Building Multiplicative Rating Relativities

From a Log-Linked Predictor to Rating Factors

A rating model with a log link is additive on the linear-predictor scale, but its fitted mean is multiplicative on the original scale. That transformation is useful when a model contains categorical rating characteristics: each category can be represented by a factor, and a risk’s fitted mean can be built by multiplying the factors for its selected levels.

This tutorial focuses on categorical predictors and comparisons with a reference risk. It assumes the log link and the meaning of the linear predictor are known. The calculations use fixed fitted coefficients; they do not address how those coefficients are estimated.

Definition: Suppose a generalized linear model has a positive conditional mean $\mu$ and uses the log link. Its linear predictor is $\eta=\log(\mu)$, so $\mu=\exp(\eta)$. For categorical rating factors represented by indicator variables, exponentiating the predictor expresses the fitted mean as a base mean multiplied by a factor for each selected nonreference level. A relativity is the ratio of a profile’s fitted mean to the fitted mean for a specified reference profile.

For example, let $x_1,\ldots,x_J$ denote categorical rating characteristics. For characteristic $j$, take one level as the reference and let $I(x_j=k)$ equal 1 when the risk has nonreference level $k$, and 0 otherwise. A main-effects-only model can be written as

\eta=\log(\mu)=\beta_0+\sum_{j=1}^{J}\sum_{k\ne r_j}\beta_{jk}I(x_j=k),

where $r_j$ is the reference level for characteristic $j$. The model assumes the specified predictor is additive in these indicators; it has no interaction terms unless such terms are explicitly included. Applying the exponential to both sides gives

\mu=\exp(\beta_0)\prod_{j=1}^{J}\prod_{k\ne r_j}\left[\exp(\beta_{jk})\right]^{I(x_j=k)}.

Because exactly one level of each characteristic applies, a reference level contributes no nonreference term, while a selected nonreference level contributes $\exp(\beta_{jk})$. Thus $\exp(\beta_0)$ is the fitted mean for the all-reference profile, and $\exp(\beta_{jk})$ is the multiplicative factor associated with level $k$, relative to $r_j$, holding the other modeled characteristics fixed.

Why the Relativity Is a Product

The product form follows directly from the link and the indicator coding. It is not an extra assumption that factors happen to multiply: multiplication results from exponentiating a sum. For a profile $x$, let $\eta(x)$ be its predictor and let $x_{\mathrm{ref}}$ be the all-reference profile. The reference profile has $\eta(x_{\mathrm{ref}})=\beta_0$.

\frac{\mu(x)}{\mu(x_{\mathrm{ref}})} =\frac{\exp(\eta(x))}{\exp(\beta_0)} =\exp\!\left(\eta(x)-\beta_0\right) =\prod_{j=1}^{J}\prod_{k\ne r_j} \left[\exp(\beta_{jk})\right]^{I(x_j=k)}.

The intercept cancels, leaving the product of the selected level factors. This also explains why a profile’s relativity can be calculated without first calculating its fitted mean, provided the reference profile and the model’s factor coding are clear. The argument is exact for the stated model. If the predictor includes interactions, the relevant interaction terms must also be included in the product.

Specify the reference levels.
For each categorical characteristic, identify the level represented by the omitted indicator.

Convert coefficients to factors.
For each nonreference level, calculate $\exp(\beta_{jk})$.

Build the profile product.
Multiply the factors corresponding to the profile’s selected levels; reference levels contribute a factor of 1.

Apply the base mean if needed.
Multiply the profile relativity by $\exp(\beta_0)$ to obtain the fitted mean on the response scale.

Worked Applications

Worked Example: Fitted Mean From Three Rating Characteristics

Method: Convert each supplied log-scale coefficient into a factor, then multiply the factors by the all-reference fitted mean. A hypothetical insurer models expected annual claim cost in dollars for a policy. The fitted log-link model has an all-reference mean of $500. Relative to the reference levels, the selected territory, vehicle class, and use category have coefficients $\log(1.15)$, $\log(0.80)$, and $\log(1.25)$, respectively. Find the policy’s fitted mean and its relativity. Assume the model has only these main effects and no interaction terms.

