Known Coefficients Versus Estimated Effects

Why Fix a Coefficient?

A predictor can enter a generalized linear model in two importantly different ways: its contribution to the linear predictor can be specified in advance, or its coefficient can be estimated from the observations. These choices may produce similar-looking equations, but they answer different scientific questions. Fixing a coefficient says its value is known and should not be adjusted to fit the current data. Estimating it asks what value best describes the association in these data.

The difference matters in applications such as exposure-adjusted rates, where the usual model assumes that doubling exposure doubles the expected count, and in calibrated models, where a previously specified effect may be carried into a new analysis. An offset represents such a fixed contribution. It does not make that contribution disappear; it places it in the predictor without giving it a parameter to estimate.

Definition: In a GLM with linear predictor $\eta_i$, an offset is a known quantity $o_i$ added to the predictor with its coefficient fixed at one: $$ \eta_i=x_i^\mathsf{T}\beta+o_i. $$ If $z_i$ is a predictor whose coefficient is fixed at one, then $o_i=z_i$. If its coefficient is fixed at a known value $c$, use $o_i=cz_i$. By contrast, estimating its coefficient means fitting $\eta_i=x_i^\mathsf{T}\beta+\gamma z_i$, where $\gamma$ is an unknown parameter.

The number one here refers to the coefficient multiplying the specified offset, not to the coefficient of an intercept or another predictor. The offset is known for every observation, just as the covariates are known; what distinguishes it is that its coefficient is not estimated. The remaining coefficients can still be estimated in the usual way.

What the Two Specifications Say

Let $g(\mu_i)=\eta_i$, where $g$ is the link function and $\mu_i=E[Y_i]$. Compare these specifications:

\begin{aligned} \text{Fixed contribution:}\quad &g(\mu_i)=x_i^\mathsf{T}\beta+z_i,\\ \text{Estimated effect:}\quad &g(\mu_i)=x_i^\mathsf{T}\beta+\gamma z_i. \end{aligned}

The first specifies in advance that a one-unit increase in $z_i$ changes the linear predictor by exactly one unit. The second lets the data determine the change, through $\gamma$. These are statements on the link scale; their implications for the mean depend on the link. For example, with a log link, a fixed unit change multiplies the mean by $e$, whereas an estimated coefficient $\gamma$ multiplies it by $e^\gamma$. With a logit link, the corresponding changes are in log odds, so exponentiating gives the odds multiplier.

Fixing a coefficient is a substantive restriction, not just a convenient way to write a model. It is appropriate when the coefficient is genuinely prescribed by the scientific setup or deliberately imposed as a modeling assumption. If the strength of the association is scientifically uncertain and the data are intended to inform it, estimate the coefficient instead. A coefficient taken from an earlier study is not automatically known without uncertainty; treating it as an offset treats its value as fixed for the current analysis.

Worked Example: Exposure and a Poisson Rate

Suppose $Y_i$ is a Poisson count observed over exposure $t_i>0$, and use a log link. The fixed-exposure model is $$ \log(\mu_i)=\beta_0+\log(t_i), \qquad \mu_i=E[Y_i]. $$

Exponentiating gives $\mu_i=e^{\beta_0}t_i$, so $$ \frac{\mu_i}{t_i}=e^{\beta_0}. $$ The model asserts a common expected count per unit exposure. The known quantity $\log(t_i)$ is an offset: its coefficient is fixed at one.

If the exposure coefficient is instead estimated, write $$ \log(\mu_i)=\beta_0+\gamma\log(t_i). $$ Then $\mu_i=e^{\beta_0}t_i^\gamma$, and the expected rate per unit exposure is $$ \frac{\mu_i}{t_i}=e^{\beta_0}t_i^{\gamma-1}. $$ Only when $\gamma=1$ is this rate constant across exposure levels. For $\gamma<1$, the expected rate decreases as exposure grows; for $\gamma>1$, it increases.

