Choosing A Link For A Ratemaking Target

Choosing a Link Means Choosing a Mean-Scale Relationship

A generalized linear model (GLM) connects predictors to a conditional mean through a link function. For ratemaking, that mean might be the expected claim count per exposure unit or the expected amount of a positive claim severity. Choosing a link is therefore not just a computational decision: it specifies how changes in rating variables are allowed to change the modeled mean.

A useful choice must meet two tests. First, its predictions must remain in the valid range for the target over the full range of predictors where the model will be used. Second, the way it translates predictor changes into mean changes should be plausible and useful for rating. A link that satisfies the first test is not automatically the best choice for the second.

Definition: Let $Y$ be a response and let $\mu=E[Y\mid\mathbf{x}]$ be its conditional mean at predictor values $\mathbf{x}$. A link function $g$ specifies the mean model $g(\mu)=\eta$, where $\eta=\beta_0+\sum_j\beta_jx_j$ is the linear predictor. The inverse link gives the modeled mean, $\mu=g^{-1}(\eta)$. The link determines the relationship between predictors and the mean; it does not, by itself, specify the response distribution or its variance.

For the three links considered here, the mean-scale forms are:

\begin{array}{lll} \text{Log:} & \log(\mu)=\eta, & \mu=\exp(\eta),\\ \text{Identity:} & \mu=\eta, & \mu=\eta,\\ \text{Reciprocal:} & 1/\mu=\eta, & \mu=1/\eta. \end{array}

The log link allows any real value of $\eta$, and its inverse always produces a strictly positive mean. The identity link allows $\eta$ to be any real number mathematically, but the resulting mean must still be in the response’s valid range. For frequency, a mean cannot be negative; for positive severity, it must be strictly positive. The reciprocal link also requires $\eta>0$: its inverse is not a valid positive mean when $\eta\leq0$.

These are constraints on the conditional mean, not on each observed response. A frequency count can be zero even when its modeled mean is positive. Similarly, a positive-severity model is intended for observations whose modeled target is positive, not for the frequency of claims.

What the Coefficients Say

Suppose one numeric predictor $x$ enters a model as $\eta=\beta_0+\beta_1x$, with other predictors held fixed. The inverse link determines what a one-unit increase in $x$ means for the expected response.

Link	Mean	Effect of increasing $x$ by one unit	Validity check
Log	$\mu=\exp(\beta_0+\beta_1x)$	The mean is multiplied by $\exp(\beta_1)$	Predicted mean is positive for every finite linear predictor
Identity	$\mu=\beta_0+\beta_1x$	The mean changes by $\beta_1$ response units	Check that the mean stays in range over the intended domain
Reciprocal	$\mu=1/(\beta_0+\beta_1x)$	The predictor changes by $\beta_1$; the mean change is not constant	Check that the linear predictor stays strictly positive

The log-link interpretation follows directly from taking the ratio of predictions. At $x$, the mean is $\exp(\beta_0+\beta_1x)$; at $x+1$, it is $\exp(\beta_0+\beta_1x+\beta_1)$. Thus:

\frac{\mu(x+1)}{\mu(x)} =\frac{\exp(\beta_0+\beta_1x+\beta_1)} {\exp(\beta_0+\beta_1x)} =\exp(\beta_1).

For the reciprocal link, the direction of the mean change is opposite the direction of the predictor when the predictor stays positive: increasing $1/\mu$ reduces $\mu$. More precisely, if $\eta=\beta_0+\beta_1x>0$, then the slope of the mean is:

\frac{d\mu}{dx} =\frac{d}{dx}\left(\frac{1}{\beta_0+\beta_1x}\right) =-\frac{\beta_1}{(\beta_0+\beta_1x)^2}.

This slope depends on $x$, unlike the constant slope under an identity link. If the reciprocal predictor approaches zero from above, the predicted mean grows without bound. That behavior can make the reciprocal link unsuitable even on a domain where it is technically valid.

