Why Standardize Territory Experience?
Territories can have different portfolios as well as different experience within comparable parts of those portfolios. For example, one territory may contain more risks in a rating class that tends to have higher losses everywhere. Its aggregate result can therefore look worse even if its results within each rating class are no worse. Standardization addresses this composition effect by comparing territories as though they had the same distribution of rating variables.
The goal is a fairer comparison of the experience measure chosen for the analysis, not an automatic conclusion about what a territory’s final rate should be. The experience measure, exposure unit, period, and treatment of losses must be comparable across territories. Standardization aligned with the CAS Exam 5 (Basic Ratemaking and Estimated Claim Liabilities) syllabus focuses here on separating geographic differences from portfolio mix.
The result \(m_{tj}\) must be defined consistently. It might be a cell-level loss measure, a frequency measure, or another appropriate outcome. This tutorial concerns how to standardize a supplied or calculated cell measure; it does not decide which experience measure is appropriate for a particular ratemaking analysis. The weights may come from a statewide, combined-territory, or explicitly selected target distribution. State the choice, because different reference mixes can produce different standardized comparisons.
Observed Results and Standardized Results
Let \(n_{tj}\) be the exposure in territory \(t\), cell \(j\), and let \(N_t=\sum_j n_{tj}\). If \(m_{tj}\) is a cell result per exposure, the territory’s observed aggregate result is $$ M_t^{\mathrm{obs}}=\sum_j \frac{n_{tj}}{N_t}m_{tj}. $$ Its weights are the territory’s own exposure shares. The standardized result replaces those territory-specific shares with the common reference weights: $$ M_t^{\mathrm{std}}=\sum_j w_jm_{tj}. $$ Thus, standardization changes the weighting, not the observed cell results. It does not assert that a territory actually has the reference mix.
Worked Example: Comparing Two Territories Across Two Classes
Suppose two hypothetical territories have the following cell results, in dollars per exposure. Class 1 and Class 2 are the rating-variable cells. The reference mix is 50% in each class. Calculate each observed aggregate result and each mix-standardized result. State the adjusted difference.
| Territory | Class 1 result | Class 2 result | Observed share in Class 1 | Observed share in Class 2 |
|---|---|---|---|---|
| A | $100 | $200 | 80% | 20% |
| B | $120 | $180 | 20% | 80% |
Method. First weight each territory’s results by its own observed shares to get its aggregate. Then apply the same 50%/50% reference weights to both territories.
For Territory A, the observed aggregate is \(0.80(100)+0.20(200)=80+40=\$120\) per exposure. For Territory B, it is \(0.20(120)+0.80(180)=24+144=\$168\) per exposure. The observed aggregate for B is therefore \(168-120=\$48\) higher.
Using the common reference mix, Territory A’s standardized result is \(0.50(100)+0.50(200)=50+100=\$150\) per exposure. Territory B’s is \(0.50(120)+0.50(180)=60+90=\$150\) per exposure. The adjusted difference is \(\$150-\$150=\$0\) per exposure.
Check. The adjusted results are equal because the two territories’ cell results offset under equal weights: A is \(\$20\) lower in Class 1 and \(\$20\) higher in Class 2. The observed difference arose from their different class mixes. Common mistake avoided: comparing \(\$120\) and \(\$168\) as though those aggregates reflected the same portfolio composition.
Why the Standardization Formula Works
The observed aggregate is a weighted average of cell results, with each weight equal to the territory’s share of exposure in that cell. Differences between observed aggregates can therefore contain two components: differences in cell results and differences in weights. Standardization holds the weights fixed so that the comparison is not driven by territories having different proportions of the cells.
For per-exposure cell results, the observed aggregate is \(M_t^{\mathrm{obs}}=\sum_j (n_{tj}/N_t)m_{tj}\), where the shares sum to 1.
The share \(n_{tj}/N_t\) can differ across territories. Even if two territories have similar cell results, their aggregate results can differ because those shares differ.
Choose \(w_j\) once, require \(w_j\ge0\) and \(\sum_j w_j=1\), then calculate \(M_t^{\mathrm{std}}=\sum_jw_jm_{tj}\) for every territory.
