Tutorials › CAS Exam 8 - Advanced Ratemaking › Centering and Scaling Continuous Rating Predictors

CAS Exam 8 · Predictor Design and Interactions · Tutorial 201 of 989

Centering and Scaling Continuous Rating Predictors

See how affine changes to mileage and age predictors alter coefficient interpretation while preserving fitted pure premiums under an unchanged model.

Intermediate 12 min read

What You'll Learn

  • Re-express a continuous predictor by subtracting a chosen center and dividing by a scale.
  • Derive how centering changes the intercept but preserves the linear predictor and fitted means.
  • Convert coefficients between per-mile, per-1,000-mile, per-year, and per-decade units.
  • Interpret centered intercepts as fitted values at specified predictor levels.
  • Identify when re-centering or re-scaling can change a model rather than merely its parameterization.

Why Center and Scale Rating Predictors?

A continuous rating variable may be measured in units that are inconvenient for interpretation. A mileage coefficient expressed per mile can be very small, while a driver-age coefficient expressed per year may not match the business question, which may concern a ten-year age difference. Centering and scaling provide a simple way to make those quantities easier to read. They are changes of coordinates, not nonlinear transformations of the predictor.

The key distinction is between changing a model's parameterization and changing its fitted predictions. With an intercept and an unconstrained coefficient, subtracting a constant from a predictor changes the intercept but leaves the linear predictor, and therefore the fitted mean, unchanged. Dividing by a scale changes the numerical coefficient to reflect the new units. These statements require an affine, one-to-one re-expression of the predictor and the same model structure.

Definition: For a continuous predictor \(x\), a centered and scaled predictor is \(z=(x-c)/s\), where \(c\) is a chosen center and \(s>0\) is a chosen scale. Centering alone uses \(s=1\); scaling alone can use \(c=0\). In a generalized linear model with an intercept, this affine re-expression changes coefficient values but can preserve every fitted mean.

For the examples, suppose the response is a hypothetical policy's positive annual pure premium \(Y\), and a Gamma GLM is fitted with a log link. Use the parameterization \(Y\mid x\sim\operatorname{Gamma}(k,\theta)\), where \(k=5\) is the shape, \(\theta=\mu/5\) is the scale, \(E[Y\mid x]=\mu\), and \(\operatorname{Var}(Y\mid x)=\mu^2/5\). The assumed dispersion is therefore \(0.20\). The response distribution and dispersion do not determine the algebra of centering or scaling; they simply make the example's GLM assumptions explicit. There is no interest or mortality assumption in these non-life pure-premium examples.

Centering: The Intercept Moves, the Fitted Mean Does Not

Start with a one-predictor GLM whose linear predictor is \(\eta=\beta_0+\beta_1x\). Under a log link, the conditional mean is \(\mu=\exp(\eta)\). Define \(x_c=x-c\), so \(x=x_c+c\). Substitution gives the centered parameterization:

$$ \eta=\beta_0+\beta_1(x_c+c) =(\beta_0+\beta_1c)+\beta_1x_c. $$

Thus the new intercept is \(\alpha_0=\beta_0+\beta_1c\), and the slope is \(\alpha_1=\beta_1\). For each observation, the new and old linear predictors are equal. Applying the same inverse link to equal linear predictors gives equal fitted means. This argument also holds for another GLM link, because it relies on preserving \(\eta\), not on the log link specifically.

Result: In an intercept-included linear predictor \(\eta=\beta_0+\beta_1x\), centering at \(c\) gives \(\eta=\alpha_0+\alpha_1(x-c)\), where \(\alpha_0=\beta_0+\beta_1c\) and \(\alpha_1=\beta_1\). The new intercept is the linear predictor at \(x=c\); it is not generally the original intercept.

Worked Example: Centering Annual Mileage

A hypothetical insurer fits the Gamma model described above to annual pure premium, using annual mileage \(m\). The fitted log-mean is \(\eta=\beta_0+0.00008m\), with \(\beta_0=\ln(500)-0.64\). Re-express the model by centering mileage at 8,000 miles, then calculate fitted means at 5,000, 8,000, and 12,000 miles. The model assumes the same Gamma shape \(k=5\) and dispersion \(0.20\) at each mileage.

Method. Let \(m_c=m-8000\). Since \(m=m_c+8000\), substitution yields \(\eta=\beta_0+0.00008(m_c+8000)=(\beta_0+0.64)+0.00008m_c\). By the given intercept, \(\beta_0+0.64=\ln(500)\). The centered model is therefore \(\eta=\ln(500)+0.00008m_c\), or \(\mu=500\exp(0.00008m_c)\).

At 5,000 miles: \(m_c=5000-8000=-3000\). The exponent adjustment is \(0.00008(-3000)=-0.24\), so the relativity is \(\exp(-0.24)=0.7866279\). The fitted mean is \(500(0.7866279)=393.3139\), or $393.31.

