Mapping Means To Linear Predictors

From a Response Mean to a Linear Predictor

A generalized linear model connects a response mean to explanatory variables in two steps. First, the model applies a link function to the mean; the resulting quantity is the linear predictor, a sum of coefficients multiplied by known covariate values. To obtain the modeled mean, apply the inverse link to that sum. Keeping these steps distinct is essential: the predictor and the response mean are generally not the same quantity.

In ratemaking, the response might represent a claim amount or a claim count, while the explanatory variables describe characteristics of a risk. This tutorial focuses on translating between the mean and predictor. The mapping itself does not require a particular probability distribution for the response; distributional assumptions are a separate part of specifying a GLM.

Definition: Let $Y$ be a response and let $\mu=E[Y\mid\mathbf{x}]$ be its conditional mean for covariate vector $\mathbf{x}$. A link function $g$ relates that mean to the linear predictor $\eta$ by $$ g(\mu)=\eta=\mathbf{x}^{\mathsf T}\boldsymbol{\beta}. $$ Here $\boldsymbol{\beta}$ is a vector of coefficients, and $\mathbf{x}$ contains the corresponding covariate values, often including a 1 for the intercept. If $g$ is invertible over the relevant mean domain, then $$ \mu=g^{-1}(\eta)=g^{-1}\!\left(\mathbf{x}^{\mathsf T}\boldsymbol{\beta}\right). $$

The term linear predictor refers to linearity in the coefficients: $\eta$ is a sum of coefficient-covariate products. It does not mean that the mean itself must be a linear function of the covariates. When the inverse link is nonlinear, a linear change in $\eta$ need not produce a linear change in $\mu$.

A link and its inverse must be used as a matched pair. For example, if $g(\mu)=\mu$, then $g^{-1}(\eta)=\eta$. If $g(\mu)=\ln(\mu)$, then $g^{-1}(\eta)=\exp(\eta)$. If $g(\mu)=1/\mu$ on positive means, then $g^{-1}(\eta)=1/\eta$, with a positive predictor required. These examples illustrate the mapping; the suitability of a link for a particular target is a separate modeling question.

Why the Inverse Link Recovers the Mean

The mapping follows directly from the defining equation for a GLM. First calculate the predictor from the covariates and coefficients. Then apply the inverse function to both sides of the link equation. This is valid when the link is one-to-one on the domain being modeled.

Calculate the predictor.
For the observed covariates, form $\eta=\mathbf{x}^{\mathsf T}\boldsymbol{\beta}$.

Start from the model definition.
The link equation states $g(\mu)=\eta$.

Apply the inverse function.
Applying $g^{-1}$ to both sides gives $g^{-1}(g(\mu))=g^{-1}(\eta)$. Since $g^{-1}(g(\mu))=\mu$, the result is $\mu=g^{-1}(\eta)$.

This derivation also identifies a frequent conceptual error. The response $Y$ is an observed random quantity; $\mu=E[Y\mid\mathbf{x}]$ is its modeled conditional mean; and $\eta$ is the transformed mean. Unless the link is the identity, $\eta$ is not the mean. Nor does the equation claim that an individual observed response equals its conditional mean.

Worked Example: Identity Link for a Hypothetical Claim Amount

A hypothetical insurer models mean claim amount in dollars for a specified policy period using an identity link. The predictor is $\eta=\beta_0+\beta_1x_1+\beta_2x_2$, where $x_1=1$ denotes a selected territory and $x_2=1$ denotes a selected vehicle class; both are 0 otherwise. The fitted coefficients are $\beta_0=\$420$, $\beta_1=\$65$, and $\beta_2=-\$25$. Find the modeled mean for a risk in the selected territory and the reference vehicle class. This is a mean-mapping calculation; no response distribution, interest rate, or mortality assumption is needed or specified.

Method. Calculate the linear predictor, then use the inverse of the identity link. This method avoids treating a coefficient as the mean without first including the intercept and the risk's covariate values.

The covariate values are $x_1=1$ and $x_2=0$. Therefore, $$ \eta=420+65(1)-25(0)=420+65=485. $$

For the identity link, $g(\mu)=\mu$, so $g^{-1}(\eta)=\eta$. The modeled mean is therefore $\mu=\$485$. The result is in dollars, matching the units of the response and the identity-link coefficients. As a check, applying the link gives $g(485)=485=\eta$.

Common mistake avoided: Reporting $\$65$, the territory coefficient, as the mean ignores both the intercept and the other covariate contribution. The full predictor is $\$485$, not $\$65$.

Using Nonidentity Links

When the link is not the identity, the predictor must be transformed before it can be reported as a mean. First combine all coefficient contributions, including the intercept. Then apply the specified inverse link once to the complete predictor. Applying an inverse link to each coefficient separately and adding the results generally does not reproduce the model equation.

