Why Center a Progression Model at Baseline?
A progression model can describe the response on its original scale, or describe how that response changes from its predicted value at baseline. These are two ways to express the same trajectory when the parameter conversion is done correctly. Centering is useful because it makes the model prediction at time zero explicit and separates the baseline response from the modeled change over time.
Here, \(R(t)\) denotes the model-predicted response at time \(t\), and \(t=0\) is the baseline time origin. The change predicted from baseline is \(R(t)-R(0)\). A baseline-centered expression does not imply that the measured baseline value is error-free or that every individual begins at the same response; it is an algebraic way to write the trajectory. The response scale and time units must be stated for the parameters to be interpreted.
The conversions below assume time is measured from baseline, the stated functional form applies over the time range of interest, and the response is modeled on an additive scale. They concern the natural-history component alone; other model components would need their own definitions and parameterization.
Linear Progression: The Intercept Is the Baseline
An absolute linear trajectory can be written \(R_{\mathrm{abs}}(t)=\beta_0+\beta_1t\). Evaluating it at baseline gives \(R_{\mathrm{abs}}(0)=\beta_0\). Subtracting that baseline prediction from the trajectory yields the centered change:
Therefore the centered form is \(R(t)=B+\beta_1t\), where \(B=\beta_0\). Only the intercept's interpretation is made explicit: it is the predicted response at baseline. The slope remains the same parameter, with units of response per unit time. At every time point, both expressions predict exactly the same response.
Worked Example: Centering a Linear Trajectory
Suppose a hypothetical response is measured in score units and time in weeks. Its absolute-response model is \(R_{\mathrm{abs}}(t)=120+2.4t\). Find the baseline-centered form and the predicted response and change from baseline at 8 weeks.
At \(t=0\), \(R_{\mathrm{abs}}(0)=120+2.4(0)=120\) score units, so \(B=120\). Subtracting that baseline gives \(R(t)-B=(120+2.4t)-120=2.4t\). The centered model is therefore \(R(t)=120+2.4t\), with a predicted change of \(2.4t\) score units.
At 8 weeks, the change is \(2.4(8)=19.2\) score units, and the predicted response is \(120+19.2=139.2\) score units. Direct substitution into the absolute form gives \(120+2.4(8)=139.2\), confirming equivalence.
Exponential Progression: The Baseline and Exponential Component Must Both Be Tracked
Consider the absolute-response form \(R_{\mathrm{abs}}(t)=C+A\exp(kt)\). Here \(C\) is a constant offset, \(A\) is the coefficient of the exponential term, and \(k\) is a rate constant with units of inverse time. At baseline, \(B=C+A\), not \(C\). Subtracting that baseline gives:
Thus the equivalent centered form is \(R(t)=B+A[\exp(kt)-1]\), with \(B=C+A\). In this conversion, \(A\) remains the coefficient of the exponential component, while the constant offset \(C\) is replaced by the predicted baseline \(B\). The inverse conversion is \(C=B-A\). The two parameterizations have identical predictions when these relationships hold.
The sign of the change is not determined by \(k\) alone. Differentiating the absolute or centered form gives \(R'(t)=Ak\exp(kt)\). Because \(\exp(kt)>0\), the derivative's sign is the sign of \(Ak\). If \(A>0\), positive \(k\) gives an increasing trajectory and negative \(k\) a decreasing one; if \(A<0\), those directions reverse. If \(A=0\), the response is constant for any \(k\). The rate \(k\) controls the exponential time scale, while \(A\) also affects the direction and magnitude of change.
Worked Example: Centering an Exponential Trajectory
Suppose a hypothetical biomarker, measured in units, follows \(R_{\mathrm{abs}}(t)=80+20\exp(0.15t)\), where time is in weeks. Derive the centered model and calculate the predicted response and change at 4 weeks.
The baseline prediction is \(B=80+20\exp(0)=100\) units. The centered form is \(R(t)=100+20[\exp(0.15t)-1]\). Equivalently, \(C=B-A=100-20=80\), so converting back recovers the original absolute model.
At 4 weeks, \(0.15(4)=0.6\) and \(\exp(0.6)\approx1.8221\). The change from baseline is \(20(1.8221-1)=16.44\) units, and the response is \(100+16.44=116.44\) units. Substitution in the absolute form gives \(80+20(1.8221)=116.44\) units. Since \(A=20>0\) and \(k=0.15>0\) per week, the derivative is positive and the trajectory increases.
Saturating Progression: Separate the Baseline From the Plateau
One saturating time-course model is \(R_{\mathrm{abs}}(t)=C+E_{\max}t/(T_{50}+t)\), for \(t\geq0\). Assume \(T_{50}>0\). The progression term equals zero at baseline, so \(R_{\mathrm{abs}}(0)=C\), and the baseline-centered form is:
The predicted change from baseline is \(E_{\max}t/(T_{50}+t)\). For positive \(E_{\max}\), this change approaches \(E_{\max}\) as time grows; accordingly, the absolute-response asymptote is \(B+E_{\max}\), not \(E_{\max}\). At \(t=T_{50}\), the change is \(E_{\max}/2\). Thus \(T_{50}\) is the time at which half of the asymptotic change has occurred, under this model. The reparameterization leaves both \(E_{\max}\) and \(T_{50}\) unchanged; it makes explicit that \(E_{\max}\) is a change from baseline rather than the absolute plateau.
