Sensitivity To Process And Between-Risk Variance

Why Variance Sensitivity Matters

A Bühlmann-Straub estimate combines a risk’s observed experience with the collective mean. The balance depends on how much of the observed variation is attributed to process variation within a risk and how much is attributed to persistent differences between risks. If either estimated variance changes, the balance can change—even when the observed losses and exposures do not.

This sensitivity matters in practice because variance components are estimated from data and may be revised as the experience period or portfolio changes. The purpose here is to isolate what those revisions do to the structural constant, credibility factor, and risk-level estimate. It is not to re-estimate the variance components from raw data.

Definition: In the Bühlmann-Straub model, let $\Theta_i$ be risk $i$'s hypothetical mean, with $E[\Theta_i]=\mu$ and $\operatorname{Var}(\Theta_i)=\tau^2$. For period $j$, let $X_{ij}$ be the observed loss cost and $w_{ij}>0$ its exposure. The model assumes $E[X_{ij}\mid\Theta_i]=\Theta_i$, $\operatorname{Var}(X_{ij}\mid\Theta_i)=\sigma^2/w_{ij}$, and conditional independence across periods given $\Theta_i$. Here $\sigma^2$ is the process variance per unit of exposure, and $\tau^2$ is the variance of hypothetical means between risks. No particular probability distribution is required for these moment assumptions.

For risk $i$, write $W_i=\sum_j w_{ij}$ for total exposure and $\bar X_i=\sum_j w_{ij}X_{ij}/W_i$ for its exposure-weighted observed mean. The structural constant $K$ and credibility factor $Z_i$ are

K=\frac{\sigma^2}{\tau^2}, \qquad Z_i=\frac{W_i}{W_i+K} =\frac{W_i\tau^2}{W_i\tau^2+\sigma^2}.

The credibility estimate of the risk’s hypothetical mean is

\widehat{\Theta}_i =Z_i\bar X_i+(1-Z_i)\mu =\mu+Z_i(\bar X_i-\mu).

These formulas use the variance components as inputs. The exposure total $W_i$ is held fixed when studying sensitivity to those inputs. As $K$ increases, credibility falls; as $W_i$ increases with $K$ fixed, credibility rises.

How the Variance Components Affect Credibility

The structural constant is a ratio, not a sum: more process variance relative to between-risk variance makes an individual risk’s experience less informative about its persistent mean. Conversely, more between-risk variance relative to process variance makes the risk-specific experience more valuable for distinguishing that risk from the collective.

\frac{\partial K}{\partial \sigma^2}=\frac{1}{\tau^2}>0, \qquad \frac{\partial K}{\partial \tau^2}=-\frac{\sigma^2}{(\tau^2)^2}<0.

For fixed positive exposure and positive variance components, the corresponding effects on credibility are

\frac{\partial Z_i}{\partial \sigma^2} =-\frac{W_i\tau^2}{(W_i\tau^2+\sigma^2)^2}<0, \qquad \frac{\partial Z_i}{\partial \tau^2} =\frac{W_i\sigma^2}{(W_i\tau^2+\sigma^2)^2}>0.

Thus, increasing $\sigma^2$ while holding $\tau^2$ fixed reduces credibility, while increasing $\tau^2$ while holding $\sigma^2$ fixed increases it. The estimate itself moves toward or away from the collective mean according to the direction of the risk’s experience: increasing $Z_i$ moves the estimate toward $\bar X_i$, not necessarily upward.

A Short Derivation of the Credibility Formula

The formula can be understood by considering the weighted observed mean. Conditional on $\Theta_i$, it has mean $\Theta_i$ and variance $\sigma^2/W_i$, because the period observations are conditionally independent and their variances are $\sigma^2/w_{ij}$. In particular,

\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}.

The best linear estimate of $\Theta_i$ from $\bar X_i$, under squared-error loss, has coefficient on $\bar X_i-\mu$ equal to the covariance of $\Theta_i$ and $\bar X_i$, divided by the variance of $\bar X_i$. The covariance is $\tau^2$, and the total variance of $\bar X_i$ is $\tau^2+\sigma^2/W_i$. Therefore the coefficient is

\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\tau^2}{W_i\tau^2+\sigma^2} =Z_i.

This also shows why the exposure total appears alongside the process variance: more exposure reduces the conditional variance of the observed mean. The derivation establishes the linear credibility coefficient under the stated moment model; it does not require a normal distribution.

