Deriving The Bühlmann-Straub Credibility Estimate

Why Derive the Credibility Estimate?

A risk’s observed experience provides information about its underlying expected cost, but that experience is subject to random variation. The Bühlmann-Straub estimate balances the risk’s own experience against the collective mean. Its weight depends on how much experience is available and on the relative sizes of process variation and variation between risks.

The goal here is to derive the best linear estimate of a risk mean under the Bühlmann-Straub assumptions. “Best” means that, among linear blends of the collective mean and the risk’s weighted observed mean, the estimate minimizes expected squared error in estimating the risk’s true mean. The derivation does not require a normal distribution.

Definition: For risk $i$, let $\Theta_i$ be its underlying expected cost per exposure unit. Assume $E[\Theta_i]=\mu$ and $\operatorname{Var}(\Theta_i)=\tau^2$. In period $j$, let $X_{ij}$ be the observed cost per exposure unit and $w_{ij}>0$ its exposure weight. Conditional on $\Theta_i$, assume $E[X_{ij}\mid\Theta_i]=\Theta_i$, $\operatorname{Var}(X_{ij}\mid\Theta_i)=\sigma^2/w_{ij}$, and that observations from different periods are conditionally uncorrelated. Write $W_i=\sum_j w_{ij}$ and $\bar X_i=\sum_j w_{ij}X_{ij}/W_i$. The parameters $\mu,\tau^2,\sigma^2$ are common to the risks.

The parameter $\mu$ is the collective mean, $\tau^2$ measures variation in underlying means across risks, and $\sigma^2$ is the process-variance parameter per unit of exposure. The model requires finite second moments and positive exposures. It specifies conditional means and variances, but not a particular probability distribution for the observations or risk means. In applications, model parameters may need to be estimated; the derivation below treats them as known.

Why the Weighted Mean Has This Variance

For a fixed risk, the weighted mean combines observations with different exposures. Its conditional expectation is $\Theta_i$, because each period has that same conditional expected cost. Conditional uncorrelatedness then gives its variance:

E[\bar X_i\mid\Theta_i] =\frac{\sum_j w_{ij}E[X_{ij}\mid\Theta_i]}{W_i} =\Theta_i, \qquad \operatorname{Var}(\bar X_i\mid\Theta_i) =\frac{\sum_j w_{ij}^2(\sigma^2/w_{ij})}{W_i^2} =\frac{\sigma^2}{W_i}.

Thus additional exposure reduces the sampling variation of the risk’s weighted mean. The division is by total exposure $W_i$, not by its square: the squared weights in the variance sum are offset by the $1/w_{ij}$ period variances.

Deriving the Best Linear Blend

Consider the estimate $T_\beta=\mu+\beta(\bar X_i-\mu)$, where $\beta$ is a constant weight. This is the same as blending the observed mean and collective mean: $T_\beta=\beta\bar X_i+(1-\beta)\mu$. It has expectation $\mu$, the population mean of $\Theta_i$. We choose $\beta$ to minimize $E[(T_\beta-\Theta_i)^2]$.

Decompose the estimation error.
Subtract $\Theta_i$ from the blend and collect the two sources of error: deviation of the risk mean from the collective mean, and sampling error in the observed mean.

Find the component variances.
The unscaled components $\mu-\Theta_i$ and $\bar X_i-\Theta_i$ have variances $\tau^2$ and $\sigma^2/W_i$, respectively. Their covariance is zero because $E[\bar X_i-\Theta_i\mid\Theta_i]=0$.

Minimize the mean squared error.
Use the zero covariance to write the error variance as a function of $\beta$, differentiate, and solve for the minimizing weight.

T_\beta-\Theta_i =(1-\beta)(\mu-\Theta_i)+\beta(\bar X_i-\Theta_i).

The covariance is zero since $E[(\Theta_i-\mu)(\bar X_i-\Theta_i)] =E[(\Theta_i-\mu)E[\bar X_i-\Theta_i\mid\Theta_i]]=0$. The unscaled components have the variances stated in Step 2. Therefore, the two scaled terms in the error decomposition have variances $(1-\beta)^2\tau^2$ and $\beta^2\sigma^2/W_i$, respectively. In particular, the scaling factors must remain in the mean squared error:

\operatorname{MSE}(T_\beta) =E[(T_\beta-\Theta_i)^2] =(1-\beta)^2\tau^2+\beta^2\frac{\sigma^2}{W_i}.

