Why the Response Target Determines the GLM
A generalized linear model is not selected simply because its name is familiar or because it produces a convenient fit. The response determines which values are possible, while the modeling target and data-generating process suggest how its mean and variance behave. A binary indicator, a count, a positive claim severity, and a policy-period total cost therefore call for different candidate families, even if all are used to predict insurance costs in a broad sense.
The earlier material on GLM mean structures and variance functions supplies the framework: predictors determine a linear predictor, a link connects that predictor to the response mean, and a response family specifies a variance relationship. This tutorial uses that framework to make a practical choice. A family-link pairing is a candidate model, not a guarantee that the assumptions are true; diagnostics and validation still matter.
The core screening questions are: (1) What values can the response take? (2) Is zero possible, and does it have a distinct meaning? (3) How does variability change with the mean? (4) Which link gives a useful mean structure without allowing impossible predictions? A correct answer starts with the response as it is actually recorded, not with the eventual business interpretation.
Candidate Families and Their Mean-Variance Patterns
The table summarizes common candidates. The variance statements refer to a single observation conditional on predictors. For binomial responses, distinguish a count of successes from a sample proportion; their variances differ by the number of trials.
| Response target | Candidate family | Mean and variance pattern | Common link choice |
|---|---|---|---|
| Binary indicator | Bernoulli | \(0\) or \(1\); variance \(\mu(1-\mu)\) | Logit |
| Success count in a fixed number \(m\) of trials | Binomial | Mean \(mp\); variance \(mp(1-p)\) | Logit for \(p\) |
| Event count in an interval or exposure unit | Poisson | Nonnegative integers; variance equals the mean | Log |
| Unbounded continuous outcome | Gaussian | Variance constant at a given dispersion | Identity |
| Strictly positive, right-skewed outcome | Gamma | Variance proportional to \(\mu^2\) | Log is common |
| Strictly positive outcome with variance growing about as \(\mu^3\) | Inverse Gaussian | Variance proportional to \(\mu^3\) | Log is one option |
| Aggregate cost with zeros and positive continuous values | Compound Poisson-Gamma Tweedie | For \(1<p<2\), variance proportional to \(\mu^p\) | Log is common |
For Bernoulli data, \(\mu\) is a probability, so it must lie strictly between zero and one for the usual finite linear predictor. The logit link \(g(\mu)=\log(\mu/(1-\mu))\) maps that interval to the real line. For a count or a positive cost, the log link \(g(\mu)=\log(\mu)\) maps a positive mean to any real linear predictor and gives \(\mu=\exp(\eta)\), which cannot be negative. An identity link does not impose a range: that can be appropriate for an unrestricted continuous response, but can predict negative values for costs or probabilities outside their permitted range.
A family’s variance pattern is a working assumption about conditional variability, not a statement that every observation will follow a perfectly deterministic curve. For example, Poisson variance equals the mean. If the observed counts vary substantially more than that relationship allows, a plain Poisson model may be inadequate. The response’s support can rule a family out immediately, while variance behavior helps discriminate between plausible families.
Worked Examples: Matching the Family and Link
Worked Example: A Binary Renewal Indicator
Scenario. In a hypothetical portfolio, \(Y=1\) indicates that a policy renews and \(Y=0\) indicates that it does not. Suppose the response is Bernoulli conditional on predictors. A baseline policy has renewal probability \(0.25\), and a binary service indicator multiplies the baseline renewal odds by \(2\). Find the fitted probability and conditional variance for a policy with the service indicator equal to \(1\), using a logit model.
Method. Use Bernoulli because the recorded response has exactly two values. The baseline odds are \(0.25/(1-0.25)=1/3\). A logit linear predictor for the baseline is \(\eta_0=\log(1/3)=-\log 3\). The service coefficient is \(\log 2\), since adding it multiplies the odds by \(2\).
For the policy with the service indicator equal to \(1\), \(\eta=-\log 3+\log 2=\log(2/3)\). Inverting the logit gives \(\mu=\exp(\eta)/(1+\exp(\eta))=(2/3)/(1+2/3)=2/5=0.40\).
The Bernoulli variance is \(\mu(1-\mu)=0.40(0.60)=0.24\). As a check, the fitted probability is between zero and one, and the fitted odds are \(0.40/0.60=2/3\), exactly twice the baseline odds \(1/3\).
Common mistake avoided: Treating renewal as an unrestricted continuous response and using an identity link could produce a fitted value below zero or above one. The Bernoulli-logit pairing respects the response support and interprets the effect as an odds multiplier.
