Why Center a Covariate?
In a covariate model, a coefficient’s numerical value is only part of its meaning. The point at which the model’s intercept is defined and the units used for a slope also matter. If weight is recorded in kilograms, an intercept in a model using raw weight describes the parameter at weight zero. That value may be irrelevant to the population being studied, even when the model fits the observed data well.
Centering replaces the covariate with its difference from a chosen reference value. Rescaling expresses that difference in a more convenient unit. These changes can make the intercept and slope easier to interpret, but they do not by themselves change the relationship the model represents. The essential condition is that the transformed coefficients are updated so that the model gives the same value at every covariate value.
Here \(c\) has the same units as \(x\), and \(a\) is the number of those units represented by one unit of \(z\). For example, if \(x\) is weight in kilograms, \(c=70\) kg, and \(a=10\) kg, then \(z\) is weight relative to 70 kg, measured in 10-kg units. The transformed intercept is the model’s value at 70 kg; the transformed slope is the change per 10 kg.
Assumptions and the Reparameterization
The algebra applies when the covariate enters the model as an intercept plus a linear term on the chosen parameter or predictor scale. The same observations, covariate values, response model, and random-effects structure must be retained. The scale constant must be nonzero, and the transformed coefficients must not be held fixed at their old numerical values. For ordinary interpretation, \(a\) is usually positive; a negative value would also be an invertible transformation, but would reverse the slope’s sign convention.
In NLME applications, \(\eta\) may be a transformed parameter, such as a parameter represented on a log scale. Centering changes the coefficients on that same scale. It does not choose a new parameter scale or change the transformation connecting \(\eta\) to the pharmacokinetic parameter. This distinction is important: covariate coding and the scale on which the parameter model is specified are separate decisions.
Thus \(\beta_0^*=\beta_0+c\beta_1\) and \(\beta_1^*=a\beta_1\). Substitution proves prediction invariance directly: \(\beta_0^*+\beta_1^*z=\beta_0+c\beta_1+a\beta_1(x-c)/a=\beta_0+\beta_1x\). If an individual-level predictor also contains a random effect, for example \(\eta_i=\beta_0+\beta_1x_i+b_i\), the same substitution leaves \(b_i\) unchanged and preserves that individual’s predictor as well.
Because the fitted predictor is identical observation by observation, the response predictions generated from it are identical too, provided the rest of the model is unchanged. In a likelihood-based NLME model, the conditional likelihood for each subject at a given random effect is therefore unchanged; integrating over the same random-effects distribution leaves the marginal likelihood unchanged. This is a re-expression of the same model, not an additional covariate effect.
Worked Example: Centering Weight for Clearance on Its Direct Scale
Suppose a hypothetical model describes clearance \(CL\), in L/h, using weight \(W\), in kg:
The intercept 2.20 L/h is the model value at zero kg. To interpret the intercept at 70 kg, take \(c=70\) kg and \(a=1\) kg. The transformed coefficients are \(\beta_0^*=2.20+(0.0400)(70)=5.00\) L/h and \(\beta_1^*=(1)(0.0400)=0.0400\) L/h per kg. The centered expression is \(CL=5.00+0.0400(W-70)\).
At 80 kg, the original expression gives \(CL=2.20+(0.0400)(80)=2.20+3.20=5.40\) L/h. The centered expression gives \(CL=5.00+(0.0400)(80-70)=5.00+0.400=5.40\) L/h. At the reference value, it gives \(CL=5.00\) L/h. Centering has made the intercept meaningful without changing any predicted clearance.
Centering on a Transformed Parameter Scale
The same coefficient transformation applies if the model is written for a transformed parameter rather than for the parameter directly. The intercept then describes the transformed parameter at the reference covariate value. To interpret a parameter on its original scale, apply the model’s existing inverse transformation; do not treat the transformed-scale intercept as though it were already on the original scale.
Worked Example: Centering Weight in a Log-Scale Volume Model
Consider the hypothetical model \(\ln(V/L)=2.6389+0.01500W\), where \(V\) is volume in litres and \(W\) is weight in kilograms. Re-express the covariate around 70 kg, using \(a=1\) kg. The centered intercept is \(2.6389+(0.01500)(70)=2.6389+1.050=3.6889\); the slope remains \(0.01500\) per kg. Therefore, \(\ln(V/L)=3.6889+0.01500(W-70)\).
At 80 kg, the original predictor is \(2.6389+(0.01500)(80)=2.6389+1.200=3.8389\). The centered predictor is \(3.6889+(0.01500)(10)=3.8389\), the same value. Applying the model’s inverse transformation gives \(V=L\exp(3.8389)\), approximately 46.48 L, either way. At 70 kg, the centered predictor is 3.6889, corresponding to approximately 40.00 L. The centered intercept is on the log scale; it is not itself a volume of 3.6889 L.