Worked Example: Exponential Inverse Link for a Hypothetical Count Mean

A hypothetical insurer models the expected number of claims for a policy over a specified policy period. The stated link is $g(\mu)=\ln(\mu)$, so the inverse is $g^{-1}(\eta)=\exp(\eta)$. The predictor is $\eta=-0.60+0.35x_1+0.15x_2$, where a risk has $x_1=1$ and $x_2=1$. Find its modeled mean. The task specifies no response distribution; no interest or mortality assumption is involved.

Method. Add the predictor terms first, then exponentiate the result. This avoids confusing a log-scale predictor with the mean on the original scale.

Substitute the covariates: $$ \eta=-0.60+0.35(1)+0.15(1)=-0.60+0.35+0.15=-0.10. $$

Now apply the inverse link: $$ \mu=\exp(-0.10)=0.904837\ldots $$

The modeled mean is $0.9048$ claims per policy period, rounded to four decimal places. As a check, $\ln(0.904837\ldots)=-0.10$, which recovers the predictor. A mean below one is permissible for a count response: it is an average, not a claim count required to be an integer.

Common mistake avoided: Reporting $-0.10$ as the expected claim count is incorrect. It is the log of the mean; the inverse link gives the mean on the original scale.

Worked Example: Reciprocal Link for a Hypothetical Mean Severity

A hypothetical model for mean claim severity in dollars uses the reciprocal link $g(\mu)=1/\mu$, defined here for $\mu>0$. The predictor is $\eta=0.0030+0.0010x$, where $x=1$ indicates a specified risk feature and $x=0$ its reference level. Find the mean severity for a risk with $x=1$. No response distribution, interest rate, or mortality assumption is specified because only the link mapping is being evaluated.

Method. Calculate the predictor and then take its reciprocal. The domain condition matters: the resulting predictor must be positive for this inverse-link calculation to produce a positive mean.

For $x=1$, $$ \eta=0.0030+0.0010(1)=0.0040. $$

Since $g^{-1}(\eta)=1/\eta$, $$ \mu=\frac{1}{0.0040}=250. $$

The modeled mean severity is $\$250.00$. As a check, $1/250=0.0040$, the calculated predictor. The coefficient and predictor units are reciprocal dollars, while the recovered mean is in dollars.

Common mistake avoided: Treating $0.0040$ as the mean would report a reciprocal-scale value as dollars. The reciprocal link requires taking $1/\eta$, not leaving the predictor unchanged.

Checks Before Reporting a Mean

In an exam calculation, writing down the link and inverse link before substituting numbers makes the transformation explicit. It also provides a quick way to catch domain errors and unit inconsistencies. Use the following checks after computing a mean:

Check the predictor: include the intercept and use the covariate values specified for the risk.
Check the inverse: apply the inverse of the stated link, not a different familiar link.
Check the domain: confirm that the predictor lies in a range where the inverse produces a valid mean.
Check the scale: report the mean in the response's units, not on the transformed predictor scale.
Check by substitution: applying $g$ to the reported mean should recover $\eta$, subject to rounding.

These checks do not determine whether the chosen link is appropriate, whether the fitted coefficients are statistically reliable, or whether the model adequately describes the data. Those are separate modeling and validation questions. Here the objective is narrower: given a link, coefficients, and covariates, correctly map between the mean and its linear predictor. This distinction is useful whenever a rating model is expressed on a transformed scale, and it is aligned with the CAS Exam 8 (Advanced Ratemaking) syllabus.

Key takeaway: A GLM specifies $g(\mu)=\eta=\mathbf{x}^{\mathsf T}\boldsymbol{\beta}$. Calculate $\eta$ from the covariates and coefficients, then recover the response mean as $\mu=g^{-1}(\eta)$. The predictor is the transformed mean, not automatically the mean itself.

Check Your Understanding

For each problem, calculate the predictor before applying the stated inverse link. Answers are rounded only where indicated.

A hypothetical mean payment model uses the identity link and $\eta=310+45x_1-20x_2$. For $x_1=1$ and $x_2=0$, find the mean in dollars. (Answer: $\$355.00$)
A hypothetical count-mean model has $g(\mu)=\ln(\mu)$ and $\eta=-0.40+0.25(1)+0.05(1)$. Find $\mu$ to four decimal places. (Answer: $0.9048$)
A hypothetical positive severity mean uses $g(\mu)=1/\mu$ and $\eta=0.003+0.001x$. For $x=1$, find the mean in dollars. (Answer: $\$250.00$)
A hypothetical model for a probability uses $g(\mu)=\ln(\mu/(1-\mu))$. If $\eta=\ln(3)$, recover $\mu$. (Answer: $0.7500$)
A hypothetical identity-link model has $\eta=500-80x_1+30x_2$, with $x_1=1$ and $x_2=1$. State both the predictor and the mean. (Answer: $\eta=450$, $\mu=450$)

What You'll Learn