Checking Collective-Model Assumptions With Risk-Level Data

Why Check the Model Before Calculating Credibility?

Bühlmann-Straub calculations use risk-level experience to estimate a premium for each risk. Their interpretation depends on how observations relate to the underlying risk and to their exposures. A table can contain many numbers and still be unsuitable: some risks may have only one observation, exposure units may be inconsistent, or groups of risks may have systematically different expected rates.

Risk-level summaries help identify these concerns before the calculations are used. They do not prove that a model is correct. In particular, observed differences can arise from sampling variation, and a small data set may not reveal a real violation. The aim is to ask whether the data and the proposed model tell a coherent story.

Definition: In the Bühlmann-Straub model, risk $i$ has an unobserved risk parameter $\Theta_i$. Observation $X_{ij}$ for period or segment $j$ has exposure $w_{ij}>0$. Conditional on $\Theta_i$, the observations are assumed to have a common risk-specific mean and an exposure-scaled process variance: $$ E[X_{ij}\mid\Theta_i]=m(\Theta_i), \qquad \operatorname{Var}(X_{ij}\mid\Theta_i)=\frac{s^2(\Theta_i)}{w_{ij}}. $$ The observations for a given risk are also assumed conditionally independent given $\Theta_i$. Across risks, the $\Theta_i$ are treated as draws from a common collective distribution. In the usual notation, $E[m(\Theta_i)]=\mu$, $\operatorname{Var}(m(\Theta_i))=\tau^2$, and $E[s^2(\Theta_i)]=\sigma^2$.

These assumptions distinguish two sources of variation. Process variation is the variation in a risk's observations around its conditional mean. Between-risk variation arises because different $\Theta_i$ can imply different conditional means. Credibility uses both sources; a diagnostic should not mistakenly treat one as the other.

What Risk-Level Summaries Can Reveal

Start by listing, for each risk, the number of observations, each exposure, and the observed values. Then calculate or inspect the risk's exposure-weighted mean rate. When the values $X_{ij}$ are rates and the model assumptions hold, this mean is

\bar{X}_i=\frac{\sum_j w_{ij}X_{ij}}{\sum_j w_{ij}}.

The weighting matters: a rate based on more exposure carries more weight in the risk's aggregate experience. Also inspect variation within each risk, not just variation across every row of the table. A risk with one observation has no observed within-risk variation to summarize. Its single value cannot, on its own, show whether repeated observations are stable or whether variance scales with exposure.

Check the observation structure.
Identify what constitutes a risk, a period, and an observation. Confirm that repeated records refer to the same underlying risk and that the exposure measure is defined consistently.

Inspect within-risk behavior.
For risks with repeated observations, compare rates around their risk-level means. Look for abrupt changes, dependence over time, or differences in data collection that a stable conditional mean may not capture.

Adjust variability for exposure.
Under the model, rate variance is proportional to $1/w_{ij}$. Compare quantities such as exposure times an observed within-risk rate variance, while allowing for considerable sampling noise in small samples.

Compare risks and groups.
Look for systematic shifts in risk-level means across identifiable groups. A persistent shift may indicate that the risks do not share one collective structure unless the relevant group differences are modeled separately.

Why Rate Variance Scales Inversely With Exposure

The scaling rule follows from the distinction between an aggregate amount and a rate. Suppose, for a fixed risk parameter $\Theta_i$, an aggregate amount $S$ over exposure $w$ has conditional mean $w m(\Theta_i)$ and conditional variance $w s^2(\Theta_i)$. The rate is $X=S/w$. Dividing an amount by $w$ divides its variance by $w^2$, so

\operatorname{Var}(X\mid\Theta_i) =\operatorname{Var}\left(\frac{S}{w}\middle|\Theta_i\right) =\frac{w s^2(\Theta_i)}{w^2} =\frac{s^2(\Theta_i)}{w}.

