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CAS Exam 8 · Link Functions and Offsets · Tutorial 78 of 989

The Log Link And Positive Means

Learn to convert a log-link predictor to a positive fitted mean and quantify the effect of an additive predictor change.

Intermediate 9 min read

What You'll Learn

  • Define the log link and its inverse for a positive conditional mean
  • Derive why the fitted mean equals the exponential of the linear predictor
  • Explain why finite predictors produce strictly positive means
  • Calculate fitted means from hypothetical claim-count predictors
  • Convert an additive predictor change into a ratio and percentage change in the mean
  • Identify common errors in reversing the log link

From a Linear Predictor to a Positive Mean

In a generalized linear model, the linear predictor is built from model terms, while the response of interest is often a conditional mean. A link function connects these two quantities. For ratemaking targets such as expected claim counts, the fitted mean must not be negative. The log link is useful because its inverse maps every finite real-valued predictor to a strictly positive mean.

This tutorial focuses on that mapping and its consequences. It does not require a particular response distribution: the log-link relationship concerns the mean, although a GLM also specifies a response distribution and its assumptions. The examples use hypothetical claim-count models with a Poisson response so that the distribution is explicit; the arithmetic of the link itself does not depend on the Poisson choice.

Definition: Let \(\mu\) be a conditional mean and \(\eta\) the linear predictor. The log link is \(g(\mu)=\ln(\mu)\), defined for \(\mu>0\). Its inverse link is \(g^{-1}(\eta)=\exp(\eta)\), so a model using the log link has fitted mean \(\mu=\exp(\eta)\). Here the mean is expressed in a stated, consistent measurement unit.

The predictor \(\eta\) may be any finite real number, including zero or a negative value. The fitted mean is not the predictor itself: the inverse link must be applied. If \(\eta=0\), then \(\mu=\exp(0)=1\); if \(\eta<0\), then \(0<\mu<1\). A negative predictor therefore does not imply a negative expected claim count.

Why the Exponential Guarantees Positivity

The log link is defined by \(\eta=\ln(\mu)\). Because the natural logarithm and exponential function are inverses, exponentiating both sides recovers \(\mu\). This is also the reason the inverse link has the required positivity property: the exponential of every finite real number is strictly positive.

$$ \eta=\ln(\mu) \quad\Longleftrightarrow\quad \exp(\eta)=\exp(\ln(\mu))=\mu \quad\Longleftrightarrow\quad \mu=\exp(\eta)>0 $$

This argument has an important boundary condition. The log link requires a strictly positive mean, because \(\ln(0)\) is undefined. A finite predictor cannot produce a fitted mean of exactly zero; as \(\eta\) decreases without bound, \(\exp(\eta)\) approaches zero but remains positive. Thus the link ensures positivity, not nonnegativity including zero.

In a model, \(\eta\) is usually a sum of terms. For example, a hypothetical model might have an intercept and one or more predictor contributions. Once those terms produce \(\eta\), the fitted mean is obtained by exponentiating the total—not by exponentiating only one term or by adding the terms to a mean on its original scale.

Worked Example: Converting a Predictor to an Expected Claim Count

A hypothetical insurer models the number of claims per policy-year for a specified risk group. Assume the claim count \(Y\) has a Poisson distribution with parameter \(\mu\), so \(Y\sim\operatorname{Poisson}(\mu)\) and \(E[Y]=\mu\). For this group, the fitted linear predictor is \(\eta=-0.2+0.9\). Find the fitted expected claim count per policy-year. No interest or mortality assumptions enter this calculation.

Method: apply the inverse log link to the complete predictor. First add the predictor terms:

$$ \eta=-0.2+0.9=0.7 $$

The model uses the log link, so the fitted mean is:

$$ \mu=\exp(\eta)=\exp(0.7)=2.0137527\ldots $$

Rounded to four decimal places, the fitted mean is 2.0138 claims per policy-year. This is a mean and need not be an integer, even though an observed claim count is an integer. Since the Poisson parameter is \(\mu\), the distribution in this example is \(\operatorname{Poisson}(2.0138)\) after rounding for reporting.