Worked Example: A Claim Event Count
Scenario. For hypothetical policies with equal observation periods, let \(Y\) be the number of roadside-assistance events. Assume \(Y\) is Poisson conditional on predictors. A baseline group has expected count \(2\), and a vehicle feature multiplies the expected count by \(1.5\). Find the fitted mean, variance, and probability of zero events for a policy with that feature.
Method. A Poisson response is a candidate because the outcome is a nonnegative integer count. Under a log link, the baseline linear predictor is \(\log 2\), and the feature coefficient is \(\log(1.5)\). Thus \(\eta=\log 2+\log(1.5)=\log 3\), so the fitted mean is \(\mu=\exp(\eta)=3\).
For a Poisson response, \(\operatorname{Var}(Y\mid\text{predictors})=\mu\), so the fitted conditional variance is \(3\). Its standard deviation is \(\sqrt{3}\), approximately \(1.732\). The probability of zero events is \(P(Y=0)=e^{-3}3^0/0!=e^{-3}\), approximately \(0.0498\).
The mean multiplier is checked directly: \(2(1.5)=3\). The zero-event probability is between zero and one, as required. A non-integer fitted mean is valid: it is an expected count, not a possible observed count.
Common mistake avoided: A predicted mean of \(3\) does not mean every policy has exactly three events. The model assigns a distribution over integer counts with mean and variance both equal to \(3\). Also, the stated equal observation periods matter; differing exposure must be represented appropriately rather than ignored.
Worked Example: A Positive Claim Severity
Scenario. Consider positive claim severities in a hypothetical insurance product, recorded only when a claim occurs. Suppose severity follows a Gamma model with variance \(\phi\mu^2\), dispersion \(\phi=0.25\), and a log link. The baseline mean severity is \(\$500\), and a risk characteristic has coefficient \(\log(1.2)\). Find the fitted mean, variance, and standard deviation for a claim with the characteristic present.
Method. Gamma is a candidate because the response is continuous and strictly positive, and its variance increases quadratically with its mean under this parameterization. The baseline linear predictor is \(\log(500)\). Adding the characteristic gives \(\eta=\log(500)+\log(1.2)=\log(600)\), hence \(\mu=\exp(\eta)=\$600\).
The conditional variance is \(\phi\mu^2=0.25(600^2)=0.25(360{,}000)=90{,}000\) dollars squared. The standard deviation is \(\sqrt{90{,}000}=\$300\). For a distributional check, a Gamma distribution with shape \(k=1/\phi=4\) and scale \(\theta=\phi\mu=150\) has mean \(k\theta=4(150)=600\) and variance \(k\theta^2=4(150^2)=90{,}000\).
The characteristic multiplies the fitted severity mean by \(1.2\), as intended. The positive mean and positive support agree with the target. Variance and standard deviation are reported in their respective units: dollars squared and dollars.
Common mistake avoided: A severity-only data set contains positive amounts, not the zero costs of policies with no claim. Fitting this Gamma model to a portfolio’s total cost, including no-claim zeros, would mismatch the observed support.
Worked Example: Aggregate Cost With Zeros
Scenario. For a hypothetical policy-period, let \(Y\) be total covered cost. It is zero when no covered event occurs and positive continuous when one or more events occur. Assume the conditional distribution is compound Poisson-Gamma, represented by a Tweedie model with power \(p=1.5\), dispersion \(\phi=0.25\), and log link. For a fitted mean of \(\$400\), calculate the conditional variance and standard deviation.
Method. The aggregate response can be zero or positive continuous. A compound Poisson-Gamma model is a candidate for this support when event frequency and positive event costs contribute to the total. Under the stated Tweedie variance function, \(\operatorname{Var}(Y\mid\text{predictors})=\phi\mu^p\).
Substitute the given values: \(\operatorname{Var}(Y\mid\text{predictors})=0.25(400^{1.5})\). Since \(400^{1.5}=400\sqrt{400}=400(20)=8{,}000\), the variance is \(0.25(8{,}000)=2{,}000\) dollars squared. The standard deviation is \(\sqrt{2{,}000}\approx \$44.72\), rounded to the nearest cent.
The fitted mean is positive, as required by the log link, and the variance calculation uses the specified power \(p=1.5\), not the Poisson power \(p=1\). This example checks the mean-variance relationship; it does not assert that the model’s assumptions have been empirically verified.
