Tutorials › CAS Exam 8 - Advanced Ratemaking › Worked Translation From GLM Output to Rating Factors

CAS Exam 8 · Estimation and Relativity Interpretation · Tutorial 195 of 989

Worked Translation From GLM Output to Rating Factors

Learn to read a fitted GLM output table, reconstruct indications from its rating factors, and quantify uncertainty for selected factors and comparisons.

Intermediate 10 min read

What You'll Learn

  • Identify reference levels and interpret the intercept in a log-link rating model
  • Convert log-scale coefficients into multiplicative rating relativities
  • Combine selected relativities with a base rate and exposure offset
  • Calculate an indication for a policy profile from fitted GLM output
  • Use coefficient variances and covariances to quantify uncertainty in a factor or contrast
  • Construct and interpret confidence intervals on both the log and multiplicative scales

From Fitted Coefficients to a Rating Indication

A fitted GLM output table reports quantities on the model’s parameter scale, but a rate indication is usually needed on the original scale and for a particular risk profile. The translation requires more than exponentiating a coefficient: first identify the response and link, the reference levels, any offset, and the coefficients that apply to the profile. Then combine the applicable effects and, when requested, carry their uncertainty through the calculation.

This tutorial uses a log-link model. Assume the conditional mean of the response is positive and has the form \(\mu=\exp(\eta)\), where \(\eta\) is a linear predictor. A log link makes additive changes in \(\eta\) multiplicative changes in the modeled mean. The calculations below interpret fitted output; they do not establish that the model is appropriate for a particular portfolio.

Definition: In a log-link GLM, a fitted coefficient \(\hat\beta_j\) for a one-unit change or a specified categorical level contributes \(\hat\beta_j\) to the linear predictor. Holding other terms fixed, its multiplicative relativitiy is \(\exp(\hat\beta_j)\). A reference level represented by the intercept conventionally has coefficient zero and relativity one. An offset is a known term in the linear predictor whose coefficient is fixed at one, not estimated.

For a categorical factor coded with a reference level, the reference level’s effect is included in the intercept rather than shown as a separately estimated coefficient. A nonreference coefficient is therefore a log-relative effect against that factor’s reference, holding other modeled terms fixed. If the model includes interactions, the relevant effects can depend on the other factor levels; the examples here explicitly use models without interactions.

For a pure-premium model with annual exposure \(e\), one common specification is \(\log E[X\mid \text{risk profile}]=\log(e)+\eta\). Here \(\log(e)\) is the offset, and \(\exp(\eta)\) is the expected amount per exposure unit. If the response is already a rate per exposure unit, the offset may not appear in the predictor. Always check whether the modeled quantity is a rate or an exposure-scaled total before producing an indication.

Why Exponentiation Produces Relativities

The central translation follows directly from the log link. Suppose a profile’s linear predictor is \(\eta\), and changing one rating characteristic adds coefficient \(\beta\), with all other terms unchanged. The original-scale means before and after the change are \(\exp(\eta)\) and \(\exp(\eta+\beta)\). Their ratio is

$$ \frac{\exp(\eta+\beta)}{\exp(\eta)} =\exp(\beta). $$

Thus the intercept and all unchanged factors cancel in the comparison. For a profile with several additive effects and no interactions, the effects sum on the log scale and multiply on the original scale:

$$ \mu=\exp\left(\beta_0+\beta_1+\cdots+\beta_k\right) =\exp(\beta_0)\prod_{j=1}^{k}\exp(\beta_j). $$

This is why the right workflow is to identify the model terms that apply, add their fitted coefficients, and exponentiate the resulting linear predictor. Multiplying rounded relativities can produce a slightly different answer from exponentiating the unrounded coefficient sum, so retain full precision until the final step.

Worked Example: Translating a Table Into a Profile Indication

A hypothetical insurer fits a log-link GLM for expected annual pure premium per policy-year. The model has no interactions and includes these fitted terms:

TermEstimateStandard error
Intercept5.200.10
Region: Coastal (reference: Inland)0.250.08
Vehicle class: Commercial (reference: Personal)0.400.10
Driver tenure: New (reference: Experienced)0.300.12

Find the fitted annual pure-premium indication for a Coastal, Commercial policy with a New driver. Identify the reference levels and show the resulting factor relativities.

