Offsets Under Noncanonical Links

Why an Offset Does Not Have One Universal Mean-Scale Effect

An offset enters a generalized linear model as a known addition to the linear predictor. That part of the model is unchanged when the link is noncanonical: the offset shifts the predictor by its specified amount. What changes with the link is how that shift is translated into the response mean. In particular, an additive shift in the predictor does not generally produce either an additive or a multiplicative change in the mean.

This distinction matters when interpreting an offset. Under a log link, adding an offset multiplies the mean by the exponential of that offset. Under other links, the same algebraic shift can have a quite different effect, depending on the starting predictor. The calculations below focus on that link-dependent translation, rather than on how an offset is used in likelihood estimation.

Definition: For observation $i$, let $x_i$ be its covariate vector, $\beta$ the regression coefficient vector, $o_i$ a known offset, $g$ the link, and $h=g^{-1}$ the inverse link. The model with offset is $$ \eta_i=x_i^{\mathsf T}\beta+o_i,\qquad g(\mu_i)=\eta_i,\qquad \mu_i=h(x_i^{\mathsf T}\beta+o_i). $$ The offset $o_i$ is a specified value, not a covariate multiplied by an additional regression coefficient. The predictor $\eta_i$ must lie in the domain of $h$, and the resulting $\mu_i$ must lie in the response-mean domain of the model.

A link is noncanonical when it is not the distribution's canonical link. For example, the probit link for a binomial mean is noncanonical; the binomial canonical link is the logit. This distinction concerns the relationship between the distribution and the link. It does not by itself determine whether an offset acts multiplicatively on the mean: that depends on the form of $h$.

From a Predictor Shift to a Change in the Mean

For a fixed value of the regression contribution, write $\eta=x^{\mathsf T}\beta$ and consider an offset value $o$. The mean without the offset is $h(\eta)$; with the offset it is $h(\eta+o)$. Their difference is therefore $h(\eta+o)-h(\eta)$, which depends on both the offset and the starting predictor unless the inverse link has a special form.

Theorem: Suppose $h$ is continuously differentiable on an open interval containing the entire line segment between $\eta$ and $\eta+o$. Then the mean as a function of the offset is differentiable, with $$ \frac{\partial}{\partial o}h(\eta+o)=h'(\eta+o), $$ and its finite change from offset zero to offset $o$ is $$ h(\eta+o)-h(\eta)=\int_0^o h'(\eta+t)\,dt. $$ For $o<0$, the integral has the usual oriented-interval meaning.

Proof. Since $h$ is continuously differentiable in a neighborhood of the points under consideration, the chain rule applied to the function $o\mapsto h(\eta+o)$ gives derivative $h'(\eta+o)$. For the finite change, the function $t\mapsto h(\eta+t)$ is continuously differentiable on the interval with endpoints $0$ and $o$. The fundamental theorem of calculus therefore gives $$ h(\eta+o)-h(\eta)=\int_0^o \frac{d}{dt}h(\eta+t)\,dt =\int_0^o h'(\eta+t)\,dt. $$ The stated domain condition ensures that every argument along the integration path is in the interval where $h$ is continuously differentiable. $\square$

The derivative describes the local mean change per unit of offset at the current predictor. It is not generally constant as the predictor changes. The finite-change formula makes the same point over an interval: the effect accumulates the inverse-link derivative along the path from $\eta$ to $\eta+o$.

Worked Example: A Probit Offset Changes a Binomial Probability

Suppose $Y$ is Bernoulli with mean $\mu=P(Y=1)$, and use the probit link $g(\mu)=\Phi^{-1}(\mu)$, where $\Phi$ is the standard normal cumulative distribution function. Thus $h(\eta)=\Phi(\eta)$. Compare a zero offset with an offset of $0.5$.

At baseline predictor $\eta=0$, the mean without an offset is $$ \mu_0=\Phi(0)=0.50000. $$ With offset $o=0.5$, the predictor is $0+0.5=0.5$, so $$ \mu_1=\Phi(0.5)\approx 0.69146. $$ The probability difference is approximately $0.69146-0.50000=0.19146$, and the ratio is approximately $0.69146/0.50000=1.38292$.

Now apply the same offset at baseline predictor $\eta=1$. The starting mean is $\Phi(1)\approx0.84134$, while the shifted mean is $\Phi(1.5)\approx0.93319$. Their ratio is approximately $0.93319/0.84134=1.10917$. The same predictor shift therefore gives different probability ratios at the two baselines. It does not correspond to a fixed multiplicative change in probability.

When Does a Predictor Shift Multiply the Mean?

The log link is a familiar special case. If $h(\eta)=\exp(\eta)$, then adding offset $o$ gives $$ \frac{h(\eta+o)}{h(\eta)} =\frac{\exp(\eta+o)}{\exp(\eta)} =\exp(o). $$ The mean is multiplied by $\exp(o)$, regardless of the baseline predictor. This property follows from the exponential form of the inverse link, not simply from calling a known quantity an offset.

Theorem: Let $I$ be a connected open interval and let $h:I\to(0,\infty)$ be continuously differentiable. Suppose that for every $\eta\in I$ and every sufficiently small $o$ for which $\eta+o\in I$, the ratio $h(\eta+o)/h(\eta)$ depends on $o$ but not on $\eta$. Then there are constants $C>0$ and $c\in\mathbb R$ such that $h(\eta)=C\exp(c\eta)$ on $I$. Conversely, every inverse link of this form has a mean ratio that depends only on $o$.