Step 1: Identify the base mean. The reference profile’s fitted mean is $\exp(\beta_0)=\$500$.

Step 2: Convert the coefficients to factors. Exponentiation reverses the logarithm: $\exp(\log(1.15))=1.15$, $\exp(\log(0.80))=0.80$, and $\exp(\log(1.25))=1.25$.

Step 3: Multiply the selected factors. The profile relativity is $1.15\times0.80\times1.25=0.92\times1.25=1.15$.

Step 4: Calculate the fitted mean. The fitted mean is $\$500\times1.15=\$575$. The policy is therefore fitted at 1.15 times the reference mean, or 15% above it.

Check: Applying the factors in sequence gives $\$500\times1.15=\$575$, then $\$575\times0.80=\$460$, then $\$460\times1.25=\$575$. The final result agrees with the product calculation. The common mistake to avoid is adding the percentage changes instead of compounding the factors.

Worked Example: Comparing Profiles and Cancelling Shared Factors

Method: Write each profile’s full factor product, then divide to cancel shared factors. Consider a hypothetical portfolio model whose all-reference fitted mean is $600. The model has an urban factor of 1.30 versus rural, a commercial-vehicle factor of 1.40 versus private vehicle, and a long-tenure factor of 0.90 versus short tenure. All terms are main effects. Find the fitted mean for an urban commercial policy with long tenure, then compare it with an urban private policy with short tenure.

Step 1: Calculate the first profile’s relativity. Its selected factors are $1.30$, $1.40$, and $0.90$. Their product is $1.30\times1.40=1.82$, then $1.82\times0.90=1.638$.

Step 2: Calculate its fitted mean. The fitted mean is $\$600\times1.638=\$982.80$.

Step 3: Calculate the second profile’s relativity and fitted mean. It is urban but private and short-tenure, so its only nonreference factor is 1.30. Its fitted mean is $\$600\times1.30=\$780$.

Step 4: Compare the profiles. The ratio is $\$982.80/\$780=1.26$. Equivalently, the shared urban factor and the common base mean cancel, leaving $1.40\times0.90=1.26$. Thus the first profile’s fitted mean is 1.26 times the second’s.

Check: $\$780\times1.26=\$982.80$. This confirms the ratio. The common mistake to avoid is including the urban factor in the comparison ratio even though it applies to both profiles.

Worked Example: Keeping Dollars and Logarithms on the Same Scale

Method: Either calculate on the response scale using factors or work entirely in dollars on the log scale. A hypothetical model predicts annual expected claim cost. Its all-reference fitted mean is $320, and the selected levels have coefficients $\log(0.90)$, $\log(1.20)$, and $\log(1.25)$. Find the profile’s fitted mean. Assume these are all the selected nonreference levels and the model contains no interactions.

Step 1: Calculate the product of relativities. The factors are $0.90$, $1.20$, and $1.25$. First, $0.90\times1.20=1.08$. Then $1.08\times1.25=1.35$.

Step 2: Apply the product to the base mean. The fitted mean is $\$320\times1.35=\$432$.

Step 3: Check using logarithms in dollars. The base intercept is $\log(320)\approx5.768320996$, not $\log(0.320)$. The three coefficient values are approximately $-0.105360516$, $0.182321557$, and $0.223143551$. Their sum with the intercept is $5.768320996-0.105360516+0.182321557+0.223143551=6.068425588$. Exponentiating gives $\exp(6.068425588)\approx432$.

