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CAS Exam 5 · Credibility Weighting Models · Tutorial 583 of 989

Unbalanced Experience Across Risk Classes

Calculate and interpret Bühlmann-Straub estimates for unbalanced risk-class experience using each class’s total exposure and weighted average.

Advanced 12 min read

What You'll Learn

  • Define unbalanced experience in the Bühlmann-Straub model
  • Calculate a class’s total exposure and exposure-weighted average
  • Apply the credibility factor using total class exposure
  • Explain why equal total exposures imply equal credibility factors
  • Identify the error in substituting period count for exposure

Why Unbalanced Experience Requires Exposure Weighting

Risk classes rarely have identical experience records. One class may have more observation periods, while another has fewer periods with more exposure in each. In the Bühlmann-Straub model, neither the number of periods nor an unweighted average alone captures the amount of information in a class. The key quantity is its total exposure, which combines period-level weights and determines the class’s credibility.

This matters when experience is summarized across different policy years, territories, or other risk groupings. A period with twice the exposure of another period contributes twice as much to the weighted average. But even when two classes have different period counts, they may have the same total exposure and therefore the same credibility factor under the model.

Definition: In the Bühlmann-Straub model, class \(i\) has observations \(X_{ij}\) with positive exposure weights \(w_{ij}\), for periods \(j=1,\ldots,n_i\). Its total exposure is \(W_i=\sum_{j=1}^{n_i}w_{ij}\), and its exposure-weighted average is \(\bar X_i=\sum_{j=1}^{n_i}w_{ij}X_{ij}/W_i\). Experience is unbalanced when classes differ in their number of periods, their period weights, or both.

The Model and the Class Estimate

For a class \(i\), let \(\Theta_i\) denote its underlying expected level. The standard Bühlmann-Straub assumptions are that \(E[X_{ij}\mid\Theta_i]=\Theta_i\) and \(\operatorname{Var}(X_{ij}\mid\Theta_i)=\sigma^2/w_{ij}\). Conditional on \(\Theta_i\), observations from distinct periods are uncorrelated. Across classes, the structural mean is \(\mu=E[\Theta_i]\), and the variance of the underlying class means is \(\tau^2=\operatorname{Var}(\Theta_i)\).

The exposure weight represents the relative amount of information in a period. A larger weight reduces the conditional variance: a period with weight \(4\) has one quarter the conditional variance of a period with weight \(1\), under these assumptions. The weights must be on a consistent scale, and the same variance parameters must apply across the classes being compared.

Definition: The expected process variance per unit of exposure is \(\sigma^2\), and the variance of hypothetical class means is \(\tau^2\). Define \(K=\sigma^2/\tau^2\), assuming \(\tau^2>0\). With known \(\mu,\sigma^2,\tau^2\), the Bühlmann-Straub estimate of class \(i\)'s underlying mean is \(Z_i\bar X_i+(1-Z_i)\mu\), where \(Z_i=W_i/(W_i+K)\).

The estimate blends the class’s own weighted average with the structural mean. A class with more total exposure receives a larger \(Z_i\), so its observed experience has greater influence. The complement \(1-Z_i=K/(W_i+K)\) is the weight on the structural mean.

Why Total Exposure Determines Credibility

The result follows by examining the weighted average for a single class. Given its underlying mean \(\Theta_i\), the weighted average is unbiased for \(\Theta_i\). Conditional independence and the model’s variance assumption give:

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

Thus the weighted average’s conditional variance depends on the period weights through their sum, not through the number of terms. Unconditionally, the variance includes both variation among class means and process variation:

$$ \operatorname{Var}(\bar X_i)=\tau^2+\frac{\sigma^2}{W_i}, \qquad \operatorname{Cov}(\Theta_i,\bar X_i)=\tau^2. $$

For a linear estimate of the form \(\mu+c(\bar X_i-\mu)\), the coefficient minimizing mean squared prediction error is the covariance of \(\Theta_i\) with \(\bar X_i\), divided by the variance of \(\bar X_i\). Therefore,

$$ c=\frac{\operatorname{Cov}(\Theta_i,\bar X_i)} {\operatorname{Var}(\bar X_i)} =\frac{\tau^2}{\tau^2+\sigma^2/W_i} =\frac{W_i}{W_i+\sigma^2/\tau^2} =\frac{W_i}{W_i+K}. $$

This establishes the credibility factor and the stated estimate. The derivation relies on the model assumptions above; it is not a general rule for arbitrary weighted data. In particular, if the weights do not represent exposure in the specified variance relationship, using their sum in this formula may not be justified.

