Why Boundary Behavior Matters
In a generalized linear model, predictors first determine a linear predictor, often written \(\eta\), and the inverse link then converts \(\eta\) to a conditional mean \(\mu\). The fitted mean is therefore not automatically an observed response, nor is it automatically allowed to take every value in the response distribution’s support. Its possible values depend on the inverse link.
This distinction matters when interpreting extreme predictions. A model for a claim-event probability may produce values very close to zero or one, while a model for an expected count may produce a value very close to zero but can never produce a negative mean under a log link. A displayed value rounded to zero or one can also conceal a fitted mean that is strictly inside its allowable range.
The distribution’s support and the mean parameter’s range are related but not identical. For example, a Poisson response takes nonnegative integer values, whereas its mean \(\mu\) can be any positive real number. A Bernoulli response takes values zero and one, whereas its conditional mean is a probability between zero and one. In each case, it is the conditional mean that the inverse link maps from \(\eta\).
Inverse Links and Their Boundaries
Several common inverse links illustrate different kinds of behavior. For a real-valued predictor \(\eta\), the identity, log, logit, and probit inverse links have the following ranges and limits.
| Link | Inverse link \(\mu=g^{-1}(\eta)\) | Range for finite \(\eta\) | Behavior at extremes |
|---|---|---|---|
| Identity | \(\mu=\eta\) | All real numbers | \(\mu\) decreases without bound as \(\eta\to-\infty\), and increases without bound as \(\eta\to+\infty\). |
| Log | \(\mu=e^\eta\) | \((0,\infty)\) | \(\mu\to0\) as \(\eta\to-\infty\), and \(\mu\to\infty\) as \(\eta\to+\infty\). |
| Logit | \(\mu=\dfrac{e^\eta}{1+e^\eta}\) | \((0,1)\) | \(\mu\to0\) as \(\eta\to-\infty\), and \(\mu\to1\) as \(\eta\to+\infty\). |
| Probit | \(\mu=\Phi(\eta)\) | \((0,1)\) | \(\mu\to0\) as \(\eta\to-\infty\), and \(\mu\to1\) as \(\eta\to+\infty\). |
Here \(\Phi\) is the cumulative distribution function of a standard normal random variable. The logit and probit inverse links are often used when the mean is a probability. The log inverse link is used when a positive mean is appropriate, as for a Poisson mean. The identity inverse link is natural for a Gaussian model whose conditional mean can be any real number; it does not, by itself, keep a fitted mean positive.
Why the Logit Mean Approaches but Does Not Reach Its Boundaries
For a Bernoulli response with success probability \(\mu\), the logit inverse link is \(\mu=e^\eta/(1+e^\eta)\). Both numerator and denominator are positive for every finite \(\eta\), and the denominator exceeds the numerator by one. Thus \(0<\mu<1\). This proves that neither endpoint is attained at a finite predictor.
To see the lower limit, rewrite the expression as \(\mu=1/(1+e^{-\eta})\). As \(\eta\to-\infty\), \(e^{-\eta}\to\infty\), so the denominator grows without bound and \(\mu\to0\). For the upper limit, as \(\eta\to+\infty\), \(e^{-\eta}\to0\), giving \(\mu\to1\). This form also avoids unnecessarily large exponentials when \(\eta\) is strongly negative.
For the log inverse link, \(\mu=e^\eta\) is positive for every finite \(\eta\). Since \(e^\eta\to0\) as \(\eta\to-\infty\), its lower boundary is approached but not reached. It grows without bound as \(\eta\to+\infty\). These statements concern the mathematical inverse link; they do not say that arbitrarily extreme fitted values are sensible for every application.
Worked Example: A Claim-Event Probability Near Zero and One
Method. Use the logit inverse link to convert each linear predictor to a conditional event probability. A hypothetical insurer models whether a small commercial vehicle has a specified event during a policy period, with \(\eta=-5+0.5x\), where \(x\) is a risk score. Find the fitted probabilities at \(x=2\), \(x=10\), and \(x=18\), and state whether any is exactly zero or one.
Step 1: Calculate each predictor. For \(x=2\), \(\eta=-5+0.5(2)=-4\). For \(x=10\), \(\eta=-5+0.5(10)=0\). For \(x=18\), \(\eta=-5+0.5(18)=4\).
Step 2: Apply the inverse link. At \(\eta=-4\), \(e^{-4}=0.0183156\), so \(\mu=0.0183156/(1+0.0183156)=0.0179862\). At \(\eta=0\), \(e^0=1\), so \(\mu=1/(1+1)=0.5\). At \(\eta=4\), \(e^4=54.5982\), so \(\mu=54.5982/(1+54.5982)=0.982014\).
Step 3: Report and check. Rounded to four decimal places, the fitted probabilities are 0.0180, 0.5000, and 0.9820. Each is strictly between zero and one, as the logit inverse link requires. The common mistake to avoid is treating \(\eta\) itself as a probability: \(-4\) and \(4\) are predictors, not fitted probabilities.
Positive Means Under the Log Inverse Link
The log inverse link transforms a predictor by exponentiation. If \(\eta\) changes by one unit, the mean is multiplied by \(e\), but the boundary question is simpler: no finite exponential is zero or negative. A very negative predictor can yield a tiny positive fitted mean, while a very positive predictor can yield a large one.
Worked Example: Expected Counts That Become Small
Method. Apply the log inverse link and distinguish an expected count from the probability of a count event. Suppose a hypothetical repair depot models the number \(Y\) of urgent service calls received per day as Poisson with conditional mean \(\mu=e^\eta\), where \(\eta=-3+0.5t\) and \(t\) is a workload index. Calculate the fitted means for \(t=0,2,6,\) and \(10\).
