Link-Scale and Response-Scale Coefficients

One Coefficient, Two Scales

A GLM coefficient describes an additive change on the linear predictor scale. The quantity a business decision-maker usually wants, however, is a change in the expected response: for example, dollars, event counts, or a probability. The link function connects those scales, so the coefficient’s response-scale meaning depends on the link and, for nonlinear links, on the starting predictor value.

This distinction matters when explaining fitted results. A coefficient of 0.4 is not automatically a change of 0.4 dollars, events, or probability points. It is a change of 0.4 in the linear predictor for a one-unit predictor increase, under the stated model and with other predictors fixed. The inverse link then determines the corresponding fitted-response change. This interpretation is aligned with the SOA Exam PA (Predictive Analytics) syllabus.

Definition: In a generalized linear model, the conditional mean of response $Y_i$, given its predictors, is $\mu_i=E[Y_i\mid \mathbf{x}_i]$. The link function $g$ relates that mean to the linear predictor $\eta_i$: $g(\mu_i)=\eta_i$, where $\eta_i=\beta_0+\beta_1x_{i1}+\cdots+\beta_px_{ip}$. The inverse link $g^{-1}$ gives the mean on the response scale: $\mu_i=g^{-1}(\eta_i)$.

For a predictor $x_j$ that enters the linear predictor without an interaction, increasing $x_j$ by $h$ while holding all other predictors fixed changes the linear predictor by $\beta_jh$. This result is additive for any link. The response-scale change is instead $g^{-1}(\eta+\beta_jh)-g^{-1}(\eta)$, which can depend on the original value $\eta$.

How the Three Common Links Translate Changes

The identity, log, and logit links illustrate three different relationships between a coefficient and the expected response. The response distribution and link must be specified as part of the model; a link is not interchangeable with the response distribution.

Link	Mean and linear predictor	Response-scale effect of increasing $x_j$ by $h$
Identity: $g(\mu)=\mu$	$\mu=\eta$	$\beta_jh$, an additive mean change
Log: $g(\mu)=\log(\mu)$	$\mu=\exp(\eta)$	$\mu\{\exp(\beta_jh)-1\}$, a change that depends on the starting mean
Logit: $g(p)=\log\{p/(1-p)\}$	$p=1/(1+\exp(-\eta))$	$p_{\text{new}}-p_{\text{old}}$, a probability change that depends on the starting probability

The identity-link formula is a direct consequence of $\mu=\eta$: changing $\eta$ by $\beta_jh$ changes $\mu$ by the same amount. For a log link, $\eta=\log(\mu)$, so applying the inverse link gives $\mu=\exp(\eta)$. This gives a useful justification for the response-scale comparison:

\frac{\mu_{\text{new}}}{\mu_{\text{old}}} = \frac{\exp(\eta+\beta_jh)}{\exp(\eta)} = \exp(\beta_jh), \qquad \mu_{\text{new}}-\mu_{\text{old}} = \mu_{\text{old}}\{\exp(\beta_jh)-1\}.

Thus, under a log link, the coefficient determines a multiplicative comparison of fitted means for a specified predictor change, while the difference in mean units depends on the baseline mean. For the logit link, applying the inverse link yields $p=\exp(\eta)/(1+\exp(\eta))$. The same additive change in $\eta$ does not produce the same change in $p$ at every starting probability. These formulas describe fitted means; they do not assert that each individual response changes deterministically.

Worked Examples

Worked Example: Identity Link and Mean Claim Cost

A hypothetical insurer models claim cost $Y$, measured in thousands of dollars, using a normal GLM with identity link. Conditional on predictors, $Y$ has distribution $N(\mu,16)$, where the second parameter is the variance, and the fitted mean is $\mu=12+1.5x$. Here $x$ is a risk score. Compare fitted mean costs at $x=4$ and $x=6$. State the coefficient’s link-scale and response-scale interpretations.