For example, compare two observations with exposures $t_1=1,t_2=4$ and counts $y_1=4,y_2=4$. Under the fixed-offset model, the Poisson log-likelihood, omitting terms constant in $\beta_0$, is $$ \ell(\beta_0)=\sum_{i=1}^2\{y_i(\beta_0+\log t_i)-e^{\beta_0}t_i\}. $$ Its derivative is $8-5e^{\beta_0}$, so the maximum occurs at $e^{\hat\beta_0}=8/5$. The fitted means are $1.6$ and $6.4$: the model allocates expected counts in proportion to exposure.

In the estimated-coefficient model, these two observations and two parameters allow the fitted means to equal the two observed counts. Setting $\hat\mu_1=\hat\mu_2=4$ gives $e^{\hat\beta_0}=4$ and $4^{\hat\gamma}=1$, hence $\hat\gamma=0$. The data in this example show no increase in expected count between the two exposure levels. The models therefore encode different assumptions, even though both include $\log(t_i)$ in the predictor.

Likelihood Consequences of the Restriction

For a fixed dataset, the fixed-offset model is the estimated-coefficient model restricted to $\gamma=1$. This simple relationship has a general likelihood consequence. Assume the same response distribution, link, and other model terms are used in both models, and that both likelihoods are defined on the same observations. Let $\ell_{\mathrm{free}}$ be the log-likelihood maximized over $(\beta,\gamma)$, and let $\ell_{\mathrm{fixed}}$ be the log-likelihood maximized over $\beta$ with $\gamma=1$.

Result: The maximized log-likelihood in the model with an estimated coefficient cannot be smaller than the maximized log-likelihood with that coefficient fixed at one: $$ \ell_{\mathrm{free}}\geq \ell_{\mathrm{fixed}}. $$

Proof. The parameter values allowed by the fixed model are exactly the subset of the free model’s parameter space with $\gamma=1$. Every parameter vector $(\beta,1)$ available to the fixed model is therefore available to the free model. Maximizing a function over a set cannot give a smaller supremum than maximizing it over a set containing it. Thus $$ \sup_{\beta,\gamma}\ell(\beta,\gamma)\geq \sup_\beta\ell(\beta,1). $$ When both maxima exist, these suprema are the stated maximized log-likelihoods. $\square$

This result does not say that the estimated-coefficient model is scientifically preferable: a more flexible model can fit at least as well simply because it has more freedom. Nor does it say that the estimated coefficient will differ from one in every dataset. It says only that fixing the coefficient imposes a restriction, and the unrestricted fit can retain that restriction if it suits the data.

Worked Example: Fixed Versus Estimated Effect in a Normal Model

Let $Y_i$ be independent normal responses with known variance one, and use the identity link. Suppose the observed pairs $(z_i,y_i)$ are $(0,1),(1,4),(0,2),(1,2)$. Compare a fixed coefficient of one with an estimated coefficient: $$ \mu_i=\alpha+z_i \qquad\text{versus}\qquad \mu_i=\alpha+\gamma z_i. $$

With variance one, maximizing the normal likelihood is equivalent to minimizing the sum of squared residuals. In the fixed model, the values $y_i-z_i$ are $1,3,2,1$. Their mean is $$ \hat\alpha_{\mathrm{fixed}}=\frac{1+3+2+1}{4}=\frac{7}{4}=1.75. $$ The fitted means are $1.75,2.75,1.75,2.75$. The residuals are $-0.75,1.25,0.25,-0.75$, so the residual sum of squares is $$ (-0.75)^2+(1.25)^2+(0.25)^2+(-0.75)^2 =0.5625+1.5625+0.0625+0.5625=2.75. $$

In the estimated model, the observations with $z=0$ have mean $(1+2)/2=1.5$, and those with $z=1$ have mean $(4+2)/2=3$. Thus $\hat\alpha=1.5$ and $\hat\gamma=3-1.5=1.5$. The fitted values are $1.5,3,1.5,3$, and the residual sum of squares is $$ (-0.5)^2+(1)^2+(0.5)^2+(-1)^2=2.5. $$ The estimated model fits better, as the general likelihood comparison predicts. It also gives a different scientific description: the mean changes by $1.5$ units per unit of $z$, rather than imposing a change of exactly one.