For territories A and B, \(M_A^{\mathrm{std}}-M_B^{\mathrm{std}}=\sum_jw_j(m_{Aj}-m_{Bj})\). The comparison is a weighted average of within-cell differences using identical weights.
The final identity gives the justification: because the same \(w_j\) multiplies each territory’s cell result, the difference is the weighted average of cell-by-cell differences. A territory’s actual distribution \(n_{tj}/N_t\) no longer appears in that comparison. The result is consequently a comparison at the chosen reference mix, not a universal or mix-free measure of geographic effect.
Standardizing Across Several Rating Variables
When the analysis involves multiple rating variables, define the cells using their joint combinations. For example, if class and vehicle use are both relevant, a cell can represent one class-and-use combination. Assigning a common weight to each joint cell preserves the specified joint distribution. Applying separate marginal distributions and multiplying them would instead impose an independence assumption; it could create a reference mix that is not the intended portfolio distribution.
Worked Example: Applying a Common Joint Mix
Two hypothetical territories are divided into four cells defined by the joint combination of driver group and vehicle use. The common joint weights are 40%, 30%, 20%, and 10%, respectively. Cell results are dollars per exposure. Calculate both standardized results and their difference. Also calculate the observed aggregate in each territory using its own mix.
| Joint cell | Common weight | Territory X result | Territory X observed share | Territory Y result | Territory Y observed share |
|---|---|---|---|---|---|
| 1 | 40% | $80 | 10% | $90 | 50% |
| 2 | 30% | $100 | 20% | $90 | 20% |
| 3 | 20% | $140 | 30% | $130 | 20% |
| 4 | 10% | $200 | 40% | $190 | 10% |
Standardized Territory X. Apply each common weight to the corresponding cell result: \(0.40(80)=32\), \(0.30(100)=30\), \(0.20(140)=28\), and \(0.10(200)=20\). The total is \(32+30+28+20=\$110\) per exposure.
Standardized Territory Y. The four weighted contributions are \(0.40(90)=36\), \(0.30(90)=27\), \(0.20(130)=26\), and \(0.10(190)=19\). The total is \(36+27+26+19=\$108\) per exposure. The standardized difference, X minus Y, is \(\$110-\$108=\$2\) per exposure.
Observed aggregates. For X, its own-mix calculation is \(0.10(80)+0.20(100)+0.30(140)+0.40(200)=8+20+42+80=\$150\) per exposure. For Y, it is \(0.50(90)+0.20(90)+0.20(130)+0.10(190)=45+18+26+19=\$108\) per exposure.
The observed aggregates differ by \(\$42\) per exposure, whereas the standardized difference is \(\$2\) per exposure. This large change reflects the different observed joint mixes, especially X’s larger share in cell 4. Common mistake avoided: replacing the four joint-cell weights with separate driver-group and vehicle-use percentages without checking whether their product gives the intended joint distribution.
From Cell Totals to Standardized Results
Cell results are often calculated from recorded totals. If \(L_{tj}\) is the relevant total amount in a cell and \(n_{tj}\) is its exposure, a per-exposure cell result is \(m_{tj}=L_{tj}/n_{tj}\), provided \(n_{tj}>0\). Calculate those cell results first; then apply the common weights. Keep the units consistent: if the cell results are dollars per exposure, the standardized result is also dollars per exposure.
Worked Example: A Mix-Adjusted Comparison Reverses the Aggregate Ranking
Two hypothetical territories each have 100 exposures divided between two classes. The experience amounts and exposure counts are shown below. Use an equal 50%/50% reference mix. Calculate the cell results, observed aggregates, and standardized results, then explain the ranking change.
| Territory | Class 1 exposures | Class 1 amount | Class 2 exposures | Class 2 amount |
|---|---|---|---|---|
| North | 90 | $9,000 | 10 | $4,000 |
| South | 20 | $2,600 | 80 | $16,000 |
Step 1: Calculate cell results. North’s Class 1 result is \(\$9{,}000/90=\$100\) per exposure, and its Class 2 result is \(\$4{,}000/10=\$400\) per exposure. South’s Class 1 result is \(\$2{,}600/20=\$130\) per exposure, and its Class 2 result is \(\$16{,}000/80=\$200\) per exposure.