Worked Example: Comparing Frequency Mean Interpretations

A hypothetical insurer models annual claim count $N$ for a fixed exposure unit. Let $z$ be a rating variable taking values 0, 1, or 2. For this illustration, assume $N\mid z$ follows a Poisson distribution with conditional mean $\mu(z)$; the Poisson variance is therefore also $\mu(z)$. Compare a log-link model with $\eta=-2.5+0.30z$ and an identity-link model with $\mu=0.08+0.03z$. Which interpretation does each model impose, and are the identity-link predictions valid on this domain?

Log-link calculations. At $z=0$, $\eta=-2.5$, so $\mu(0)=\exp(-2.5)=0.082085$. At $z=1$, $\eta=-2.2$, so $\mu(1)=\exp(-2.2)=0.110803$. At $z=2$, $\eta=-1.9$, so $\mu(2)=\exp(-1.9)=0.149569$. These are all positive means.

Each one-unit increase in $z$ multiplies the mean by $\exp(0.30)=1.3499$, or about a 35.0% increase. This is a proportional interpretation: the same relativity applies from $z=0$ to 1 and from $z=1$ to 2.

Identity-link calculations. The three means are $0.08+0.03(0)=0.08$, $0.08+0.03(1)=0.11$, and $0.08+0.03(2)=0.14$. The mean increases by 0.03 expected claims for each unit increase in $z$. Since the model is linear and increasing, its minimum on $0\leq z\leq2$ is 0.08, so it predicts no negative means on this domain.

Conclusion and method. Both candidates give valid means on the stated domain, but they impose different effects: a constant relativity for the log link and a constant additive change for the identity link. The log link is a natural candidate when proportional changes are intended and strictly positive frequency means are required. The identity link is not ruled out here by validity, but its predictions should be checked over the model’s full intended domain. The common mistake is to reject an identity link merely because it can produce negative values somewhere; the relevant check is whether it does so in the domain where the model is to be used.

Check. The Poisson mean must be nonnegative, and all six calculated means are positive. The log-link ratio is the same at both steps: $0.110803/0.082085\approx1.3499$ and $0.149569/0.110803\approx1.3499$.

Worked Example: A Reciprocal Link for Positive Severity

Worked Example: Checking a Reciprocal Severity Model

A hypothetical portfolio models positive claim severity in dollars using predictor $x$, which is known to lie between 0 and 10. Assume the positive-severity response has a Gamma distribution with conditional mean $\mu(x)$ and variance $\phi\mu(x)^2$, where $\phi=0.25$. This distribution assumption is separate from the link choice. Consider the reciprocal-link predictor $1/\mu=0.0025+0.00005x$, where its coefficients have units of inverse dollars. Calculate the means at $x=0,5,10$, and describe the effect of increasing $x$.

At $x=0$, the reciprocal predictor is $0.0025+0.00005(0)=0.0025$ per dollar. Therefore, $\mu(0)=1/0.0025=\$400.00$.

At $x=5$, the predictor is $0.0025+0.00005(5)=0.00275$ per dollar. Therefore, $\mu(5)=1/0.00275=\$363.636\ldots$, or $\$363.64$ to the nearest cent.

At $x=10$, the predictor is $0.0025+0.00005(10)=0.0030$ per dollar. Therefore, $\mu(10)=1/0.0030=\$333.333\ldots$, or $\$333.33$ to the nearest cent.

The reciprocal predictor increases by $0.00005$ per dollar of severity for each one-unit increase in $x$, while the mean decreases. The mean does not decrease by a constant dollar amount: the changes are $\$363.64-\$400.00=-\$36.36$ from 0 to 5 and $\$333.33-\$363.64=-\$30.31$ from 5 to 10, using displayed rounded values. This model is valid throughout $0\leq x\leq10$, because its predictor is at least $0.0025>0$. The method is to check positivity of the reciprocal predictor before taking its inverse; the common mistake is to treat the reciprocal coefficient as a constant dollar change in the mean.

