Centering Predictors and Interpreting Intercepts

Why Center a Predictor?

A numerical predictor is often measured from a convenient zero that is not typical of the observations. For example, age may be recorded in years even when the observed ages are all near 50. In a model with an intercept, centering age at its sample mean changes the reference point for the intercept: it describes the linear predictor at the average age rather than at age zero. The central algebraic fact is that this change of reference point does not change the fitted linear predictors.

Centering is a change in how the same linear predictor is represented, not a new predictor effect. We will derive the exact coefficient transformation, check it numerically, and distinguish the intercept’s linear-predictor meaning from a response-scale interpretation.

One Numerical Predictor

Suppose a model includes an intercept and one numerical predictor, with linear predictor \(\eta_i=\beta_0+\beta_1x_i\). Write \(x_i=x_{ci}+\bar{x}\) and substitute:

Thus the centered representation has the same slope and a changed intercept. The new intercept is the old linear predictor evaluated at the predictor mean, because when \(x_i=\bar{x}\), the centered value is zero.

Several Numerical Predictors

The same operation applies when the model contains several numerical predictors. Let \(\eta_i=\beta_0+\sum_{j=1}^{p}\beta_jx_{ij}\), where \(x_{ij}\) is predictor \(j\) for observation \(i\). Center each predictor at its own sample mean, \(x_{cij}=x_{ij}-\bar{x}_j\). Substituting \(x_{ij}=x_{cij}+\bar{x}_j\) gives:

All slopes remain unchanged; the intercept increases by the sum of each slope multiplied by its predictor mean. The new intercept is the linear predictor at the point where every centered predictor is zero, meaning the vector of predictor values is \((\bar{x}_1,\ldots,\bar{x}_p)\). That point need not correspond to an actual observation.

Why the Fitted Linear Predictors Are Preserved

The preceding substitution gives the transformation directly. The next result makes its scope explicit: for a model with an intercept, centering numerical predictors changes the coefficient coordinates but leaves every possible vector of linear predictors available in the model unchanged.

Proof. Define \(\beta_{jc}=\beta_j\) for \(j=1,\ldots,p\), and define \(\beta_{0c}=\beta_0+\sum_{j=1}^{p}\beta_j\bar{x}_j\). Since \(x_{ij}=x_{cij}+\bar{x}_j\), for every observation \(i\),

This proves that the transformed coefficients yield the same linear predictors. To show the correspondence is reversible, start with any centered coefficients and set \(\beta_j=\beta_{jc}\) for each \(j\), and \(\beta_0=\beta_{0c}-\sum_{j=1}^{p}\beta_{jc}\bar{x}_j\). Substitution recovers the centered coefficients. The forward and reverse transformations are therefore inverses, so the correspondence is one-to-one. \(\square\)

This result is about the linear predictor. In a GLM, the model’s other ingredients act on that predictor, so a re-expression that gives the same \(\eta_i\) for every observation also preserves whatever model quantities are determined by those linear predictors. In particular, centering alone does not change the model’s fitted linear predictors.

What Centering Does—and Does Not—Say

Centering is useful because an intercept at a typical predictor value can be easier to describe than one at zero, particularly when zero is far outside the observed range. It can also make the intercept’s reference point explicit in a model with several numerical predictors: when all are centered, the intercept is the linear predictor at their joint sample-mean point, with any other predictors held at their specified reference values.

The intercept remains a value on the linear-predictor scale. It should not automatically be described as a mean response. Its relation to the response depends on the model’s link and other specifications. Likewise, the joint mean point can be mathematically well-defined even if no individual in the data has exactly that combination of predictor values.

Centering is defined relative to a chosen reference value. The sample mean is common, but the same algebra works if a different constant \(c\) is chosen and \(x_{ci}=x_i-c\): the centered intercept becomes \(\beta_0+\beta_1c\). The choice of reference affects the interpretation of the intercept, not the slope or the represented linear predictors. This is centering, not a change in the predictor’s units.