Why Compare Severity Across Periods?
A change in average claim severity can reflect several different things: a shift in typical claim costs, a small number of unusually large claims, or a change in the kinds of claims being observed. Period claim counts matter too. A mean based on a few claims is generally less stable than one based on many claims, and period means cannot be combined correctly by simply giving every period equal weight.
The purpose of this analysis is to diagnose what the observed data say before treating a difference as a persistent change. Comparisons are most interpretable when the periods use the same claim definition, coverage, deductible and limit basis, valuation basis, and currency. If these differ, the figures may not be measuring comparable severities. The earlier material on defining severity and matching frequency and severity data provides the relevant claim-level setup.
In this tutorial, the claim amounts in worked examples are hypothetical observed losses measured on a common basis and at a common valuation date. There is no discounting, so the effective annual interest rate is \(i=0\%\); mortality is not applicable. The examples use the empirical observations given, not an assumed parametric severity distribution. Where a probability model is useful for interpretation, its assumptions are stated separately.
Period Means, Pooled Means, and Unequal Counts
A period mean answers, “What was the average severity among claims in this period?” A pooled mean answers, “What was the average severity among all claims in these periods combined?” When periods have different claim counts, the pooled mean weights each period mean in proportion to its claim count:
The identity follows directly from the definition of each period mean: \(L_j=n_j\bar{X}_j\). Adding those period totals gives \(\sum_j L_j=\sum_j n_j\bar{X}_j\); dividing by the total claim count gives the pooled formula. An unweighted average of period means instead gives each period equal influence, regardless of how many claims it contains. That may be a deliberate period-level summary, but it is not the average severity per claim across the combined data.
Worked Example: Combining Periods With Different Claim Counts
A hypothetical insurer records four claims in Period A, with severities of $1,000, $1,200, $1,400, and $1,600, and two claims in Period B, with severities of $1,800 and $2,200. Find each period mean, the simple average of the two period means, and the pooled mean. The data are exact observed amounts; no parametric distribution is assumed. The interest rate is 0%, and mortality is not applicable.
Period A. Its total loss is \(1{,}000+1{,}200+1{,}400+1{,}600=\$5{,}200\). With \(n_A=4\), its mean severity is \(5{,}200/4=\$1{,}300\).
Period B. Its total loss is \(1{,}800+2{,}200=\$4{,}000\). With \(n_B=2\), its mean severity is \(4{,}000/2=\$2{,}000\).
The unweighted average of the two period means is \((1{,}300+2{,}000)/2=\$1{,}650\). This is the average of the two period-level results, not the average amount per claim.
For the pooled mean, total losses are \(5{,}200+4{,}000=\$9{,}200\), and the total claim count is \(4+2=6\). Therefore, the pooled mean is \(9{,}200/6=\$1{,}533.33\), rounded to the nearest cent. Equivalently, \((4(1{,}300)+2(2{,}000))/6=9{,}200/6\). The answer checks against the total loss divided by total claims.
Method and trap. Use claim-count weights when the question asks for average severity across all claims. Treating the periods equally overweights Period B relative to its two claims; the unweighted $1,650 is not the pooled per-claim average.
Volatility, Small Counts, and Large Claims
A period mean is a summary, not a full description of the severity data. The sample standard deviation describes the spread of observed severities around their period mean. For observations \(X_1,\ldots,X_n\), with \(n>1\), the sample variance is \(s^2=\sum_i(X_i-\bar{X})^2/(n-1)\), and \(s\) is its square root. The coefficient of variation is \(s/\bar{X}\) when \(\bar{X}>0\); it expresses spread relative to the mean and is unit-free.
These are descriptive measures of the observed data. If the claims are treated as independent observations from a stable severity distribution with finite variance, \(s/\sqrt{n}\) is the estimated standard error of the sample mean. That interpretation depends on the model: it does not account for changes in claim mix, dependence among claims, or a changing severity process. In particular, a small count often makes a mean sensitive to individual claims, but a large count alone does not guarantee comparable data.
Worked Example: Diagnosing a Period Mean Affected by a Large Claim
In a hypothetical period, five claim severities are $800, $900, $1,000, $1,100, and $6,200. In a comparison period, four severities are $900, $950, $1,050, and $1,100. Calculate the means and medians, and quantify how unusual the first period’s spread is relative to its mean. These are observed amounts, with no assumed parametric distribution; \(i=0\%\), and mortality is not applicable.
