Estimating Process Variance From Replicates

Why Replicates Identify Process Variation

In a credibility model, an observed risk outcome reflects both a persistent feature of the risk and random variation around that feature. Replicates—repeated observations on the same risk under comparable conditions—let us focus on the random, within-risk part. Centering each risk’s observations around its own average removes its persistent level, so the remaining variation can be used to estimate expected process variance.

This tutorial considers a balanced dataset: each of $I$ risks has the same number, $T$, of observations. The target is the average conditional variance of an observation, not the variance of the risk averages. No interest or mortality assumptions are relevant here. The method does not require a particular parametric distribution, but it does require finite conditional variances and suitable conditional independence.

Definition: Let $\Theta_i$ denote the persistent risk characteristic for risk $i$, and let $Y_{ij}$ be its $j$th observation, for $j=1,\ldots,T$. The expected process variance (EPV) is $$ \operatorname{EPV}=E[\operatorname{Var}(Y_{ij}\mid\Theta_i)]. $$ It is the average, over the risk population, of the variance remaining after the risk characteristic is given.

Assume observations from a risk have a common conditional mean $m(\Theta_i)$ and variance $v(\Theta_i)$. Given $\Theta_i$, the $T$ observations are independent and have the same distribution. Across risks, the $\Theta_i$ are sampled from the risk population represented by the target EPV. The conditional variance may differ by risk; it need not be a single constant for all risks.

Estimator: For each risk, calculate its sample variance $$ s_i^2=\frac{1}{T-1}\sum_{j=1}^{T}(Y_{ij}-\bar{Y}_{i\cdot})^2, \qquad \bar{Y}_{i\cdot}=\frac{1}{T}\sum_{j=1}^{T}Y_{ij}. $$ For a balanced dataset, estimate EPV by the average of the $I$ within-risk sample variances: $$ \widehat{\operatorname{EPV}}=\frac{1}{I}\sum_{i=1}^{I}s_i^2 =\frac{\displaystyle\sum_{i=1}^{I}\sum_{j=1}^{T}(Y_{ij}-\bar{Y}_{i\cdot})^2}{I(T-1)}. $$

Each risk receives equal weight because each has the same number of replicates and the target is the average process variance across risks. In the balanced setting, averaging the separate sample variances is equivalent to pooling their within-risk squared deviations and dividing by the total within-risk degrees of freedom, $I(T-1)$.

Why the Within-Risk Estimator Works

The key result is that, conditional on a risk’s persistent characteristic, its sample variance is unbiased for that risk’s conditional variance. To see why, write $\mu_i=m(\Theta_i)$ and $\sigma_i^2=v(\Theta_i)$. The identity $$ \sum_{j=1}^{T}(Y_{ij}-\bar{Y}_{i\cdot})^2 = \sum_{j=1}^{T}(Y_{ij}-\mu_i)^2 - T(\bar{Y}_{i\cdot}-\mu_i)^2 $$ holds because the cross-product term vanishes: $\sum_j(Y_{ij}-\bar{Y}_{i\cdot})=0$. Conditional on $\Theta_i$, the expected value of the first sum on the right is $T\sigma_i^2$. The conditional variance of the average is $\operatorname{Var}(\bar{Y}_{i\cdot}\mid\Theta_i)=\sigma_i^2/T$, so the expected value of the second term is also $T(\sigma_i^2/T)=\sigma_i^2$. Therefore, $$ E\left[\sum_{j=1}^{T}(Y_{ij}-\bar{Y}_{i\cdot})^2\mid\Theta_i\right] =(T-1)\sigma_i^2, \qquad E[s_i^2\mid\Theta_i]=\sigma_i^2. $$ Averaging over risks gives $E[s_i^2]=E[\sigma_i^2]=\operatorname{EPV}$. Thus the average within-risk sample variance is unbiased for EPV under the stated sampling assumptions. This result depends on conditional independence and a common conditional mean across replicates for a given risk; it does not require normality.

The centering step explains the practical separation. If every observation for risk $i$ includes the same persistent level $\mu_i$, subtracting $\bar{Y}_{i\cdot}$ removes that level. Differences between risk levels therefore do not directly inflate the within-risk squared deviations. By contrast, a variance calculated across all observations generally contains both process variation and persistent differences across risks.