Worked Example: Increasing Process Variance

A hypothetical portfolio has estimated process variance $\sigma^2=900$ and between-risk variance $\tau^2=100$, in squared loss-cost units. One risk has total exposure $W_i=12$, collective mean $\mu=\$600$, and observed weighted mean $\bar X_i=\$700$. Calculate the estimate, then recalculate it if process variance increases by 25% while all other inputs stay fixed.

Method and assumptions. Apply the Bühlmann-Straub moment model above, treating the given variance components as inputs. No interest or mortality assumptions apply, and no parametric distribution is assumed. The estimate uses the risk’s total exposure, not the number of periods.

Baseline. First calculate $K=900/100=9$. The credibility factor is $Z_i=12/(12+9)=12/21=0.5714$. The risk’s experience exceeds the collective mean by $\$700-\$600=\$100$. Thus,

\widehat{\Theta}_i =\$600+0.5714286(\$100) =\$657.1429 \approx \$657.14.

Revised process variance. A 25% increase gives $\sigma^2=900(1.25)=1{,}125$. With $\tau^2=100$, the revised constant is $K=1{,}125/100=11.25$. Therefore $Z_i=12/(12+11.25)=12/23.25=0.5161$. The revised estimate is

\widehat{\Theta}_i =\$600+\frac{12}{23.25}(\$100) =\$651.6129 \approx \$651.61.

The estimate falls by $\$657.14-\$651.61=\$5.53$, moving toward the collective mean because the risk’s observed mean is above it. This method avoids treating a higher process variance as evidence that the risk’s expected loss increased: it changes the weight on experience, not the observed mean. Dollar amounts are rounded to the nearest cent only at the end.

From Credibility Changes to Estimate Changes

With $\mu$, $\bar X_i$, and $W_i$ fixed, all changes in the estimate come through $Z_i$. Subtracting the original estimate from a revised estimate gives a useful exact relationship:

\Delta\widehat{\Theta}_i =(Z_{i,\mathrm{new}}-Z_{i,\mathrm{old}})(\bar X_i-\mu).

If $\bar X_i>\mu$, a reduction in credibility lowers the estimate. If $\bar X_i<\mu$, a reduction in credibility raises it. If $\bar X_i=\mu$, changing credibility has no effect on the estimate. These statements assume the collective mean and observed risk mean do not themselves change.

For small relative changes, the local sensitivity can also be expressed in elasticity form. Since $Z_i=W_i\tau^2/(W_i\tau^2+\sigma^2)$, its proportional response to either variance component depends on the amount of credibility not already assigned to the risk:

\frac{\partial \ln Z_i}{\partial \ln \sigma^2}=-(1-Z_i), \qquad \frac{\partial \ln Z_i}{\partial \ln \tau^2}=1-Z_i.

These are local approximations for small changes, not substitutes for recalculating $K$ and $Z_i$ after a substantial change. In particular, one should not interpret a 10% increase in process variance as necessarily producing a 10% decrease in credibility.

Worked Example: Changing Between-Risk Variance

For a hypothetical class, $\sigma^2=800$, $\tau^2=100$, and a risk has total exposure $W_i=5$, observed weighted mean $\bar X_i=\$572$, and collective mean $\mu=\$500$. Recalculate the estimate if the estimated between-risk variance rises to 160, with the other inputs unchanged.

Method and assumptions. Use the same conditional moment assumptions, with the revised between-risk variance treated as an input. No interest or mortality assumptions apply, and the model specifies moments rather than a distribution. Keep the risk’s experience and exposure fixed.

Original estimate. $K=800/100=8$, so $Z_i=5/(5+8)=5/13=0.3846154$. The observed mean is $\$72$ above the collective mean. Thus,

\widehat{\Theta}_{i,\mathrm{old}} =\$500+\frac{5}{13}(\$72) =\$527.6923 \approx \$527.69.

Revised estimate. With $\tau^2=160$, $K=800/160=5$. The new credibility is $Z_i=5/(5+5)=0.5$. Therefore,

\widehat{\Theta}_{i,\mathrm{new}} =\$500+0.5(\$72) =\$536.00.

Credibility rises by $0.5-0.3846154=0.1153846$, and the estimate rises by $0.1153846(\$72)=\$8.3077$, or $\$8.31$ to the nearest cent. The exact change identity checks the recalculation. The increase occurs because this risk’s observed mean is above the collective mean; for a below-average risk, the same credibility increase would lower its estimate. The common error is to say that higher between-risk variance always raises the risk estimate. It raises credibility, but the estimate’s direction depends on $\bar X_i-\mu$.