First-period mean. The total is \(800+900+1{,}000+1{,}100+6{,}200=\$10{,}000\). Dividing by five gives a mean of \(\$2{,}000\). Its sorted observations are already shown in increasing order, so the median is the third observation, \(\$1{,}000\).
The $6,200 claim is \(6{,}200/10{,}000=0.62\), or 62% of the period’s total. If it is set aside solely as a diagnostic check, the other four claims total \(800+900+1{,}000+1{,}100=\$3{,}800\), for a mean of \(3{,}800/4=\$950\). This calculation shows the large claim’s influence; it does not establish that the claim should be excluded from a rate or estimate.
For the first period, deviations from the $2,000 mean are \(-1{,}200\), \(-1{,}100\), \(-1{,}000\), \(-900\), and \(4{,}200\). Their squared deviations sum to \(1{,}440{,}000+1{,}210{,}000+1{,}000{,}000+810{,}000+17{,}640{,}000=22{,}100{,}000\). Thus \(s^2=22{,}100{,}000/(5-1)=5{,}525{,}000\), and \(s=\sqrt{5{,}525{,}000}=\$2{,}350.53\), rounded to cents. Its coefficient of variation is \(2{,}350.53/2{,}000=1.1753\), or 117.53%.
Comparison-period checks. Its total is \(900+950+1{,}050+1{,}100=\$4{,}000\), so its mean is \(4{,}000/4=\$1{,}000\). The median is \((950+1{,}050)/2=\$1{,}000\). Its deviations from the mean are \(-100,-50,50,100\). The squared deviations sum to \(10{,}000+2{,}500+2{,}500+10{,}000=25{,}000\), giving \(s^2=25{,}000/3=8{,}333.33\) and \(s=\$91.29\). Its coefficient of variation is \(91.29/1{,}000=9.13\%\).
The first period’s mean is twice its median, and its sample spread is much larger relative to its mean. The large claim is a clear diagnostic flag. The correct conclusion is that the average is highly influenced by one observation—not that the claim is invalid or should automatically be removed.
Method and trap. Compare the mean with a robust center such as the median and inspect the claim amounts. Do not rely on the mean alone, and do not silently delete a genuine claim to make the periods appear more alike.
Separating Severity Changes From Claim-Mix Changes
A period-wide average can change even when the severity within each claim class changes only modestly. For example, a portfolio with a greater share of complex equipment claims may have higher average severity than one with more minor property claims. To diagnose this, calculate counts, proportions, and mean severities within meaningful classes, then compare those class summaries across periods.
A useful check is to apply the later period’s within-class means to the earlier period’s mix. This standardized comparison answers: “What would the later class severities average to if the earlier claim mix had remained in place?” It is a diagnostic decomposition, not by itself a projection or a selected trend.
Worked Example: Distinguishing Within-Class Changes From Mix Shifts
A hypothetical portfolio has two claim classes, routine and complex. In Period 1, it has eight routine claims averaging $1,000 and two complex claims averaging $5,000. In Period 2, it has four routine claims averaging $1,100 and six complex claims averaging $5,500. Calculate each period’s total mean and separate the change into a mix component and a within-class component, using Period 1 severities as the mix-effect base. The given class means are exact summaries of observed claims; no parametric distribution is assumed. The interest rate is 0%, and mortality is not applicable.
Period 1. Its total losses by class are \(8(1{,}000)=\$8{,}000\) and \(2(5{,}000)=\$10{,}000\). Total losses are $18,000 over 10 claims, so the overall mean is \(18{,}000/10=\$1{,}800\). The class proportions are 80% routine and 20% complex.
Period 2. Its class totals are \(4(1{,}100)=\$4{,}400\) and \(6(5{,}500)=\$33{,}000\). Total losses are $37,400 over 10 claims, giving an overall mean of \(37{,}400/10=\$3{,}740\). Its proportions are 40% routine and 60% complex. The observed overall increase is \(3{,}740-1{,}800=\$1{,}940\) per claim.
First hold Period 1’s mix fixed and substitute Period 2’s class means. The standardized mean is \(0.8(1{,}100)+0.2(5{,}500)=880+1{,}100=\$1{,}980\). The within-class severity component, evaluated at Period 1’s mix, is \(1{,}980-1{,}800=\$180\).