Check. The unrounded means are $400$, $1/0.00275=363.636\ldots$, and $333.333\ldots$ dollars. Rounding only the final means to cents gives $\$400.00$, $\$363.64$, and $\$333.33$, all positive as required.

Validity Must Hold Across the Intended Domain

A link can produce sensible predictions at a reference level and still fail elsewhere. For an identity model, check the minimum and maximum predicted means over the domain. For a reciprocal model, check that its linear predictor remains strictly positive everywhere; checking only the reciprocal prediction at one rating level is not enough. A log link avoids nonpositive mean predictions, although positive predictions alone do not establish that its pattern is plausible.

For a linear predictor in one variable, $\eta=\beta_0+\beta_1x$, its minimum and maximum on a closed interval occur at an endpoint. With several predictors, the relevant check is over the combinations of predictor values the model may actually encounter. If a model is intended only for a restricted range, that range should be stated and enforced in its use; extrapolating beyond it can invalidate an otherwise acceptable mean model.

Worked Example: Finding an Invalid Reciprocal Prediction

A hypothetical positive-severity model uses $x$ between 0 and 14 and the reciprocal-link predictor $1/\mu=0.004-0.00035x$, with predictor units of inverse dollars. Determine whether it gives valid means throughout the domain. For comparison, consider the candidate log-link model $\log(\mu)=\log(250)+0.04x$, where $\mu$ is in dollars.

Reciprocal-link check. At $x=0$, the predictor is $0.004$, giving $\mu(0)=1/0.004=\$250.00$. The predictor reaches zero when $0.004-0.00035x=0$, so $x=0.004/0.00035=80/7\approx11.4286$. This value is inside the stated domain. At $x=10$, the predictor is $0.004-0.0035=0.0005$, giving $\mu(10)=\$2{,}000.00$. At $x=12$, the predictor is $0.004-0.0042=-0.0002$, which cannot be inverted to obtain a valid positive mean under this link. At $x=14$, it is $0.004-0.0049=-0.0009$, also invalid. Thus the reciprocal model is not valid throughout $0\leq x\leq14$; it becomes undefined at $x=80/7$ and has a negative predictor beyond that point.

Log-link comparison. Its mean at $x=14$ is $\exp(\log(250)+0.04(14))=250\exp(0.56)\approx\$437.67$, which is positive. The log link avoids the reciprocal model’s zero crossing, but whether its relatively modest increase is plausible must be judged against the intended application. The common mistake is to inspect only $x=0$ and conclude that the reciprocal model is valid because its reference mean is positive. The method is to check the predictor across the entire specified range.

Check. The reciprocal predictor is positive at $x=10$, zero at approximately 11.4286, and negative at $x=12$ and 14; the domain therefore includes invalid values. The log-link prediction at 14 is approximately $\$437.67$.

A Practical Selection Sequence

Link choice should be made in context, with the response’s valid range and the intended use of the model made explicit. A candidate link is not selected solely because it is familiar or because it fits the observed data. A clear sequence helps separate mathematical validity from the actuarial meaning of the modeled effect.

State the target.
Identify whether the modeled conditional mean is a frequency, a positive severity, or another quantity, and state its valid range.

Translate the link to the mean scale.
Write the inverse-link equation and identify the restrictions it places on the linear predictor and the mean.

Interpret predictor changes.
Decide whether additive changes, proportional changes, or the nonlinear pattern implied by a reciprocal link is credible for the rating variable.

Check the whole intended range.
Evaluate boundary values and any combinations that could make an identity mean nonpositive or a reciprocal predictor nonpositive.

Keep the distribution choice distinct.
Specify an appropriate response distribution and variance structure separately. A link does not determine either one.

For frequency, a log link is often attractive because it ensures a positive mean and expresses rating effects as relativities. An identity link can be meaningful when additive changes are appropriate, provided its mean remains nonnegative over the relevant range. For positive severity, all three links may be candidates, but the identity mean must stay strictly positive and the reciprocal predictor must stay strictly positive. The choice depends on the target, predictor range, and plausible effect—not on a universal ranking of links.