1
Find total exposure.
Add the period weights: \(W_i=\sum_j w_{ij}\). Do not substitute the number of periods unless every period has weight \(1\).
2
Find the weighted average.
Multiply each observation by its weight, sum those products, and divide by \(W_i\).
3
Calculate credibility.
Use \(Z_i=W_i/(W_i+K)\), with \(K=\sigma^2/\tau^2\) in the same exposure units as \(W_i\).
4
Blend with the structural mean.
Calculate \(Z_i\bar X_i+(1-Z_i)\mu\), then check that the result lies between \(\bar X_i\) and \(\mu\).

Worked Example: Three Classes With Different Period Counts and Weights

A hypothetical portfolio has structural mean \(\mu=100\), expected process variance \(\sigma^2=900\), and variance of class means \(\tau^2=100\). Class A has observations \(96,104,110\), each with weight \(1\). Class B has observations \(120,90\), with weights \(2,1\), respectively. Class C has observations \(130,110,90,120\), each with weight \(0.5\). Calculate each class estimate. The variance parameters and exposures use consistent units.

Method and parameter. Use Bühlmann-Straub credibility. First, \(K=\sigma^2/\tau^2=900/100=9\).

Class A. Its total exposure is \(W_A=1+1+1=3\). Its weighted average is \(\bar X_A=(96+104+110)/3=310/3=103.3333\). Its credibility is \(Z_A=3/(3+9)=0.25\). Thus its estimate is \(0.25(103.3333)+0.75(100)=25.8333+75=100.8333\).

Class B. Its total exposure is \(W_B=2+1=3\). Its weighted average is \(\bar X_B=(2(120)+1(90))/3=(240+90)/3=110\). Its credibility is \(Z_B=3/(3+9)=0.25\). Its estimate is \(0.25(110)+0.75(100)=27.5+75=102.5\).

Class C. Its total exposure is \(W_C=4(0.5)=2\). Its weighted average is \(\bar X_C=0.5(130+110+90+120)/2=225/2=112.5\). Its credibility is \(Z_C=2/(2+9)=2/11=0.1818\) to four decimals. Its estimate is \((2/11)(112.5)+(9/11)(100)=20.4545+81.8182=102.2727\) to four decimals.

Check and common mistake. Each estimate lies between its class average and \(100\), as a credibility-weighted blend should. Classes A and B have the same total exposure and therefore the same credibility, despite having different period counts. Using B’s period count \(2\) in place of its total exposure \(3\) would incorrectly give \(Z_B=2/(2+9)\) and an estimate of about \(101.8182\).

Equal Total Exposure, Different Period Schedules

Two classes can have different period schedules yet the same weighted average and total exposure. In that case, they have the same estimate under this model. This is a consequence of the model’s variance structure: once the total weight and weighted average are known, the details of how the exposure was divided among periods do not change the formula.

Worked Example: Same Exposure Split Across Periods

For a hypothetical book, let \(\mu=50\), \(\sigma^2=400\), and \(\tau^2=100\). Portfolio A has two observations, \(60\) and \(80\), each with weight \(4\). Portfolio B has four observations, \(55,65,75,85\), each with weight \(2\). Find both estimates and explain whether their different period counts change credibility.

Method and parameter. \(K=400/100=4\). For A, \(W_A=4+4=8\), and \(\bar X_A=(4(60)+4(80))/8=(240+320)/8=70\). Thus \(Z_A=8/(8+4)=2/3\), and the estimate is \((2/3)(70)+(1/3)(50)=46.6667+16.6667=63.3333\), rounded to four decimals.

For B, \(W_B=2+2+2+2=8\). Its weighted average is \(\bar X_B=2(55+65+75+85)/8=560/8=70\). Therefore \(Z_B=8/(8+4)=2/3\), and the estimate is again \((2/3)(70)+(1/3)(50)=63.3333\).

Check and common mistake. Both schedules have total exposure \(8\) and weighted average \(70\), so their estimates agree. Using period counts as exposures would give different and incorrect credibility factors: \(2/(2+4)=1/3\) for A and \(4/(4+4)=1/2\) for B. The relevant input is total exposure, not the number of records.

Unbalanced Classes and the Period-Count Trap

When class schedules are unbalanced, the period count can still be useful for describing the data, but it is not a substitute for \(W_i\). A class with only a few heavily weighted periods may have as much credibility as a class with many lightly weighted periods. Conversely, many periods need not imply high credibility if their total exposure is small.

