Partial Histories And Missing Risk-Periods

Why Partial Histories Need Care

In a credibility data set, risks may have different amounts of observed experience. A risk might enter the portfolio late, leave before the end of the study, or have a period for which no usable record exists. The Bühlmann-Straub method accommodates unequal observed exposure by weighting each observed risk-period appropriately. It does not treat an unobserved period as a period with zero losses.

That distinction matters twice: observed exposure contributes to the risk’s weighted average and to the credibility assigned to that average. A missing record contributes neither. By contrast, a recorded zero loss is an observation: it contributes zero to the loss numerator but its exposure still contributes to the denominator and the credibility weight.

Definition: For risk $i$, let $X_{ij}$ be its observed loss cost per exposure unit in period $j$, and let $w_{ij}>0$ be the exposure weight for that period. Sum only over the periods with observed data. The total observed weight and weighted risk mean are $$ W_i=\sum_{j\in O_i}w_{ij}, \qquad \bar X_i=\frac{\sum_{j\in O_i}w_{ij}X_{ij}}{W_i}, $$ where $O_i$ is the set of observed periods for risk $i$. A period that is missing is not in $O_i$; a period with an observed loss cost of zero is in $O_i$, with its actual positive exposure weight.

The Bühlmann-Straub Estimate With Incomplete Histories

Assume the standard Bühlmann-Straub hierarchical model. Conditional on a risk’s underlying mean $\Theta_i$, observations from its periods are independent, have conditional mean $\Theta_i$, and have conditional variance $\sigma^2/w_{ij}$. Across risks, the underlying means have common mean $m$ and variance $\tau^2$. Thus exposure increases the information in a period: a larger $w_{ij}$ gives a smaller conditional variance.

The calculation uses only observed periods. Write $K=\sigma^2/\tau^2$, the structural constant, when $\tau^2>0$. For a risk with positive total observed weight, the credibility factor and credibility estimate are

Z_i=\frac{W_i}{W_i+K}, \qquad \widehat{\Theta}_i=Z_i\bar X_i+(1-Z_i)m.

The estimate blends the risk’s observed weighted mean with the collective mean. More observed exposure increases $W_i$ and therefore increases $Z_i$, holding the model parameters fixed. A zero-loss observation can lower $\bar X_i$, but it does not erase its exposure or reduce $W_i$.

These formulas presume that the recorded periods and their exposure weights are correctly identified, and that the model assumptions apply to the observed experience. In particular, a missing period must not be silently coded as zero. If missingness is systematically related to losses or to the underlying risk, the observed history may be unrepresentative; the standard calculation alone does not correct that selection effect. The model parameters may be specified or estimated from a broader data set, but the examples below treat them as given.

Worked Example: A Risk With One Missing Period

A hypothetical portfolio has collective mean $m=100$, process variance parameter $\sigma^2=400$, and between-risk variance $\tau^2=100$. For one risk, three annual records are available: loss cost 70 with exposure weight 1, loss cost 100 with exposure weight 2, and loss cost 150 with exposure weight 1. A fourth annual record is missing. Calculate its Bühlmann-Straub estimate.

Method and observed history. Use the Bühlmann-Straub estimate with the observed periods only. The fourth period has no record, so it contributes neither a loss-cost amount nor an exposure weight.

Step 1: Calculate total observed weight. $W_i=1+2+1=4$.

Step 2: Calculate the weighted mean. The weighted numerator is $1(70)+2(100)+1(150)=70+200+150=420$. Therefore, $\bar X_i=420/4=105$.

Step 3: Calculate the structural constant and credibility. $K=\sigma^2/\tau^2=400/100=4$, so $Z_i=4/(4+4)=1/2=0.5$.

Step 4: Blend the risk mean and collective mean. $\widehat{\Theta}_i=0.5(105)+0.5(100)=52.5+50=102.5$.

Check. The estimate is halfway between 105 and 100, as required when $Z_i=0.5$. The missing fourth period has not been treated as zero exposure or zero loss. The common mistake avoided is adding a zero-loss record for that unobserved year, which would incorrectly change both the weighted mean and the credibility.

Why the Exposure-Weighted Formula Works

The weighted mean is the natural summary because, conditional on $\Theta_i$, more heavily exposed periods are more precise. Its variance explains how the total observed weight enters the credibility factor.