Calculating the Estimate

Worked Example: Three Risks With Three Replicates

A hypothetical portfolio has three sampled risks, each observed in three comparable periods. The observations, in claim-count units per period, are: Risk A: 4, 6, 8; Risk B: 7, 8, 9; Risk C: 2, 5, 8. Estimate EPV using the balanced within-risk estimator. Assume observations are conditionally independent and identically distributed within each risk, with no period trend. No distributional form, interest rate, or mortality assumption is used.

Method. Calculate each risk’s sample variance using denominator $T-1=2$, then average the three variances.

For Risk A, $\bar{Y}_{A\cdot}=(4+6+8)/3=6$. Its squared deviations are $(4-6)^2=4$, $(6-6)^2=0$, and $(8-6)^2=4$. Hence $s_A^2=(4+0+4)/2=4$.

For Risk B, $\bar{Y}_{B\cdot}=(7+8+9)/3=8$. Its squared deviations are $1,0,1$, so $s_B^2=(1+0+1)/2=1$.

For Risk C, $\bar{Y}_{C\cdot}=(2+5+8)/3=5$. Its squared deviations are $9,0,9$, so $s_C^2=(9+0+9)/2=9$.

The estimate is $$ \widehat{\operatorname{EPV}}=\frac{4+1+9}{3}=\frac{14}{3}=4.6667. $$ As a check, the total within-risk sum of squared deviations is $8+2+18=28$, and $I(T-1)=3(2)=6$. Thus $28/6=4.6667$, matching the average of the three sample variances. The estimate is $4.67$ claim-count-squared units per period, rounded to two decimals. The common mistake avoided is pooling deviations from the overall mean instead of centering each risk separately.

Worked Example: Two Replicates Per Risk

A hypothetical service-contract portfolio records a repair count in two comparable periods for each of four risks. The paired observations are $(3,5)$, $(8,12)$, $(1,1)$, and $(6,12)$. Estimate EPV. Assume the two observations within each risk are conditionally independent with the same conditional mean and variance. No parametric distribution, interest rate, or mortality assumption is needed.

Method. With $T=2$, the sample variance for a pair $(a,b)$ is $((a-\bar{Y})^2+(b-\bar{Y})^2)/(2-1)=(a-b)^2/2$. This follows because each observation is one-half of the pair’s difference from the pair mean.

The first pair has difference $5-3=2$, so $s_1^2=2^2/2=2$. The second pair has difference $12-8=4$, so $s_2^2=4^2/2=8$. The third pair has difference $1-1=0$, so $s_3^2=0^2/2=0$. The fourth pair has difference $12-6=6$, so $s_4^2=6^2/2=18$.

Therefore, $$ \widehat{\operatorname{EPV}}=\frac{2+8+0+18}{4}=\frac{28}{4}=7. $$ A pooled check gives the same result: the within-pair squared-deviation totals are $2,8,0,18$, totaling $28$, and the denominator is $I(T-1)=4(1)=4$, giving $28/4=7$. The estimate is $7$ repair-count-squared units per period. The calculation uses the pair difference squared divided by two, not the pair difference squared by itself; omitting the divisor would double every pair’s sample variance.

Worked Example: Unequal Within-Risk Variation

A hypothetical group of three insured locations has four annual observations per location, all measured in the same exposure and currency units. The observations are Location A: 10, 12, 14, 16; Location B: 20, 20, 22, 22; Location C: 5, 8, 11, 14. Estimate EPV. Assume a stable conditional mean and conditionally independent observations within each location. No distributional form, interest rate, or mortality assumption is imposed.

Method. Compute each sample variance with denominator $T-1=3$, then average them. Different sample variances across locations are permitted; the estimator targets their average over the sampled risk population.

For Location A, the mean is $(10+12+14+16)/4=13$. The squared deviations are $9,1,1,9$, whose sum is $20$. Thus $s_A^2=20/3$.

For Location B, the mean is $(20+20+22+22)/4=21$. The squared deviations are $1,1,1,1$, whose sum is $4$. Thus $s_B^2=4/3$.

