Why Cross-Classified Tables Matter
A rating plan may classify a risk on more than one characteristic, such as territory and vehicle use. A cross-classified relativity table displays the factor for each combination of levels, so the combined effect is visible rather than being hidden in separate lists. The table also makes the exposure mix explicit: a factor applying to many exposures can matter more to the portfolio average than a factor applying to only a few.
The construction described here assumes that the two classification effects combine multiplicatively, without an additional cell-specific adjustment. That is a modeling assumption, not a consequence of arranging factors in a table. This tutorial is aligned with the CAS Exam 5 (Basic Ratemaking and Estimated Claim Liabilities) syllabus.
Constructing and Normalizing the Table
First identify the two classification variables and their levels. Put one variable in rows and the other in columns. Multiply the corresponding row and column relativities to obtain each raw cell factor. If the factor table must have an exposure-weighted average of 1, divide every raw cell factor by the weighted average across the portfolio exposures being rated.
Here, \(C_{ij}\) is the raw combined factor, \(\overline{C}_e\) is its exposure-weighted average, and \(C^{*}_{ij}\) is the normalized factor. The weights must match the intended rating basis: for example, use policy exposures if the target average is over policy exposures. A cell with zero exposure contributes zero to both weighted sums, although its factor may still be calculable from the multiplicative assumption.
Confirm how the row and column relativities are based. Their product determines the scale of the raw table; a different common scale does not change the relative comparisons between cells.
For each row-column combination, multiply its two factors. Do not add the factors or average them.
Multiply each cell factor by its exposure count, add those products, and divide by total exposure.
Divide every cell factor by the same weighted average. Recalculate the weighted average to confirm it is 1, subject only to rounding.
Why the Weighted Normalization Works
The normalization is chosen to make the exposure-weighted average of the revised factors equal to 1. Because the same divisor is applied to every cell, the relative factor between any two cells is preserved. Substituting \(C^{*}_{ij}=C_{ij}/\overline{C}_e\) into the weighted average verifies the result directly:
This calculation explains why the denominator must be the total exposure and why the weighted average must use the exposure in each particular cell. An unweighted average gives each cell equal influence, even when the cells contain very different numbers of exposures.
Worked Example: Building and Normalizing a Two-by-Two Table
A hypothetical insurer uses two classification variables. The row relativities are 0.80 for standard territory and 1.20 for remote territory. The column relativities are 0.90 for personal use and 1.10 for business use. The respective cell exposures are 50, 30, 20, and 100, arranged in that same row-by-row order. Construct the raw table and normalize it to an exposure-weighted average of 1.
Model and assumptions. Use the multiplicative, separable structure \(C_{ij}=A_iB_j\). Exposures are deterministic weights for this calculation; no interest rate, mortality assumption, or probability distribution is assumed.
Step 1: Calculate each raw cell factor. Standard territory and personal use give \(0.80(0.90)=0.72\). Standard territory and business use give \(0.80(1.10)=0.88\). Remote territory and personal use give \(1.20(0.90)=1.08\). Remote territory and business use give \(1.20(1.10)=1.32\).
| Territory / use | Personal use | Business use |
|---|---|---|
| Standard territory | 0.72 (50 exposures) | 0.88 (30 exposures) |
| Remote territory | 1.08 (20 exposures) | 1.32 (100 exposures) |
Step 2: Calculate the exposure-weighted average. The weighted factor total is \(50(0.72)+30(0.88)+20(1.08)+100(1.32)=36+26.4+21.6+132=216\). Total exposure is \(50+30+20+100=200\), so the weighted average is \(216/200=1.08\).
Step 3: Normalize every cell. Divide each raw factor by 1.08. The normalized factors are \(0.72/1.08=0.6667\), \(0.88/1.08=0.8148\), \(1.08/1.08=1.0000\), and \(1.32/1.08=1.2222\), rounded to four decimal places.
Check. Using the unrounded values, the weighted total of normalized factors is \(216/1.08=200\). Dividing by 200 exposures gives 1. The normalization changes the scale of the table but preserves every ratio between cells.
Exposure Weights and Empty Cells
An empty cell is a combination with no observed exposure in the data being used. It is important to distinguish two statements: the multiplicative formula can produce a factor for that combination, but the empty cell has no direct experience with which to validate that factor. The factor is therefore a model-implied value under the separability assumption, not an empirical estimate from that cell’s own observations.
An empty cell receives zero weight in the portfolio average when its exposure is zero. Do not treat the missing experience as a zero loss cost or as evidence that the cell has no risk. Also, do not silently include it as an observed cell when calculating an exposure-weighted average. Whether the resulting modeled factor is appropriate for rating requires judgment beyond the arithmetic of constructing the table.
