Why Interactions Matter
A classification factor is intended to reflect how expected cost varies among risks. With two classification variables, a simple rating structure may apply one factor for each variable and multiply them. That structure assumes the effect of one variable is consistent across levels of the other. If, for example, the cost difference associated with vehicle use is proportionally larger in one territory than another, a single use factor may not describe all the cross-classified experience. This is an interaction.
Interactions matter because the cross-classified table can reveal patterns hidden by one-way summaries. A marginal factor averages over the mix of the other variable’s levels; it can therefore conceal a different pattern within particular groups. The focus here is on identifying and describing those patterns, not on fitting a statistical model or deciding how to smooth or select final filed factors.
Cell Loss Costs and Multiplicative Structure
For a cross-classified cell, a basic experience measure is the cell loss cost: losses divided by the corresponding earned exposures. The losses, exposures, and time basis must be consistent across cells. For instance, losses should be on a comparable basis with respect to development and trend, and every cell must use the same exposure unit. The calculation describes observed average cost per exposure; it does not, by itself, establish that the experience is sufficiently credible or that a classification difference is causal.
Here \(L_{ij}\) is the observed loss cost for level \(i\) of variable A and level \(j\) of variable B. A multiplicative structure represents the expected cost as a base amount multiplied by a factor for each variable. In a two-variable table, write this structure as \(L_{ij}=K A_i B_j\), where \(K\) is a base cost and \(A_i\) and \(B_j\) are the respective class factors. The particular factor values depend on the chosen reference levels or normalization, but the interaction test does not.
Under this structure, compare the relative cost for two levels of A within each level of B. The ratio \(L_{2j}/L_{1j}\) equals \(A_2/A_1\), because the base amount and B factor cancel. It does not depend on \(j\). The same reasoning applies when comparing levels of B. Thus, the multiplicative structure requires consistent relative effects across the table.
The ratio of ratios is a useful way to express interaction. It compares the effect of moving from level 1 to level 2 of A at level 2 of B with the same relative effect at level 1 of B. A value above 1 means the A-level ratio is larger at B level 2; a value below 1 means it is smaller. The value alone does not identify a cause.
Worked Example: Detecting an Interaction in a Two-by-Two Table
A hypothetical insurer cross-classifies policies by territory (suburban or central) and annual mileage (lower or higher). For this illustration, all losses and exposures are assumed to be on the same comparable basis. Calculate the cell loss costs and assess whether a multiplicative structure describes them.
| Territory | Mileage | Earned exposures | Losses |
|---|---|---|---|
| Suburban | Lower | 200 | $120,000 |
| Suburban | Higher | 100 | $90,000 |
| Central | Lower | 100 | $90,000 |
| Central | Higher | 50 | $90,000 |
Step 1: Calculate each cell loss cost. Suburban/lower is \(120{,}000/200=\$600\). Suburban/higher is \(90{,}000/100=\$900\). Central/lower is \(90{,}000/100=\$900\). Central/higher is \(90{,}000/50=\$1{,}800\).
Step 2: Compare mileage relativities across territories. The higher-to-lower mileage ratio is \(900/600=1.5\) in the suburban territory and \(1{,}800/900=2.0\) in the central territory. These ratios differ, so the cell costs do not satisfy a multiplicative structure without an interaction.
Step 3: Quantify the interaction. The ratio of ratios is \(2.0/1.5=1.3333\), or \(4/3\). Thus, the higher-mileage relativity is one-third larger in the central territory than in the suburban territory, as measured by this cell experience.
Step 4: Check against a multiplicative prediction. Use suburban/lower as the reference cost, \(\$600\). The central/lower factor is \(900/600=1.5\), and the suburban/higher factor is \(900/600=1.5\). Without an interaction, central/higher would be \(600(1.5)(1.5)=\$1{,}350\). The observed cell cost is \(\$1{,}800\), consistent with the \(4/3\) interaction multiplier: \(1{,}350(4/3)=\$1{,}800\).