Method. The identity link means $g(\mu)=\mu$, so $\eta=\mu$. The slope coefficient is $1.5$ thousand dollars per one-unit increase in risk score. The normal distribution’s variance is $16$ thousand-dollars-squared; it does not change the calculation of the fitted mean.

At $x=4$, the linear predictor and fitted mean are

\eta_4=12+1.5(4)=12+6=18, \qquad \mu_4=18.

At $x=6$, they are

\eta_6=12+1.5(6)=12+9=21, \qquad \mu_6=21.

The predictor increases by $6-4=2$ units, so the linear predictor increases by $1.5(2)=3$. Because the link is identity, the fitted mean also increases by 3 thousand dollars, or $3,000. The fitted costs are $18,000 and $21,000, respectively. The direct check is $21-18=3$ thousand dollars.

Common mistake avoided: Do not report 1.5 as the change for a two-unit increase. It is the change per one-unit increase; the two-unit change is $1.5(2)=3$.

Worked Example: Log Link and Poisson Event Counts

For a hypothetical policy group, the annual event count $Y$, conditional on its predictor $x$, follows a Poisson distribution with mean $\mu$. A log-link model has $\log(\mu)=\log(4)+0.30x$. Calculate the fitted mean at $x=0$ and $x=2$, then give both the mean difference and the multiplicative comparison. Hold any other predictors fixed.

Method. Under a Poisson model, the conditional mean is the Poisson parameter $\mu$. With the log link, exponentiate the linear predictor to obtain the fitted mean.

At $x=0$,

\eta_0=\log(4)+0.30(0)=\log(4), \qquad \mu_0=\exp\{\log(4)\}=4.

At $x=2$,

\eta_2=\log(4)+0.30(2)=\log(4)+0.60,

\mu_2=\exp\{\log(4)+0.60\} =4\exp(0.60) \approx 4(1.8221) \approx 7.288.

The multiplicative comparison is $\mu_2/\mu_0=\exp(0.60)\approx1.8221$. The fitted mean therefore increases by about $7.288-4=3.288$ events per policy group per year. These are expected counts, so a non-integer fitted mean is valid. As a check, applying the factor to the baseline gives $4(1.8221)\approx7.288$.

Common mistake avoided: The coefficient 0.30 is not an increase of 0.30 expected events per one-unit increase. It changes the log mean by 0.30. For the two-unit change here, the fitted mean is multiplied by $\exp(0.60)$; its absolute increase depends on the starting mean.

Worked Example: Logit Link and Claim Probability

A hypothetical model for a binary claim indicator $Y$ uses a Bernoulli response with conditional probability $p=P(Y=1\mid x)$ and a logit link. The linear predictor is $\eta=-\log(4)+(\log 2)x$, where $x$ is either 0 or 1. Calculate the fitted probability at each value and the probability-point change.

Method. For a Bernoulli response and logit link, $\log\{p/(1-p)\}=\eta$, so $p=1/(1+\exp(-\eta))$. The coefficient is $\log 2$, approximately 0.6931, on the linear-predictor scale.

At $x=0$, $\eta_0=-\log(4)$. Thus $\exp(-\eta_0)=4$, and

p_0=\frac{1}{1+\exp(-\eta_0)} =\frac{1}{1+4} =\frac{1}{5} =0.2000.

At $x=1$,

\eta_1=-\log(4)+\log(2) =\log(2/4) =-\log(2).

Therefore $\exp(-\eta_1)=2$, and

p_1=\frac{1}{1+\exp(-\eta_1)} =\frac{1}{1+2} =\frac{1}{3} \approx0.3333.

The fitted probability increases by $0.3333-0.2000=0.1333$, or about 13.33 percentage points. The coefficient itself is approximately 0.6931 on the logit scale, not a 0.6931 increase in probability. The direct check is that both probabilities lie between 0 and 1 and their difference is positive.

Common mistake avoided: Do not read a logit coefficient as a probability-point change. The change from 0.2000 to 0.3333 is determined by applying the inverse link to both linear predictors.

