Translating Linear Predictors Through GLM Links

From a Linear Predictor to a Fitted Mean

A generalized linear model (GLM) connects predictors to an outcome’s conditional mean through a link function. This means a coefficient first changes a quantity on the link scale; the corresponding change in the fitted mean depends on the inverse link. For the logit, log, and identity links, that distinction is essential: the same numerical coefficient does not have the same response-scale meaning.

For an observation with predictor vector $\mathbf{x}$ and coefficient vector $\boldsymbol{\beta}$, the linear predictor is $\eta=\mathbf{x}^{\mathsf T}\boldsymbol{\beta}$. It includes the intercept when the model has one. A GLM specifies a link $g$ so that the conditional mean $\mu=E[Y\mid\mathbf{x}]$ satisfies $g(\mu)=\eta$. Therefore, to obtain a fitted mean, calculate $\eta$ first and then apply the inverse link $g^{-1}$.

Definition: In a GLM, the linear predictor is $\eta=\mathbf{x}^{\mathsf T}\boldsymbol{\beta}$, and the link function relates the conditional mean to that predictor through $g(\mu)=\eta$. The fitted conditional mean is $\hat{\mu}=g^{-1}(\hat{\eta})$. The inverse link is applied to the entire linear predictor, including the intercept and all predictor terms.

The link does not by itself specify the full probability distribution of $Y$; the response distribution and its parameters must also be specified. The examples below state those assumptions explicitly. No interest-rate or mortality assumption is needed for these statistical models.

Three Links and Their Inverse Calculations

The logit link is commonly used when $Y$ is Bernoulli, taking values 0 and 1. Its conditional mean is a probability, so it must be between 0 and 1. The logit link and inverse-logit transformation are

g(\mu)=\log\left(\frac{\mu}{1-\mu}\right), \qquad g^{-1}(\eta)=\frac{e^\eta}{1+e^\eta} =\frac{1}{1+e^{-\eta}}.

The log link is used when a positive conditional mean is appropriate. Its inverse is the exponential function:

g(\mu)=\log(\mu), \qquad g^{-1}(\eta)=e^\eta.

The identity link leaves the mean on its original scale:

g(\mu)=\mu, \qquad g^{-1}(\eta)=\eta.

An identity link does not automatically keep fitted means within a desired range. The model must be appropriate for the response, and its fitted values should be checked for plausibility. In contrast, the inverse-logit always gives a value strictly between 0 and 1, and the inverse-log always gives a positive value.

Worked Example: Converting Logit Predictors to Probabilities

A hypothetical insurer models whether an applicant completes a voluntary wellness assessment. Let $Y=1$ indicate completion and $Y=0$ otherwise. Assume $Y\mid x$ is Bernoulli with probability $p(x)$, and use the logit model $\log(p(x)/(1-p(x)))=-\log 3+x\log 3$, where $x$ is a score taking values 0, 1, or 2. Find the fitted probability for each score. No interest-rate or mortality assumption is involved.

Method. First evaluate the linear predictor, then apply the inverse-logit. The intercept is $-\log 3$, and the coefficient of $x$ is $\log 3$.

For $x=0$, $\hat{\eta}=-\log 3+0=-\log 3$. Since $e^{-\log 3}=1/3$,

\hat{p}(0)=\frac{1/3}{1+1/3} =\frac{1/3}{4/3} =\frac14=0.2500.

For $x=1$, $\hat{\eta}=-\log 3+\log 3=0$, so

\hat{p}(1)=\frac{e^0}{1+e^0}=\frac12=0.5000.

For $x=2$, $\hat{\eta}=-\log 3+2\log 3=\log 3$, and $e^{\log 3}=3$. Thus

\hat{p}(2)=\frac{3}{1+3}=\frac34=0.7500.

Check. All three fitted probabilities lie between 0 and 1, as required for a Bernoulli mean. The coefficient $\log 3$ is an increment in the logit predictor per score unit; the calculated probabilities show the response-scale values at the specified scores. The common mistake avoided here is reporting $\hat{\eta}$ itself as a probability.

Worked Example: Converting a Log Predictor to a Positive Mean

Suppose a hypothetical service team models the number $Y$ of follow-up requests per account. Assume $Y\mid x$ has a Poisson distribution with mean $\lambda(x)$, and the fitted log-link model is $\log(\lambda(x))=\log 4+x\log 2$. Calculate the fitted mean number of requests for $x=0,1,2$. No interest-rate or mortality assumption is involved.

Method. Evaluate the predictor and exponentiate it, because the inverse of the log link is $e^\eta$.

At $x=0$, $\hat{\eta}=\log 4$, so $\hat{\lambda}(0)=e^{\log 4}=4$.

At $x=1$, $\hat{\eta}=\log 4+\log 2=\log 8$, so $\hat{\lambda}(1)=e^{\log 8}=8$.

At $x=2$, $\hat{\eta}=\log 4+2\log 2=\log 4+\log 4=\log 16$, so $\hat{\lambda}(2)=e^{\log 16}=16$.

Check. Each fitted mean is positive, as required by the log link, and the values are 4, 8, and 16 requests per account. Do not report $\log 16$ as the mean at $x=2$; it is the predictor on the link scale. The Poisson assumption specifies the conditional distribution here, with its parameter equal to the conditional mean.

Worked Example: Interpreting an Identity-Link Predictor

A hypothetical study models annual claim cost $Y$, in dollars, by a normal conditional distribution $Y\mid x\sim N(\mu(x),150^2)$. Here the second parameter is the variance, so the conditional standard deviation is $150. The fitted identity-link model is $\mu(x)=1200-150x$, where $x$ is a nonnegative risk score. Find the fitted mean at $x=0,1,2$, and compare the changes. No interest-rate or mortality assumption is involved.