For Location C, the mean is $(5+8+11+14)/4=9.5$. The squared deviations are $20.25,2.25,2.25,20.25$, whose sum is $45$. Thus $s_C^2=45/3=15$.

The estimate is $$ \widehat{\operatorname{EPV}}=\frac{20/3+4/3+15}{3} =\frac{69/3}{3} =\frac{23}{3} =7.6667. $$ For a pooled check, the within-risk squared-deviation totals sum to $20+4+45=69$, while $I(T-1)=3(3)=9$; therefore $69/9=7.6667$. The estimate is $7.67$ squared measurement units, rounded to two decimals. The check confirms that the low variation at Location B does not justify discarding it: the stipulated target averages process variance across the three risks.

Assumptions, Interpretation, and Exam Traps

The estimator is useful because it separates repeat-to-repeat noise from persistent differences in risk levels. Its interpretation nevertheless depends on how the observations were collected. In particular:

Comparable replicates: All observations for a risk must refer to the same scale and comparable exposure conditions. A systematic change in exposure or a period trend can create within-risk variation that is not the process variance being targeted.
Conditional independence: After conditioning on the risk characteristic, observations must be independent for the sample-variance justification above. Serial dependence can change the expected within-risk sample variance.
Stable conditional mean: The conditional mean must remain the same across replicates for a given risk. If it changes systematically by period, centering at one risk average does not remove that trend.
Relevant risk sample: Averaging sample variances estimates the average process variance for the population represented by the sampled risks. A deliberately selected or unrepresentative set of risks may not estimate the target portfolio’s EPV.
Enough replication: At least two observations per risk are needed to calculate a sample variance. With few replicates or few risks, the estimator may be noisy even when its assumptions hold.

A frequent exam trap is to calculate one sample variance from all $IT$ observations and call it EPV. In the first worked example, the overall mean is $57/9=6.3333$. The sum of all observations squared is $403$, so the overall sum of squared deviations is $403-57^2/9=403-361=42$. The overall sample variance is $42/(9-1)=5.25$, not the within-risk estimate $4.6667$. Its deviations include differences between risks as well as fluctuations within each risk. For EPV, calculate deviations from each risk’s own replicate average.

The estimate is not a claim that every risk has exactly the same process variance. It estimates an average of conditional variances; individual $s_i^2$ values can be quite different. Nor does the calculation estimate the variability of the hypothetical means themselves. That is a separate component, obtained from variation across risk averages.

Key takeaway: In a balanced repeated-observation dataset, estimate EPV by averaging the within-risk sample variances. Center each risk’s observations on that risk’s own mean, use $T-1$ degrees of freedom per risk, and interpret the result under conditional independence and stable, comparable replicates. This treatment is aligned with the CAS Exam 8 (Advanced Ratemaking) syllabus.

Check Your Understanding

For each problem, use within-risk sample variances and state the estimate in the units implied by the observations.

Two risks each have three observations: Risk A has $1,3,5$, and Risk B has $4,4,7$. Find both sample variances and the EPV estimate. (Answer: $s_A^2=4$, $s_B^2=3$, and $\widehat{\operatorname{EPV}}=3.5$.)
Three risks each have two observations, with pair differences $2,6,4$. Find the EPV estimate using the pair-difference shortcut, and state the denominator used to pool within-risk sums of squares. (Answer: $9.3333$; the pooled denominator is $3(2-1)=3$.)
Two risks each have four observations: Risk A has $2,4,6,8$, and Risk B has $10,10,14,14$. Calculate each sample variance and their average. (Answer: $s_A^2=20/3$, $s_B^2=16/3$, and $\widehat{\operatorname{EPV}}=6$.)
A researcher calculates the variance of all observations around one overall mean, rather than centering within each risk. What variation can this calculation include that the EPV estimator is designed to exclude? (Answer: It can include persistent differences between risk levels, in addition to within-risk process variation.)
Each of five sampled risks has three observations, and the within-risk sums of squared deviations total $80$. Calculate the balanced EPV estimate and identify the assumption needed about replicate dependence. (Answer: $80/[5(3-1)]=8$ squared measurement units; observations must be conditionally independent within risk, along with a stable conditional mean.)

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