Why Organize Repeated Observations by Risk?
Bühlmann variance components describe two distinct sources of variation: variation from one observation period to another for the same risk, and variation in underlying expected outcomes across risks. A table that keeps risk and period separate makes these sources easier to distinguish. Simply pooling all observations into one list discards the repeated-risk structure that carries this information.
The discussion here concerns equal-exposure observations on a common scale. For example, a hypothetical insurer might compare annual claim counts for risks observed over the same length of time. Before using repeated observations to study components, check that the observations are organized consistently and that the assumptions supporting the comparison are reasonable.
For the component interpretation, regard risks as drawn from a population of risks. Let \(\Theta_i\) represent the underlying characteristics of risk \(i\), and define its hypothetical mean as \(m_i=E[X_{ij}\mid\Theta_i]\). The observation for a risk in a particular period can differ from this mean because of process variation. The hypothetical mean can differ across risks because of between-risk variation.
Arrange the Data and Identify the Summaries
Put one row per risk and one column per period. Three summaries are particularly useful for describing the data:
- The risk average \(\bar X_i=\frac{1}{n}\sum_{j=1}^{n}X_{ij}\) summarizes the observed experience for risk \(i\).
- The grand average \(\bar X=\frac{1}{r}\sum_{i=1}^{r}\bar X_i\) is also the average of all \(rn\) observations when the data are balanced.
- The within-risk deviations \(X_{ij}-\bar X_i\) show how each period’s observation differs from its own risk average.
The deviations within each row describe period-to-period fluctuation around that risk’s observed average. Differences among the row averages describe how the observed averages vary across risks. Neither summary alone identifies a variance component: a row average still contains process noise, and within-risk deviations are observed data rather than the unobservable true process distribution.
Worked Example: Building Risk and Overall Summaries
A hypothetical portfolio records equal-exposure annual claim counts for three risks over three periods. The observations are shown below. Calculate each risk average, the grand average, and the within-risk deviations. Explain which summaries describe within-risk and between-risk information.
| Risk | Period 1 | Period 2 | Period 3 |
|---|---|---|---|
| A | 8 | 10 | 12 |
| B | 14 | 16 | 18 |
| C | 10 | 12 | 14 |
Method. First average across periods within each risk; then average the risk averages. This preserves the repeated-risk structure instead of treating the nine observations as unrelated.
For risk A, the total is \(8+10+12=30\), so \(\bar X_A=30/3=10\). Its deviations from its own average are \(8-10=-2\), \(10-10=0\), and \(12-10=2\).
For risk B, the total is \(14+16+18=48\), so \(\bar X_B=48/3=16\). Its deviations are \(14-16=-2\), \(16-16=0\), and \(18-16=2\).
For risk C, the total is \(10+12+14=36\), so \(\bar X_C=36/3=12\). Its deviations are \(10-12=-2\), \(12-12=0\), and \(14-12=2\).
The grand average of the three risk averages is \((10+16+12)/3=38/3=12.6667\), rounded to four decimal places. As a check, the total of all observations is \(30+48+36=114\), and \(114/9=12.6667\) to four decimal places.
Interpretation. Each row’s deviations show within-risk period fluctuation. The risk averages 10, 16, and 12 show between-risk differences in observed averages. They are descriptive summaries; they do not, by themselves, separate process noise in the averages from true variation in hypothetical means. The method avoids the common mistake of interpreting differences among observed risk averages as pure between-risk variation.
How Repeated Observations Carry Component Information
Let \(\operatorname{EPV}=E[\operatorname{Var}(X_{ij}\mid\Theta_i)]\) denote the expected process variance, and let \(\operatorname{VHM}=\operatorname{Var}(m_i)\) denote the variance of hypothetical means. These are population quantities, not statistics calculated directly by taking one particular sum of squares. The variance decomposition gives \(\operatorname{Var}(X_{ij})=\operatorname{EPV}+\operatorname{VHM}\) under the model assumptions.
The covariance between two distinct-period observations on the same risk has a different interpretation. Conditional independence says that, after the risk characteristics are fixed, the two observations do not share process fluctuation. Unconditionally, however, both are associated with the same risk-specific mean. The following derivation shows why their covariance reflects VHM.
For distinct periods \(j\ne k\), conditional independence gives \(\operatorname{Cov}(X_{ij},X_{ik}\mid\Theta_i)=0\). Both conditional means equal \(m_i\).
\(\operatorname{Cov}(X_{ij},X_{ik})=E[\operatorname{Cov}(X_{ij},X_{ik}\mid\Theta_i)]+\operatorname{Cov}(E[X_{ij}\mid\Theta_i],E[X_{ik}\mid\Theta_i])\).
The first term is zero by conditional independence. The second is \(\operatorname{Cov}(m_i,m_i)=\operatorname{Var}(m_i)=\operatorname{VHM}\). Thus \(\operatorname{Cov}(X_{ij},X_{ik})=\operatorname{VHM}\) for distinct periods under these assumptions.
This result explains why repeated observations on the same risk are informative: shared risk characteristics induce association across periods. By contrast, variation among observations in the same row also includes independent process fluctuations. A risk average based on \(n\) conditionally independent observations has variance \(\operatorname{VHM}+\operatorname{EPV}/n\) under the common-component model. To see this, write \(\bar X_i=m_i+\frac{1}{n}\sum_j(X_{ij}-m_i)\). The second term has conditional mean zero and conditional variance \(\operatorname{Var}(X_{ij}\mid\Theta_i)/n\); averaging over risks gives variance \(\operatorname{EPV}/n\), while the first term contributes VHM. The cross term is zero because the conditional process deviations have mean zero. Therefore variation among observed risk averages is not identical to VHM.