Worked Example: Scaling Both Variances Equally

A hypothetical risk has $W_i=6$, $\mu=\$480$, and $\bar X_i=\$560$. The variance estimates are $\sigma^2=720$ and $\tau^2=120$. An analyst considers replacing both estimates by values 10% higher. Determine the original and revised credibility and risk estimate.

Method and assumptions. Apply the Bühlmann-Straub formulas under the stated moment assumptions, holding exposure and both means fixed. No interest or mortality assumptions or parametric distribution are needed.

Original values. $K=720/120=6$, so $Z_i=6/(6+6)=0.5$. The observed mean is $\$80$ above the collective mean, giving

\widehat{\Theta}_{i,\mathrm{old}} =\$480+0.5(\$80) =\$520.00.

Increase both variances. The revised values are $\sigma^2=720(1.10)=792$ and $\tau^2=120(1.10)=132$. Their ratio remains $K=792/132=6$, so $Z_i=6/(6+6)=0.5$ and

\widehat{\Theta}_{i,\mathrm{new}} =\$480+0.5(\$80) =\$520.00.

There is no change in credibility or estimate because the variance ratio is unchanged. The method avoids the trap of assuming that any increase in either variance must change credibility: the structural constant depends on their ratio. Money is reported to the nearest cent.

Exam Traps and Practical Interpretation

Sensitivity calculations are straightforward when each input is kept distinct. Most errors come from using the right formula with the wrong interpretation or from changing more than one input without checking the ratio.

Reversing the ratio. The structural constant is process variance divided by between-risk variance, $K=\sigma^2/\tau^2$, not the reverse. If process variance rises with between-risk variance fixed, $K$ rises and $Z_i$ falls.
Confusing exposure with a period count. Use total exposure $W_i$. For unequal periods, it is the sum of the exposure weights, not the number of observations.
Assuming the estimate must move in one direction. A change in $Z_i$ pulls the estimate toward or away from $\bar X_i$; whether the dollar estimate rises or falls depends on whether $\bar X_i$ is above or below $\mu$.
Changing both variances without recomputing the ratio. Multiplying both by the same positive constant leaves $K$, $Z_i$, and the estimate unchanged when other inputs are fixed. Changing only one generally does not.

These calculations are useful for assessing whether a selected variance estimate materially affects an experience-rated result. They do not, by themselves, establish whether the assumed process-variance structure is appropriate or whether observations from different periods are dependent. Those are separate model questions.

Key takeaway: With exposure fixed, credibility depends on the ratio $\sigma^2/\tau^2$: increasing process variance reduces $Z_i$, while increasing between-risk variance raises it. The resulting estimate change equals the credibility change times the risk’s deviation from the collective mean.

Check Your Understanding

For each hypothetical problem, apply the Bühlmann-Straub moment formulas. Unless stated otherwise, hold all unmentioned inputs fixed; no parametric distribution, interest rate, or mortality assumption is needed.

A risk has $W_i=8$, $\sigma^2=640$, and $\tau^2=80$. Calculate $K$ and $Z_i$. (Answer: $K=8$, $Z_i=0.5000$.)
A risk has $W_i=3$, $\sigma^2=900$, $\tau^2=150$, $\mu=\$700$, and $\bar X_i=\$820$. Calculate the credibility estimate. (Answer: $K=6$, $Z_i=3/9=1/3$, and $\widehat{\Theta}_i=\$740.00$.)
A risk has $W_i=5$, $\sigma^2=500$, $\tau^2=100$, $\mu=\$400$, and $\bar X_i=\$430$. Recalculate the estimate if $\sigma^2$ increases by 20%. (Answer: Revised $\sigma^2=\$600$ squared loss-cost units, $K=6$, $Z_i=5/11$, and $\widehat{\Theta}_i=\$413.64$ to the nearest cent.)
A risk has $W_i=4$, $\sigma^2=400$, $\tau^2=100$, $\mu=\$250$, and $\bar X_i=\$310$. Both variance components are doubled. Compare the original and revised estimates. (Answer: Both cases have $K=4$ and $Z_i=0.5000$; the estimate is $\$280.00$ in both cases.)
A risk has $W_i=9$ and $\sigma^2=600$. What value of $\tau^2$ gives credibility $Z_i=0.6000$? Verify the resulting structural constant. (Answer: $\tau^2=100$, so $K=600/100=6$ and $Z_i=9/(9+6)=0.6000$.)

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