Why Functional Form Matters
An untreated natural-history model describes how response would change over time without active drug treatment. The choice of trajectory is consequential: models can describe the observed follow-up similarly yet imply very different future responses. A linear trajectory continues at a constant rate, an exponential trajectory accelerates, and a saturating trajectory approaches a limiting change. Comparing them requires attention both to fit within observed follow-up and to the assumptions built into extrapolation.
For this comparison, let \(Y(t)\) denote the response at time \(t\), measured in response units, and let \(Y_0=Y(0)\) denote baseline response. Define the baseline-centered change as \(D(t)=Y(t)-Y_0\). We take increasing response to mean worsening in the examples; if a clinical endpoint is oriented so that increasing values mean improvement, the signs and clinical interpretation must be adjusted. The data are hypothetical and represent an untreated group with no active drug effect.
The three candidate forms below assume time \(t\) is measured from baseline in years. Their parameters have different units and meanings, so values should not be compared across models as if they were interchangeable.
For increasing worsening, the linear rate has \(\beta>0\); the exponential form has \(A>0\) and \(k>0\); and the saturating form has \(M>0\) and \(k>0\). Here \(\beta\) is in response units per year, \(A\) and \(M\) are in response units, and \(k\) is in inverse years. \(A\) scales exponential change, while \(M\) is the limiting change under the saturating model. In the exponential model, the initial rate is \(Ak\); in the saturating model, the initial rate is \(Mk\).
What the Shapes Assume
The linear model assumes a constant rate of change: its slope is \(\beta\) at every time. The exponential form assumes an increasing rate for \(A,k>0\), because its derivative is \(Ak e^{kt}\). It has no finite upper asymptote. The saturating model assumes a decreasing rate: its derivative is \(Mk e^{-kt}\), which tends toward zero, and its predicted change approaches \(M\) as time increases. These are substantive assumptions about disease progression, not merely different curve-drawing choices.
Each expression is baseline-centered because substituting \(t=0\) gives \(D(0)=0\). This is a structural property of the trajectory. It does not mean a measured baseline value is error-free, nor that every individual starts at the same response. In an analysis with subject-specific baselines or measurement error, those features must be represented separately.
A comparison is meaningful only when the models are evaluated on the same response scale, time origin, untreated population, and observations. The measured follow-up must also be adequate to reveal curvature: a short, sparse series may not distinguish a nearly linear segment from part of an exponential or saturating curve. For a basic least-squares comparison, one may minimize the sum of squared residuals, but the residual pattern and scientific plausibility remain important; a smaller in-sample sum alone does not establish the correct long-term form.
Worked Example: Comparing Candidate Curves on Untreated Data
Suppose a hypothetical untreated cohort has mean baseline-centered worsening of 0 response units at baseline, 2.0 at year 1, 3.2 at year 2, and 3.9 at year 3. Consider the following candidate parameterizations: a linear least-squares fit through baseline, \(D(t)=1.4357t\); a saturating curve, \(D(t)=5(1-e^{-0.5t})\); and an illustrative exponential curve, \(D(t)=14.4(e^{0.1t}-1)\). The latter two are parameterizations for demonstrating shape comparison, not claims that all three parameters were estimated by the same optimization procedure.
For the linear fit through baseline, the least-squares slope is \(\sum t_iD_i/\sum t_i^2\). Using the three post-baseline observations gives \((1(2.0)+2(3.2)+3(3.9))/(1^2+2^2+3^2)=20.1/14=1.4357\) response units per year. For example, the year-2 prediction is \(1.4357(2)=2.8714\) response units.
The saturating curve predicts \(5(1-e^{-0.5})=1.967\) at year 1, \(5(1-e^{-1})=3.161\) at year 2, and \(5(1-e^{-1.5})=3.884\) at year 3. For the exponential curve, the corresponding predictions are \(14.4(e^{0.1}-1)=1.514\), \(14.4(e^{0.2}-1)=3.188\), and \(14.4(e^{0.3}-1)=5.038\) response units.
Residuals are defined here as observed minus fitted: \(e_i=D_i-\widehat{D}_i\). The table shows the observed values, predictions, and resulting residuals. Small differences in the final digit can result from rounding.
| Time (years) | Observed change | Linear prediction | Linear residual | Saturating prediction | Saturating residual | Exponential prediction | Exponential residual |
|---|---|---|---|---|---|---|---|
| 1 | 2.000 | 1.436 | 0.564 | 1.967 | 0.033 | 1.514 | 0.486 |
| 2 | 3.200 | 2.871 | 0.329 | 3.161 | 0.039 | 3.188 | 0.012 |
| 3 | 3.900 | 4.307 | -0.407 | 3.884 | 0.016 | 5.038 | -1.138 |
For the linear fit, the sum of squared post-baseline residuals is approximately \(0.564^2+0.329^2+(-0.407)^2=0.592\) response-units squared. The saturating candidate gives approximately \(0.033^2+0.039^2+0.016^2=0.0029\) response-units squared, while the illustrative exponential candidate gives approximately \(0.486^2+0.012^2+(-1.138)^2=1.531\) response-units squared. On these observations, the stated saturating curve is closest. This comparison does not prove that the disease truly has a plateau: it compares particular parameterizations over only three post-baseline times.
Reading Residual Patterns
A residual is a signed difference between an observation and its fitted value. Under the convention \(e_i=D_i-\widehat{D}_i\), a positive residual means the curve is below the observation, and a negative residual means it is above. Plotting residuals against time can reveal a pattern that a single summary of squared error hides. In the example, the linear residuals are positive at years 1 and 2 and negative at year 3. That sequence suggests the linear curve is too low relative to the earlier observations and too high relative to the later one. It is consistent with a curve that bends downward, but does not prove the functional form is wrong: measurement noise, changing conditions, or other features could also contribute.