Check and method. The cross-product check gives \(1{,}800(600)=1{,}080{,}000\), while \(900(900)=810{,}000\); their ratio is \(4/3\), not 1. This method avoids mistaking a large dollar difference for a multiplicative interaction. It identifies a pattern in the observed costs; it does not establish that the pattern is stable or sufficiently supported by the cell exposures.
Interactions in Tables with More Than Two Levels
When a variable has more than two levels, compare the same pair of levels across every level of the other variable. A common relative effect across those comparisons supports a multiplicative structure for that pair. If one comparison differs, inspect the size and pattern of the differences rather than forcing a single factor to describe them all. A table can show an interaction in only some levels; it need not have one uniform interaction multiplier for every cell.
The calculations are always relative to a stated comparison. If the ratio for one pair of levels changes, that is evidence of interaction for that comparison. Changing which level is used as the reference changes how the individual ratios are written, but it does not turn a genuine inconsistency into a consistent set of ratios.
Worked Example: Locating an Interaction Across Three Usage Tiers
A hypothetical insurer examines a vehicle classification with standard and specialty vehicle groups and three annual usage tiers. Losses are divided by earned exposures to obtain these observed cell costs:
| Usage tier | Standard vehicle cost | Specialty vehicle cost | Specialty-to-standard ratio |
|---|---|---|---|
| Light | $400 | $600 | 1.50 |
| Moderate | $600 | $900 | 1.50 |
| Heavy | $800 | $1,400 | 1.75 |
Step 1: Compare the vehicle-group relativity at each usage tier. At light usage, the ratio is \(600/400=1.50\). At moderate usage, it is \(900/600=1.50\). At heavy usage, it is \(1{,}400/800=1.75\). The first two comparisons agree, but the heavy-usage comparison does not.
Step 2: Form a multiplicative benchmark. Relative to the light/standard cell, the moderate/standard cost factor is \(600/400=1.5\), and the heavy/standard factor is \(800/400=2.0\). The specialty/standard factor at light usage is \(600/400=1.5\). A structure using these factors predicts a heavy/specialty cost of \(400(2.0)(1.5)=\$1{,}200\).
Step 3: Measure the departure. The observed heavy/specialty cost is \(\$1{,}400\). The ratio of its observed specialty-to-standard effect to the light-tier effect is \(1.75/1.50=1.1667\), or \(7/6\). Equivalently, \(1{,}400/1{,}200=7/6\). In this table, the specialty-vehicle relativity at heavy usage is \(7/6\) times the light-usage relativity.
Check and interpretation. The moderate-tier ratio remains \(1.50\), so the evidence does not indicate the same departure there. The \(7/6\) multiplier describes the heavy-versus-light difference in observed relative effects. It is a comparison of cell experience, not a universal factor to apply without considering the reliability and intended use of that experience. This check avoids assuming that an interaction must affect every level equally.
Using an Interaction Factor in a Rating Illustration
If an interaction is represented explicitly, the joint class receives an additional multiplier relative to the stated reference structure. This makes the departure from a simple product visible: the two main classification factors still describe their reference comparisons, while the interaction multiplier adjusts the combination whose relative effect differs. The following illustration is a loss-cost-based calculation, not a complete premium indication; it omits expenses, profit provisions, and other rating components.
Worked Example: Applying a Joint-Class Adjustment
Suppose comparable hypothetical cell experience gives these loss costs for two binary variables, territory and usage:
| Territory | Usage | Observed loss cost |
|---|---|---|
| Reference | Reference | $600 |
| Higher-cost | Reference | $720 |
| Reference | Higher | $900 |
| Higher-cost | Higher | $1,440 |
Step 1: Calculate the two reference factors. The territory factor is \(720/600=1.20\). The usage factor is \(900/600=1.50\).
