Response Scale Versus Link Scale

Why Two Scales Appear in a Generalized Linear Model

A generalized linear model (GLM) can describe a mean that is restricted—for example, a probability must lie between 0 and 1, while a count mean must be positive. The model handles such restrictions by relating the mean to a predictor through a link function. This creates two scales for the same fitted quantity: the response scale, where the mean has the units and interpretation of the outcome, and the link scale, where the model’s predictor is expressed.

The distinction matters whenever a fitted value is reported or interpreted. A number on the link scale is generally not itself a fitted response mean. It must be converted using the inverse link. The correct conversion depends on the selected link, so a value such as \(0.5\) can represent a mean of \(0.5\) under an identity link but a different mean under a log or logit link.

Definition: Let \(Y\) be a response and let \(\mu=E[Y\mid x]\) denote its conditional mean at predictor information \(x\). The response scale is the scale on which \(\mu\) is expressed, in the response’s units or as a response quantity such as a probability. A link function \(g\) maps the mean to the link scale: \(\eta=g(\mu)\). The value \(\eta\) is the linear predictor on that scale. To return to the response scale, use the inverse link: \(\mu=g^{-1}(\eta)\).

The inverse link must be applied within its valid domain. For a particular model, the chosen link is intended to map allowable mean values to link-scale values that can be used by the model. Familiar examples include the identity link for an unrestricted real-valued mean, the log link for a positive mean, and the logit link for a probability strictly between 0 and 1. The formula \(\mu=g^{-1}(\eta)\) is a conversion rule, not a claim that every link is appropriate for every response.

The response scale is about a model’s mean, not a guarantee about an individual observation. For instance, a fitted mean count of 2.01 does not mean that an observed count must equal 2.01; an observed count is an outcome, while the fitted mean summarizes the model’s expected response at the specified predictors. This distinction holds even when the response itself is numeric.

Three Common Link-Scale Conversions

For the identity link, \(g(\mu)=\mu\), so the link-scale value and the mean have the same numerical value. For the log link, \(g(\mu)=\log(\mu)\), and the inverse link is the exponential function. For the logit link, \(g(\mu)=\log\!\left(\mu/(1-\mu)\right)\), where \(\mu\) is a probability; its inverse is the logistic function.

\begin{aligned} \text{Identity:}\quad &\eta=\mu, & \mu&=\eta,\\ \text{Log:}\quad &\eta=\log(\mu), & \mu&=\exp(\eta),\\ \text{Logit:}\quad &\eta=\log\!\left(\frac{\mu}{1-\mu}\right), & \mu&=\frac{\exp(\eta)}{1+\exp(\eta)} =\frac{1}{1+\exp(-\eta)}. \end{aligned}

These equations show why the scale label is essential. With a log link, a predictor value of \(0.7\) is not a mean of \(0.7\); it is the logarithm of the mean. With a logit link, the predictor is the logarithm of the odds, not the probability. The inverse link is what restores the response-scale interpretation.

Worked Example: Converting a Log-Scale Prediction to a Mean Count

A hypothetical count model uses a log link for its conditional mean number of service requests per location per month. At one location, the fitted link-scale value is \(\eta=0.7\). Find the fitted mean count and interpret it. No count distribution is needed for this scale conversion.

Method. For a log link, \(\eta=\log(\mu)\), so apply the inverse link, \(\mu=\exp(\eta)\). Since a count mean must be positive, this inverse transformation also gives a positive fitted mean.

Calculation.

\begin{aligned} \mu &=\exp(0.7)\\ &=2.0137527\ldots\\ &\approx 2.0138. \end{aligned}

The fitted mean is approximately 2.0138 service requests per location per month. It is an expected count, not a claim that a particular location must have a fractional observed count.

Check and mistake avoided. The result is positive, as required for a log-link mean. Reporting \(0.7\) as the fitted mean would confuse the link scale with the response scale; \(0.7\) is the log of the fitted mean.

Worked Example: Converting a Logit Prediction to a Probability

In a hypothetical study of whether a sensor flags a maintenance issue, the model gives a logit-scale prediction \(\eta=0.5\) for one sensor. Calculate the fitted probability and the corresponding odds. Report the probability to four decimal places.

Method. Under the logit link, \(\eta=\log(\mu/(1-\mu))\). First exponentiate to obtain the odds, then convert odds to probability using \(\mu=\exp(\eta)/(1+\exp(\eta))\).

Odds calculation.

\frac{\mu}{1-\mu}=\exp(0.5)=1.6487213\ldots

Probability calculation.

\begin{aligned} \mu &=\frac{\exp(0.5)}{1+\exp(0.5)}\\ &=\frac{1.6487213\ldots}{1+1.6487213\ldots}\\ &=\frac{1.6487213\ldots}{2.6487213\ldots}\\ &=0.6224593\ldots\\ &\approx 0.6225. \end{aligned}

The fitted probability is 0.6225, or about 62.25%, and the corresponding odds are approximately 1.6487 to 1. As a check, the odds exceed 1, so the probability must exceed 0.5; 0.6225 does.

Common mistake avoided. Treating \(\eta=0.5\) as a probability gives 0.5, which is wrong here. The value \(0.5\) is the log-odds; applying the inverse logit gives the probability.

Why the Logit Inverse Has This Form

The inverse-logit formula follows directly from the definition of the logit link. Let \(\mu\) be a probability strictly between 0 and 1, and suppose \(\eta=\log(\mu/(1-\mu))\). Exponentiating both sides removes the logarithm and gives the odds. Solving that equation for \(\mu\) produces the inverse link.