Check: The logarithmic calculation agrees with the direct calculation, $\$320\times1.35=\$432$. The common mistake to avoid is using $\log(0.320)\approx-1.139434283$ while labeling the base mean as 320 dollars. That logarithm is for 0.320, not 320. Working in thousands is valid only if the base mean and final fitted mean are both expressed in thousands; in that case, the base is 0.320 thousand dollars and the final result is 0.432 thousand dollars, or $432.

Reference Coding and Model Limits

A reference level is the level whose indicator is omitted from the predictor. Under this coding, its factor is implicitly 1, because it contributes no coefficient term. The choice of reference affects the individual coefficients and factors reported by the model, but it does not change fitted means when the model is represented equivalently. For a comparison, use factors defined against the same reference coding, or calculate each profile’s predictor consistently before taking the ratio.

The product formula also depends on which terms the model contains. In a main-effects-only model, the factor associated with one characteristic does not vary with the level of another characteristic. If the predictor includes an interaction, such as a term for a particular combination of territory and vehicle class, that term contributes an additional multiplicative factor for that combination. Leaving it out would not reproduce the model’s fitted mean.

A fitted mean is the model’s conditional expected response for the stated profile, not a guarantee of an individual policy’s realized claim cost. The arithmetic explains how the model combines its terms; whether the model is appropriate for a rating decision is a separate question.

Common Exam Traps

Adding percentage changes. A wrong calculation for factors 1.15, 0.80, and 1.25 is $1+0.15-0.20+0.25=1.20$. The correct calculation is $1.15\times0.80\times1.25=1.15$. Percentage changes compound through multiplication.
Forgetting the reference profile. A factor of 1.30 means 1.30 times the specified reference level for that characteristic, not 1.30 times an arbitrary profile. State the reference levels before multiplying.
Mixing a scale with its logarithm. $\log(320)$ corresponds to 320 dollars, while $\log(0.320)$ corresponds to 0.320 thousand dollars. Keep the response units consistent from the base mean through the final answer.
Failing to cancel shared terms in a comparison. When two profiles share a factor, it appears in both fitted means and cancels in their ratio. Including it only once produces the wrong comparison.
Ignoring modeled interactions. The product must include every term in the predictor that applies to the profile. A main-effects-only product is insufficient when the fitted model contains interaction terms.

Key takeaway: With a log link, exponentiating an additive categorical predictor yields a product of positive factors. Use the omitted reference levels to define the base profile, multiply the selected factors to obtain a relativity, and multiply by the base mean only when a fitted mean is required.

Check Your Understanding

For each problem, state the factor product before giving the requested fitted mean or comparison.

A hypothetical model has an all-reference mean of $720. A selected region factor is 0.85 and a selected equipment factor is 1.10. What are the profile relativity and fitted mean? (Answer: Relativity $0.85\times1.10=0.935$; fitted mean $\$673.20$.)
A model’s nonreference factors are 1.25 for coastal territory and 0.80 for an upgraded safety class. The reference mean is $900. Find the fitted mean for a coastal risk with the standard safety class, and for a noncoastal risk with the upgraded class. (Answer: $\$900\times1.25=\$1,125$ and $\$900\times0.80=\$720$, respectively.)
Two hypothetical profiles share a factor of 1.20. Profile A also has a factor of 1.35, while Profile B has a factor of 0.90. What is A’s fitted mean relative to B’s? (Answer: $1.35/0.90=1.50$; A’s fitted mean is 1.50 times B’s.)
A reference mean is $400, and a profile’s selected coefficients are $\log(1.10)$ and $\log(0.75)$. Calculate the fitted mean and show the factor product. (Answer: $1.10\times0.75=0.825$; $\$400\times0.825=\$330$.)
A model uses a reference mean of $250 and has two selected factors, 1.40 and 1.15. A candidate adds their percentage changes to claim the profile is 55% above reference. Correct the calculation and give the fitted mean. (Answer: Correct relativity $1.40\times1.15=1.61$, or 61% above reference; fitted mean $\$250\times1.61=\$402.50$.)

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