Diagnosing Disease-Progression Functional-Form Errors

Why Progression-Shape Diagnostics Matter

A disease-progression model describes how an outcome would change over time in the absence of active drug, after baseline and any modeled placebo response have been accounted for. Choosing its functional form affects more than the fit at observed visits: it also determines what the model predicts between visits and beyond the follow-up period. A straight-line model continues at a constant rate; an exponential model accelerates or decelerates according to its rate parameter; a saturating model approaches a limiting value. These distinct implications make untreated follow-up especially useful for diagnosing progression-shape errors.

The aim is not to select a curve from one striking residual or to assume that the most flexible model is best. Instead, inspect whether residuals show a systematic pattern over time, check whether observed changes are consistent with the proposed curvature, and assess whether the implied extrapolation is scientifically plausible. These checks presume that baseline and placebo components have been handled adequately; an unmodeled time-varying placebo response can look like misspecified disease progression.

Definition: Let \(Y_{ij}\) be the outcome for individual \(i\) at time \(t_{ij}\), and let \(m(t)\) be the model-predicted untreated mean outcome at time \(t\), after accounting for the specified baseline and placebo components. The residual is \(r_{ij}=Y_{ij}-m(t_{ij})\). A positive residual means the observed outcome is above its model prediction; a negative residual means it is below.

Expected Shapes and Their Diagnostic Implications

For illustration, suppose the outcome increases as disease progresses and time is measured from baseline. Three common mean trajectories are:

\begin{aligned} \text{Linear:}\quad &m(t)=B+\beta t,\\ \text{Exponential:}\quad &m(t)=B+A\{\exp(kt)-1\},\\ \text{Saturating:}\quad &m(t)=B+S\{1-\exp(-kt)\}. \end{aligned}

Here \(B\) is the baseline mean on the outcome scale. In the exponential form, \(A>0\) and \(k>0\) give an increasing trajectory whose rate of increase grows over time. In the saturating form, \(S>0\) and \(k>0\) give an increasing trajectory that approaches \(B+S\). The linear slope \(\beta\) is the constant change in outcome per unit time. These parameter restrictions describe the examples below; other clinical settings may require decreasing trajectories or other parameterizations.

The linear trajectory has zero curvature. The increasing exponential trajectory is convex: its slope increases with time. The increasing saturating trajectory is concave: its slope decreases toward zero. If a straight reference is the chord joining the endpoints of an interval, it lies above a convex curve and below a concave curve between those endpoints. This chord relationship helps explain expected deviations from linear interpolation, but it is not a universal rule for a least-squares fitted line: the latter depends on the observation times, weights, and fitting procedure.

Using Residuals and Successive Differences

Plot residuals against time, preferably also showing individual trajectories or subject-level summaries when repeated measurements are available. A pattern of positive residuals at early and late times with negative residuals in the middle can be consistent with a convex trajectory fitted by a straight line. The reverse pattern can be consistent with concavity. Look for persistence across subjects or visits, not just one isolated deviation. Residuals should fluctuate around zero without a systematic time trend if the mean functional form is adequate, although random effects and correlated errors may create additional structure.

Successive differences provide a related check when visits are equally spaced. Define the first difference as the change between adjacent outcomes and the second difference as the change between adjacent first differences. A linear mean has constant first differences and zero second differences. For an increasing convex mean, second differences tend to be positive; for an increasing concave mean, they tend to be negative. With irregular visit spacing, raw differences are not directly comparable; use changes per unit time or model-based predictions at a common time grid.

Justification: For spacing \(h\), define \(\Delta_h^2m(t)=m(t+h)-2m(t)+m(t-h)\). If \(m\) has continuous fourth derivatives near \(t\), Taylor expansion at \(t\) gives \(m(t\pm h)=m(t)\pm hm'(t)+\frac{h^2}{2}m''(t)\pm\frac{h^3}{6}m'''(t)+\frac{h^4}{24}m''''(t)+o(h^4)\). Adding the two expansions cancels the odd-order terms, so \(\Delta_h^2m(t)=h^2m''(t)+O(h^4)\). Thus, for sufficiently small equal spacing, the sign of the second difference reflects the sign of curvature. In observed data, measurement error can change that sign, so several consistent differences are more informative than one.

