qquad W_i=3+4+5=12. $$ The original weighted total and mean are $$ (3)(100)+(4)(120)+(5)(90)=300+480+450=1{,}230, \qquad \bar X_i=\frac{1{,}230}{12}=102.5. $$

2. Calculate the original estimate. $$ Z_i=\frac{12}{12+12}=0.5, \qquad \widehat{\mu}_i=(0.5)(102.5)+(0.5)(90)=51.25+45=96.25. $$ The original estimate is $96.25 per exposure.

3. Add the new experience. The new block contributes weighted amount $(24)(110)=2{,}640$. The updated weighted total and exposure are $$ 1{,}230+2{,}640=3{,}870, \qquad W_i^{\text{new}}=12+24=36. $$ Therefore, $$ \bar X_i^{\text{new}}=\frac{3{,}870}{36}=107.5, \qquad Z_i^{\text{new}}=\frac{36}{36+12}=0.75. $$

4. Calculate the updated estimate. $$ \widehat{\mu}_i^{\text{new}} =(0.75)(107.5)+(0.25)(90) =80.625+22.5 =103.125. $$ Rounded to the nearest cent, the updated estimate is $103.13 per exposure. Its increase reflects both the higher new experience and the fact that the added weight raises credibility from $0.50$ to $0.75$.

Common Calculation Traps

Taking an ordinary mean. With unequal weights, $\sum_j X_{ij}/n$ is generally not the class mean used in the model. Calculate $\sum_j w_{ij}X_{ij}/\sum_jw_{ij}$.
Using the number of observations as total weight. $W_i$ is the sum of exposure weights, not the count of periods, unless every observation has weight 1.
Using $K$ as the credibility factor. $K=\phi/a$ is a structural ratio; the factor is $Z_i=W_i/(W_i+K)$.
Reversing the blend. $Z_i$ multiplies the class weighted mean, while $1-Z_i$ multiplies the collective mean.
Combining mismatched quantities. A loss total cannot be averaged directly with a loss cost per exposure. Put observations on a consistent rate basis and pair them with their corresponding weights.

Key takeaway: For unequal-weight experience, compute the weighted class mean and total weight separately. Use total weight in $Z_i=W_i/(W_i+K)$, then apply that credibility to the class mean and the remaining credibility to the collective mean.

Check Your Understanding

For each problem, calculate the requested quantities before checking the answer.

A hypothetical class has loss costs $40 and $60 with weights 1 and 3. The collective mean is $45, $\phi=400$, and $a=100$. Find $W_i$, $\bar X_i$, $Z_i$, and the estimate. (Answer: $W_i=4$, $\bar X_i=\$55$, $Z_i=0.5000$, estimate $=\$50.00$ per exposure.)
A class has loss costs $30, $50, and $70 with weights 2, 1, and 1. The collective mean is $40, $\phi=600$, and $a=100$. Find its weighted mean and estimate. (Answer: $W_i=4$, $\bar X_i=\$45$, $Z_i=0.4000$, estimate $=\$42.00$ per exposure.)
A class has loss costs $80 and $100 with weights 2 and 4. The collective mean is $70, $\phi=800$, and $a=100$. Calculate its weighted mean, total weight, and estimate. (Answer: $W_i=6$, $\bar X_i=\$93.33$, $Z_i=6/14=0.4286$, estimate $=\$80.00$ per exposure.)
A class has $K=10$ and currently has total weight 20. How much additional weight is needed for credibility to reach 0.75? (Answer: target total weight $=0.75(10)/(1-0.75)=30$; additional weight $=30-20=10$.)
A class has loss costs $50 and $80 with weights 1 and 3. Its collective mean is $60, $\phi=900$, and $a=100$. Calculate its weighted mean and credibility-weighted estimate, and state why the ordinary mean is inappropriate. (Answer: $W_i=4$, $\bar X_i=(50+240)/4=\$72.50$, $K=9$, $Z_i=4/13=0.3077$, estimate $=(4/13)(72.50)+(9/13)(60)=\$63.85$ per exposure; the ordinary mean assigns equal influence to observations with unequal weights.)

What You'll Learn