At 8,000 miles: \(m_c=0\), so the relativity is \(\exp(0)=1\), and the fitted mean is \(500(1)=\) $500.00.

At 12,000 miles: \(m_c=4000\). The exponent adjustment is \(0.00008(4000)=0.32\), so the relativity is \(\exp(0.32)=1.3771278\). The fitted mean is \(500(1.3771278)=688.5639\), or $688.56.

Check. Each result uses the same coefficient and mean as the original model, because the substitution preserved \(\eta\) exactly. The centered intercept has a direct interpretation: the fitted mean at 8,000 miles is \(\exp(\ln 500)=$500.00\).

Scaling: The Coefficient Changes Units

Suppose the predictor is scaled as \(z=x/s\), without centering. Then \(x=sz\), and the linear predictor becomes \(\eta=\beta_0+(s\beta_1)z\). The new slope is \(s\beta_1\), not \(\beta_1/s\). If \(z=m/1000\), a one-unit increase in \(z\) is an increase of 1,000 miles, so its coefficient is 1,000 times the coefficient per mile. The fitted mean remains unchanged because the product of coefficient and predictor is unchanged.

More generally, if \(z=(x-c)/s\), then \(x=c+sz\), and \(\eta=(\beta_0+\beta_1c)+(\beta_1s)z\). Thus the transformed intercept is \(\beta_0+\beta_1c\), and the transformed slope is \(s\beta_1\). Under a log link, a change of \(\Delta z\) has relativity \(\exp((s\beta_1)\Delta z)\), equal to the original relativity \(\exp(\beta_1(s\Delta z))\).

Worked Example: Expressing Mileage Per 1,000 Miles

A second hypothetical Gamma pure-premium model has mean \(\mu=400\exp(0.00006(m-7000))\), where \(m\) is annual mileage. It uses shape \(k=5\), scale \(\mu/5\), and dispersion \(0.20\). Express the slope per 1,000 miles and calculate the fitted mean at 9,000 miles, as well as the relativity for a 3,000-mile increase.

Method. Define \(z=(m-7000)/1000\). Then \(m-7000=1000z\), so \(0.00006(m-7000)=0.00006(1000z)=0.06z\). The equivalent model is \(\mu=400\exp(0.06z)\). Its coefficient is \(0.06\) per 1,000 miles, while the fitted mean at \(z=0\) remains $400.00.

Fitted mean at 9,000 miles. Here \(z=(9000-7000)/1000=2\). The exponent is \(0.06(2)=0.12\); the relativity is \(\exp(0.12)=1.1274969\). Therefore \(\mu=400(1.1274969)=450.9987\), or $451.00.

Relativity for 3,000 miles. A 3,000-mile increase is three units of \(z\). Its log-mean change is \(0.06(3)=0.18\), and its relativity is \(\exp(0.18)=1.1972174\), or 1.197 to three decimal places. This is the same result as \(\exp(0.00006(3000))=\exp(0.18)\) in the original units.

Check. The transformed coefficient describes a 1,000-mile change, not a one-mile change. The model gives the same fitted mean at 9,000 miles whether evaluated with the per-mile coefficient or the per-1,000-mile coefficient.

Centering and Scaling Together: Driver Age

A useful center is often a meaningful or typical rating value. After centering, scaling can put a coefficient on a useful comparison unit. For age, \(z=(a-40)/10\) makes \(z=0\) correspond to age 40 and makes a one-unit increase correspond to ten years. Under a log link, the coefficient on \(z\) is the log relativity for a ten-year age increase, with other predictors held fixed.

Worked Example: Centering Age and Scaling to Decades

A hypothetical Gamma pure-premium model for driver age \(a\) and annual mileage \(m\) is \(\eta=\ln(600)+0.02(a-40)+0.00004(m-8000)\). Assume shape \(k=5\), scale \(\mu/5\), and dispersion \(0.20\). Re-express age in decades from age 40, state the age relativity per decade, and calculate the fitted mean for a 45-year-old driver with 9,000 annual miles.

Method. Set \(z=(a-40)/10\), so \(a-40=10z\). Then \(0.02(a-40)=0.02(10z)=0.20z\). The equivalent model is \(\eta=\ln(600)+0.20z+0.00004(m-8000)\). The age coefficient is now \(0.20\) per decade. The relativity for a ten-year increase is \(\exp(0.20)=1.2214028\), or 1.221 to three decimal places.

Fitted mean. For age 45, \(z=(45-40)/10=0.5\), giving an age contribution \(0.20(0.5)=0.10\). For mileage 9,000, the centered mileage is \(9000-8000=1000\), giving a mileage contribution \(0.00004(1000)=0.04\). The total adjustment to \(\ln(600)\) is \(0.10+0.04=0.14\). Hence \(\mu=600\exp(0.14)\), where \(\exp(0.14)=1.1502738\), and \(\mu=600(1.1502738)=690.1643\), or $690.16.