Worked Example: Interpreting a Risk Average’s Variation
Suppose a hypothetical equal-exposure portfolio is described by the common-component model, with \(\operatorname{EPV}=25\) and \(\operatorname{VHM}=9\), in squared units of the observation. Each risk is observed in four conditionally independent periods. Find the variance of a risk average and the covariance between two distinct-period observations on the same risk.
Method. Use the repeated-observation variance identity for the average, and the conditional-independence result for the covariance. These are theoretical quantities under the stated model, not estimates from a particular data table.
The process contribution to the average’s variance is \(\operatorname{EPV}/n=25/4=6.25\). The between-risk contribution is \(\operatorname{VHM}=9\). Thus \(\operatorname{Var}(\bar X_i)=9+6.25=15.25\) squared units.
For two observations from distinct periods on the same risk, their covariance is \(\operatorname{VHM}=9\) squared units. As a check, the average’s variance exceeds VHM by 6.25, exactly the remaining process contribution \(25/4\). The common mistake is to report 9 as the variance of the risk average; 9 is only its between-risk contribution, not the total variance.
Read the Table Before Calculating
Balanced data require the same number of comparable observations for every risk. This is more than a convenient layout: equal numbers of observations ensure that the grand average of risk averages also equals the average across all individual observations. If a row is incomplete, these two averages can differ. Unequal exposure can create another problem: counts recorded over different exposure amounts are not directly comparable as equal-exposure observations.
Worked Example: A Missing Period Breaks the Balanced Setup
A hypothetical data set records equal-exposure annual claim counts for four risks over three periods, but risk D has no observation in period 3. Calculate the available risk averages and both natural overall averages, then explain why the balanced-data interpretation cannot be applied unchanged.
| Risk | Period 1 | Period 2 | Period 3 |
|---|---|---|---|
| A | 3 | 5 | 7 |
| B | 5 | 7 | 9 |
| C | 7 | 9 | 11 |
| D | 9 | 11 | Missing |
Method. Calculate each available row average using its observed periods, and distinguish an average across observations from an equally weighted average across risks.
The risk averages are \(\bar X_A=(3+5+7)/3=5\), \(\bar X_B=(5+7+9)/3=7\), and \(\bar X_C=(7+9+11)/3=9\). For risk D, \(\bar X_D=(9+11)/2=10\). There are 11 observations, with total \(15+21+27+20=83\), so the observation-weighted average is \(83/11=7.5455\), rounded to four decimal places.
The equally weighted average of the four risk averages is \((5+7+9+10)/4=31/4=7.75\). The two overall averages differ because risk D has only two observations while the other risks have three. This data set is not balanced, so the balanced-data equivalence between the two averages does not hold. Do not silently treat the missing value as zero or claim the balanced summaries apply unchanged. The method avoids confusing a missing observation with an observed claim count of zero.
Common Mistakes and Practical Checks
- Pooling too early. A grand average alone hides which observations belong to the same risk. Keep the risk-by-period organization and calculate row averages first.
- Calling the spread of risk averages VHM. Each observed average contains process fluctuation as well as a risk-specific mean. Under the model, its variance includes both VHM and \(\operatorname{EPV}/n\).
- Ignoring unequal row sizes. With unequal numbers of observations, the average across risks need not equal the average across individual observations.
- Overlooking period comparability. A common exposure basis and absence of unmodeled period effects are needed for the standard interpretation. A change affecting all risks in one period can resemble a risk-level pattern if the period structure is ignored.
- Confusing observed zero with missing data. A recorded zero is an observation; a missing period supplies no observed value. They must not be interchanged.
Check Your Understanding
For each problem, keep the risk and period structure visible and distinguish descriptive summaries from model quantities.
- Three risks have two equal-exposure observations each: A has 4 and 6, B has 6 and 8, and C has 10 and 12. Find the three risk averages and the grand average. (Answer: Risk averages 5, 7, and 11; grand average \(46/6=7.6667\), to four decimal places.)
- Two risks have observations A: 2, 4, 6 and B: 8, 10, 12. Find the grand average and the total of the squared within-risk deviations. (Answer: Grand average 7; total squared within-risk deviations \(8+8=16\).)
- Under the common-component model, \(\operatorname{EPV}=25\), \(\operatorname{VHM}=9\), and each risk has four conditionally independent observations. Find the risk-average variance and the covariance of two distinct-period observations on the same risk. (Answer: Risk-average variance \(9+25/4=15.25\); covariance \(9\), in squared observation units.)
- Three risks have three observations each with averages 4, 6, and 8. A fourth risk has only two observations, 10 and 12. Find that fourth risk’s average, the equally weighted average of the four risk averages, and explain whether the balanced-data identity applies. (Answer: Fourth risk average 11; equally weighted risk-average mean \((4+6+8+11)/4=7.25\); the data are unbalanced, so the balanced-data identity does not apply unchanged.)
- For distinct periods on the same risk, conditional independence holds and both conditional means equal \(m_i\), where \(\operatorname{Var}(m_i)=14\). What is the unconditional covariance of those observations under the model? (Answer: 14, since the conditional covariance is zero and the covariance of the conditional means is \(\operatorname{Var}(m_i)\).)