Canonical Links Are About the Distribution
A link function transforms a model’s mean onto a scale where predictors can be added. For ratemaking, the log link is attractive because it produces positive fitted means and supports multiplicative effects. But these features do not make it canonical for every positive-valued response. Canonical status is determined by the response distribution’s exponential-family form, not by whether the link respects the range of the mean.
This distinction is especially useful when comparing Poisson claim counts with Gamma claim severities. Both distributions can be used in ratemaking, and both can be paired with a log link. The log link is canonical for Poisson counts. For Gamma severity, it is a useful noncanonical choice; the canonical link is the negative inverse mean under the natural-parameter convention used here. This topic is aligned with the CAS Exam 8 (Advanced Ratemaking) syllabus.
The predictor \(\eta\) is typically a linear combination of model terms and may include a known offset. An offset changes the value of the predictor; it does not change which link is canonical for a given response distribution. The earlier material on offsets distinguishes such a known predictor component from an estimated coefficient.
Why the Poisson Log Link Is Canonical
For a Poisson response \(Y\) with mean \(\mu>0\), the probability mass function is \(P(Y=y)=e^{-\mu}\mu^y/y!\), for nonnegative integer \(y\). Taking logarithms and collecting the terms involving \(y\) gives \(y\log\mu-\mu-\log(y!)\). Comparing this with the exponential-family form identifies \(\theta=\log\mu\), \(b(\theta)=e^\theta\), and \(a(\phi)=1\). Therefore the canonical link is \(g(\mu)=\log\mu\).
In a frequency model, if the count mean is exposure \(e\) times a rate \(\lambda\), then \(\log\mu=\log e+\log\lambda\). Thus the exposure offset \(\log e\) appears on the canonical link scale. The rate may itself be modeled through covariates, for example \(\log\lambda=\mathbf{x}^{\mathsf T}\boldsymbol{\beta}\). The resulting count mean remains positive because it is the exponential of its predictor.
Worked Example: Poisson Count Mean With an Exposure Offset
Method. Use the Poisson natural parameter, which is the log of the count mean, and include the known exposure on that same scale. A hypothetical policy has exposure \(e=2.5\) years, and its modeled claim rate is \(\lambda=\exp(-1.2)\) claims per year. Find the rate, expected count, and canonical predictor. The common mistake to avoid is treating the rate as the count mean and omitting exposure.
Step 1: Calculate the rate. Using the unrounded exponential, \(\lambda=\exp(-1.2)=0.3011942\) claims per year. To four decimal places, the rate is \(0.3012\).
Step 2: Calculate the count mean. The expected count is exposure multiplied by rate: \( \mu=e\lambda=2.5\exp(-1.2)=0.7529855 \) claims. To four decimal places, \(\mu=0.7530\).
Step 3: Find the canonical predictor. For a Poisson mean, \(\eta=\log\mu\). Equivalently, \(\eta=\log e+\log\lambda=\log(2.5)-1.2=0.9162907-1.2=-0.2837093\). Thus \(\exp(\eta)=0.7529855\), agreeing with the count mean calculated in Step 2.
Check and trap. The offset is \(\log(2.5)\), so the count mean is not just \(\exp(-1.2)\). Multiplying the rate by exposure gives \(0.7529855\), as required. The displayed rate rounded to four decimals should not be used to claim more precision in the count mean.
Why the Gamma Log Link Is Noncanonical
Consider a Gamma response \(Y>0\) with mean \(\mu>0\) and variance \(\phi\mu^2\), where \(\phi>0\). One corresponding density is \( f(y;\mu,\phi)=y^{1/\phi-1}\exp[-y/(\phi\mu)]/\{\Gamma(1/\phi)(\phi\mu)^{1/\phi}\}\). Taking its logarithm, the terms involving \(y\) and \(\mu\) are \(-y/(\phi\mu)-(1/\phi)\log\mu\). In particular, the log-mean term has a negative sign.