Next measure the mix effect using Period 1’s class means but Period 2’s proportions. That mean is \(0.4(1{,}000)+0.6(5{,}000)=400+3{,}000=\$3{,}400\). The mix component is \(3{,}400-1{,}800=\$1{,}600\). Finally, the within-class component at Period 2’s mix is \(0.4(1{,}100-1{,}000)+0.6(5{,}500-5{,}000)=40+300=\$340\). Therefore, \(1{,}600+340=\$1{,}940\), exactly the observed overall increase.
This decomposition assigns the mix effect first and the within-class effect second. A different order of standardization can allocate the interaction differently, so state the convention used. Here the arithmetic checks: the two components sum to the observed overall change.
Method and trap. Compare class-level severities as well as the overall averages. Attributing the entire $1,940 increase to higher costs within each class would miss that the complex-claim share rose from 20% to 60%.
A Practical Diagnostic Sequence
Check that claim definitions, coverage, limits, valuation basis, and currency align. A difference in basis can mimic a severity change.
For each period, record \(n_j\), \(L_j\), and \(\bar{X}_j=L_j/n_j\). Counts indicate how much data support each mean; totals allow a pooled check.
Use the range, sample standard deviation, coefficient of variation, and median as appropriate. Review unusually large claims rather than assuming they should be excluded.
Compare the counts and proportions of relevant claim classes, and compare within-class averages where the data permit.
Distinguish an observed difference from an established change in the underlying process. Small counts, outliers, and changing mix limit what a simple average can establish.
Common Exam Traps and Interpretation
- Taking an unweighted average of period means. That describes an equally weighted average across periods. Use claim-count weights to find average severity across all claims.
- Reading a small-count mean as stable. A few claims can produce a volatile average. Report the count and inspect the individual losses where possible.
- Treating an outlier as an error without evidence. A large claim can strongly affect the mean and still be a valid observation. Diagnose its influence separately from any decision about treatment.
- Confusing mix shifts with within-class severity changes. An overall mean may rise because the portfolio contains more high-severity claim types, even if each class changes little.
- Comparing unlike bases. Differences in limits, deductibles, claim definitions, or valuation dates can make a raw comparison misleading.
These diagnostics help determine what a period-level average does and does not represent. They do not, on their own, select a future severity trend or adjust losses to a different time period; those require additional assumptions and methods. The central habit is to pair each average with its count, inspect variation and influential observations, and ask whether the claim mix stayed comparable.
Check Your Understanding
For each hypothetical problem, calculate the requested summaries and state what they reveal. Assume amounts are on a common basis and valuation date, with no discounting (\(i=0\%\)); mortality is not applicable. Treat the listed data as empirical observations unless a problem states otherwise.
- Period A has three severities: $900, $1,200, and $1,500. Period B has six severities: $1,000, $1,200, $1,400, $1,600, $1,800, and $2,000. Find both period means and the pooled mean. (Answer: Period A mean = $1,200; Period B mean = $1,500; pooled mean = $1,400.)
- Two periods each have mean severity $1,000. Period X has four claims totaling $4,000; Period Y has twelve claims totaling $12,000. Find the combined mean and explain why equal period means do not imply equal precision. (Answer: Combined mean = $16,000/16 = $1,000; Period X’s mean is based on one-third as many claims as Period Y’s, so it may be less stable.)
- Period A has claims of $700, $900, $1,100, and $1,300. Period B has claims of $950 and $1,050. Find each mean and the pooled mean, and compare the counts. (Answer: Both period means are $1,000; pooled mean = $6,000/6 = $1,000; Period A has four claims and Period B has two.)
- A period has severities of $600, $800, $1,000, $1,200, and $6,400. Calculate the mean, median, and the largest claim’s share of total losses. What is the appropriate diagnostic interpretation? (Answer: Mean = $10,000/5 = $2,000; median = $1,000; largest claim share = $6,400/$10,000 = 64%; the mean is strongly influenced by a large observation, which should be investigated rather than automatically removed.)
- Period 1 has 9 routine claims averaging $800 and 1 complex claim averaging $4,000. Period 2 has 5 routine claims averaging $880 and 5 complex claims averaging $4,400. Find both overall means and the change. Then calculate the Period 2 severity standardized to Period 1’s mix. (Answer: Period 1 mean = \((9(800)+1(4,000))/10=\$1,120\); Period 2 mean = \((5(880)+5(4,400))/10=\$2,640\); increase = $1,520; Period 1 mix standardized to Period 2 class means = \(0.9(880)+0.1(4,400)=\$1,232\).)