Why Territory Interactions Need Careful Treatment
A territory factor is intended to represent differences in expected cost across territories. A rating plan may also distinguish classes of insureds, vehicles, or risks. If the relative difference between territories changes from one class to another, the data suggest a territory-by-class interaction: one set of territory factors may not describe every class equally well.
An interaction indication from a finely divided table is not automatically a reliable basis for a rating distinction. Some combinations of territory and class may have very few exposures. Their observed loss costs can move sharply when one claim changes, even if the underlying risk pattern is similar to the pattern elsewhere. The task is to diagnose the interaction, decide whether to retain the cell distinction, and choose a method that reflects the intended rating structure.
For this tutorial, a cell’s loss cost is its adjusted losses divided by its exposures. Assume that the losses and exposures being compared have already been placed on a consistent basis for the rating analysis, including any required treatment of mix, trend, and loss development. No probability distribution or interest assumption is needed for these deterministic calculations. The methods below address the structure of the indications; they do not replace the earlier material on standardizing territory experience for mix or separating territory effects from class effects.
Diagnosing an Interaction
A useful first check is to compare the territorial relativity within each class. If Territory B has a loss cost 20% above Territory A in one class and 40% above Territory A in another, the territorial relationship is not constant across those classes. The size of the difference must be considered alongside the exposure supporting each cell.
Worked Example: A Large Indication in a Sparse Cell
A hypothetical insurer has two territories and two vehicle classes. Adjusted losses and earned exposures are as follows. Determine the cell loss costs, compare the territorial relativities by class, and assess what the table does and does not establish.
| Cell | Exposures | Adjusted losses |
|---|---|---|
| Territory A, Class 1 | 100 | $90,000 |
| Territory A, Class 2 | 50 | $75,000 |
| Territory B, Class 1 | 100 | $108,000 |
| Territory B, Class 2 | 10 | $21,000 |
Step 1: Calculate each cell loss cost. Divide adjusted losses by exposures:
Step 2: Compare territory relativities within each class. In Class 1, B relative to A is \(1{,}080/900=1.20\). In Class 2, it is \(2{,}100/1{,}500=1.40\). The indicated relative difference is therefore not the same across classes.
Step 3: Quantify the interaction contrast. The Class 2 territorial relativity divided by the Class 1 territorial relativity is \(1.40/1.20=1.1667\). The observed Class 2 relationship is about 16.67% higher than the Class 1 relationship. Under a no-interaction structure that uses the Class 1 territorial relativity, the corresponding B, Class 2 loss cost would be \(1.20(1{,}500)=\$1{,}800\), rather than the observed \(\$2{,}100\).
Interpretation and check. The numerical table indicates a possible interaction, but the B, Class 2 indication is based on only 10 exposures. The computation does not establish that the difference will persist. The method in this example is a within-class relativity comparison; its main trap is treating the largest cell indication as conclusive without checking the exposure supporting it.
A useful diagnostic is to ask whether the interaction is visible in more than one comparison and whether it is consistent with a plausible risk distinction. A single dramatic cell may be driven by one claim or by an unusual mix within the cell. Conversely, a small number of exposures does not prove that the cell is wrong; it means the indication should be treated cautiously and the structure should be selected deliberately.
Pooling Cells and Simplifying the Rating Structure
Pooling combines experience so that cells share an estimate. For a group of cells, the pooled loss cost is total adjusted losses divided by total exposures. It is not generally the unweighted average of the cell loss costs. Pooling can be appropriate when the distinctions being removed are not needed in the rating plan or when the data do not support separate indications.
Simplifying the structure is related but not identical. A plan can retain territory and class main effects while removing the interaction, so that every class uses the same territorial relationship. This is a structural assumption: it says the plan will not assign separate territory factors by class. It does not say that the underlying risks are identical in every respect.
Worked Example: Pooling a Sparse Work-Van Cell
A hypothetical insurer is deciding whether to use separate territory factors for work vans. The regular-vehicle cells each have a loss cost of $1,000. For work vans, Territory East has 6 exposures and $18,000 in adjusted losses; Territory West has 54 exposures and $81,000 in adjusted losses. Compare separate indications with one pooled work-van indication, using the regular-vehicle loss cost as the reference.
Step 1: Calculate the separate work-van loss costs. East is \(18{,}000/6=\$3{,}000\) per exposure. West is \(81{,}000/54=\$1{,}500\) per exposure. Relative to the $1,000 reference, the separate factors would be 3.00 and 1.50.
Step 2: Pool work-van experience. Total losses are \(18{,}000+81{,}000=\$99{,}000\), and total exposures are \(6+54=60\). Thus the pooled loss cost is \(99{,}000/60=\$1{,}650\) per exposure, corresponding to one work-van factor of \(1{,}650/1{,}000=1.65\).
Step 3: Check the weighted calculation. The pooled value equals \((6(3{,}000)+54(1{,}500))/60=\$1{,}650\). The unweighted average, \((3{,}000+1{,}500)/2=\$2{,}250\), is not the pooled loss cost because it gives the six-exposure East cell the same weight as the 54-exposure West cell.
Interpretation. Pooling removes the separate East and West work-van indications; it does not prove that the two territories have identical underlying work-van risk. The example uses the pooled loss cost as a common work-van indication and assumes the reference loss cost is appropriate for expressing that indication as a factor. The method avoids allowing a small cell to determine a sharply different factor solely through an unweighted average.