For positive $\tau^2,\sigma^2,W_i$, differentiate with respect to $\beta$, set the derivative to zero, and solve:

\frac{d}{d\beta}\operatorname{MSE}(T_\beta) =-2(1-\beta)\tau^2+2\beta\frac{\sigma^2}{W_i}=0, \qquad \beta^*=\frac{\tau^2}{\tau^2+\sigma^2/W_i}.

The second derivative is $2\tau^2+2\sigma^2/W_i>0$, so this stationary point is the minimum. The resulting estimate is the best linear credibility estimate under the stated model:

Theorem: Given known $\mu,\tau^2,\sigma^2$ and total exposure $W_i$, the minimum-mean-squared-error linear blend of $\bar X_i$ and $\mu$ is $T_{\beta^*}=\beta^*\bar X_i+(1-\beta^*)\mu$, where $\beta^*=\tau^2/(\tau^2+\sigma^2/W_i)$. This result is a best linear estimate; without additional distributional assumptions, it need not be the full conditional expectation $E[\Theta_i\mid\text{observations}]$.

Worked Applications

Worked Example: Unequal Period Exposures and a Credibility Blend

A hypothetical equipment portfolio has a risk with two observed periods. Its cost per exposure unit is $X_{i1}=\$1{,}100$ at exposure $w_{i1}=2$, and $X_{i2}=\$1{,}300$ at exposure $w_{i2}=4$. The collective mean is $\mu=\$1{,}000$, the between-risk variance is $\tau^2=400$, and the process parameter is $\sigma^2=900$, in squared dollars per exposure unit. Find the best linear estimate of this risk’s mean. No interest rate or mortality assumption applies. No distributional form is assumed; use the stated model parameters.

Method and common mistake avoided. Calculate the exposure-weighted mean, then apply the derived blend. Do not use the simple average of the two period costs: their exposures differ.

Step 1: Total exposure. $W_i=2+4=6$.

Step 2: Weighted observed mean. The weighted total is $2(\$1{,}100)+4(\$1{,}300)=\$2{,}200+\$5{,}200=\$7{,}400$. Hence $\bar X_i=\$7{,}400/6=\$1{,}233.333\ldots$.

Step 3: Credibility weight. The process variance of the weighted mean is $\sigma^2/W_i=900/6=150$. Thus $\beta^*=400/(400+150)=400/550=8/11$.

Step 4: Estimate. The observed mean is $\$233.333\ldots$ above the collective mean, so $T_{\beta^*}=\$1{,}000+(8/11)(\$233.333\ldots)=\$1{,}169.696\ldots$. Rounded to the nearest cent, the estimate is \$1,169.70.

Check. The weight $8/11$ lies between zero and one, and the estimate lies between $\$1{,}000$ and $\$1{,}233.33$, as a convex blend should.

Worked Example: Verifying the Minimum Mean Squared Error

For a hypothetical risk, $\tau^2=225$, $\sigma^2=900$, and $W_i=4$. The collective mean is $\$1{,}000$, and the weighted observed mean is $\$850$. Calculate the best linear estimate and compare its mean squared error with the error from using $\beta=0.75$. No interest rate or mortality assumption applies. No distributional form is assumed; the stated parameters define the model.

Method and common mistake avoided. First minimize the model’s mean squared error, then evaluate both candidate weights using the full formula. Do not omit the squared factors on either component.

Step 1: Process variance of the mean. $\sigma^2/W_i=900/4=225$.

Step 2: Best weight and estimate. The two variance components are equal, so $\beta^*=225/(225+225)=0.5$. The estimate is $\$1{,}000+0.5(\$850-\$1{,}000)=\$925$.

Step 3: Minimum mean squared error. At $\beta=0.5$, $\operatorname{MSE}=(0.5)^2(225)+(0.5)^2(225)=56.25+56.25=112.5$ squared dollars.

Step 4: Compare the alternative. At $\beta=0.75$, $\operatorname{MSE}=(0.25)^2(225)+(0.75)^2(225)=14.0625+126.5625=140.625$ squared dollars. The alternative has greater error.

Check. The quadratic error function is strictly convex, and $112.5<140.625$. The derived weight therefore performs better than the stated alternative under the model.