\operatorname{Var}(\bar X_i\mid\Theta_i) =\operatorname{Var}\left(\frac{\sum_{j\in O_i}w_{ij}X_{ij}}{W_i}\,\middle|\,\Theta_i\right) =\frac{1}{W_i^2}\sum_{j\in O_i}w_{ij}^2\frac{\sigma^2}{w_{ij}} =\frac{\sigma^2}{W_i}.

The middle equality uses conditional independence, so the conditional variance of the weighted sum is the sum of the weighted conditional variances. The final equality follows because $\sum_{j\in O_i}w_{ij}=W_i$. Thus, although the number of observed periods may differ between risks, their information under this model is summarized by total observed weight.

To see the credibility blend, consider a linear estimate $a\bar X_i+(1-a)m$. Its squared-error risk, averaged over the model, is

a^2\frac{\sigma^2}{W_i}+(1-a)^2\tau^2.

The first term is the conditional process variance contribution; the second reflects uncertainty in the risk mean when the estimate shrinks toward $m$. Differentiating with respect to $a$ and setting the derivative to zero gives

2a\frac{\sigma^2}{W_i}-2(1-a)\tau^2=0, \qquad a\left(\frac{\sigma^2}{W_i}+\tau^2\right)=\tau^2, \qquad a=\frac{W_i\tau^2}{\sigma^2+W_i\tau^2} =\frac{W_i}{W_i+\sigma^2/\tau^2}.

This is $Z_i=W_i/(W_i+K)$. The result depends on the stated model assumptions and on using the observed weights correctly; it is not a rule for imputing unobserved losses.

Worked Example: A Recorded Zero Loss Is Still an Observation

A hypothetical risk has three fully observed periods, each with exposure weight 1. Its loss costs are 0, 150, and 150. The collective mean is $m=120$, with $\sigma^2=900$ and $\tau^2=225$. Calculate the estimate, then quantify the effect of incorrectly dropping the zero-loss period.

Correct method. The zero is an observed outcome, so include its period and weight.

Step 1: Calculate total weight and weighted mean. $W_i=1+1+1=3$. The weighted numerator is $1(0)+1(150)+1(150)=300$, hence $\bar X_i=300/3=100$.

Step 2: Calculate credibility. $K=900/225=4$, so $Z_i=3/(3+4)=3/7$.

Step 3: Calculate the estimate. $\widehat{\Theta}_i=120+\frac{3}{7}(100-120)=120+\frac{3}{7}(-20)=120-\frac{60}{7}=\frac{780}{7}\approx111.43$.

Incorrect calculation and correction. If the zero is incorrectly treated as missing, the weighted mean becomes $(150+150)/2=150$, and the weight becomes 2. That produces $Z_i=2/(2+4)=1/3$ and an incorrect estimate $120+\frac13(150-120)=130$. The correction is to keep the observed zero: it contributes zero to the numerator, but its weight of 1 belongs in $W_i$. The correct estimate is approximately 111.43.

Check. The correctly calculated mean of 100 is below the collective mean of 120, so the estimate must also be below 120. The result $111.43$ passes this check; the incorrect result $130$ does not, given that the observed weighted mean is below 120.

Applying the Rule to Different Exposure Weights

The weight need not be one for every observed period. If a period’s loss cost is loss amount divided by exposure, its weight may be that period’s exposure. The weighted mean then combines the period-specific loss costs in proportion to their exposure. A missing period is excluded even if other periods for the same risk are observed.

Worked Example: Unequal Exposures in a Partial History

For a hypothetical risk, the collective mean is $m=80$, $\sigma^2=360$, and $\tau^2=90$. In the first observed period, exposure is 2 and total loss is 100. The second period is missing. In the third period, exposure is 1 and total loss is 110. Treat each observed period’s loss cost as total loss divided by exposure. Find the credibility estimate.

Method. Convert each observed period to loss cost per exposure unit, retain its exposure as the weight, and sum only the observed periods.

Step 1: Calculate observed loss costs. In period 1, $X_{i1}=100/2=50$, with $w_{i1}=2$. In period 3, $X_{i3}=110/1=110$, with $w_{i3}=1$. The missing second period supplies no term.