Step 1: Calculate the predictors. The four predictor values are \(-3+0.5(0)=-3\), \(-3+0.5(2)=-2\), \(-3+0.5(6)=0\), and \(-3+0.5(10)=2\).
Step 2: Exponentiate. At \(\eta=-3\), \(\mu=e^{-3}=0.0497871\). At \(\eta=-2\), \(\mu=e^{-2}=0.1353353\). At \(\eta=0\), \(\mu=e^0=1\). At \(\eta=2\), \(\mu=e^2=7.3890561\).
Step 3: Report and check. To four decimal places, the fitted means are 0.0498, 0.1353, 1.0000, and 7.3891 calls per day. All are positive, including the two small means. A mean of 0.0498 does not mean that the probability of a call is 0.0498; it is the expected count. The common mistake is to interpret the mean as a probability simply because it is small.
When a Link Can Reach a Boundary at a Finite Predictor
Not every inverse link behaves like the log or logit. Under the identity link, \(\mu=\eta\). A finite predictor of zero gives a mean of zero, and a negative predictor gives a negative mean. That is acceptable for a Gaussian response with real-valued support, but it may be unsuitable when the response and its mean must be positive, such as a cost or count.
The inverse link must therefore be considered together with the response distribution and the model’s purpose. A fitted mean of zero from an identity link is not an asymptotic limit; it is attained at \(\eta=0\). Conversely, a log-link mean can be arbitrarily close to zero as the predictor decreases, but cannot equal zero for any finite predictor.
Worked Example: Checking a Positive Cost Mean
Method. Compare the fitted means produced by identity and log inverse links at the same predictor values. A hypothetical model considers the mean of a nonnegative repair cost, in thousands of dollars, at three risk profiles with predictors \(\eta=-1,0,1\). Determine which inverse-link predictions respect a positive-mean requirement.
Step 1: Calculate identity-link means. Since \(\mu=\eta\), the predictions are \(-1\), \(0\), and \(1\) thousand dollars. The first is negative and the second is zero. Neither is a positive mean for a cost, so the identity link does not preserve the required range for these profiles.
Step 2: Calculate log-link means. Since \(\mu=e^\eta\), the predictions are \(e^{-1}=0.367879\), \(e^0=1\), and \(e^1=2.718282\) thousand dollars. Rounded to four decimal places, these are 0.3679, 1.0000, and 2.7183 thousand dollars.
Step 3: Interpret and check. All three log-link means are positive, as required, while the identity-link means include values outside that range. The log-link prediction at \(\eta=-1\) is below one thousand dollars but is not zero. The common mistake to avoid is assuming that a link appropriate for a real-valued Gaussian mean automatically preserves positivity for a cost mean.
Numerical Rounding and Exam Traps
The distinction between a limit and an attained value can disappear in a calculator display. For example, with a logit predictor of \(\eta=-10\), \(\mu=1/(1+e^{10})\approx0.0000454\). Rounded to four decimal places this is 0.0000, even though the exact fitted probability is positive. At \(\eta=10\), the probability is approximately 0.9999546, which rounds to 1.0000 but is still less than one.
A rounded display is not evidence that the inverse link reached a boundary. If the question asks whether the fitted mean equals a boundary, use the formula and its range, not only the rounded number. Conversely, if a model uses the identity link, a zero fitted mean at a finite predictor is possible and is not merely a rounding artifact.
Write the formula that maps \(\eta\) to \(\mu\); do not substitute the predictor directly unless the inverse link is identity.
Check whether the formula can equal the proposed boundary at any allowed, finite \(\eta\).
Examine \(\eta\to-\infty\) and \(\eta\to+\infty\) separately; the two directions need not have symmetric behavior.
State whether the result is a probability, an expected count, or another conditional mean, and distinguish the exact value from a rounded display.
Another important check is the inverse link’s domain. The logit and log inverse links accept every real \(\eta\). Other inverse links can have restricted predictor domains. In that case, a candidate predictor outside the domain does not produce a valid fitted mean by that inverse link; it is not appropriate to extend the formula without checking the model’s definition.
Check Your Understanding
For each problem, identify the inverse-link behavior before interpreting the numerical result.
- A Bernoulli model has logit predictor \(\eta=-\ln 3\). Calculate the fitted probability and state whether it is a boundary value. (Answer: \(\mu=0.2500\), strictly between zero and one.)
- A Poisson model uses the log link and has \(\eta=\ln(0.2)\). Find its fitted mean and explain whether it can equal zero at this predictor. (Answer: \(\mu=0.2000\); it is positive and does not equal zero.)
- A logit model has \(\eta=-8\). Calculate the fitted probability to four decimal places, then state whether the exact value is zero. (Answer: \(\mu=1/(1+e^8)\approx0.0003\); the exact value is positive, not zero.)
- A Gaussian identity-link model has predictors \(-0.6\) and \(0.6\). Find both fitted means and explain why the first is not a boundary-behavior contradiction. (Answer: \(-0.6\) and \(0.6\); the identity inverse link permits all real means, which is consistent with a real-valued Gaussian response.)
- A probit model has predictor \(\eta=0.5\), and the standard normal cumulative distribution function satisfies \(\Phi(0.5)\approx0.6915\). Give the fitted mean and state its limiting values as the predictor tends to negative and positive infinity. (Answer: \(\mu\approx0.6915\); the limits are zero and one, neither attained at a finite predictor.)