Worked Example: Centering a Saturating Trajectory
Suppose a hypothetical clinical score has the absolute-response model \(R_{\mathrm{abs}}(t)=50+36t/(6+t)\), with response in score units and time in weeks. Write its centered form and calculate the response at 6 and 18 weeks.
At time zero, \(R_{\mathrm{abs}}(0)=50+36(0)/(6+0)=50\) score units. Therefore \(B=50\), and the centered model is \(R(t)=50+36t/(6+t)\). The baseline-centered change is \(36t/(6+t)\); \(T_{50}=6\) weeks and the asymptotic change is 36 score units.
At 6 weeks, the change is \(36(6)/(6+6)=216/12=18\) score units, so the predicted response is \(50+18=68\) score units. At 18 weeks, the change is \(36(18)/(6+18)=648/24=27\) score units, so the response is \(50+27=77\) score units. The model's absolute-response asymptote is \(50+36=86\) score units.
Parameter Mapping and Equivalence
A reliable conversion starts by evaluating the absolute-response model at \(t=0\), then subtracting that value from the full expression. This order prevents the baseline from being confused with an offset, coefficient, or asymptote.
Calculate \(B=R_{\mathrm{abs}}(0)\) using the absolute-response model.
Simplify \(R_{\mathrm{abs}}(t)-B\) to obtain the modeled change.
State which parameter is relabeled or transformed, and check that the centered model returns \(B\) at \(t=0\).
The equivalence is direct: if \(g(t)=R_{\mathrm{abs}}(t)-R_{\mathrm{abs}}(0)\), then \(R_{\mathrm{abs}}(t)=R_{\mathrm{abs}}(0)+g(t)\) by addition. Conversely, if \(R(t)=B+g(t)\) and \(g(0)=0\), then \(R(0)=B\), and subtracting \(R(0)\) recovers \(g(t)\). This verifies that centering changes representation, not the trajectory, provided the parameters obey the stated mapping.
| Absolute-response form | Baseline-centered form | Parameter interpretation |
|---|---|---|
| \(\beta_0+\beta_1t\) | \(B+\beta_1t\), \(B=\beta_0\) | The intercept is explicitly the baseline; slope is unchanged. |
| \(C+A\exp(kt)\) | \(B+A[\exp(kt)-1]\), \(B=C+A\) | The offset maps to \(B-A\); \(A\) and \(k\) retain their roles. |
| \(C+E_{\max}t/(T_{50}+t)\) | \(B+E_{\max}t/(T_{50}+t)\), \(B=C\) | \(E_{\max}\) is the asymptotic change; the absolute plateau is \(B+E_{\max}\). |
Checks, Common Errors, and Limitations
A first diagnostic is the baseline identity: substitute \(t=0\) into the centered expression and verify that it returns \(B\). A second is a numerical check at a nonzero time: calculate the prediction from both parameterizations after applying the mapping. Disagreement usually indicates an incorrect baseline, an omitted offset conversion, or a unit error. Also check whether the modeled shape is plausible over the observed time range; algebraic equivalence cannot make a chosen functional form appropriate for the data.
A common mistake is to replace the intercept or offset with the baseline while leaving the rest of an exponential expression untouched. For example, writing \(B+A\exp(kt)\) after defining \(B=C+A\) predicts \(B+A\) at \(t=0\), not \(B\). The correct centered expression is \(B+A[\exp(kt)-1]\). Likewise, in the saturating model, calling \(E_{\max}\) the absolute plateau is wrong: the plateau is \(B+E_{\max}\). These are parameter-interpretation errors, even though the equations may look similar.
Baseline-centering is an algebraic reparameterization, not a diagnostic test of the linear, exponential, or saturating shape. It does not establish that the baseline measurement equals the latent model value, resolve uncertainty in estimated parameters, or guarantee predictions remain sensible outside the time range where the form is useful. The main alternative is to retain the absolute-response form, which may be convenient when the offset or absolute asymptote is the scientific quantity of interest. The centered form is often clearer when the target is change from baseline. Either is suitable if the mapping is explicit and interpretations are consistent.
Check Your Understanding
For each question, use time zero as baseline and keep units attached to interpreted quantities. Solutions follow each question.
- A linear model is \(R_{\mathrm{abs}}(t)=75-1.2t\), with time in months and response in units. What is its centered form and its predicted change at 5 months?
- For \(R_{\mathrm{abs}}(t)=40+10\exp(0.2t)\), what is the baseline-centered form and the predicted response at 3 time units?
- For \(R_{\mathrm{abs}}(t)=12+24t/(8+t)\), identify the baseline, \(T_{50}\), asymptotic change, and absolute plateau.
- In \(R(t)=B+A[\exp(kt)-1]\), what is the sign of the initial direction of change when \(A<0\) and \(k>0\)? What if \(A=0\)?
Solutions. (1) \(B=75\), so \(R(t)=75-1.2t\) and the change at 5 months is \((-1.2)(5)=-6.0\) units. (2) \(B=40+10=50\), so \(R(t)=50+10[\exp(0.2t)-1]\). At \(t=3\), this is \(50+10[\exp(0.6)-1]\approx58.22\) response units. (3) \(B=12\) units, \(T_{50}=8\) time units, and \(E_{\max}=24\) units; the absolute plateau is \(12+24=36\) units. (4) \(R'(t)=Ak\exp(kt)\), so with \(A<0\) and \(k>0\) the derivative is negative and the trajectory decreases. If \(A=0\), the trajectory is constant for any \(k\).