Set \(\theta=-1/\mu\), so that \(-\theta=1/\mu>0\), and let \(b(\theta)=-\log(-\theta)\). Then \(\log\mu=-\log(-\theta)\), and \(y\theta-b(\theta)=-y/\mu-\log\mu\). Consequently the terms just identified equal \([y\theta-b(\theta)]/\phi\). The natural parameter is \(\theta=-1/\mu\), and differentiating \(b\) verifies the mean and variance relationship: \(b'(\theta)=-1/\theta=\mu\) and \(b''(\theta)=1/\theta^2=\mu^2\). With dispersion \(\phi\), the variance is \(\phi b''(\theta)=\phi\mu^2\), as specified. Hence the canonical link in this convention is \(g(\mu)=-1/\mu\).
The log link \(g(\mu)=\log\mu\) is not this natural parameter. It is often chosen for Gamma severity because any real-valued predictor maps to a positive mean, and predictor differences correspond to multiplicative mean changes. Some presentations refer to the Gamma canonical link as the inverse link; sign conventions can differ. In the natural-parameter representation above, the canonical parameter is explicitly \(-1/\mu\). A positive inverse link \(1/\mu\) is its negative, and the corresponding coefficients must be sign-reversed to represent the same linear-predictor structure.
Worked Example: A Gamma Severity Model With a Log Link
Method. Calculate the mean and variance from a log-link predictor, then compare that predictor with the Gamma natural parameter. A hypothetical severity model has \(\phi=0.25\) and predictor \(\eta=-0.4\) under the log link. The common mistake to avoid is calling the link canonical simply because its fitted mean is positive.
Step 1: Calculate the mean. Under the log link, \(\eta=\log\mu\), so \(\mu=\exp(-0.4)=0.6703200\). The fitted mean is positive, as the Gamma response requires.
Step 2: Calculate the variance. The Gamma variance assumption gives \(\operatorname{Var}(Y)=\phi\mu^2=0.25[\exp(-0.4)]^2=0.25\exp(-0.8)=0.1123324\). Thus the variance is \(0.1123\) to four decimal places.
Step 3: Calculate the natural parameter. The canonical parameter is \(\theta=-1/\mu=-1/\exp(-0.4)=-\exp(0.4)=-1.4918247\), or \(-1.4918\) to four decimal places. It is not equal to the log-link predictor \(-0.4\).
Check and trap. If the predictor \(-0.4\) were instead interpreted as the canonical natural parameter, the implied mean would be \(-1/(-0.4)=2.5\), not \(0.6703200\). The two links therefore define different mean models for the same numerical predictor. The correct variance calculation uses the mean under the stated log link.
Canonical Status Does Not Mean “Best” or “Positive”
Canonical links have a special relationship to the likelihood because they use the distribution’s natural parameter directly. That relationship does not make every noncanonical link unsuitable. A modeler may prefer a noncanonical link for interpretability, multiplicative rating effects, or a convenient mean domain. The model must still use a response distribution and mean structure appropriate to the data, and its fitted means must satisfy the response’s requirements.
For Gamma severity, the log link guarantees positive means for all real predictors. The canonical link also produces positive means only on its valid predictor domain: since \(\mu=-1/\eta\), it requires \(\eta<0\). A canonical Gamma model with an unrestricted linear predictor can therefore produce invalid values unless its structure or fitting procedure keeps predictors in the negative domain. This is a practical reason a log link is often attractive; it is not evidence that the log link is canonical.
Worked Example: Checking the Domain of a Canonical Gamma Predictor
Method. For a Gamma canonical-link model, transform each predictor using \(\mu=-1/\eta\), check its domain, and then apply the assumed variance function. Suppose a hypothetical insurer models average repair severity in thousands of dollars for two vehicle groups. Their canonical predictors are \(\eta_A=-0.5\) and \(\eta_B=-0.25\), and \(\phi=0.20\). The common mistake to avoid is applying the canonical inverse formula without checking whether the predictor is negative.