Check: Taking the natural logarithm of the unrounded fitted mean gives \(\ln(2.0137527\ldots)=0.7\), recovering the specified predictor. The common mistake avoided is reporting \(0.7\) itself as the expected count.

Additive Predictor Changes and Multiplicative Mean Changes

The exponential function converts addition in the predictor into multiplication on the mean scale. Suppose two fitted predictors differ by an amount \(\Delta\). Write the second predictor as \(\eta_2=\eta_1+\Delta\). Their fitted means are \(\mu_1=\exp(\eta_1)\) and \(\mu_2=\exp(\eta_2)\). Dividing the second mean by the first gives:

$$ \frac{\mu_2}{\mu_1} = \frac{\exp(\eta_1+\Delta)}{\exp(\eta_1)} = \frac{\exp(\eta_1)\exp(\Delta)}{\exp(\eta_1)} = \exp(\Delta) $$

This derivation shows that the multiplicative change in the fitted mean depends only on the predictor difference \(\Delta\), not on the starting predictor \(\eta_1\). The corresponding relative change is \(\exp(\Delta)-1\). If \(\Delta>0\), the mean increases; if \(\Delta<0\), it decreases; and if \(\Delta=0\), the fitted means are equal. The result describes a difference between predictors and does not, by itself, require assigning a particular interpretation to any individual model term.

Worked Example: Measuring the Effect of a Predictor Increase

For two hypothetical groups, suppose claim counts per policy-year follow Poisson distributions with respective means \(\mu_A\) and \(\mu_B\). Their fitted predictors are \(\eta_A=0.4\) and \(\eta_B=0.4+0.25\). Find both fitted means, their ratio, and the percentage change from group A to group B. No interest or mortality assumptions are used.

Method: exponentiate each predictor, then use the predictor difference to check the ratio. The predictors are:

$$ \eta_A=0.4, \qquad \eta_B=0.4+0.25=0.65 $$

Apply the inverse link to each:

$$ \mu_A=\exp(0.4)=1.4918247\ldots, \qquad \mu_B=\exp(0.65)=1.9155408\ldots $$

The ratio calculated from the means is:

$$ \frac{\mu_B}{\mu_A} = \frac{1.9155408\ldots}{1.4918247\ldots} = 1.2840254\ldots $$

Independently, the predictor difference is \(0.65-0.4=0.25\), so the ratio must be \(\exp(0.25)=1.2840254\ldots\), which agrees. The relative increase is:

$$ 1.2840254\ldots-1=0.2840254\ldots $$

Rounded to four decimal places for the means and ratio, \(\mu_A=1.4918\), \(\mu_B=1.9155\), and \(\mu_B/\mu_A=1.2840\). The percentage increase, rounded to two decimal places, is 28.40%. The common mistake avoided is treating a predictor increase of \(0.25\) as a \(0.25\)-claim increase; the change is additive on the predictor scale but multiplicative on the mean scale.

Negative Predictors, Scale, and Model Boundaries

A negative predictor is entirely compatible with a positive fitted mean. It often corresponds to a mean below one when the modeled response is a count per stated unit, but it never makes the mean negative. The predictor and mean have different roles and may also be expressed on different scales: \(\eta\) is a logarithmic quantity, while \(\mu\) is in the response’s measurement units.

The unit used for the response must be consistent. If a count mean is stated per policy-year, the model output is interpreted per policy-year. Changing the response’s unit changes the numerical mean and requires a corresponding change in the model’s predictor representation. The log-link equation does not make units irrelevant; it specifies how the numerical mean in the chosen unit relates to \(\eta\).

The positivity guarantee is a property of the inverse link, not a guarantee that the model is otherwise appropriate. A log-link GLM still requires a suitable response distribution, a defensible predictor structure, and appropriate estimation and validation. In a Poisson model, for example, the conditional distribution is Poisson with mean \(\mu\) and variance \(\mu\). If that distributional assumption is unsuitable for the data, the fact that \(\exp(\eta)\) is positive does not repair the mismatch.

Worked Example: A Negative Predictor Still Gives a Positive Mean

A hypothetical portfolio model uses a Poisson distribution for claim counts per policy-year, \(Y\sim\operatorname{Poisson}(\mu)\), with the predictor for a risk group equal to \(\eta=-1.5\). A proposed adjustment lowers the predictor by a further \(0.7\). Calculate the original and adjusted means, the ratio of adjusted to original mean, and the percentage change. No interest or mortality assumptions enter.