Method and references. The table identifies Inland, Personal, and Experienced as the reference levels. For a profile at all three references, the linear predictor is the intercept, \(5.20\). The profile in the question uses all three displayed nonreference levels, so its linear predictor is

$$ \hat\eta=5.20+0.25+0.40+0.30=6.15. $$

Convert each coefficient. The Coastal relativity is \(\exp(0.25)=1.2840\); the Commercial relativity is \(\exp(0.40)=1.4918\); and the New-driver relativity is \(\exp(0.30)=1.3499\). The reference relativity for each factor is \(1.0000\).

Calculate the indication. The reference-profile base indication is \(\exp(5.20)=\$181.27\) per policy-year. The selected profile’s combined relativity is \(\exp(0.25+0.40+0.30)=\exp(0.95)=2.5857\). Therefore its indication is

$$ \exp(6.15)=\$468.72\text{ per policy-year}. $$

As a check, \(\$181.27\times2.5857\) is approximately \(\$468.72\), subject to rounding in the displayed intermediate values. The calculation avoids the mistake of treating each coefficient as a dollar amount: the coefficients are log-scale effects, and their exponentials are the multiplicative factors.

Offsets and the Quantity Being Indicated

An offset changes the linear predictor but is not an estimated rating effect. For an exposure-scaled total, an exposure of \(e\) contributes \(\log(e)\), so the fitted expected total is \(e\exp(\eta)\). The per-unit rate remains \(\exp(\eta)\). A policy with twice the exposure has twice the modeled expected total under this specification, but not twice the expected rate per exposure unit.

A practical translation should therefore distinguish three quantities: the base rate implied by the intercept, the multiplicative rating factors implied by the other terms, and the exposure amount used to scale a total. The relevant output table may also show a response-specific scale or other model information; do not mistake such a quantity for an additional rating coefficient.

Worked Example: A Relativity and Its Confidence Interval

Suppose a hypothetical fitted log-link model has a reference-profile indication of \(\$400\) per exposure unit. A selected territory coefficient is \(\hat\beta=0.40\), and its estimated variance is \(0.006\). Find the point indication and a 95% Wald confidence interval for the indication, using the normal critical value \(1.96\). Assume the model specification and reference profile are held fixed.

Point indication. The territory relativity is \(\exp(0.40)=1.4918\), or \(1.492\) to four significant figures. The point indication is \(\$400\exp(0.40)=\$596.73\), rounded to cents.

Log-scale uncertainty. The standard error of the coefficient is \(\sqrt{0.006}=0.0774597\). A 95% Wald interval for the log relativity is

$$ 0.40\pm1.96\sqrt{0.006} =0.40\pm0.15182095, $$

so its endpoints are \(0.24817905\) and \(0.55182095\). Exponentiating gives a relativity interval of approximately \((1.2817,\ 1.7364)\). Multiplying the unrounded endpoints by the \(\$400\) reference indication gives the 95% interval \((\$512.68,\ \$694.56)\), rounded to cents.

This interval is asymmetric around the point indication in dollars because it is symmetric on the log scale, not on the original scale. The method avoids constructing a symmetric dollar interval by adding and subtracting \(1.96\) times a dollar-scale standard error, which would not represent this log-link Wald interval.

Comparing Two Levels and Using Covariance

When comparing two nonreference levels of the same factor, subtract their fitted coefficients on the log scale. If the estimates are \(\hat\beta_A\) and \(\hat\beta_B\), the fitted relativity of level A to level B is \(\exp(\hat\beta_A-\hat\beta_B)\). The uncertainty calculation must use the variance of the difference, including covariance:

$$ \operatorname{Var}(\hat\beta_A-\hat\beta_B) =\operatorname{Var}(\hat\beta_A)+\operatorname{Var}(\hat\beta_B) -2\operatorname{Cov}(\hat\beta_A,\hat\beta_B). $$

For a combined log-scale effect \(c^\mathsf{T}\hat\beta\), where \(c\) selects and signs the coefficients in the contrast, its estimated variance is \(c^\mathsf{T}\widehat{\operatorname{Var}}(\hat\beta)c\). The standard error is the square root of that variance. A Wald interval is formed on the log scale and then exponentiated to obtain an interval for the multiplicative comparison. This procedure requires the relevant covariance estimates; adding standard errors or ignoring covariance is generally incorrect.

Worked Example: Comparing Two Fitted Levels With Covariance

A hypothetical log-link model estimates coefficients \(0.32\) for level A and \(0.11\) for level B relative to the same reference level. Their standard errors are \(0.09\) and \(0.07\), respectively, and their estimated covariance is \(0.002\). Find the fitted relativity of A to B and its 95% Wald interval.