Worked Example: Exposure Weighting with Three Periods

A hypothetical commercial risk has period costs per exposure unit of $\$700,\$1{,}100,$ and $\$900$, with exposures $1,3,$ and $2$, respectively. Let $\mu=\$900$, $\tau^2=100$, and $\sigma^2=200$. Find the best linear estimate and its mean squared error. No interest rate or mortality assumption applies. No distributional form is assumed; the stated model parameters are used.

Method and common mistake avoided. Use each exposure both in the weighted mean and in the total exposure controlling process variance. Treating the three periods as equally exposed would misstate both.

Step 1: Total exposure and weighted mean. $W_i=1+3+2=6$. The weighted total is $1(\$700)+3(\$1{,}100)+2(\$900)=\$700+\$3{,}300+\$1{,}800=\$5{,}800$. Therefore, $\bar X_i=\$5{,}800/6=\$966.666\ldots$.

Step 2: Best weight. The process variance of the weighted mean is $200/6=33.333\ldots$. Thus $\beta^*=100/(100+200/6)=100/(400/3)=0.75$.

Step 3: Estimate. $T_{\beta^*}=\$900+0.75(\$966.666\ldots-\$900)=\$900+\$50=\$950$.

Step 4: Mean squared error. $(1-0.75)^2(100)+(0.75)^2(200/6)=6.25+18.75=25$ squared dollars.

Check. The estimate of $\$950$ lies between the collective mean and weighted observed mean, and the calculated error includes both between-risk and process components.

Interpretation and Exam Traps

The derivation shows why experience is blended rather than adopted without adjustment. A larger $\tau^2$ makes risks’ underlying means more variable, so individual experience is more informative about a particular risk. A larger $\sigma^2/W_i$ means the observed mean is noisier, so the estimate places relatively more weight on the collective mean. These statements follow directly from $\beta^*=\tau^2/(\tau^2+\sigma^2/W_i)$.

A frequent error is to write the correct error decomposition but then report the variances of its scaled terms as $\tau^2$ and $\sigma^2/W_i$. Those are the variances of the unscaled components. The factors $(1-\beta)^2$ and $\beta^2$ must be included when calculating mean squared error. For example, if $\beta=1/2$, the scaled component variances are $\tau^2/4$ and $\sigma^2/(4W_i)$, not $\tau^2$ and $\sigma^2/W_i$.

Other checks help catch mistakes:

Use $W_i=\sum_j w_{ij}$, not the number of periods, when exposures differ.
Use the variance of the weighted mean, $\sigma^2/W_i$, rather than $\sigma^2/W_i^2$.
Keep the collective mean and observed mean in the same units before blending.
Check that the weight is between zero and one and that the estimate lies between its two inputs.

The result is specific to a linear blend under the Bühlmann-Straub assumptions. If period observations are conditionally dependent, have different process-variance parameters, or do not share the same conditional mean, the variance calculation and resulting estimate may no longer apply as stated.

Check Your Understanding

For each problem, apply the model and show the weighted mean and variance calculation before checking the answer.

A risk has $\mu=\$40$, $\tau^2=64$, $\sigma^2=144$, $W_i=4$, and $\bar X_i=\$52$. Find the best linear estimate. (Answer: $\$47.68$)
A hypothetical risk has $\mu=\$2{,}000$, $\tau^2=300$, $\sigma^2=600$, $W_i=5$, and $\bar X_i=\$1{,}850$. Calculate the best linear weight and estimate, rounding the estimate to cents. (Answer: $\beta^*=5/7\approx0.7143$; estimate $\$1{,}892.86$)
For $\mu=\$60$, $\tau^2=80$, $\sigma^2=320$, $W_i=8$, and $\bar X_i=\$75$, find the estimate and its minimum mean squared error. (Answer: estimate $\$70$; MSE $26.6667$ squared dollars)
Suppose $\tau^2=50$, $\sigma^2=200$, and $W_i=2$. Find the optimal weight, then compare its MSE with the MSE at $\beta=0.4$. (Answer: $\beta^*=1/3$; optimal MSE $33.3333$, versus $34$ squared units at $\beta=0.4$)
A risk’s period costs are $\$620,\$560,$ and $\$480$, at exposures $1,2,$ and $3$. If $\mu=\$500$, $\tau^2=150$, and $\sigma^2=300$, calculate its weighted observed mean, credibility weight, and estimate. (Answer: weighted mean $\$530$; $\beta^*=0.75$; estimate $\$522.50$)

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