An alternative simplification is to retain territory factors and class factors as separate components, but omit territory-by-class factors. The resulting structure assumes that the territory relativity is common across classes. This can be easier to explain and maintain than a large table of cell factors. Its trade-off is that it cannot reproduce every observed cell indication when the data show different territorial relationships by class.
Constrained Factor Fitting
A constrained approach starts with preliminary factors and chooses adjusted factors that satisfy a stated condition, such as a weighted average factor of 1. The objective function matters: two procedures can both satisfy the balance condition yet make very different adjustments. Therefore, state both the constraint and the quantity being minimized before calculating fitted factors.
When the objective weights and constraint weights are the same, \(w_i=a_i\), the adjustment has a particularly simple form. The fitted factors all move by the same absolute amount. This result follows by introducing a multiplier \(\lambda\) for the constraint. For the objective \(Q=\sum_i w_i(f_i-g_i)^2\), the first-order condition for each factor is \(2w_i(f_i-g_i)+\lambda a_i=0\). If \(a_i=w_i\), then \(f_i-g_i=-\lambda/2\), which is constant across \(i\). The balance condition determines that common change as \(1-\sum_i w_i g_i\), when the weights sum to 1. This derivation assumes no additional bounds or constraints; adding them can change which factors move and by how much.
Worked Example: Balancing Preliminary Factors by Minimum Weighted Change
A hypothetical two-territory plan has preliminary factors 0.90 and 1.20. The exposure weights are 25% and 75%. Fit factors with an exposure-weighted average of 1 by minimizing exposure-weighted squared changes:
Step 1: Calculate the preliminary weighted average.
Step 2: Apply the minimum-change result. The objective weights equal the balance weights, so the minimizing adjustment is the same for both factors. The preliminary weighted average must decrease by \(1.125-1=0.125\). Subtract 0.125 from each factor: \(f_1=0.90-0.125=0.775\) and \(f_2=1.20-0.125=1.075\).
Step 3: Verify the constraint and objective. The weighted average is \(0.25(0.775)+0.75(1.075)=0.19375+0.80625=1.00000\). The objective is \(0.25(0.775-0.90)^2+0.75(1.075-1.20)^2 =0.25(0.015625)+0.75(0.015625)=0.015625\).
Method and trap. This is a constrained weighted least-squares fit. Multiplying both preliminary factors by \(1/1.125\) gives 0.8000 and approximately 1.0667, which also satisfies the balance constraint. But its objective is approximately \(0.25(0.1000)^2+0.75(0.1333)^2=0.01583\), which is larger than 0.015625. Balance alone does not identify the least-squares solution; the specified objective does.
In practice, a constrained fit may also be required to respect approved ranges, preserve a selected relationship, or keep changes within operational limits. Such conditions must be written explicitly. A fitted set should then be checked against every constraint, and the objective value should be compared when the question asks for a minimum. A result that is merely balanced is not automatically the minimizing result.
Choosing Among the Approaches
| Approach | What changes | Key consideration |
|---|---|---|
| Pool cells | Cells share one experience estimate | Combine losses and exposures, not cell rates with equal weights |
| Simplify the structure | Remove an interaction while retaining selected main effects | Recognize the resulting assumption of a common relationship across classes |
| Constrained fitting | Adjust preliminary factors to satisfy defined conditions | Specify the objective and constraints; balance alone is insufficient |
These approaches answer different questions. Pooling specifies which observations contribute to a common estimate. Simplification specifies which distinctions the rating plan will represent. Constrained fitting specifies how factors should be adjusted to meet mathematical or operational requirements. They can be used together, but one does not automatically imply another.
A common exam trap is to treat a noisy cell indication as a proven interaction, or to report a balanced factor set as the unique answer without stating the adjustment objective. First calculate the cell indications and compare the relationships across classes. Then state what structure is retained or removed. If factors are fitted, write the objective and constraints, solve them, and verify the result.
Check Your Understanding
For each problem, calculate the requested indication and state what the calculation implies about the rating structure.
- In Class A, Territory North has a loss cost of $800 and South has $1,000. In Class B, North is $1,200 and South is $1,500. Calculate South relative to North in each class and the interaction contrast. (Answer: Relativities are 1.25 in both classes; interaction contrast is 1.00, so these indications show no interaction.)
- A cell has 20 exposures and $16,000 of adjusted losses; another has 80 exposures and $96,000. Calculate the pooled loss cost and explain why the unweighted average of cell rates is not the pooled answer. (Answer: Cell rates are $800 and $1,200; pooled loss cost is \((16{,}000+96{,}000)/(20+80)=\$1{,}120\) per exposure, not the unweighted average of $1,000.)
- Preliminary factors are 0.95 and 1.15 with weights 0.40 and 0.60. Under the objective \(\sum_i w_i(f_i-g_i)^2\) and constraint \(\sum_i w_i f_i=1\), find the fitted factors. (Answer: Preliminary weighted average is 1.07; subtract 0.07 from both factors, giving 0.88 and 1.08. Their weighted average is 1.00.)
- A class has a sharply different territory relativity in one cell based on 4 exposures, while the other cells each have more than 100 exposures. What can be concluded from the indication, and name two possible structural responses? (Answer: The table signals a possible interaction but does not establish a persistent difference; possible responses include pooling the sparse cell with an appropriate group or simplifying the structure to remove the interaction.)
- A factor adjustment meets the required weighted-average constraint, but no objective function is given. Is it possible to conclude that the adjustment minimizes weighted squared changes? (Answer: No. The constraint establishes balance only; a minimization conclusion requires a specified objective, its weights, and any additional constraints.)