Reading and Reporting a Coefficient Correctly

A reliable interpretation separates three questions: what predictor change is being considered, what happens to the linear predictor, and what happens to the fitted response after applying the inverse link. State the units and hold other predictors fixed. If the predictor changes by more than one unit, multiply its coefficient by that change before translating through the link.

Identify the link and response.
Determine whether the model uses identity, log, logit, or another link, and identify the response quantity being modeled.

Calculate the link-scale change.
For a change of $h$ in one additive predictor, calculate $\beta_jh$, holding other predictors fixed.

Translate to the response scale.
Use the inverse link for the relevant starting and ending linear predictors; do not label the coefficient itself as a mean or probability change.

Report context and units.
Give the fitted values or their difference with appropriate units, and distinguish an expected response from an individual outcome.

For a log-link model, a tempting shortcut is to multiply a baseline mean by $1+\beta_j$. That is generally not exact: the multiplier for a one-unit change is $\exp(\beta_j)$. For example, if $\beta_j=0.30$, the exact proportional change is $\exp(0.30)-1\approx0.3499$, or about 35.0%, rather than 30.0%. The approximation $\exp(\beta_j)-1\approx\beta_j$ can be useful for a small coefficient, but it is an approximation, not the exact interpretation.

A related logit-scale trap is to multiply a probability by a coefficient or add the coefficient directly to it. A logit coefficient changes $\log\{p/(1-p)\}$, not $p$. To obtain a probability change, calculate the fitted probability at each relevant predictor value using the inverse logit, as in the worked example.

Key takeaway: A coefficient is additive on the linear-predictor scale. With an identity link, that additive change carries directly to the mean. With log and logit links, use the inverse link: the resulting response-scale difference depends on the starting fitted mean or probability.

Check Your Understanding

For each problem, identify the link-scale change before reporting the fitted response-scale result.

A normal GLM with identity link has fitted mean $\mu=8+2x$, measured in thousands of dollars. Find the fitted means at $x=1$ and $x=3$, and their difference. (Answer: 10 and 14 thousand dollars; increase of 4 thousand dollars.)
A Poisson GLM with log link has $\log(\mu)=\log(5)+0.20x$. Compare fitted means at $x=0$ and $x=1$, rounding means to three decimals. (Answer: 5.000 and 6.107; increase of 1.107 events; multiplier $\exp(0.20)\approx1.2214$.)
A Bernoulli GLM with logit link has $\eta=-\log(3)+(\log 3)x$, with $x=0$ or 1. Find both fitted probabilities and the probability-point change. (Answer: 0.2500 and 0.5000; increase of 0.2500, or 25.00 percentage points.)
A log-link model has coefficient $0.10$ for a predictor. For a two-unit increase, find the exact multiplier and exact proportional increase, rounding the latter to one decimal percentage point. (Answer: multiplier $\exp(0.20)\approx1.2214$; proportional increase approximately 22.1%, not 20.0%.)
A logit model gives a fitted probability of 0.40 at a starting predictor value. The predictor increases by one unit and its coefficient is $\log(2)$. Use the odds at the starting value and the inverse logit to find the new probability and the probability-point change. (Answer: starting odds $2/3$, new odds $4/3$, new probability $4/7\approx0.5714$; increase approximately 0.1714, or 17.14 percentage points.)

Link	Mean and linear predictor	Response-scale effect of increasing \(x_j\) by \(h\)
Identity: \(g(\mu)=\mu\)	\(\mu=\eta\)	\(\beta_jh\), an additive mean change
Log: \(g(\mu)=\log(\mu)\)	\(\mu=\exp(\eta)\)	\(\mu\{\exp(\beta_jh)-1\}\), a change that depends on the starting mean
Logit: \(g(p)=\log\{p/(1-p)\}\)	\(p=1/(1+\exp(-\eta))\)	\(p_{\text{new}}-p_{\text{old}}\), a probability change that depends on the starting probability

What You'll Learn