Combining Several Years of Equal-Exposure Experience

Why Combine Several Years of Experience?

A risk’s experience in one year can be unusually high or low because of random variation. Several years of comparable experience provide a more stable basis for estimating the risk’s expected outcome. In the equal-exposure Bühlmann model, the years are combined through their arithmetic mean, while the total number of observations determines how much credibility that mean receives.

The essential conditions are that each observation represents the same exposure unit and that observations are independent and identically distributed conditional on the risk’s underlying parameter. Equal exposure does not mean that observed outcomes must be equal; it means that they are measured on the same basis.

Definition: Let $\Theta$ denote a risk’s unknown parameter. Let $X_1,\ldots,X_n,X_{n+1}$ be observations from $n$ past periods and one future period. Assume they are conditionally independent and identically distributed given $\Theta$, with $$ E[X_j\mid\Theta]=\mu(\Theta), \qquad \operatorname{Var}(X_j\mid\Theta)=\sigma^2(\Theta). $$ The collective mean is $m=E[\mu(\Theta)]$. The expected process variance (EPV) is $v=E[\sigma^2(\Theta)]$, and the variance of hypothetical means (VHM) is $a=\operatorname{Var}(\mu(\Theta))$. For $a>0$, define the Bühlmann credibility constant $K=v/a$.

The Bühlmann premium is the best affine linear predictor of the future observation based on the past observations, under squared-error loss. It has the form $m+Z(\overline X-m)$, where $\overline X$ is the arithmetic mean of the $n$ past observations. When $a>0$ and $v>0$, the credibility factor is $Z=n/(n+K)$. The premium is a linear predictor; it need not equal the full conditional expectation $E[X_{n+1}\mid X_1,\ldots,X_n]$ for every possible model.

Deriving the Equal-Exposure Formula

The formula follows from the covariance structure created by conditional independence. Write $X=(X_1,\ldots,X_n)^\mathsf{T}$, and let $\mathbf{1}$ be the $n$-vector of ones. By the law of total variance, $\operatorname{Var}(X_j)=v+a$. For two different periods $j\ne k$, conditional independence gives $\operatorname{Cov}(X_j,X_k)=a$. The same conditioning argument gives $\operatorname{Cov}(X_j,X_{n+1})=a$.

Thus, if $v>0$, the covariance matrix of the observations is nonsingular:

\Sigma=\operatorname{Cov}(X)=vI+a\mathbf{1}\mathbf{1}^{\mathsf{T}}, \qquad \operatorname{Cov}(X,X_{n+1})=a\mathbf{1}.

Consider the general affine linear predictor $c+\boldsymbol{\beta}^{\mathsf{T}}X$, allowing a separate coefficient for every observation. The least-squares normal equations require its residual to have mean zero and to be uncorrelated with each component of $X$. Since $E[X_j]=E[X_{n+1}]=m$, these equations give $c=m-\boldsymbol{\beta}^{\mathsf{T}}m\mathbf{1}$ and $\Sigma\boldsymbol{\beta}=a\mathbf{1}$. This full-vector calculation is important: it establishes optimality among affine predictors based on all the observations, not only among predictors that were assumed in advance to depend on their mean.

Because $\Sigma\mathbf{1}=(v+na)\mathbf{1}$, the unique solution when $v>0$ is $\boldsymbol{\beta}=\frac{a}{v+na}\mathbf{1}$. Therefore all observations receive equal coefficients. Rewriting the predictor in terms of $\overline X$ gives the result:

\widehat X_{n+1} =m+\frac{na}{v+na}(\overline X-m) =m+Z(\overline X-m), \qquad Z=\frac{na}{v+na}=\frac{n}{n+K}.

The same result can be checked from the variance of the sample mean. Its covariance with the future observation is $a$, and its variance is $a+v/n$. The regression slope of the future observation on the mean is therefore $a/(a+v/n)=na/(v+na)$, exactly the stated $Z$.

Result: Under the equal-exposure Bühlmann assumptions, with $a>0$ and $v>0$, use $Z=n/(n+K)$, where $K=v/a$, and calculate the premium as $m+Z(\overline X-m)$. The count $n$ is the number of past observations, and $\overline X$ is their arithmetic mean.

There are boundary cases. If $a=0$, there is no variation in risk-specific means, so past experience does not change the linear predictor: the premium is $m$. If $v=0$ and $a>0$, each observation equals its risk-specific mean with probability one. For any $n\ge1$, the observations reveal that mean exactly and the future observation equals it; the premium is $\overline X$. In this case the covariance matrix is singular and the unique-coefficient argument above does not apply. If $a=v=0$, all observations are constant at $m$.

Worked Example: Four Years of Equal-Exposure Loss Experience

A hypothetical portfolio uses annual losses per identical exposure unit. The collective mean is $m=100$ thousand dollars, the EPV is $v=144$ (thousand dollars)$^2$, and the VHM is $a=36$ (thousand dollars)$^2$. The risk’s four observed annual losses, in thousands of dollars, are 90, 110, 130, and 150. Calculate the Bühlmann premium for the next year.

Method and assumptions. Treat the observations as conditionally independent and identically distributed given the risk parameter, with equal annual exposure. Since $v>0$ and $a>0$, the usual Bühlmann formula applies.

First calculate the experience mean: $\overline X=(90+110+130+150)/4=480/4=120$ thousand dollars. The credibility constant is $K=v/a=144/36=4$. With $n=4$, the factor is $Z=4/(4+4)=1/2$.