Rescaling to Clinically Useful Units
Centering and rescaling can be done together. A useful choice is often a reference value that has a clear interpretation and a scale unit that makes the coefficient readable. The choice does not need to be the sample mean. A clinically meaningful value can be preferable, especially when it is well represented in the data. If the reference is far outside the observed range, the intercept is still defined algebraically but may describe an extrapolation rather than a well-supported value.
Worked Example: Expressing an Age Slope per Decade
Suppose a hypothetical model for an absorption parameter \(P\), in hours, is \(P=0.30+0.0120A\), with age \(A\) measured in years. Its raw intercept is the model value at age zero. Let the reference be 50 years and choose \(a=10\) years, so \(z=(A-50)/10\) measures decades from age 50.
The centered intercept is \(0.30+(0.0120)(50)=0.90\) h. The slope per decade is \((10)(0.0120)=0.120\) h per decade. Thus \(P=0.90+0.120z\). At age 65, \(z=(65-50)/10=1.5\), and the centered prediction is \(0.90+(0.120)(1.5)=0.90+0.180=1.08\) h. The original expression gives \(0.30+(0.0120)(65)=0.30+0.780=1.08\) h. The slope’s numerical value changed because its units changed; the fitted values did not.
Coefficient Uncertainty, Diagnostics, and Limitations
The coefficient estimates are re-expressed along with their uncertainty. For centering alone, where \(z=x-c\), the transformed intercept is \(\beta_0^*=\beta_0+c\beta_1\). Its estimated variance is therefore
This follows from the variance of a linear combination. The covariance term matters: simply carrying over the old intercept’s standard error is generally wrong. Rescaling also changes the slope’s units and standard error by the same factor \(a\), for positive \(a\). An interval for the transformed coefficient should be calculated from the transformed estimates and their covariance, or by transforming the full covariance matrix.
A practical check is to compute predictions before and after recoding at several covariate values, including the reference value. They should agree up to rounding. If they do not, check whether the intercept was updated, whether the same units were used in \(c\) and \(x\), and whether the scale factor was applied to the slope in the correct direction. Also check that the reference value lies in a substantively useful part of the covariate range and that the stated slope units match the coding.
Centering can sometimes improve numerical conditioning when an intercept and slope are strongly correlated, especially when the covariate’s zero is far from its observed values. It does not guarantee better convergence, remove confounding, create information about poorly sampled covariate values, or correct a misspecified covariate relationship. It also does not change the model’s substantive fit when carried out as an exact reparameterization.
The main alternative is to retain the raw covariate. That coding is valid and may be convenient when zero has a meaningful interpretation, but its intercept is then tied to zero. Standardizing by a sample standard deviation is another rescaling choice; it expresses the slope per standard-deviation unit, which can aid numerical comparison but may be less clinically intuitive. Neither choice is inherently superior. Select \(c\) and \(a\) to serve interpretation and computation while keeping the model and units explicit.
Set \(c\) to an interpretable covariate value and \(a\) to the number of original units represented by one rescaled unit.
Use \(\beta_0^*=\beta_0+c\beta_1\) and \(\beta_1^*=a\beta_1\); retain the same parameter scale and remaining model components.
Check equality of predictions at representative covariate values, then state the intercept at \(x=c\) and the slope per \(a\) original units.
Check Your Understanding
Answer each question, then compare with the solutions below.
- A model is \(\eta=1.5+0.20x\). Recenter at \(c=4\), without rescaling. Give both transformed coefficients.
- For \(\eta=3.0-0.05x\), define \(z=(x-20)/10\). Give the transformed model and calculate its prediction at \(x=30\).
- In a log-parameter model, what does the centered intercept represent, and what extra step is needed to report the parameter on its original scale?
- Why is it generally incorrect to keep the raw intercept after replacing \(x\) with \(x-c\)?
Solutions. (1) The intercept is \(1.5+(0.20)(4)=2.3\), and the slope remains \(0.20\), so \(\eta=2.3+0.20(x-4)\). (2) The intercept is \(3.0+(-0.05)(20)=2.0\), and the slope per 10 units is \(10(-0.05)=-0.50\). Thus \(\eta=2.0-0.50z\); at \(x=30\), \(z=(30-20)/10=1\), so \(\eta=2.0-0.50=1.5\). (3) It is the transformed parameter’s model value at \(x=c\); apply the specified inverse transformation to express the value on the original parameter scale. (4) The intercept must absorb \(c\beta_1\); otherwise the predictor changes by \(c\beta_1\), altering fitted predictions.