Method: exponentiate both finite predictors and compare their means. The adjusted predictor is:

$$ \eta_{\mathrm{adjusted}}=-1.5-0.7=-2.2 $$

The original and adjusted fitted means are:

$$ \mu_{\mathrm{original}}=\exp(-1.5)=0.2231302\ldots, \qquad \mu_{\mathrm{adjusted}}=\exp(-2.2)=0.1108032\ldots $$

Both means are positive, despite both predictors being negative. The ratio can be checked directly from the predictor difference:

$$ \frac{\mu_{\mathrm{adjusted}}}{\mu_{\mathrm{original}}} = \exp(-2.2-(-1.5)) = \exp(-0.7) = 0.4965853\ldots $$

Thus the adjusted mean is \(0.4965853\ldots\) times the original mean. Its percentage change is \(0.4965853\ldots-1=-0.5034147\ldots\), or a 50.34% decrease, rounded to two decimal places. The common mistake avoided is interpreting \(-1.5\) as a negative expected count; it is the predictor, and the fitted count mean is its exponential.

Common Exam Trap: Using the Link Instead of Its Inverse

The most direct error is to compute the log of the predictor or to use the predictor as the fitted mean. With the log link, the model statement is \(\ln(\mu)=\eta\), not \(\mu=\ln(\eta)\) and not \(\mu=\eta\). The correct operation is to exponentiate \(\eta\). A quick reasonableness check is that a finite predictor must produce a positive fitted mean.

1
Find the complete predictor.
Add the specified predictor contributions before applying the link transformation.
2
Apply the inverse link.
For the log link, calculate \(\mu=\exp(\eta)\), retaining the response’s stated measurement unit.
3
For a predictor difference, use the ratio result.
If \(\eta_2-\eta_1=\Delta\), then \(\mu_2/\mu_1=\exp(\Delta)\); convert to a percentage change by subtracting one.
4
Check scale and sign.
A finite predictor must yield a positive mean. Do not report the predictor itself as a mean or claim the fitted mean is exactly zero.

This distinction matters in ratemaking because fitted means are used as estimates of expected response on the original scale, while the model is constructed on the predictor scale. The log link supports positive fitted means and gives a precise relationship between predictor differences and mean ratios. These are core properties aligned with the CAS Exam 8 (Advanced Ratemaking) syllabus; they do not replace the need to assess whether the chosen response family and model structure suit the intended application.

Check Your Understanding

For each problem, identify the relevant transformation before calculating. Answers are rounded as stated.

  1. A hypothetical claim-count model has \(Y\sim\operatorname{Poisson}(\mu)\) and predictor \(\eta=0.2\) per policy-year. What is the fitted mean, rounded to four decimal places? (Answer: \(1.2214\) claims per policy-year.)
  2. Two hypothetical groups have Poisson claim-count means linked to predictors \(\eta_1=-0.3\) and \(\eta_2=-0.1\). Find \(\mu_1\), \(\mu_2\), and the ratio \(\mu_2/\mu_1\), to four decimal places. (Answer: \(\mu_1=0.7408\), \(\mu_2=0.9048\), and \(\mu_2/\mu_1=1.2214\).)
  3. A Poisson claim-count model’s predictor falls by \(0.4\) between two risk profiles. What is the ratio of the second fitted mean to the first, and what is the percentage change, rounded to two decimal places? (Answer: ratio \(=\exp(-0.4)=0.6703\); a \(32.97\%\) decrease.)
  4. A finite log-link predictor is \(-3\). A candidate says the fitted mean is \(-3\), while another says it is exactly zero because the predictor is negative. What is the fitted mean to four decimal places, and are either statements correct? (Answer: \(\exp(-3)=0.0498\); neither statement is correct.)
  5. For a hypothetical Poisson count model, group A has predictor \(0.1\), and group B has predictor \(0.1+\Delta\). The fitted mean for B is \(1.5\) times the fitted mean for A. Find \(\Delta\), rounded to four decimal places. (Answer: \(\Delta=\ln(1.5)=0.4055\).)
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