Point comparison. The log-scale contrast is \(0.32-0.11=0.21\), so the fitted A-to-B relativity is \(\exp(0.21)=1.2337\). In this fitted model, level A’s indicated mean is about 1.2337 times level B’s, with other terms held fixed.

Contrast variance. The coefficient variances are \(0.09^2=0.0081\) and \(0.07^2=0.0049\). Therefore,

$$ \widehat{\operatorname{Var}}(\hat\beta_A-\hat\beta_B) =0.0081+0.0049-2(0.002)=0.0090, $$

and the standard error is \(\sqrt{0.0090}=0.09487\). The log-scale 95% interval is \(0.21\pm1.96(0.09487)\), or approximately \((0.02406,\ 0.39594)\). Exponentiating the endpoints gives the relativity interval \((1.0244,\ 1.4858)\), to four decimals.

The interval is consistent with the direction of the point estimate: its endpoints are above one. The calculation avoids the common error of using \(0.0081+0.0049\) as the contrast variance and thereby omitting the covariance term.

A Reliable Translation Process

When given a compact GLM output table, use a fixed sequence rather than jumping directly to exponentiation. This keeps the model interpretation, the selected profile, and the uncertainty calculation aligned.

1
Read the model statement.
Confirm the response, link, unit of the modeled mean, and whether an offset is present. Establish whether the requested result is a rate or an exposure-scaled total.
2
Map the profile to the table.
Identify the reference level for each factor and select only the coefficients that apply to the risk profile. Include interactions only when they are part of the stated model and apply to that profile.
3
Build the linear predictor.
Add the intercept, applicable coefficient estimates, and any known offset. Keep the offset separate conceptually because its coefficient is fixed rather than estimated.
4
Translate to the requested scale.
Exponentiate the relevant log-scale sum. For a comparison, use the log-scale contrast; for a total with an exposure offset, apply exposure to the per-unit fitted mean.
5
Carry uncertainty through correctly.
Use the variance of the coefficient or contrast, including covariance terms. Construct the Wald interval on the log scale, then exponentiate its endpoints.

These translations matter because a rating indication is often presented as a base rate multiplied by a set of selected relativities, while the fitted model output is expressed in coefficient form. The conversion makes the implied comparison and the modeled unit visible. It also makes assumptions easier to audit: a missing reference level, an omitted offset, or an unsupported independence assumption can materially change the result.

For standard errors, the displayed coefficient standard error alone is enough for a single coefficient’s Wald interval, but not necessarily for a combined profile or level-to-level contrast. Those require the estimated covariance matrix, and a confidence interval for a fitted mean can require uncertainty in the intercept as well as in the factor coefficients. The uncertainty examples above are specifically for the stated coefficient or contrast, not a full allowance for model selection, process variation, or uncertainty in the model specification.

Key takeaway: A fitted log-link table becomes a rating indication by selecting the correct coefficients, adding them with the intercept and any applicable offset, and exponentiating. For uncertainty, calculate the variance of the relevant log-scale quantity—including covariance where applicable—before exponentiating interval endpoints. This translation is aligned with the CAS Exam 8 (Advanced Ratemaking) syllabus.

Check Your Understanding

For each problem, retain precision through the calculation and round the final quantities as requested.

  1. A fitted log-link model reports a coefficient of \(0.18\) for a selected territory versus its reference. What is the fitted territory relativity, to four decimals? (Answer: \(1.1972\).)
  2. A reference-profile rate is \(\$250\) per policy-year. A profile has two applicable coefficients, \(0.20\) and \(-0.10\), with no interaction or offset. Find its indicated rate to cents. (Answer: \(\$276.29\).)
  3. Two level coefficients relative to one reference are \(0.32\) and \(0.11\), with variances \(0.0081\) and \(0.0049\), and covariance \(0.002\). Find the A-to-B relativity and its 95% Wald interval, to four decimals. (Answer: relativity \(1.2337\), interval \((1.0244,\ 1.4858)\).)
  4. A model estimates a rate of \(\$120\) per policy-year, and the stated exposure for a hypothetical account is \(1.5\) policy-years. Under a log-exposure offset, with no additional factor effects, what is the expected total? (Answer: \(\$180.00\).)
  5. A combined log-scale factor estimate is \(0.30\). Its estimated variance, already calculated from the coefficient covariance matrix, is \(0.0224\). Find the factor relativity and its 95% Wald interval, to four decimals. (Answer: relativity \(1.3499\), interval \((1.0067,\ 1.8100)\).)
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