The model also distinguishes a weighted average from an ordinary average. If period weights differ, an ordinary average gives each observation equal influence, even though the model assigns different information to those observations. Using an ordinary average together with exposure-based credibility can therefore produce an estimate inconsistent with the model.

Worked Example: Avoiding the Count-Based Credibility Error

A hypothetical insurer studies two classes with \(\mu=80\), \(\sigma^2=600\), and \(\tau^2=100\). Class R has two observations, \(100\) and \(40\), with weights \(5\) and \(1\). Class S has four observations, \(120,100,80,100\), each with weight \(1.5\). Calculate both estimates, and show the consequence of using the number of periods instead of total exposure.

Method and parameter. \(K=600/100=6\). For R, \(W_R=5+1=6\), and \(\bar X_R=(5(100)+1(40))/6=540/6=90\). Therefore \(Z_R=6/(6+6)=0.5\), and the estimate is \(0.5(90)+0.5(80)=85\).

For S, \(W_S=4(1.5)=6\). Its weighted average is \(\bar X_S=1.5(120+100+80+100)/6=600/6=100\). Therefore \(Z_S=6/(6+6)=0.5\), and the estimate is \(0.5(100)+0.5(80)=90\).

Check and common mistake. Both classes have total exposure \(6\), so both have credibility \(0.5\), even though R has two periods and S has four. Substituting the counts gives incorrect factors \(2/(2+6)=0.25\) and \(4/(4+6)=0.4\), producing incorrect estimates \(82.5\) and \(88\). The correct estimates, \(85\) and \(90\), lie between each class’s weighted average and the structural mean.

Interpretation and Practical Checks

Total exposure has a direct interpretation in the model. As \(W_i\) increases, the conditional variance \(\sigma^2/W_i\) of the class weighted average decreases. The class experience is then more informative about \(\Theta_i\), and \(Z_i\) rises toward \(1\). As \(W_i\) becomes small relative to \(K\), the estimate places more weight on \(\mu\).

Some useful checks are:

  • Exposure check: confirm that each \(W_i\) is the sum of the period weights, not the period count.
  • Average check: verify that the weighted products sum to \(W_i\bar X_i\).
  • Parameter check: calculate \(K=\sigma^2/\tau^2\), and ensure its exposure scale matches the weights.
  • Blend check: for \(0\leq Z_i\leq1\), the estimate must lie between \(\bar X_i\) and \(\mu\).

The credibility formula treats \(\mu,\sigma^2,\tau^2\) as specified inputs; uncertainty from estimating those parameters is not included in this class-level calculation. If \(W_i=0\), there is no class experience to average, and the formula gives \(Z_i=0\), so the estimate is \(\mu\). These conclusions are tied to the Bühlmann-Straub assumptions, especially the inverse relationship between conditional variance and exposure weight.

Key takeaway: For unbalanced Bühlmann-Straub experience, calculate both the exposure-weighted class average and the sum of its weights. Total exposure \(W_i\), not the number of periods, determines \(Z_i=W_i/(W_i+K)\).

Check Your Understanding

For each hypothetical problem, use the stated Bühlmann-Straub parameters and exposure weights.

  1. A class has \(\mu=120\), \(K=8\), total exposure \(W=4\), and weighted average \(\bar X=132\). Find its credibility and estimate. (Answer: \(Z=1/3\); estimate \(124\).)
  2. Let \(\mu=60\), \(\sigma^2=500\), and \(\tau^2=100\). Two observations are \(70\) and \(90\), with weights \(2\) and \(3\). Find the total exposure, weighted average, and estimate. (Answer: \(W=5\), \(\bar X=82\), estimate \(71\).)
  3. For a class, \(\mu=200\), \(K=10\), \(W=4\), and \(\bar X=230\). Calculate its credibility-weighted estimate. (Answer: \(Z=2/7\); estimate \(208.5714\) to four decimals.)
  4. A class has observations \(100,120,90\) with weights \(0.5,1.5,2\). Let \(\mu=100\), \(\sigma^2=800\), and \(\tau^2=100\). Find \(W\), the weighted average, and the estimate. (Answer: \(W=4\), \(\bar X=102.5\), estimate \(100.8333\) to four decimals.)
  5. Two classes share \(\mu=75\) and \(K=5\). Class A has three periods, each with weight \(1\), and weighted average \(90\). Class B has two periods with weights \(2\) and \(1\), and weighted average \(60\). Find each credibility and estimate, and state why the factors agree. (Answer: both have \(W=3\) and \(Z=3/8\); A’s estimate is \(80.625\) and B’s is \(69.375\).)
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