Worked Example: Convex Progression Fitted With a Linear Trend

Consider hypothetical untreated mean outcomes generated by \(m(t)=5+2\{\exp(0.2t)-1\}\), in outcome points, with time in years. Measurements are available at years 0, 2, and 4. Assume baseline and placebo effects have already been handled and that these values represent the mean trajectory, not noisy individual observations.

At year 0, \(m(0)=5.000\). At year 2, \(m(2)=5+2\{\exp(0.4)-1\}\approx5.983649\). At year 4, \(\exp(0.8)=2.225541\), so \(m(4)=5+2(2.225541-1)=7.451082\). The two successive changes are \(5.983649-5.000=0.983649\) points and \(7.451082-5.983649=1.467433\) points over equal two-year intervals. The increase in the successive change is \(1.467433-0.983649=0.483784\) points, consistent with convex progression.

Fit an ordinary least-squares line to these three equally spaced mean values. Its slope is \(0.612770\) points per year, and its intercept is \(4.919369\) points. The fitted values at years 0, 2, and 4 are \(4.919369\), \(6.144910\), and \(7.370451\) points. Observed-minus-fitted residuals are therefore \(+0.080631\), \(-0.161261\), and \(+0.080631\) points. The positive endpoint residuals and negative middle residual are a curved pattern consistent with the convex trajectory that generated the data.

A line through just the first and last points would have slope \((7.451082-5.000)/4=0.612770\) points per year and would predict \(8.676623\) points at year 6. The exponential model instead predicts \(5+2\{\exp(1.2)-1\}=9.640234\) points at year 6. The difference, about \(0.963611\) points, illustrates how a linear extrapolation can understate an accelerating trajectory. It does not establish that the exponential form is correct; it shows why extrapolation depends strongly on the assumed form.

Worked Example: Concave Saturating Progression and a Linear Chord

Suppose hypothetical untreated outcomes follow \(m(t)=10+8\{1-\exp(-0.5t)\}\), in points, with time in years. The baseline is 10 points and the trajectory approaches 18 points. Assume the stated curve describes the mean and that the placebo component has been separated.

At year 0, \(m(0)=10.000\) points. At year 2, \(m(2)=10+8(1-\exp(-1))=15.056964\) points. At year 4, \(m(4)=10+8(1-\exp(-2))=16.917318\) points. The increases over the two equal intervals are \(5.056964\) and \(1.860353\) points. The smaller later increase is consistent with a concave trajectory that is approaching its upper limit.

The endpoint chord from year 0 to year 4 has slope \((16.917317734-10)/4=1.729329434\) points per year. At year 2, that chord predicts \(10+1.729329434(2)=13.458658868\) points. The true year-2 outcome is therefore \(15.056964471-13.458658867=1.598305604\), or approximately \(1.598306\) points, above the chord. This direction is expected for a concave curve. The chord is a visual reference, not necessarily the least-squares line that would be fitted to a larger, unevenly sampled dataset.

If a linear model were extended beyond the observed interval, it would continue increasing at its constant slope. The saturating model instead predicts progressively smaller increases and approaches 18 points. When predictions beyond year 4 matter, this difference is clinically important, but the observed follow-up alone may not determine which long-term behavior is true.

Check Whether Apparent Curvature Is More Than Noise

Residual shape is evidence to investigate, not a stand-alone test of the mean model. Under independent measurement errors with mean zero, a single observed second difference includes measurement error from three visits; correlated errors change its variability and may create smooth-looking patterns. Unequal visit timing, missing follow-up, informative dropout, and heterogeneity in individual progression rates can also distort aggregate curves. Plotting subject-level trajectories alongside the population fit can reveal whether an apparent shape is broadly shared or driven by a few individuals.