Interpreting a Known Effect on the Link Scale

Offsets can encode known contributions other than exposure. For instance, in a logistic regression a known log-odds contribution $o_i$ gives $$ \operatorname{logit}(\mu_i)=x_i^\mathsf{T}\beta+o_i. $$ The model uses that contribution as specified; it does not estimate an effect multiplying $o_i$. If instead the contribution is $\gamma z_i$, the odds multiplier for a one-unit increase in $z_i$, holding the other predictors fixed, is $e^\gamma$. Fixing its coefficient at one asserts the multiplier $e$, if $z_i$ is measured on the log-odds scale itself.

The units and construction of the predictor matter. If a covariate is measured in arbitrary units, fixing its coefficient at one may impose a very different substantive change than intended. A known effect $c$ multiplying a covariate $z_i$ is represented by the offset $cz_i$, not by $z_i$ alone unless $c=1$. Scaling or transforming a predictor changes the meaning of a coefficient fixed at one.

Common Mistakes and Practical Consequences

Calling an offset “a coefficient estimated as one.” An offset coefficient is not estimated. It is fixed at one by construction. A freely estimated coefficient that happens to have fitted value one is a different model.
Assuming the fixed coefficient is harmless notation. Fixing $\gamma=1$ removes possible values from the model and can change the fitted means and the other coefficient estimates.
Interpreting the coefficient on the mean scale without applying the link. The fixed unit change is on the linear-predictor scale. Its mean-scale implication follows from the particular inverse link.
Treating an uncertain external estimate as known. Using it as an offset does not account for uncertainty in its value. This is justified only when treating that value as fixed is appropriate for the analysis.

In practice, ask what is known and what is being learned from the current data. If a model needs the exposure-proportionality assumption, the log exposure belongs in the predictor as an offset. If the data should assess whether counts rise proportionally with exposure, estimate its coefficient instead. Neither choice is universally correct; each corresponds to a distinct scientific claim.

Worked Example: Choosing the Form for a Calibrated Predictor

Suppose a logistic model has predictor $\eta_i=\alpha+x_i^\mathsf{T}\beta+o_i$, where $o_i$ is a prespecified log-odds contribution for each observation. The scientific assumption is that this contribution is already calibrated and must enter without adjustment. Then it is appropriate to use $o_i$ as an offset, and the model estimates $\alpha$ and $\beta$, not a coefficient on $o_i$.

If instead the calibration may be too strong or too weak, introduce a known-scale predictor $z_i=o_i$ and estimate $$ \eta_i=\alpha+x_i^\mathsf{T}\beta+\gamma o_i. $$ The fitted odds multiplier associated with a one-unit increase in $o_i$, with the other predictors held fixed, is $e^\gamma$. The offset specification corresponds to imposing $\gamma=1$; it is not evidence from the current data that $\gamma=1$.

Thus the choice follows the question. “Use the calibrated contribution exactly” calls for an offset. “Check how strongly the contribution is associated with the outcome” calls for an estimated coefficient. The two specifications can be compared as a restricted and an unrestricted model, but their scientific interpretations should be stated before interpreting fit.

Key takeaway: An offset is a known contribution to the linear predictor, with coefficient fixed at one. Estimating the corresponding coefficient relaxes that restriction and lets the data determine the link-scale effect. The choice changes both the model’s assumptions and the interpretation of its fitted effects.

Check Your Understanding

Use the distinction between a fixed predictor contribution and an estimated coefficient to answer these questions.

Write a linear predictor with known coefficient $c$ multiplying $z_i$ as a model with an offset.
In a Poisson model with a log link, what does fixing the coefficient of $\log(t_i)$ at one assert about the expected rate?
If the coefficient of $z_i$ is estimated as $\gamma$ in a log-link model, what multiplicative change in the mean corresponds to a one-unit increase in $z_i$?
Why can the maximized log-likelihood of the model with an estimated coefficient not be lower than that of the model fixing the coefficient at one?
A coefficient from an earlier study is inserted as an offset. What uncertainty assumption is being made about that coefficient?