Why Link Choice Changes the Model
A Poisson regression models a count through its conditional mean. The link function determines how a linear predictor changes that mean. Choosing a link is therefore not just a matter of notation: it determines whether a covariate acts additively, multiplicatively, or through a reciprocal relationship, and whether the model can produce invalid means.
For example, a change in operating conditions might add a roughly constant number of expected events, or it might multiply the expected number by a constant factor. A third possibility is that the reciprocal of the expected count changes linearly. These are distinct assumptions about the mean scale. A link should be selected to reflect a reasonable mean relationship and to preserve valid predictions over the predictor values where the model will be used.
The three links compared here are the log link \(g(\mu)=\log\mu\), the identity link \(g(\mu)=\mu\), and the reciprocal link \(g(\mu)=1/\mu\). The reciprocal link is also called the inverse link. Under each link, the linear predictor is linear in the model parameters, but the implied mean-scale relationship differs.
| Link | Mean as a function of predictor | Mean-scale effect of a predictor change | Main constraint |
|---|---|---|---|
| Log: \(g(\mu)=\log\mu\) | \(\mu=\exp(\eta)\) | A fixed predictor change gives a fixed multiplicative factor | Any finite \(\eta\) gives a positive mean |
| Identity: \(g(\mu)=\mu\) | \(\mu=\eta\) | A fixed predictor change gives a fixed additive change | Fitted means must be positive over the relevant covariate range |
| Reciprocal: \(g(\mu)=1/\mu\) | \(\mu=1/\eta\) | A predictor change alters the reciprocal mean linearly | The predictor must be positive wherever the model is used |
How the Three Links Translate Changes
Suppose a single numeric covariate \(x\) enters the predictor with coefficient \(\beta\), and compare \(x\) with \(x+\Delta x\), holding other predictors fixed. Under the log link, the predictor changes by \(\beta\Delta x\), so the mean is multiplied by \(\exp(\beta\Delta x)\). Under the identity link, the mean changes by \(\beta\Delta x\). Under the reciprocal link, it is the reciprocal mean that changes by \(\beta\Delta x\).
These statements assume that the covariate change is within the range where the fitted model is valid. In particular, identity and reciprocal models require feasibility checks; a linear predictor that gives a nonpositive mean, or a nonpositive reciprocal-link predictor, cannot be used as a Poisson mean.
Worked Link Interpretations
Worked Example: Comparing Additive and Multiplicative Count Effects
A hypothetical monitoring system records the number of alarms during equal-length operating periods. For a baseline setting \(x=0\), suppose two candidate Poisson models both give a mean of 4 alarms. In a log-link model, \(\log\mu=\log 4+x\log(1.5)\). In an identity-link model, \(\mu=4+2x\). Compare their predictions at \(x=1\) and \(x=2\), and state the assumed effect of increasing \(x\) by one unit.
Method. In the log model, exponentiate the linear predictor. In the identity model, substitute directly into the mean equation.
Log link at \(x=1\): \(\mu=\exp\{\log 4+\log(1.5)\}=4(1.5)=6\) alarms.
Identity link at \(x=1\): \(\mu=4+2(1)=6\) alarms.
The models agree at \(x=1\), but for different reasons. The log model multiplies the mean by 1.5 per one-unit increase. The identity model adds 2 alarms per one-unit increase.
Log link at \(x=2\): \(\mu=\exp\{\log 4+2\log(1.5)\}=4(1.5)^2=4(2.25)=9\) alarms.
Identity link at \(x=2\): \(\mu=4+2(2)=8\) alarms.
The predictions differ because the models encode different patterns of change. The log model predicts a 50% increase at each step: \(4\) to \(6\), then \(6\) to \(9\). The identity model predicts an increase of 2 alarms at each step: \(4\) to \(6\), then \(6\) to \(8\). The calculations are internally consistent with the shared baseline and each stated model.
Common mistake avoided: Matching predictions at one covariate value does not make the links equivalent. Their effects must be compared across relevant covariate values and on the scale specified by the link.
Worked Example: Checking Identity-Link Feasibility
A hypothetical facility models the number of minor equipment faults per fixed inspection period as \(Y\mid x\sim\operatorname{Poisson}(\mu)\), with an identity-link mean \(\mu=2-0.7x\). Here \(x\) is a numeric operating-condition score observed from 0 through 4. Calculate the means at \(x=0,2,4\), and determine whether this model is feasible over the full stated range.
Method. Under the identity link, the linear predictor equals the mean. Substitute each score and check that each resulting Poisson mean is positive.
At \(x=0\), \(\mu=2-0.7(0)=2\). At \(x=2\), \(\mu=2-0.7(2)=2-1.4=0.6\). At \(x=4\), \(\mu=2-0.7(4)=2-2.8=-0.8\).
The first two means are positive, but the prediction at \(x=4\) is negative. A Poisson distribution cannot have a negative mean, so the model as written is not feasible over the full stated range. In fact, positivity requires \(2-0.7x>0\), or \(x<2/0.7\approx2.8571\). The observed range extends beyond this boundary.
A fitted identity-link model can be useful when an additive mean effect is scientifically appropriate, but its predictions must be checked throughout the domain of use. Restricting use to a smaller range is defensible only if that restriction matches the intended application; silently replacing negative predictions with zero changes the model.