The premium is $m+Z(\overline X-m)=100+(1/2)(120-100)=100+10=110$ thousand dollars, or $110,000.

Check. The experience mean is 20 thousand dollars above the collective mean, and the premium moves halfway toward it, by 10 thousand dollars. The common mistake avoided here is using $n=1$ because the data are presented as one list; there are four separate annual observations.

Worked Examples

Worked Example: Updating a Premium After Adding a Year

A hypothetical insurer models annual claim counts per identical policy unit. The collective mean is $m=1{,}200$ claims per unit, with $v=90{,}000$ claims$^2$ and $a=30{,}000$ claims$^2$. Five observed annual counts are 900, 1,100, 1,200, 1,300, and 1,500. Find the premium for the next year.

Method and assumptions. Apply the equal-exposure Bühlmann model: the counts are conditionally independent and identically distributed given the risk parameter, and the observations have the same exposure basis. Both structural variances are positive.

The total observed count is $900+1{,}100+1{,}200+1{,}300+1{,}500=6{,}000$. Hence $\overline X=6{,}000/5=1{,}200$. The constant is $K=90{,}000/30{,}000=3$, so $Z=5/(5+3)=5/8$.

The experience mean equals the collective mean. Thus $m+Z(\overline X-m)=1{,}200+(5/8)(1{,}200-1{,}200)=1{,}200$ claims per unit.

Check. With no difference between the experience mean and collective mean, credibility weighting cannot move the premium. The common mistake avoided is applying the credibility factor directly to $\overline X$; it weights the deviation $\overline X-m$, not the entire experience mean.

Worked Example: Deriving the Premium From the Full Observation Vector

For a hypothetical risk, let $m=100$, $v=36$, and $a=16$, with all outcomes measured in thousands of dollars and variances in (thousand dollars)$^2$. Two past equal-exposure observations are $X_1=110$ and $X_2=130$. Calculate the least-squares affine linear predictor of the next observation using both past observations.

Method and assumptions. Assume conditional independence and identical conditional distributions across the two past periods and the future period. Form the covariance matrix and solve the full-vector normal equations.

Each observation has variance $v+a=36+16=52$, and the covariance between the two past observations is $a=16$. Each past observation also has covariance $16$ with the future observation. Therefore $\Sigma=\begin{pmatrix}52&16\\16&52\end{pmatrix}$, and the covariance vector with the future outcome is $(16,16)^\mathsf{T}$.

The normal equations are $52\beta_1+16\beta_2=16$ and $16\beta_1+52\beta_2=16$. Subtracting the equations gives $36(\beta_1-\beta_2)=0$, so $\beta_1=\beta_2$. Substitution gives $68\beta_1=16$, hence $\beta_1=\beta_2=4/17$. Their sum is $8/17$.

The intercept is $c=m-m(\beta_1+\beta_2)=100(1-8/17)=900/17$. The prediction is $(900+4(110)+4(130))/17=1{,}860/17=109.4118$ thousand dollars, rounded to four decimal places in thousands. The requested premium is $109,411.76, rounded to the nearest cent.

Check. Here $K=36/16=9/4$ and $Z=2/(2+9/4)=8/17$. The experience mean is $(110+130)/2=120$, giving $100+(8/17)(20)=109.4118$ thousand dollars, in agreement with the vector calculation. The common mistake avoided is assuming equal coefficients without checking the full normal equations.

Interpreting the Credibility Factor

The factor $Z$ lies between zero and one when $a>0$ and $v>0$. Increasing the number of equal-exposure observations raises $Z$, because the sample mean becomes more informative about the risk’s underlying mean. Increasing the EPV relative to the VHM lowers $Z$: more within-risk random variation makes the experience mean less revealing of the risk’s true level.

The formula is especially useful because all $n$ observations enter through two summaries: their arithmetic mean and their count. That compression relies on the model’s equal-exposure assumptions. If periods have different exposure amounts, simply taking an unweighted arithmetic mean and using the number of periods is generally not justified; the equal-exposure formula should not be carried over unchanged.

Common exam trap: Suppose $K=4$, $n=3$, $m=80$, and $\overline X=100$. The wrong calculation $Z=K/(n+K)=4/7$ gives $80+(4/7)(20)=91.4286$. The correction is $Z=n/(n+K)=3/7$, so the premium is $80+(3/7)(20)=88.5714$. The experience mean receives credibility $n/(n+K)$; $K$ is not the numerator.

Check Your Understanding

For each problem, assume the stated equal-exposure observations are conditionally independent and identically distributed given the risk parameter, unless the problem asks about the model conditions.

A hypothetical risk has $m=60$, $v=75$, $a=25$, and three past observations 50, 65, and 70. Find its Bühlmann premium. (Answer: 60.8333)
A hypothetical book has $m=1{,}000$, $v=64$, and $a=16$. Four annual observations are 900, 1,000, 1,100, and 1,200. Calculate $K$, $Z$, and the premium. (Answer: $K=4$, $Z=1/2$, premium $=1{,}025$)
A risk has $K=6$. What is the smallest integer number of equal-exposure observations needed for $Z\ge0.8$? (Answer: 24 observations)
For a risk with $m=100$, $n=5$, and $\overline X=140$, the observed premium is 120. Find $Z$ and $K$. (Answer: $Z=1/2$, $K=5$)

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