Step 2: Calculate total weight and weighted mean. $W_i=2+1=3$. The weighted numerator is $2(50)+1(110)=100+110=210$. Thus, $\bar X_i=210/3=70$.

Step 3: Calculate the credibility factor. $K=360/90=4$, so $Z_i=3/(3+4)=3/7$.

Step 4: Calculate the estimate. $\widehat{\Theta}_i=80+\frac{3}{7}(70-80)=80+\frac{3}{7}(-10)=80-\frac{30}{7}=\frac{530}{7}\approx75.71$.

Check. The weighted mean of 70 is below the collective mean of 80, so the estimate of approximately 75.71 lies between them. The common mistake avoided is using the unweighted average $(50+110)/2=80$: the first period has twice the exposure of the third and must receive twice its weight.

Common Errors and Practical Checks

Unequal histories are manageable once the data status of every period is explicit. Before calculating, classify each period as observed or missing and confirm its exposure weight. Then compute the numerator, denominator, and credibility from the same set of observed periods.

Filling a missing period with zero. This incorrectly treats absence of data as an observed no-loss result. Exclude the period from both weighted sums.
Dropping an observed zero. This removes genuine evidence and reduces the weight. Keep the period in the denominator and in $W_i$; its numerator contribution is zero.
Using a simple mean with unequal exposure. Unless all observed weights are equal, use $\sum w_{ij}X_{ij}/W_i$, not the unweighted average.
Using different period sets in the mean and credibility factor. The same observed periods must define both $\bar X_i$ and $W_i$.
Ignoring the reason data are missing. If high-loss periods are more likely to be missing, the observed mean may be biased. The ordinary Bühlmann-Straub calculation does not repair informative missingness.

A useful final check is that $0\leq Z_i\leq1$, and the estimate lies between $\bar X_i$ and $m$. If the observed risk mean equals the collective mean, the estimate equals both regardless of the credibility factor. If a missing period has been inadvertently included as a zero, these checks may reveal the error, but careful period-by-period data handling is the primary safeguard.

Key takeaway: For a partial history, calculate $W_i$ and $\bar X_i$ using observed periods only. Missing periods contribute neither exposure nor loss; observed zero-loss periods contribute their exposure and a zero loss-cost amount.

Check Your Understanding

For each problem, use the observed periods only and show how the exposure weights affect the estimate.

A risk has $m=50$, $K=2$, and two observed periods with weights 1 and 1 and loss costs 40 and 60. A third period is missing. Find $W_i$, $\bar X_i$, $Z_i$, and the estimate. (Answer: $W_i=2,\ \bar X_i=50,\ Z_i=0.5000,\ \widehat{\Theta}_i=50$.)
A risk has $m=90$, $\sigma^2=400$, and $\tau^2=100$. Its observed records are loss cost 70 with weight 1 and loss cost 110 with weight 3; one other period is missing. Find its credibility estimate. (Answer: $W_i=4,\ \bar X_i=100,\ K=4,\ Z_i=0.5000,\ \widehat{\Theta}_i=95$.)
A risk has $m=100$, $K=3$, and three observed periods, each with weight 1 and loss costs 0, 120, and 150. Find the estimate. (Answer: $W_i=3,\ \bar X_i=90,\ Z_i=0.5000,\ \widehat{\Theta}_i=95$.)
A risk has $m=60$, $\sigma^2=250$, and $\tau^2=50$. Its observed periods have exposures 2 and 1, and total losses 80 and 100, respectively; a third period is missing. Find the exposure-weighted loss mean and credibility estimate. (Answer: $\bar X_i=60,\ K=5,\ W_i=3,\ Z_i=0.3750,\ \widehat{\Theta}_i=60$.)
A risk has $m=75$, $K=4$, and two observed periods: loss cost 0 with weight 2 and loss cost 120 with weight 1. A third period is missing. Calculate the correct estimate and compare it with the estimate obtained by wrongly treating the observed zero as missing. (Answer: Correctly, $W_i=3,\ \bar X_i=40,\ Z_i=3/7,\ \widehat{\Theta}_i=60$. If the zero is wrongly dropped, $W_i=1,\ \bar X_i=120,\ Z_i=1/5$, giving the incorrect estimate $84$.)

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