Residuals that fluctuate around zero without an obvious time trend are more compatible with an adequate mean trajectory than a systematic sequence, although a small dataset gives limited diagnostic power. A smooth residual curve, long runs of one sign, or curvature in residuals can indicate that the chosen trajectory misses structure. Repeated measures from one participant are not independent observations; a residual display can still be informative, but formal uncertainty calculations must account for the data structure rather than treating every measurement as independent.
Worked Example: Interpreting the Sign of a Residual Sequence
A model predicts 4.0 response units at an early visit when the observed change is 4.6, and predicts 7.2 units at a later visit when the observed change is 6.5. Calculate and interpret the residuals using observed minus fitted.
At the early visit, \(e=4.6-4.0=+0.6\) response units. The model underpredicts the observation by 0.6 units. At the later visit, \(e=6.5-7.2=-0.7\) response units. The model overpredicts the observation by 0.7 units. This positive-early, negative-late pattern suggests a trajectory that is too low early and too high later, as may occur when a straight line is imposed on a flattening trend. It is a diagnostic clue, not conclusive evidence about the biological mechanism.
The residuals should also be examined in relation to baseline value, study visit, and relevant grouping variables when those are part of the dataset. A time pattern could reflect a misspecified time course, but it could also arise from systematic changes in measurement, dropout, or cohort composition. Residuals help identify questions to investigate; they do not by themselves select a causal explanation.
Why Long-Term Predictions Separate the Models
Observed fit is local to the times sampled. Long-term predictions expose the consequences of the trajectory's shape. The linear and increasing exponential forms have no upper limit for positive parameters, whereas the saturating form is bounded above by \(M\). This difference may matter clinically if the response has a known scale or if forecasts extend beyond the observed period. Extrapolation is especially uncertain when the data contain little information about curvature or an eventual plateau.
Worked Example: Projecting the Same Curves Beyond Follow-Up
Using the three candidate trajectories from the first example, calculate predicted change at year 10. For the linear model, \(D(10)=1.4357(10)=14.357\) response units. For the exponential model, \(D(10)=14.4(e^1-1)\approx24.743\) response units. For the saturating model, \(D(10)=5(1-e^{-5})\approx5(0.993262)\approx4.966\) response units.
At year 10, therefore, the same short follow-up supports candidate forecasts ranging from about 5.0 to 24.7 response units. The saturating forecast is close to its assumed 5-unit limit; the linear and exponential forecasts continue upward, with the exponential increasing most rapidly. These are consequences of the stated mathematical forms, not independently verified predictions. If the outcome has a possible range incompatible with a forecast, that is a reason to question the model or its extrapolation, not a reason to silently cap the prediction.
Comparing long-range forecasts is a sensitivity analysis: it shows how conclusions depend on functional form. It does not mean that a model should be selected solely because its forecast seems intuitively reasonable. Conversely, a slightly better short-term fit is not sufficient justification for distant extrapolation. The observed time span, measurement precision, plausibility of rate changes, and intended use of the model all matter.
Choosing and Checking a Trajectory
The main alternatives differ in both flexibility and extrapolation. The linear model is parsimonious and interpretable when a constant rate is credible over the period of interest. The exponential model is appropriate when accelerating change is scientifically defensible and supported by observations spanning enough time to reveal acceleration. The saturating model is useful when a slowing trajectory and finite limiting change are plausible, but it assumes that such a limit is meaningful. These models are not interchangeable approximations at all time horizons.
- Check the time scale and baseline anchor. Confirm that time zero corresponds to the intended baseline and that predicted change equals zero there.
- Inspect observed and fitted trajectories. A curve should represent the direction and curvature supported by the observations, not merely pass near one or two points.
- Plot observed-minus-fitted residuals against time. Look for sign runs, curvature, or other systematic structure, interpreting positive and negative signs according to the stated convention.
- Compare forecasts over a scientifically relevant horizon. Report how conclusions change across plausible forms instead of presenting a distant projection as if it were determined by the observed data.
A common mistake is to choose the model with the smallest residual sum of squares and then treat its long-term forecast as established. The correction is to distinguish in-sample fit from extrapolation: residual error describes agreement at observed times, while the long-term forecast follows from assumptions about how the trajectory continues. A second common mistake is to reverse residual signs. With observed minus fitted, a positive residual means the model is too low, not too high, relative to the observation.
Check Your Understanding
Answer each question, then compare with the brief solutions below.
- For \(D(t)=M(1-e^{-kt})\), with \(M,k>0\), what change does the model predict at baseline, and what does it approach as time increases?
- A model predicts 5.2 response units and the observed change is 4.7 units. What is the residual under the observed-minus-fitted convention, and what does its sign mean?
- Why can two models with similar residual error over three years give very different predictions at year 10?
- State one assumption of the linear form and one assumption of the exponential form for increasing progression.
- A fitted curve has positive early residuals and negative later residuals. State the direction of its mismatch relative to the observations, and give one reason this pattern is not definitive proof of misspecification.
Solutions. (1) At \(t=0\), \(D(0)=M(1-1)=0\); as \(t\) increases without bound, \(e^{-kt}\) tends to zero and \(D(t)\) tends to \(M\). (2) The residual is \(4.7-5.2=-0.5\) response units, so the fitted value is 0.5 units higher than observed. (3) Their mathematical forms impose different rates and limits outside the observed interval; a nearly straight segment does not determine which form continues. (4) Linear progression assumes a constant rate; increasing exponential progression assumes a rate that grows over time for positive parameters. (5) The fitted curve is too low relative to early observations and too high relative to later observations. Noise, changing conditions, or other data features could also produce a pattern, so the residual sequence is suggestive rather than conclusive.