Thus, if the same risk and process structure are being observed, a rate based on four times as much exposure should have one-quarter the conditional process variance, not the same variance. This is a model implication to examine, not a guarantee that small observed samples will display the ratio exactly.

Worked Example: Separating Repeated Experience From Risk Differences

A hypothetical service plan records annual claim rates for three risks, using the same exposure unit in each year. Risk A has rates 90, 100, and 110; Risk B has rates 145, 155, and 165; Risk C has one recorded rate, 130. Assess what the summaries say about repeated observations and within-risk variation.

Method. Calculate each risk's mean and, where possible, its sample variance. The sample variance for $n$ observations is the sum of squared deviations from the sample mean divided by $n-1$.

For Risk A, the mean is $(90+100+110)/3=100$. The squared deviations are $100,0,100$, so its sample variance is $(100+0+100)/(3-1)=100$.

For Risk B, the mean is $(145+155+165)/3=155$. Its squared deviations are also $100,0,100$, giving sample variance $200/2=100$.

Risk C has just one observation. Its mean is 130, but a sample variance cannot be calculated because there are no deviations from that single value with which to estimate within-risk variation.

Risks A and B each have repeated observations, with equal observed sample variances, while their means differ by $155-100=55$. This is consistent with substantial differences between these two risks' average experience, but the summaries alone do not establish why they differ or whether one common collective distribution is appropriate. Risk C provides no within-risk variance check.

Check and common mistake avoided. The two calculated variances both equal 100, and both use the sample-variance denominator $n-1=2$. A common mistake is to treat the three values for each risk as if they were three different risks. That would mix within-risk process variation with between-risk differences; calculate deviations within each risk first.

Checking the Exposure-Scaling Pattern

When repeated rates are available at differing exposures, a simple diagnostic is to compare $w_{ij}$ times an estimate of the rate variance. If the assumed process structure is stable, these products should be broadly similar for observations governed by the same conditional variance parameter. This is only an approximate check: a sample variance based on a few observations can be highly unstable, and observations from different risks can have different $s^2(\Theta_i)$.

Worked Example: Comparing Variability at Different Exposures

For one hypothetical risk, three pairs of annual rate observations have exposure levels $w=1$, $w=2$, and $w=4$, respectively. The observed pairs are $(8,12)$, $(9,11)$, and $(9.5,10.5)$. Check whether these small samples are consistent with exposure-scaled process variance.

Method. Calculate the sample variance within each pair, then multiply it by that pair's exposure. For two observations equidistant by $a$ from their mean, the sample variance is $(a^2+a^2)/(2-1)=2a^2$.

At $w=1$, the mean is 10 and $a=2$, so the variance is $2(2^2)=8$, and $w s^2=1(8)=8$.

At $w=2$, the mean is 10 and $a=1$, so the variance is $2(1^2)=2$, and $w s^2=2(2)=4$.

At $w=4$, the mean is 10 and $a=0.5$, so the variance is $2(0.5^2)=0.5$, and $w s^2=4(0.5)=2$.

These products are 8, 4, and 2, rather than approximately equal. The observed data do not display the expected scaling particularly closely. This is a warning to check the exposure definition, possible changes in the process, and sampling noise; with only two observations per exposure level, it is not a conclusive rejection of the model.

Check and common mistake avoided. The unadjusted rate variances are 8, 2, and 0.5; the exposure-adjusted values are 8, 4, and 2. Comparing the raw variances alone would overlook the model's inverse-exposure prediction. Conversely, declaring the model false solely because three noisy products differ would overstate what these data establish.

Assessing a Common Collective Structure

The common collective assumption concerns the distribution of risk parameters, not the requirement that every risk have the same observed mean. Risk-level means naturally vary, and some variation is expected when $\tau^2>0$. A more concerning pattern is a systematic difference tied to a known feature, such as a product type or measurement regime, especially when that shift is large relative to variation within the groups.