Common mistake avoided: Do not use a severity-only Gamma model just because positive claim costs look skewed. The target here is total policy-period cost, and it includes zero outcomes; its frequency-plus-severity structure motivates a different candidate.
Why the Link and Variance Must Agree With the Target
The link controls the mean scale; the family controls the response’s support and conditional variance. This distinction explains why selecting only one of them is not enough. A log link is useful for positive means and multiplicative effects, but it does not turn a Gaussian response into a count or make a Gamma response permit zero observations. Conversely, choosing a family with appropriate support does not ensure that an arbitrary link yields valid fitted means.
The logit link follows directly from the odds representation. If \(Y\) is Bernoulli with success probability \(\mu\), then its odds are \(\mu/(1-\mu)\). Taking logarithms gives \(g(\mu)=\log(\mu/(1-\mu))\). Solving for \(\mu\) gives \(\mu=\exp(\eta)/(1+\exp(\eta))\). Because \(\exp(\eta)>0\), the denominator exceeds the numerator, so every finite \(\eta\) produces \(0<\mu<1\). The link therefore enforces the probability range without separately constraining every linear predictor.
A similar range check supports the log link for positive conditional means. If \(\eta\) is any real number, then \(\mu=\exp(\eta)>0\). In contrast, under the identity link, \(\mu=\eta\); the linear predictor can be negative, so an identity-link model for cost needs extra scrutiny. A link may be mathematically possible yet unsuitable in practice if it generates implausible means or makes effects difficult to interpret.
Decide whether each row is a binary outcome, a count, a positive amount conditional on an event, or a total that can include zeros.
Do not use a strictly positive-only family for a response whose recorded values include zero, or a continuous family for an integer count without a compelling approximation.
Ask whether a constant, mean-proportional, quadratic, cubic, or other variance pattern is plausible for the response at its conditional mean.
Confirm that fitted means stay in the response’s valid range, then interpret the link-scale effects in context.
These checks identify defensible starting models; fit assessment, diagnostics, and the predictive objective determine whether a candidate is useful.
Common pairings are not universal rules. A Gaussian model can be reasonable for a continuous response with approximately constant conditional variance, but its unrestricted support may be awkward for nonnegative costs. Gamma and inverse Gaussian models both describe positive continuous outcomes, but their variance functions differ: \(\phi\mu^2\) versus \(\phi\mu^3\). The observed support alone cannot decide between them. Likewise, Poisson variance equaling the mean is a specific assumption, not a consequence of the word “count.”
For an aggregate cost with zeros and positive continuous values, the compound Poisson-Gamma Tweedie model provides one candidate because its distribution has a point mass at zero and a continuous positive component. For \(1<p<2\), its variance function is \(\phi\mu^p\). The choice still depends on whether that variance behavior and the assumed process are plausible. Using a different family simply because it is easy to fit can lead to invalid means, poor uncertainty representation, or predictions that do not match the intended target.
These family-link choices are aligned with the SOA Exam PA (Predictive Analytics) syllabus. In a written justification, name the response target, state its support, identify the relevant variance assumption, and explain why the selected link respects the mean’s range. That reasoning is more defensible than naming a family without connecting it to the data.
Check Your Understanding
For each original scenario, choose a candidate family and link, then check the stated calculation or support issue.
- A hypothetical warranty dataset records whether each device requires a repair during its first year. The baseline repair probability is \(0.10\), and a feature doubles the odds. Under a Bernoulli-logit model, what is the feature-group probability? (Answer: \(2/11\approx0.1818\).)
- In a hypothetical equal-exposure count model, the log-link predictor for a segment is \(\eta=\log 4+\log 1.25\). Under a Poisson model, give the fitted mean and variance. (Answer: Mean \(=5\); variance \(=5\).)
- A positive-only severity model has Gamma variance \(\phi\mu^2\), with \(\phi=0.20\) and fitted mean \(\$250\). Find the variance and standard deviation. (Answer: Variance \(=\$12{,}500\) squared; standard deviation \(\approx\$111.80\).)
- A hypothetical policy-period total is zero with positive probability and otherwise positive continuous. Which is the more directly matched candidate: a Gamma severity model or a compound Poisson-Gamma Tweedie model? (Answer: Compound Poisson-Gamma Tweedie, because it can represent both zero totals and positive continuous totals.)
- A model uses an identity link for a binary outcome and gives a fitted mean of \(1.08\) for one record. What is the problem, and what common family-link candidate avoids this range violation? (Answer: \(1.08\) is not a valid probability; a Bernoulli model with a logit link constrains finite fitted probabilities to between zero and one.)