Method. With the identity link, the inverse link is the identity function. The fitted mean is the linear predictor itself.

At $x=0$, $\hat{\eta}=1200-150(0)=1200$, so $\hat{\mu}(0)=\$1{,}200$.

At $x=1$, $\hat{\eta}=1200-150(1)=1200-150=1050$, so $\hat{\mu}(1)=\$1{,}050$.

At $x=2$, $\hat{\eta}=1200-150(2)=1200-300=900$, so $\hat{\mu}(2)=\$900$.

Check. The fitted means are positive and decrease by $150 for each one-unit increase in the score. Here the coefficient is a direct change in the fitted mean because the link is identity. In another link model, a coefficient generally cannot be read as a fixed dollar change. The conditional variance is $150^2=22{,}500$ dollars squared at each score, as specified by the example’s normal model.

Why the Response-Scale Effect Depends on the Link

The distinction between coefficient and response-scale effect follows directly from the inverse-link relationship. If predictor $x_j$ enters linearly with coefficient $\beta_j$, then $\eta=\beta_0+\beta_jx_j+\cdots$. A one-unit increase in $x_j$, holding the other predictors fixed, changes $\eta$ by $\beta_j$. But the resulting change in $\mu=g^{-1}(\eta)$ depends on where $\eta$ starts and on the inverse link’s shape.

For a continuous predictor, the local response-scale change can be obtained by differentiating $\mu=g^{-1}(\eta)$ with respect to $x_j$. The chain rule gives the following justification:

\frac{\partial\mu}{\partial x_j} =\frac{d\,g^{-1}(\eta)}{d\eta} \frac{\partial\eta}{\partial x_j} =(g^{-1})'(\eta)\beta_j.

Thus the coefficient is multiplied by the slope of the inverse link at the current predictor value. For the three links in this tutorial, the derivatives are:

\begin{array}{c|c} \text{Link} & \displaystyle\frac{\partial\mu}{\partial x_j}\\ \hline \text{Logit} & \beta_j\mu(1-\mu)\\ \text{Log} & \beta_j e^\eta=\beta_j\mu\\ \text{Identity} & \beta_j \end{array}

For example, under a logit link the local change depends on $\mu$, so it varies with the starting predictor even when $\beta_j$ is fixed. Under a log link the local change also depends on the current mean. Under the identity link it is constant and equals the coefficient. For a finite one-unit change, the exact response-scale difference is $g^{-1}(\eta+\beta_j)-g^{-1}(\eta)$; the derivative describes a local rate of change, not necessarily that entire finite difference.

Link	Fitted mean	Meaning of a coefficient increment on the link scale	Response-scale caution
Logit	$1/(1+e^{-\eta})$	Adds $\beta_j$ to the linear predictor for a one-unit predictor increase.	The corresponding probability change depends on the starting predictor.
Log	$e^\eta$	Adds $\beta_j$ to the log of the mean.	The corresponding mean change depends on the starting mean.
Identity	$\eta$	Adds $\beta_j$ to the predictor.	The same increment is a direct mean change, subject to plausibility of fitted values.

Common Exam Trap: Inverting the Wrong Quantity

A frequent error is to stop after computing $\eta$, or to apply an inverse link only to part of the predictor. For the logit model in the first example, at $x=0$ the predictor is $-\log 3$. Treating that value as the probability gives the wrong calculation $\hat{p}=-\log 3$, which is negative and cannot be a probability. The correction is to transform the complete predictor:

\hat{p}=\frac{e^{-\log 3}}{1+e^{-\log 3}} =\frac{1/3}{1+1/3} =0.2500.

A related log-link error is to report $\eta$ as the fitted mean. If $\eta=\log 16$, the fitted mean is $e^{\log 16}=16$, not $\log 16$. For an identity link, by contrast, no exponentiation or inverse-logit is required. Always identify the stated link before converting the predictor.

Key takeaway: Calculate the complete linear predictor first, then apply the inverse of the specified link. A coefficient always describes an increment on the link scale; whether that becomes a probability change, a change in a positive mean, or a direct mean change depends on the link and the starting predictor.

Check Your Understanding

For each problem, identify the relevant inverse-link calculation before giving the fitted mean or probability.

A Bernoulli response has logit predictor $\eta=-\log 4+x\log 2$. Find the fitted probability at $x=1$. (Answer: $1/3=0.3333$.)
A Poisson response has log-link predictor $\eta=\log 5+x\log 2$. Find the fitted mean at $x=2$. (Answer: $20$.)
A normal response has identity-link conditional mean $\mu=70+6x$ and conditional variance $9$. Find the fitted mean at $x=3$, and state the conditional standard deviation. (Answer: fitted mean $88$; standard deviation $3$.)
In a logit model, the current fitted probability is $0.5$, and a one-unit increase in $x$ changes the predictor by $\beta_x=0.8$. Find the exact new probability and the exact probability-point change. (Answer: new probability $1/(1+e^{-0.8})=0.6900$; increase $0.1900$.)
A log-link model has current fitted mean $12$, and increasing a predictor by one unit changes the linear predictor by $\log(1.5)$. Find the new fitted mean and the absolute increase. (Answer: new mean $18$; increase $6$.)

Link	Fitted mean	Meaning of a coefficient increment on the link scale	Response-scale caution
Logit	\(1/(1+e^{-\eta})\)	Adds \(\beta_j\) to the linear predictor for a one-unit predictor increase.	The corresponding probability change depends on the starting predictor.
Log	\(e^\eta\)	Adds \(\beta_j\) to the log of the mean.	The corresponding mean change depends on the starting mean.
Identity	\(\eta\)	Adds \(\beta_j\) to the predictor.	The same increment is a direct mean change, subject to plausibility of fitted values.

What You'll Learn