Step 2: Calculate observed aggregates. North’s total amount is \(\$9{,}000+\$4{,}000=\$13{,}000\), and its total exposure is \(90+10=100\). Its observed aggregate is \(\$13{,}000/100=\$130\) per exposure. South’s total amount is \(\$2{,}600+\$16{,}000=\$18{,}600\), over \(20+80=100\) exposures, so its observed aggregate is \(\$18{,}600/100=\$186\) per exposure.
Step 3: Standardize. North’s result at the common mix is \(0.50(100)+0.50(400)=50+200=\$250\) per exposure. South’s is \(0.50(130)+0.50(200)=65+100=\$165\) per exposure. The standardized difference, North minus South, is \(\$250-\$165=\$85\) per exposure.
Check and interpretation. North has the lower observed aggregate (\(\$130\) versus \(\$186\)), but the higher standardized result (\(\$250\) versus \(\$165\)). North has 90% of its exposures in the lower-result Class 1, while South has 80% in Class 2. Standardizing removes that mix difference from the comparison. Common mistake avoided: assuming the lower observed aggregate means North has lower results at a common mix.
Conditions, Limits, and Exam Traps
Standardization is useful only when its inputs and reference distribution are defensible. Before interpreting a result, check the following:
- Comparable cell definitions: a cell must represent the same combination of rating variables in every territory. A “Class 1” definition that changes by territory cannot support a direct cell comparison.
- Comparable measurements: use the same exposure unit, experience period, and outcome definition across territories. Otherwise, differences may reflect measurement or timing rather than geography or mix.
- Common support: each cell with positive reference weight needs an estimable result in every territory being compared. If a territory has no observations in a weighted cell, its cell result is not observed there. Standardization does not create that missing information.
- Reference-mix relevance: the weights should match the question. A statewide reference mix answers a comparison at that statewide mix; a target portfolio mix answers a comparison at that target. Neither choice is automatically correct for every purpose.
- Cell detail: overly fine cells may be sparse or empty, while overly broad cells can conceal important differences in mix. Cell definition therefore affects what the adjusted comparison can reveal.
If a needed cell is empty or too thin to support a result, do not silently assign it zero and proceed. That would treat “not observed” as “no experience,” which is a different claim. The analysis may need a different cell definition or a clearly stated estimation approach; the standardized result should be limited to what its inputs support.
A frequent exam trap is to use each territory’s own exposure shares in the purported adjusted result. That simply reproduces the observed aggregate:
The correction is to select one common set of weights, verify that it sums to 1, and use it unchanged for all territories. Also avoid treating a standardized comparison as a final territorial indication: it compares experience at a specified mix, but does not by itself select a rating factor or resolve other ratemaking considerations.
Check Your Understanding
For each problem, use the stated common weights for every territory and keep the units as given.
- Two territories have results of \($90\) and \($210\) in Classes 1 and 2 for Territory A, and \($110\) and \($170\) for Territory B. The common weights are 70% in Class 1 and 30% in Class 2. Calculate both standardized results and the difference, A minus B. (Answer: A = $126; B = $128; A minus B = -$2 per exposure.)
- Three cells have common weights 20%, 50%, and 30%. Territory C’s cell results are \($60\), \($120\), and \($240\); Territory D’s are \($80\), \($100\), and \($200\). Calculate both standardized results. (Answer: C = $144; D = $126 per exposure.)
- Territory E has exposures of 40 and 60 and amounts of \($4,000\) and \($12,000\) in Classes 1 and 2. Territory F has exposures of 70 and 30 and amounts of \($9,100\) and \($4,500\). With common weights of 25% and 75%, calculate the cell results and standardized result for each territory. (Answer: E’s cell results are $100 and $200, with standardized result $175; F’s are $130 and $150, with standardized result $145 per exposure.)
- Four joint cells have equal common weights. Territory G’s results are \($80\), \($120\), \($160\), and \($240\); Territory H’s are \($100\), \($100\), \($180\), and \($220\). Calculate the standardized difference, G minus H. (Answer: G = $150; H = $150; difference = $0 per exposure.)
- A territory has no observed exposures in a cell assigned a positive common weight, while the other territories do have observations there. Can its standardized result be calculated from observed cell results alone? (Answer: No. The missing cell result must not be treated as zero; the comparison requires a defensible way to obtain a result or a revised analysis that addresses the unsupported cell.)