A practical check is to summarize risk-level means by relevant groups and compare their centers and spreads. If groups show distinct levels, ask whether the groups are genuinely comparable for the intended portfolio, whether a group-specific model or adjustment is needed, or whether the apparent difference could plausibly be sampling variation. This is a diagnostic judgment, not a formal hypothesis test.

Worked Example: Flagging Group Differences in Risk Means

A hypothetical administrator compares annual rates for six risks, all with equal exposure and three repeated observations per risk. The three risk means in Group A are 10, 12, and 14. The three risk means in Group B are 20, 22, and 24. Assess whether a single common collective structure merits review.

Method. Calculate the average risk mean in each group, then compare the difference with the variation of risk means within the groups.

The Group A average is $(10+12+14)/3=36/3=12$. The Group B average is $(20+22+24)/3=66/3=22$. The gap is $22-12=10$.

In Group A, deviations from 12 are $-2,0,2$, so the sample variance of risk means is $(4+0+4)/(3-1)=4$. Group B has the same deviations from 22 and the same sample variance, 4. Thus each group has a sample standard deviation of risk means $\sqrt{4}=2$, while the gap between group averages is 10, or five times that within-group standard deviation.

The systematic separation is a strong reason to investigate whether the groups share one collective distribution. It does not by itself prove that the model is invalid: the data are small, and the group labels or other differences may explain the shift. If the groups represent distinct risk populations, pooling them without adjustment may obscure important structure.

Check and common mistake avoided. Both group averages and variances have been calculated from the risk means, not from the 18 individual observations. Comparing one group's average with the overall average alone, or treating each repeated period as a new independent risk, would answer a different question and could misrepresent between-risk structure.

Common Traps and Limits of the Checks

Assuming one observation establishes a risk's variability. One recorded value provides no within-risk sample variance. It may contribute to a larger analysis, but cannot independently show how that risk varies over repeated periods.
Ignoring exposure when comparing rates. The model implies rate variance proportional to $1/w$. Similar raw variances at very different exposures may be inconsistent with that implication.
Confusing variation within and between risks. Pooling all observations can make systematic differences between risk means look like process noise. Summarize within each risk first.
Treating an imperfect pattern as a definitive failure. Small samples, random fluctuation, changing risk composition, and data errors can all affect diagnostics. Investigate a warning rather than converting it automatically into a model decision.
Assuming “common” means identical observations. The model allows distinct risk parameters and therefore different risk means. The concern is a systematic collective difference not represented by the chosen model.

Key takeaway: Before applying Bühlmann-Straub calculations, check that the records represent repeated observations of stable risks, that exposure is meaningful and consistently measured, that rate variability is broadly compatible with inverse-exposure scaling, and that risks plausibly belong to a common collective. These checks support model judgment; they do not replace it.

Check Your Understanding

Use the stated model assumptions and show the intermediate calculations before comparing your result with each answer.

A risk has four equally exposed annual rates: 7, 9, 11, and 13. Calculate its mean and sample variance. (Answer: mean $=10$; sample variance $=20/3\approx6.6667$.)
A set of rate observations has exposure $w=5$ and observed within-risk sample variance $1.6$. Calculate the exposure-adjusted variance $w s^2$. (Answer: $8$.)
Two exposure levels for the same process have $w=1$ with observed rate variance 12 and $w=4$ with observed rate variance 3. Compare their exposure-adjusted variances. (Answer: both equal 12, so the summaries match the exposure-scaling pattern.)
Two groups have risk means 6 and 8 in Group A, and 14 and 16 in Group B. Calculate the difference between group averages and state the diagnostic implication. (Answer: averages are 7 and 15, a difference of 8; this is a group-separation warning to investigate, not proof of model failure.)
Suppose the process variance parameter is $\sigma^2=10$. At exposure $w=2$, calculate the conditional variance of the rate and the conditional variance of the aggregate amount. (Answer: rate variance $=10/2=5$; aggregate variance $=2(10)=20$.)

What You'll Learn