Check. Direct substitution into the original model gives the same total adjustment: \(0.02(45-40)+0.00004(9000-8000)=0.10+0.04=0.14\). The transformed coefficient changes age units; it does not change this fitted mean.

What Is Preserved, and What Can Go Wrong?

For an ordinary GLM with an intercept and no coefficient constraints, an affine re-expression of a predictor preserves the model's set of possible linear predictors. Each original coefficient pair maps to a transformed pair by the equations above, and the inverse mapping exists when \(s>0\). Consequently, for the same observations and response distribution, the likelihood as a function of fitted means is unchanged; fitted means and residual-based summaries are unchanged as well. The individual coefficient values and their units do change, so coefficient interpretation must match the predictor's current coding.

The intercept's meaning is especially easy to misread. With several predictors, the intercept is the linear predictor when all predictors equal zero. After centering, zero corresponds to the chosen center for each centered variable. Under a log link, exponentiating the intercept gives the fitted mean at those reference predictor values, provided the other terms in the model are also evaluated at their zero values.

Scaling also changes the coefficient's standard error in the same units. If \(z=x/s\) and the transformed slope is \(\alpha_1=s\beta_1\), then its standard error is \(s\) times the standard error of \(\beta_1\). The ratio of estimate to standard error, and thus the Wald test for a zero slope, is unchanged for a positive scale. A confidence interval must be converted using the same scale factor if it is reported in the original units.

The preservation result has limits. If a model has no intercept, centering generally adds a constant contribution that the model cannot absorb, so fitted values may change. Constraints on coefficients or some penalized fitting procedures can also make the transformed fitting problem differ unless the constraints or penalty are adjusted consistently. Finally, centering and scaling only re-express a linear effect; they do not make a linear predictor flexible enough to capture a nonlinear relationship.

Common exam trap: If \(z=m/1000\), it is incorrect to divide a per-mile coefficient by 1,000. For example, \(0.00008m\) becomes \(0.00008(1000z)=0.08z\), not \(0.00000008z\). The wrong calculation gives a log-mean change of \(0.00000008\) for a 1,000-mile increase; the correct change is \(0.08\), with relativity \(\exp(0.08)=1.0833\) to four decimal places. Check the units by substituting \(m=1000z\) into the original linear predictor.

For practical rating work, centered predictors can give a useful reference profile and more readable intercepts. Scaled predictors can make coefficient tables easier to compare and communicate. Neither operation improves fit by itself or changes a fitted indication under the stated conditions. Record the center and scale wherever coefficients are reported; without them, a coefficient's unit and reference point are ambiguous.

Check Your Understanding

For each problem, assume a log-link Gamma pure-premium GLM with shape \(k=5\), scale \(\mu/5\), and dispersion \(0.20\). Round monetary answers to cents and relativities to four decimal places.

  1. A fitted model is \(\mu=250\exp(0.00005(m-6000))\), with annual mileage \(m\). Write the coefficient per 1,000 miles and find the relativity for an increase from 6,000 to 8,000 miles. (Answer: coefficient \(0.05\) per 1,000 miles; relativity \(\exp(0.10)=1.1052\).)
  2. A driver-age coefficient is \(0.012\) per year. Define \(z=(a-35)/5\). What is the coefficient per five years, and what is the relativity for a ten-year increase? (Answer: coefficient \(0.06\) per five years; relativity \(\exp(0.12)=1.1275\).)
  3. A mileage model is \(\eta=\ln(350)+0.00004(m-9000)\). Re-center mileage at 12,000 miles and calculate the fitted mean at that mileage. (Answer: new intercept \(\ln(350)+0.12\), slope \(0.00004\); fitted mean \(350\exp(0.12)=$394.62\).)
  4. A model has age coefficient \(0.01\) per year and mileage coefficient \(0.00002\) per mile. Use \(z_a=(a-50)/10\) and \(z_m=(m-5000)/1000\). Find both transformed coefficients and the multiplicative change for a simultaneous increase of 10 years and 3,000 miles. (Answer: coefficients \(0.10\) per decade and \(0.02\) per 1,000 miles; multiplicative change \(\exp(0.10+0.06)=1.1735\).)
  5. A model is fitted without an intercept as \(\eta=\beta_1x\). Explain whether replacing \(x\) with \(x-c\) is guaranteed to preserve fitted means. (Answer: No. The original term becomes \(\beta_1((x-c)+c)=\beta_1(x-c)+\beta_1c\); without an intercept, the constant \(\beta_1c\) generally cannot be absorbed, so the model's fitted means need not be preserved.)
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