The following hypothetical group means illustrate why mixed second differences should temper a strong curvature claim. Assume visits are equally spaced by one year, the outcome is in points, and baseline and placebo effects have been addressed.

Worked Example: Noisy Follow-Up Does Not Show Stable Curvature

Suppose the observed mean outcomes at years 0 through 4 are 30.0, 32.2, 34.1, 35.5, and 37.2 points. The first differences are \(32.2-30.0=2.2\), \(34.1-32.2=1.9\), \(35.5-34.1=1.4\), and \(37.2-35.5=1.7\) points per year. The corresponding second differences are \(1.9-2.2=-0.3\), \(1.4-1.9=-0.5\), and \(1.7-1.4=+0.3\) points per year.

The first two second differences are negative, but the last is positive. These few means do not show a consistent curvature direction: the pattern could reflect modest concavity, measurement variability, or both. It would be premature to conclude that the progression saturates. A reasonable next step is to inspect individual trajectories and residuals under candidate models, consider the error structure and follow-up completeness, and compare fit over the observed time range. Extrapolation should remain cautious because these data do not establish a stable long-term shape.

Comparing Candidate Forms and Recognizing Failure Modes

Candidate mean form	Expected progression pattern	Extrapolation implication	Diagnostic warning
Linear	Constant change per unit time; zero mean curvature	Continues at the same rate without bound	Persistent curved residuals or changing increments
Exponential	For the increasing form shown, change accelerates	Can grow rapidly at long times	Residual pattern inconsistent with convexity or implausible growth
Saturating	Increases by progressively smaller amounts toward a limit	Approaches a finite asymptote	Residuals or later increments indicate continued acceleration or a different trend

A useful comparison starts with plots and residuals, then asks whether each model's assumptions and extrapolation are plausible. Candidate forms can also be compared by likelihood-based criteria or predictive performance, provided they are fitted to the same outcome data with a suitable error model. A better in-sample fit is not, by itself, proof of the correct biological mechanism. Flexible models can follow noise; simple models can miss real curvature. Prefer conclusions that are stable to reasonable modeling choices and supported by follow-up spanning the times at which predictions are needed.

Common pitfalls include treating a curved residual pattern as proof of one particular mechanism, using a chord as though it were always the fitted regression line, and extrapolating a short linear trend far beyond observed follow-up. Another frequent error is attributing all residual structure to disease progression when baseline, placebo response, covariates, or individual effects have been misspecified. The correction is to diagnose the progression component conditionally on those other components, examine both population and individual patterns, and state the range over which the data support the chosen form.

Key takeaway: Linear, increasing exponential, and increasing saturating models imply zero, positive, and negative curvature, respectively. Residual patterns and successive differences can reveal a mismatch, but they must be interpreted alongside noise, visit structure, placebo adjustment, and the limits of the observed follow-up.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions.

For a model with residuals mostly positive at the first and last visits and negative in the middle, what progression shape might be missing from a linear fit?
For equally spaced observations, the first differences are 1.0, 1.0, and 1.0 outcome points. What does this suggest about the mean trajectory, and what does it not prove?
An increasing trajectory has consistently shrinking first differences. Which of the three forms is most consistent with that pattern?
Why should an endpoint chord not automatically be treated as the residual reference from a least-squares fit?
Name two reasons residual patterns in untreated follow-up might not be caused by progression functional-form error.

Solutions. (1) Convex progression, such as an increasing exponential trajectory, is a plausible mismatch; the pattern alone does not identify a unique model. (2) It is consistent with a linear mean over those visits, but finite noisy observations cannot prove that the underlying trajectory is exactly linear or that it will remain linear outside the observed period. (3) A saturating increasing trajectory is consistent with shrinking increments toward a limit. (4) A chord joins two selected endpoints, whereas a least-squares line depends on all observations, their timing, and any weights; their predictions need not match. (5) Examples include an unmodeled placebo time course, measurement error or serial correlation, heterogeneous individual trajectories, irregular visit timing, or informative dropout.