Worked Example: An Empty Cell and the Portfolio Average
A hypothetical rating table has row factors of 0.80 for group A and 1.20 for group B, and column factors of 0.75 for use X, 1.00 for use Y, and 1.25 for use Z. Exposures and raw combined factors are needed for the occupied cells; group A with use Z has no exposure. Calculate the table, the exposure-weighted average of the occupied cells, and the normalized factor implied for the empty cell.
Model and assumptions. Assume \(C_{ij}=A_iB_j\). The exposure counts below are the weights. As in the prior example, no interest rate, mortality assumption, or probability distribution is used.
Step 1: Form the raw factors. Group A has factors \(0.80(0.75)=0.60\), \(0.80(1.00)=0.80\), and \(0.80(1.25)=1.00\). Group B has factors \(1.20(0.75)=0.90\), \(1.20(1.00)=1.20\), and \(1.20(1.25)=1.50\).
| Group / use | X | Y | Z |
|---|---|---|---|
| A | 0.60 (100 exposures) | 0.80 (50 exposures) | 1.00 (0 exposures) |
| B | 0.90 (40 exposures) | 1.20 (60 exposures) | 1.50 (50 exposures) |
Step 2: Weight only by observed exposure. The weighted factor total is \(100(0.60)+50(0.80)+40(0.90)+60(1.20)+50(1.50)=60+40+36+72+75=283\). Total observed exposure is \(100+50+40+60+50=300\). Thus the average is \(283/300=0.9433\), rounded to four decimal places.
Step 3: Normalize the occupied factors and the model-implied empty-cell factor. Divide each raw factor by \(283/300\). For example, the occupied A-X factor becomes \(0.60/(283/300)=180/283=0.6360\), and the empty A-Z cell has the model-implied normalized factor \(1.00/(283/300)=300/283=1.0601\). These values are rounded to four decimal places.
Check and interpretation. The empty cell contributes no amount to the weighted total because its exposure is zero. Using the unrounded normalized factors, the occupied cells’ weighted factor total is \(283/(283/300)=300\); divided by 300 exposures, the weighted average is 1. The factor 1.0601 for A-Z follows from the assumed multiplication, but it is not supported by observed A-Z experience in this data set.
Consistency Checks and Common Traps
A table built from row and column factors has useful arithmetic checks. First, each raw cell must equal its row factor multiplied by its column factor. Second, the ratio between two rows should be the same in every column, provided the same separable factors are used. For example, with row factors \(A_1,A_2\), the ratio \(C_{1j}/C_{2j}\) equals \(A_1/A_2\) in each column \(j\). Such checks can reveal a transcription or multiplication error. If rounded values are used, small differences can result from rounding.
The most common weighting error is using an unweighted mean of the cell factors. Consider three candidate cell factors, 0.75, 1.10, and 1.35, with corresponding exposures 40, 40, and 20. The unweighted mean is \((0.75+1.10+1.35)/3=1.0667\), but that gives each cell equal weight. The correct exposure-weighted mean is
The correction is to weight each factor by its corresponding exposure before averaging. Normalizing by 1.01, rather than 1.0667, makes the exposure-weighted average equal to 1. Other frequent errors include adding the row and column relativities instead of multiplying them, using a different divisor for different cells, and treating an empty cell as an observed zero-risk cell.
Check Your Understanding
For each problem, use the stated multiplicative structure where applicable. Round numerical factors to four decimal places only at the final step.
- A two-variable rating plan has row factors 0.90 and 1.30, and column factors 0.80 and 1.10. What are the four raw combined factors, in row-by-row order? (Answer: 0.72, 0.99, 1.04, 1.43.)
- Two occupied cells have raw factors 0.80 and 1.20, with 20 and 30 exposures, respectively. Find their exposure-weighted average. (Answer: 1.04.)
- In a separable table, row factors are 0.90 and 1.20, while column factors are 0.80 and 1.25. What is the raw factor for the second row and second column, and what is the ratio of the second-row factor to the first-row factor in either column? (Answer: The raw factor is 1.50, and the row ratio is \(1.20/0.90=1.3333\).)
- Three occupied cells have raw factors 0.70, 0.90, and 1.10, with 10, 20, and 70 exposures. Find the weighted average and the normalized factor for the third cell. (Answer: The weighted average is \(102/100=1.02\); the third normalized factor is \(1.10/1.02=1.0784\).)
- A cell has no exposure, but its row and column factors are 1.10 and 0.80. Under the multiplicative assumption, state its raw factor and whether that factor is directly supported by experience in the cell. (Answer: The model-implied raw factor is 0.88; it is not directly supported by that cell’s experience because the cell is empty.)