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Method-of-Moments Estimation for Bühlmann Models

Estimate Bühlmann variance components from repeated observations in equal-sized risk classes, and apply the nonnegative boundary correction when needed.

Advanced 10 min read

What You'll Learn

  • Define the Bühlmann EPV and VHM in terms of conditional means and variances
  • Estimate EPV from pooled within-class variation
  • Correct between-class variation for the sampling noise in class means
  • Derive why the correction recovers the VHM in expectation
  • Apply a nonnegative estimate when the raw VHM is below zero
  • Recognize when the balanced-class formulas do not apply

Why Estimate the Bühlmann Variance Components?

Bühlmann credibility separates variation within a risk class from variation between risk classes. The within-class component measures how much an individual class’s observations fluctuate around its own expected value. The between-class component measures how much those expected values differ across classes. Estimating these components from data is essential because the credibility assigned to observed experience depends on their relative sizes.

A key detail is that observed class averages do not reveal only genuine differences among classes. Each average also contains sampling variation. A method-of-moments estimate of the between-class variance must remove that contribution; otherwise, it tends to overstate the variation among the underlying class means.

Definition: Let \(\Theta_i\) denote the risk characteristic of class \(i\), and let \(X_{ij}\) be observation \(j\) from that class. The Bühlmann expected process variance is \(\operatorname{EPV}=E[\operatorname{Var}(X_{ij}\mid\Theta_i)]\). The variance of hypothetical means is \(\operatorname{VHM}=\operatorname{Var}(E[X_{ij}\mid\Theta_i])\). EPV describes average variation within a class; VHM describes variation in the underlying class means.

Model and Balanced-Class Estimators

Use the classical Bühlmann model with \(r\) independent risk classes and \(n\) observations in every class. Conditional on \(\Theta_i\), the observations in class \(i\) are independent and identically distributed, with conditional mean \(\mu_i=E[X_{ij}\mid\Theta_i]\) and conditional variance \(\sigma_i^2=\operatorname{Var}(X_{ij}\mid\Theta_i)\). The classes are drawn from the same population of risks, and the relevant means and variances are finite.

Write \(\bar X_i=\frac{1}{n}\sum_{j=1}^n X_{ij}\) for class \(i\)'s sample mean and \(\bar X=\frac{1}{r}\sum_{i=1}^r\bar X_i\) for the grand mean. The within-class sample variance is \(s_i^2=\frac{1}{n-1}\sum_{j=1}^n(X_{ij}-\bar X_i)^2\). The balanced method-of-moments estimates are:

$$ \widehat{\operatorname{EPV}}= \frac{\displaystyle\sum_{i=1}^r\sum_{j=1}^n(X_{ij}-\bar X_i)^2}{r(n-1)} =\frac{1}{r}\sum_{i=1}^r s_i^2 $$
$$ \widehat{\operatorname{VHM}}_{\rm raw} = \frac{\displaystyle\sum_{i=1}^r(\bar X_i-\bar X)^2}{r-1} -\frac{\widehat{\operatorname{EPV}}}{n} $$

The first estimate pools the within-class sums of squares and divides by their total degrees of freedom. The first term in the second estimate is the sample variance of the class averages. The subtraction removes the average sampling variance of those averages. Because a variance component cannot be negative, a common constrained estimate is \(\widehat{\operatorname{VHM}}=\max(0,\widehat{\operatorname{VHM}}_{\rm raw})\). This nonnegative estimate is a boundary adjustment to the raw method-of-moments result.

Why the Between-Class Correction Works

The within-class estimate targets EPV because, conditional on a class, the usual sample variance is unbiased for that class’s conditional variance. Averaging over the class population gives \(E[s_i^2]=E[\sigma_i^2]=\operatorname{EPV}\). Thus, averaging the \(r\) sample variances estimates EPV.