Step 2: Predict the joint cost without interaction. The multiplicative prediction is \(600(1.20)(1.50)=\$1{,}080\).
Step 3: Calculate the joint adjustment. Divide the observed joint cost by the no-interaction prediction: \(1{,}440/1{,}080=4/3\). The corresponding ratio-of-ratios gives the same result: the territory ratio at higher usage is \(1{,}440/900=1.60\), compared with \(720/600=1.20\) at reference usage, and \(1.60/1.20=4/3\).
Step 4: Verify the adjusted calculation. The reference cost with both main factors and the joint adjustment is \(600(1.20)(1.50)(4/3)=\$1{,}440\), matching the observed joint cell cost. Therefore, the interaction multiplier is \(4/3\) for this joint comparison. The arithmetic check confirms that the adjustment reproduces the cell cost; it does not show whether a separate joint factor is appropriate in a final rating plan.
Common Exam Trap and Practical Limits
A frequent mistake is to compare differences in dollars when the question concerns a multiplicative rating structure. In the first example, the mileage cost increase is \(\$900-\$600=\$300\) in the suburban territory and \(\$1{,}800-\$900=\$900\) in the central territory. Those dollar differences are not the multiplicative interaction test. The relevant ratios are \(900/600=1.5\) and \(1{,}800/900=2.0\), which show that the relative mileage effect differs. A structure based on additive dollar increments would ask a different question.
Observed differences also require careful interpretation. Unequal exposures, random variation, changes in mix, or imperfect comparability of losses can create unstable cell costs. A ratio far from 1 can be a useful signal to investigate, but it is not by itself proof that a permanent rating distinction is warranted. Check that cells are defined consistently, use aligned exposure and loss bases, and interpret the result in the context of the available experience.
Divide comparable losses by corresponding earned exposures using the stated exposure unit.
State which levels of one variable are being compared and at which levels of the other variable.
Calculate the within-level ratios. Equal ratios are consistent with no interaction for that comparison.
Calculate the ratio of ratios, check the multiplicative prediction, and avoid treating the observed result as proof of causation or stability.
Check Your Understanding
For each problem, use the stated cell costs and compare relative effects rather than dollar differences.
- A two-by-two table has costs of $500 and $750 for the two levels of variable B when variable A is at its reference level, and $700 and $1,120 when A is at its second level. Calculate the interaction ratio for the B effect and state whether the table is exactly multiplicative. (Answer: \( (1{,}120/700)/(750/500)=16/15=1.0667\); no, the B relativity is about 6.67% larger at the second A level.)
- For two class variables, the four costs are $800 at reference/reference, $960 at second-A/reference, $1,200 at reference/second-B, and $1,440 at second-A/second-B. Is there a multiplicative interaction? Show the ratio check. (Answer: No interaction; the B ratios are \(1{,}200/800=1.5\) and \(1{,}440/960=1.5\), and the ratio of ratios is 1.)
- For a base cost of $400, the first level of A has cost $480, the second level of B has cost $600, and the joint cell costs $864. Find the no-interaction prediction and the joint interaction multiplier. (Answer: The prediction is \(400(480/400)(600/400)=\$720\); the multiplier is \(864/720=1.20\).)
- In a three-tier classification, the second-to-first level cost ratios for another variable are 1.25, 1.25, and 1.50 across the tiers. What is the ratio-of-ratios comparing the third tier with the first, and what does it mean? (Answer: \(1.50/1.25=1.20\); the relative effect is 20% larger in tier three than in tier one.)
- Two cells for a classification comparison have losses of $84,000 on 120 exposures and $96,000 on 100 exposures. Calculate each loss cost and the second-to-first relativity. If the same relativity is 1.20 in another level of the other variable, is there an interaction between these two comparisons? (Answer: The costs are \(\$700\) and \(\$960\), and the relativity is \(960/700=1.3714\); compared with 1.20, the ratio-of-ratios is about 1.1429, so the relative effect differs.)