Step 1: Check the domain and calculate means. Both predictors are negative, so both yield valid positive means. For group A, \(\mu_A=-1/(-0.5)=2\) thousand dollars. For group B, \(\mu_B=-1/(-0.25)=4\) thousand dollars.
Step 2: Calculate variances. For group A, \(\operatorname{Var}(Y_A)=0.20(2^2)=0.80\) thousand-dollars-squared. For group B, \(\operatorname{Var}(Y_B)=0.20(4^2)=3.20\) thousand-dollars-squared.
Step 3: Test a predictor outside the domain. If a third group had \(\eta_C=0.2\), then \(\mu_C=-1/0.2=-5\) thousand dollars. This is not a valid Gamma mean, so that predictor must be rejected or the model structure must be changed.
Check and trap. The two valid canonical predictors yield positive means, but not because positivity defines canonical status: the natural-parameter derivation does. The positive predictor \(0.2\) exposes the canonical model’s domain restriction. A log-link predictor of \(0.2\), by contrast, would imply \(\exp(0.2)>0\); it would be a different, noncanonical Gamma mean model.
A Reliable Way to Classify a Link
When asked whether a link is canonical, do not start by asking whether its fitted means are positive or whether it is commonly used in insurance. Start from the response distribution’s exponential-family representation and identify its natural parameter. Then compare that parameter with the link applied to the mean. The following sequence keeps the two questions separate.
State the assumed distribution and its mean and variance. A link cannot be classified as canonical without reference to a distributional model.
Write the density or probability mass function in exponential-family form and read off \(\theta\), checking the sign of every term involving the mean.
The link is canonical when \(g(\mu)=\theta\) under the stated parameterization. Otherwise, it is noncanonical.
Confirm that the implied means are valid for the response distribution. This is a separate check from canonical status.
A common exam trap is to reason that the log link must be canonical whenever the mean is positive. That conclusion confuses a range property with a distributional property. Another trap is to lose a negative sign while deriving the Gamma natural parameter. The Gamma terms are \(-y/(\phi\mu)-(1/\phi)\log\mu\), not \(-y/(\phi\mu)+(1/\phi)\log\mu\). With the correct signs, the natural parameter is \(-1/\mu\). For Poisson counts, the corresponding natural parameter is \(\log\mu\), so the log link is canonical there.
Check Your Understanding
For each problem, identify the distributional link relationship and check any relevant mean or predictor calculation.
- A hypothetical Poisson count model has mean \(\mu=\exp(0.3)\). What is its canonical predictor, and why? (Answer: \(\eta=\log\mu=0.3\); the Poisson canonical link is the log link.)
- A Gamma model has \(\mu=5\) and dispersion \(\phi=0.10\). Find its canonical natural parameter and variance. (Answer: \(\theta=-1/5=-0.2\), and \(\operatorname{Var}(Y)=0.10(5^2)=2.5\).)
- A Gamma severity model uses a log link with predictor \(0.2\) and \(\phi=0.30\). Find its mean and variance, and state whether the link is canonical. (Answer: \(\mu=\exp(0.2)=1.2214\), \(\operatorname{Var}(Y)=0.30\exp(0.4)=0.4476\), and the log link is noncanonical for Gamma.)
- A Gamma model uses the canonical predictor \(\eta=-0.4\), with \(\phi=0.25\). Find the mean and variance. (Answer: \(\mu=-1/(-0.4)=2.5\), and \(\operatorname{Var}(Y)=0.25(2.5^2)=1.5625\).)
- A proposed canonical Gamma model assigns predictor \(\eta=0.1\) to a hypothetical claim class. Is the implied mean valid? Give the implied value and explain. (Answer: \(\mu=-1/0.1=-10\), which is invalid because a Gamma mean must be positive; the canonical predictor must be negative.)