For the between-class estimate, decompose a class average as \(\bar X_i=\mu_i+(\bar X_i-\mu_i)\). The second term is the sampling error in that average. Conditional on \(\Theta_i\), it has mean zero and variance \(\sigma_i^2/n\). Consequently, the law of total variance gives:

$$ \begin{aligned} \operatorname{Var}(\bar X_i) &=\operatorname{Var}(E[\bar X_i\mid\Theta_i]) +E[\operatorname{Var}(\bar X_i\mid\Theta_i)]\\ &=\operatorname{Var}(\mu_i)+E\left[\frac{\sigma_i^2}{n}\right]\\ &=\operatorname{VHM}+\frac{\operatorname{EPV}}{n}. \end{aligned} $$

Since the class averages are independent and identically distributed, their sample variance has expectation \(\operatorname{Var}(\bar X_i)\). Subtracting \(\operatorname{EPV}/n\) therefore leaves VHM in expectation. In practice EPV is unknown, so its estimate is used in the correction.

Important condition: These formulas assume the same number \(n\) of observations in every class. They are not the appropriate unadjusted formulas when class sizes differ. Also, the method requires at least two classes to estimate between-class variation and at least two observations per class to estimate within-class variation.

Worked Calculations

Worked Example: Estimating Both Components From Three Classes

A hypothetical insurer groups equal-exposure risks into three classes and records three annual loss costs per class. The observations are \(8,10,12\), \(13,15,17\), and \(18,20,22\), respectively. Estimate EPV and VHM using the balanced method of moments, then check whether the adjustment is needed.

Step 1: Find each class mean and within-class sum of squares. The means are \(10,15,20\). In each class the deviations from its mean are \(-2,0,2\), so each within-class sum of squares is \(4+0+4=8\). The pooled within-class sum of squares is \(8+8+8=24\).

Step 2: Estimate EPV. There are \(r=3\) classes and \(n=3\) observations per class, so the denominator is \(r(n-1)=3(2)=6\). Hence \(\widehat{\operatorname{EPV}}=24/6=4\).

Step 3: Find the sample variance of class means. The grand mean is \((10+15+20)/3=15\). The squared deviations are \(25,0,25\), giving a between-class sample variance of \((25+0+25)/(3-1)=25\).

Step 4: Correct for sampling variation. The correction is \(\widehat{\operatorname{EPV}}/n=4/3\). Thus \(\widehat{\operatorname{VHM}}_{\rm raw}=25-4/3=71/3\), or \(23.6667\) to four decimal places. This is positive, so no boundary adjustment is needed.

Check. The within-class estimate is \(4\), and the estimated sampling contribution to the variance of a class mean is \(4/3\). Adding that contribution back to the raw VHM estimate gives \(71/3+4/3=25\), the observed sample variance of the class means.

Method and trap avoided. This is the balanced method-of-moments calculation. The check confirms that the raw VHM is the between-class sample variance after removing the class-mean sampling contribution.

Worked Example: Why the EPV Correction Matters

A hypothetical portfolio has three equally sized classes, each with three observations. The observations are \(9,10,11\), \(10,11,12\), and \(11,12,13\). Estimate both variance components. A reviewer proposes using the sample variance of the class means as the VHM estimate without adjustment; assess that proposal.

Step 1: Estimate EPV. Each class mean is \(10,11,12\), respectively. Within each class the deviations are \(-1,0,1\), so each within-class sum of squares is \(1+0+1=2\). The pooled sum is \(6\), and the total within-class degrees of freedom are \(3(3-1)=6\). Therefore \(\widehat{\operatorname{EPV}}=6/6=1\).

Step 2: Calculate the between-class sample variance. The grand mean of the class means is \((10+11+12)/3=11\). The squared deviations are \(1,0,1\), so the sample variance is \(2/(3-1)=1\).

Step 3: Apply the correction. The estimated sampling variance of each class mean is \(1/3\). Thus \(\widehat{\operatorname{VHM}}_{\rm raw}=1-1/3=2/3\), or \(0.6667\) to four decimal places. It is positive, so the nonnegative estimate is also \(2/3\).

Incorrect calculation: Setting VHM equal to the between-class sample variance gives \(1\). That counts both genuine class-to-class variation and the sampling noise in the class means. The correction removes \(1/3\), leaving \(2/3\).

Check. The estimated VHM plus the estimated sampling variance is \(2/3+1/3=1\), exactly the observed variance of the class means. The reviewer’s unadjusted value is therefore too large by the estimated sampling contribution.

Worked Example: Handling a Negative Raw VHM

For a hypothetical coverage, three equal-sized classes each have two observations: \(7,11\), \(8,12\), and \(9,13\). Estimate EPV and the raw VHM, then report the constrained nonnegative VHM estimate.