Common Exam Traps

Checking only a base risk. A valid reference prediction does not ensure the model remains valid for all rating-variable values. Check the full intended domain.
Calling a log coefficient a dollar change. Under a log link, $\beta_1$ changes the log mean; the one-unit mean relativity is $\exp(\beta_1)$.
Treating the reciprocal coefficient as a mean-scale slope. It changes $1/\mu$, not $\mu$ by a fixed amount. The mean-scale slope varies with the predictor.
Confusing the link with the response distribution. A log link does not by itself imply a Poisson or Gamma response, and a distribution choice does not settle whether the link’s mean predictions are valid.
Assuming any positive prediction is plausible. A reciprocal predictor just above zero can generate an extremely large mean. Check whether the resulting pattern makes sense as well as whether it is mathematically defined.

Key takeaway: Choose a link by checking both its mean constraints and its implied predictor effect. Log links give positive means and multiplicative effects; identity links give additive effects but require range checks; reciprocal links require a strictly positive predictor and imply a nonlinear mean response.

Check Your Understanding

For each problem, check the link’s implications and the stated predictor range before interpreting the result.

A hypothetical claim-count model has $\log(\mu)=-1.6+0.20x$. Calculate the mean at $x=3$ and the one-unit relativity. (Answer: $\mu=e^{-1}\approx0.3679$; relativity $e^{0.20}\approx1.2214$.)
A positive-severity model has identity-link mean $\mu=180+12x$, with $0\leq x\leq8$. Find the endpoint means and state whether they are positive throughout the domain. (Answer: $\mu(0)=\$180$, $\mu(8)=\$276$; yes, the increasing linear mean stays positive throughout.)
A reciprocal severity model has $1/\mu=0.005+0.0001x$, with $0\leq x\leq20$, and the predictor is in inverse dollars. Find the endpoint means and describe the direction of change. (Answer: $\mu(0)=\$200.00$, $\mu(20)=1/0.007=\$142.86$; the mean decreases and remains positive throughout.)
A reciprocal model has $1/\mu=0.003-0.0002x$ over $0\leq x\leq20$. Find where its predictor reaches zero and decide whether the model is valid throughout the domain. (Answer: It reaches zero at $x=0.003/0.0002=15$; no, it is undefined at 15 and invalid for $x>15$.)
For a claim-count target on $0\leq x\leq3$, compare $\log(\mu)=-2+0.5x$ with identity mean $\mu=0.1+0.15x$. Calculate both means at $x=2$, and state the one-unit effect under each model. (Answer: Log-link mean $e^{-1}\approx0.3679$, with one-unit relativity $e^{0.5}\approx1.6487$; identity mean $0.40$, with an additive increase of $0.15$ claim-count units per unit of $x$.)

Link	Mean	Effect of increasing \(x\) by one unit	Validity check
Log	\(\mu=\exp(\beta_0+\beta_1x)\)	The mean is multiplied by \(\exp(\beta_1)\)	Predicted mean is positive for every finite linear predictor
Identity	\(\mu=\beta_0+\beta_1x\)	The mean changes by \(\beta_1\) response units	Check that the mean stays in range over the intended domain
Reciprocal	\(\mu=1/(\beta_0+\beta_1x)\)	The predictor changes by \(\beta_1\); the mean change is not constant	Check that the linear predictor stays strictly positive

What You'll Learn

Choosing a Link Means Choosing a Mean-Scale Relationship

What the Coefficients Say

Worked Example: Comparing Frequency Mean Interpretations

Worked Example: Comparing Frequency Mean Interpretations

Worked Example: A Reciprocal Link for Positive Severity

Worked Example: Checking a Reciprocal Severity Model

Validity Must Hold Across the Intended Domain

Worked Example: Finding an Invalid Reciprocal Prediction

A Practical Selection Sequence

Common Exam Traps

Check Your Understanding

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