Proof. Denote the assumed ratio, for small offsets, by $R(o)$. Fix any interior point $\eta_0\in I$. Since $h$ is positive and continuously differentiable, the expression $h(\eta_0+o)/h(\eta_0)$ is differentiable for $o$ near zero. It equals $R(o)$, so $R$ is differentiable there. Also $R(0)=1$. For any $\eta\in I$, differentiating the ratio at $o=0$ gives $$ \left.\frac{d}{do}\frac{h(\eta+o)}{h(\eta)}\right|_{o=0} =\frac{h'(\eta)}{h(\eta)} =R'(0). $$ Thus $h'(\eta)/h(\eta)=c$ throughout $I$, where $c=R'(0)$ is constant. Because $h(\eta)>0$, the logarithm is defined and $$ \frac{d}{d\eta}\log h(\eta)=c. $$ Integration on the connected interval gives $\log h(\eta)=c\eta+k$ for a constant $k$, hence $h(\eta)=C\exp(c\eta)$ with $C=\exp(k)>0$. Conversely, if $h(\eta)=C\exp(c\eta)$, then $$ \frac{h(\eta+o)}{h(\eta)} =\frac{C\exp(c(\eta+o))}{C\exp(c\eta)} =\exp(co), $$ which is independent of $\eta$. $\square$

This result characterizes a strong property: the offset must give the same mean ratio at every baseline predictor. The theorem does not say that every noncanonical link fails this property. It says the property depends on the inverse-link form; an exponential inverse link has it, while links such as probit and complementary log-log do not.

Worked Example: An Identity-Link Offset Adds to a Poisson Mean

Let $Y$ have a Poisson distribution with mean $\mu>0$, and use the identity link $g(\mu)=\mu$, which is noncanonical for the Poisson distribution. Then $h(\eta)=\eta$, on the valid predictor range where the resulting mean is positive. Suppose the regression contribution is $\eta=0.8$ and the known offset is $o=0.3$.

Without the offset, $\mu_0=h(0.8)=0.8$. With the offset, the predictor is $0.8+0.3=1.1$, so $\mu_1=h(1.1)=1.1$. Both values are positive, so they are valid Poisson means in this example. The change is $1.1-0.8=0.3$, exactly the offset. The ratio, however, is $1.1/0.8=1.375$; it is not a fixed multiplier determined by the offset alone.

More generally, for the identity inverse link, $h'(\eta)=1$, so the derivative result gives a mean change of $o$ wherever the entire predictor path remains in the valid range. An identity-link model can require particular care to keep its fitted means in the response-mean domain.

Worked Example: A Complementary Log-Log Offset Has Baseline-Dependent Ratios

For a binomial mean, consider the complementary log-log link $g(\mu)=\log[-\log(1-\mu)]$. Solving for the inverse link gives $h(\eta)=1-\exp[-\exp(\eta)]$, which maps real predictors to probabilities strictly between zero and one. This is a noncanonical binomial link.

At $\eta=0$, the mean is $$ h(0)=1-\exp(-1)\approx0.63212. $$ Choose the offset $o=\log 2$. The shifted predictor is $\log 2$, and the shifted mean is $$ h(\log 2)=1-\exp(-2)\approx0.86466. $$ The ratio of shifted to unshifted means is approximately $0.86466/0.63212=1.36788$.

At baseline $\eta=\log 2$, the mean is $1-\exp(-2)\approx0.86466$. Applying the same offset gives predictor $2\log 2=\log 4$ and mean $1-\exp(-4)\approx0.98168$. This second ratio is approximately $0.98168/0.86466=1.13533$. Since the two ratios differ, this offset does not multiply the probability by a baseline-independent factor.

Interpreting and Checking a Noncanonical Offset

An offset can represent a known contribution that the model includes on the linear-predictor scale. Its numerical value is added to the regression contribution before applying the inverse link. To explain what that means for the response, calculate $h(\eta+o)$ and compare it with $h(\eta)$; do not infer the mean-scale effect from the word “offset” alone.

For a small offset, the local approximation $$ h(\eta+o)-h(\eta)\approx h'(\eta)o $$ follows from differentiability. It can be useful for describing a small change near a specified predictor, but it is local: if $o$ is not small, the derivative may vary substantially along the path. The finite-change integral gives the exact expression under the continuously differentiable condition stated above.

Common pitfall: “An offset of $o$ multiplies the mean by $\exp(o)$, because offsets are exponentiated.” This reasoning incorrectly assumes a log link. The model adds $o$ to the predictor, then applies its inverse link. Only when that inverse link is exponential does the mean ratio reduce to $\exp(o)$ (or, more generally, to $\exp(co)$ for an inverse link $C\exp(c\eta)$). With a probit link, for example, the mean is $\Phi(\eta+o)$, and the ratio $\Phi(\eta+o)/\Phi(\eta)$ generally depends on $\eta$.

A practical interpretation should therefore name both scales: the offset shifts the linear predictor by its known amount, while the inverse link determines the resulting mean-scale change. When interpreting probabilities or other constrained means, also check that the shifted predictor maps to an admissible mean. This is especially important for links whose mean effect changes rapidly over some parts of the predictor range.

Check Your Understanding

Use the inverse link and the baseline predictor in each answer.

For a model with inverse link $h(\eta)=\Phi(\eta)$, write the mean with offset $o$ and its difference from the mean with zero offset.
For an identity-link model, what is the mean change from offset zero to $o$, provided the predictor path remains in the valid mean range?
Why does an offset under a log link multiply the mean by a factor independent of the baseline predictor?
What regularity condition in the finite-change result justifies applying the fundamental theorem of calculus?
For an inverse link $h(\eta)=C\exp(c\eta)$, find the mean ratio after adding offset $o$.

Offsets Under Noncanonical Links

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