Exponentiate.
\(\exp(\eta)=\mu/(1-\mu)\).

Clear the denominator.
\(\exp(\eta)(1-\mu)=\mu\), so \(\exp(\eta)-\exp(\eta)\mu=\mu\).

Collect the terms involving \(\mu\).
\(\exp(\eta)=\mu(1+\exp(\eta))\).

Isolate the probability.
\(\mu=\exp(\eta)/(1+\exp(\eta))\). The denominator is greater than the numerator, and both are positive, so the result is strictly between 0 and 1.

This derivation also explains the roles of the scales: exponentiating a logit-scale value yields odds, and one further conversion yields probability. It is not correct to describe \(\exp(\eta)\) itself as a probability.

How Link-Scale Changes Affect the Mean

A change on the link scale does not have the same meaning under every link. Under the identity link, adding a fixed amount to \(\eta\) adds that same amount to \(\mu\). Under the log link, adding a fixed amount to \(\eta\) multiplies \(\mu\) by a fixed factor. Under the logit link, adding a fixed amount changes the log-odds by that amount, or equivalently multiplies the odds by a fixed factor; the corresponding change in probability depends on its starting value.

Worked Example: Comparing Equal Predictor Changes Under Two Links

A hypothetical analyst compares the effect of a predictor change that raises the link-scale value from \(1.0\) to \(1.4\). Calculate the two fitted means under (a) an identity link and (b) a log link, and describe the change on each response scale. For the log-link model, round means and the difference to four decimal places.

(a) Identity link. Since \(\mu=\eta\), the fitted means are

\mu_1=1.0,\qquad \mu_2=1.4,\qquad \mu_2-\mu_1=1.4-1.0=0.4.

Thus, the response-scale mean increases by 0.4 response units.

(b) Log link. Since \(\mu=\exp(\eta)\), the fitted means are

\begin{aligned} \mu_1&=\exp(1.0)=2.7182818\ldots\approx 2.7183,\\ \mu_2&=\exp(1.4)=4.0551999\ldots\approx 4.0552,\\ \mu_2-\mu_1&=4.0551999\ldots-2.7182818\ldots\\ &=1.3369181\ldots\approx 1.3369. \end{aligned}

Here the response-scale mean increases by about 1.3369, not by 0.4. The ratio of the new mean to the old mean is

\frac{\mu_2}{\mu_1} =\frac{\exp(1.4)}{\exp(1.0)} =\exp(0.4) =1.4918247\ldots \approx 1.4918.

So a 0.4 increase on the log scale multiplies the mean by about 1.4918. The calculations check because \(2.7183\times1.4918\) is approximately \(4.0552\).

Common mistake avoided. Applying the identity-link interpretation to the log-link model would incorrectly report an increase of 0.4 response units. Under the log link, the predictor difference determines a ratio of means, not an additive difference.

Interpreting and Reporting Fitted Values

A reliable interpretation begins by identifying which scale a number belongs to. If a question gives \(\eta\), it gives the link-scale value. If it asks for the fitted conditional mean, apply \(g^{-1}\). If it asks for odds under a logit link, exponentiate \(\eta\); if it asks for probability, use the inverse-logit formula. These are related but distinct quantities.

Link	Link-scale meaning	Response-scale conversion	Interpretation of a link-scale increase \(c\)
Identity	The mean itself	\(\mu=\eta\)	The mean increases by \(c\)
Log	The log of the mean	\(\mu=\exp(\eta)\)	The mean is multiplied by \(\exp(c)\)
Logit	The log of the odds	\(\mu=1/(1+\exp(-\eta))\)	The odds are multiplied by \(\exp(c)\)

The last column does not say that a fixed link-scale change creates a fixed response-scale difference. For the log link, the absolute difference depends on the starting mean, even though the ratio is fixed. For the logit link, the probability difference depends on the starting probability, even though the odds ratio is fixed. Keeping these statements separate prevents overinterpreting a coefficient or fitted predictor.

A useful reporting habit is to state both the scale and the requested quantity. For example: “The fitted logit-scale value is 0.5; the corresponding fitted probability is 0.6225.” This makes clear that the probability was obtained by transformation rather than read directly from the predictor. The same habit is useful for any link: name \(\eta\), name the inverse link, and report the response-scale mean with its units or interpretation.

Key takeaway: A model’s link-scale predictor \(\eta\) and its response-scale fitted mean \(\mu\) are connected by \(\eta=g(\mu)\) and \(\mu=g^{-1}(\eta)\). Apply the inverse link before interpreting a predictor as a mean or probability, and interpret changes according to the selected link.

Check Your Understanding

For each item, identify the relevant scale and convert only as requested.

A positive mean is modeled with a log link, and its fitted link-scale value is \(0.25\). Find the fitted mean to four decimal places. (Answer: \(1.2840\).)
A probability model uses a logit link and has fitted value \(\eta=\log(2/5)\). Find the fitted probability. (Answer: \(2/7\), or approximately \(0.2857\).)
An identity-link model has fitted value \(\eta=12.6\). What is the fitted mean, and what is the response-scale interpretation? (Answer: \(\mu=12.6\), in the response’s units.)
Under a log link, a predictor change raises \(\eta\) by \(0.6\). By what factor does the fitted mean change? Give four decimals. (Answer: \(\exp(0.6)=1.8221\); the mean is multiplied by \(1.8221\).)
A logit-link model has fitted value \(\eta=\log(3)\). Find the odds and the probability. (Answer: odds \(=3\); probability \(=3/(1+3)=0.7500\).)

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