Step 1: Calculate class means and within-class sums of squares. The means are \(9,10,11\). In each class, the deviations are \(-2,2\), so the sum of squared deviations is \(4+4=8\). The pooled within-class sum is \(24\).

Step 2: Estimate EPV. There are \(3(2-1)=3\) within-class degrees of freedom. Therefore \(\widehat{\operatorname{EPV}}=24/3=8\).

Step 3: Calculate the raw VHM. The grand mean is \(10\). The squared deviations of the class means are \(1,0,1\), so their sample variance is \(2/(3-1)=1\). The sampling correction is \(8/2=4\). Hence \(\widehat{\operatorname{VHM}}_{\rm raw}=1-4=-3\).

Step 4: Apply the nonnegative boundary. A variance component is constrained to be nonnegative, so \(\widehat{\operatorname{VHM}}=\max(0,-3)=0\).

Check and interpretation. The observed variance of the class means is only \(1\), while the estimated sampling variance of each mean is \(4\). The raw subtraction is negative because the data do not support a positive VHM estimate under this calculation. The constrained estimate is zero; it does not mean the raw calculation should be reported as \(-3\).

Method and trap avoided. The calculation uses the same moment correction as before, followed by a nonnegative boundary adjustment. Replacing the raw estimate with its absolute value would give \(3\), which is not the constrained method-of-moments estimate.

Interpretation and Common Exam Traps

The EPV estimate is measured in squared units of the observations, as is the VHM estimate. Their magnitudes should therefore be interpreted on the same variance scale. The first estimates average conditional variation; the second estimates the spread of conditional means. Neither estimate is itself a credibility factor.

  • Omitting the correction. The variance of class averages includes both VHM and \(\operatorname{EPV}/n\). Using it directly overstates VHM unless the sampling contribution is zero.
  • Using the wrong divisor. The within-class pooled sum of squares is divided by \(r(n-1)\); the sample variance of the \(r\) class means is divided by \(r-1\).
  • Subtracting the wrong amount. For \(n\) observations per class, the sampling contribution is \(\widehat{\operatorname{EPV}}/n\), not \(\widehat{\operatorname{EPV}}\) and not \(n\widehat{\operatorname{EPV}}\).
  • Reporting a negative variance component. Keep the raw result visible if useful for diagnosing the data, but report the nonnegative constrained estimate as zero when the requested method uses a nonnegative component.
  • Applying a balanced formula to unequal class sizes. The correction and pooling structure depend on equal \(n\). Unequal class sizes require an appropriate method rather than mechanically inserting an average class size.

The central reasoning is simple but easy to miss: variation among observed class averages is not all variation among hypothetical means. First estimate the within-class process variance; then remove its contribution to the sampling variance of each average. If the resulting raw VHM falls below zero, enforce the nonnegative parameter constraint rather than interpreting a negative value as a real variance.

Check Your Understanding

For each balanced data set, estimate EPV first, then calculate the raw VHM and apply a nonnegative boundary adjustment if needed. Round non-integer results to four decimal places.

  1. Three classes each have two observations: \(4,6\); \(6,8\); and \(8,10\). Find the class means, EPV, raw VHM, and constrained VHM. (Answer: means \(5,7,9\); EPV \(=2\); raw VHM \(=3\); constrained VHM \(=3\).)
  2. Four classes each have two observations: \(12,14\); \(14,16\); \(16,18\); and \(18,20\). Find EPV and constrained VHM. (Answer: EPV \(=2\); raw and constrained VHM \(=17/3=5.6667\).)
  3. Three classes each have two observations: \(17,23\); \(19,25\); and \(21,27\). Calculate the within-class estimate and the between-class correction before applying the boundary. (Answer: EPV \(=18\); sample variance of class means \(=4\); correction \(=9\); raw VHM \(=-5\); constrained VHM \(=0\).)
  4. Four classes each have three observations: \(29,30,31\); \(31,32,33\); \(33,34,35\); and \(35,36,37\). Estimate both components and verify the between-class identity by adding the estimated sampling contribution back to VHM. (Answer: EPV \(=1\); raw and constrained VHM \(=19/3=6.3333\); sampling contribution \(=1/3\); their sum is \(20/3\), the sample variance of the class means.)

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