Common mistake avoided: A positive fitted mean at the average covariate value does not guarantee valid means at all observed or forecast covariate values.
Worked Example: Interpreting a Reciprocal-Link Model
A hypothetical count model for events per fixed observation period uses the reciprocal link \(1/\mu=0.25+0.05x\), where \(x\) is a numeric score between 0 and 4. Find the predicted means at \(x=0,2,4\), and describe how the mean changes as \(x\) increases.
Method. Calculate the predictor \(1/\mu\) at each score, then take its reciprocal. Check that the predictor remains positive.
At \(x=0\), \(1/\mu=0.25+0.05(0)=0.25\), so \(\mu=1/0.25=4\).
At \(x=2\), \(1/\mu=0.25+0.05(2)=0.35\), so \(\mu=1/0.35\approx2.8571\).
At \(x=4\), \(1/\mu=0.25+0.05(4)=0.45\), so \(\mu=1/0.45\approx2.2222\).
The predictor is positive over \(0\le x\le4\), so all three predictions are positive. The mean decreases as \(x\) increases because the reciprocal mean increases. The mean-scale changes are not constant: the decrease is approximately \(4-2.8571=1.1429\) from \(x=0\) to \(x=2\), and \(2.8571-2.2222=0.6349\) from \(x=2\) to \(x=4\).
Common mistake avoided: The coefficient \(0.05\) is an increase in \(1/\mu\) per unit of \(x\), not an additive increase in \(\mu\), nor a fixed percentage change in \(\mu\).
Positivity, Exposure, and Model Suitability
The log link is often convenient for count means because \(\exp(\eta)\) is positive for every finite real \(\eta\). It also represents a multiplicative effect directly. If a predictor coefficient is \(\beta\), a one-unit increase in that predictor multiplies the mean by \(\exp(\beta)\); a change of \(\Delta x\) multiplies it by \(\exp(\beta\Delta x)\). This interpretation applies with other predictors held fixed.
For an identity link, \(\mu=\eta\), so a coefficient gives an additive change in the conditional mean per unit of its covariate. This can be a natural description when expected counts change by roughly the same amount across the relevant range. The cost is that a linear predictor can become zero or negative. Checking only the values used to estimate coefficients is not sufficient if predictions will be made at new covariate values.
For the reciprocal link, \(\mu=1/\eta\), so the reciprocal mean is linear in the predictors. A valid model requires \(\eta>0\) wherever predictions are required. Although a positive \(\eta\) guarantees a positive mean, an unconstrained linear predictor may become nonpositive outside the observed region. The mean also changes nonlinearly with the predictor: when \(\eta\) is near zero from above, \(1/\eta\) becomes large.
Exposure requires particular care. With the log link, if a count has mean equal to exposure \(e_i\) times a rate \(\lambda_i\), then \(\mu_i=e_i\lambda_i\) implies \(\log\mu_i=\log e_i+\log\lambda_i\). Thus \(\log e_i\) enters as a known offset when the rate has a log-linear model. The offset is not an estimated covariate coefficient. Under identity or reciprocal links, adding \(\log e_i\) to the predictor does not in general express \(\mu_i=e_i\lambda_i\). A model using those links must specify its exposure dependence on the mean scale rather than copying the log-link offset mechanically.
Common Interpretation Trap
Suppose a log-link model has baseline mean 5 and a coefficient of \(0.4\) for a covariate that increases by one unit. An incorrect calculation treats \(0.4\) as an additive change and gives \(5+0.4=5.4\). This is the identity-link interpretation, not the log-link interpretation.
For the log link, the mean is multiplied by \(\exp(0.4)\). Since \(\exp(0.4)\approx1.4918\), the new mean is \(5(1.4918)\approx7.4591\). The coefficient corresponds to an increase of about 49.18% in the mean for this one-unit change, not an increase of 0.4 events. The calculation assumes all other predictors are held fixed and the covariate change is one unit.
Check Your Understanding
For each problem, identify the link-scale calculation and check that the resulting mean is valid.
- A log-link model has \(\log\mu=\log 3+x\log 2\). Find the mean when \(x=1\), and give the multiplicative effect of a one-unit increase in \(x\). (Answer: \(\mu=6\); the mean is multiplied by 2 per unit.)
- An identity-link model has \(\mu=1.8+0.6x\). Find the predicted mean at \(x=3\), and state the additive effect of a one-unit increase. (Answer: \(\mu=3.6\); the mean increases by 0.6 per unit.)
- A reciprocal-link model has \(1/\mu=0.20+0.04x\). Find the predicted mean at \(x=5\). (Answer: \(1/\mu=0.40\), so \(\mu=2.5\).)
- An identity-link model has \(\mu=4-1.2x\), and predictions are required at \(x=3.5\). Is the prediction valid? (Answer: \(\mu=4-1.2(3.5)=4-4.2=-0.2\), so it is invalid as a Poisson mean.)
- At a baseline covariate value, two models both have mean 10. For a change \(\Delta x=0.5\), a log-link coefficient is \(0.4\), while an identity-link coefficient is 4. Calculate both new means and compare the assumed effects. (Answer: log link \(10\exp(0.4(0.5))=10\exp(0.2)\approx12.2140\), a multiplicative change; identity link \(10+4(0.5)=12\), an additive increase of 2.)