Why the Order of Residuals Matters
A residual is not merely a measure of how far a prediction misses an observation. In an indirect-response model, the order and timing of the misses can matter: errors that cluster during onset, peak effect, or washout may indicate that the fitted model is not representing the response dynamics adequately. A residual pattern can raise a specific question about mechanism or turnover, but it does not, by itself, establish the cause.
This tutorial considers two diagnostic situations: production inhibition fitted to data generated by loss inhibition, and a fitted turnover rate that is too large or too small. The interpretation depends on the direction of response change, the phase of the profile, the residual definition, and what other parts of the model were estimated.
Models and Assumptions Behind the Diagnostic
Let \(R(t)\) denote a response, with baseline \(R_0\). In the usual turnover formulation, baseline equilibrium satisfies \(k_{\mathrm{in}}=k_{\mathrm{out}}R_0\), where \(k_{\mathrm{in}}\) is the production rate and \(k_{\mathrm{out}}\) is the first-order loss-rate constant. For a concentration-driven inhibitory effect \(I(C)\), with \(0\leq I(C)\leq1\), the two mechanisms are:
These equations assume that production and loss follow the stated forms, the response is described by one turnover pool, and drug concentration \(C(t)\) is supplied by an adequate PK model. They also assume that observation error is represented appropriately when residuals are assessed. The concentration-effect function might, for example, be an inhibitory \(E_{\max}\) function, but residuals alone do not identify that function.
At baseline, when \(I(C)=0\), both equations give zero net change at \(R_0\). During drug effect, however, the two mechanisms alter different terms. If \(I(C)>0\) and the response starts at baseline, production inhibition initially lowers the response, whereas loss inhibition initially raises it. A fitted model may partly absorb the discrepancy by changing baseline, effect magnitude, turnover, or other parameters. Consequently, the resulting residuals need not retain a simple sign throughout the profile.
Worked Example: Reading Residuals Separately by Phase
Worked Example: Reading Residuals Separately by Phase
Suppose a hypothetical response has baseline 100 response units. A fitted indirect model produces the predictions below. The observations are invented for illustration; the table is not a real study. Calculate residuals and interpret the onset and washout phases separately.
| Phase | Time (hours) | Observed (response units) | Predicted (response units) | Residual (response units) |
|---|---|---|---|---|
| Onset | 1 | 88 | 82 | 6 |
| Onset | 2 | 78 | 72 | 6 |
| Washout | 5 | 86 | 80 | 6 |
| Washout | 7 | 94 | 91 | 3 |
| Washout | 9 | 99 | 98 | 1 |
For example, at 1 hour, \(e=88-82=6\) response units. At 7 hours, \(e=94-91=3\) response units. All listed residuals are positive, but their phase-specific meaning differs.
During onset, the observations are above predictions: the fitted model predicts a response decline that is too rapid or too large at those times. During washout, predictions are below observations: the fitted model predicts recovery that is too slow. The shrinking positive washout residuals suggest that the discrepancy is becoming smaller by 9 hours; they do not reverse its direction at the earlier washout times.
These signs are diagnostic clues, not a unique explanation. A different mechanism, an incorrect concentration-time profile, an unsuitable error model, or measurement variability could also produce such a sequence. In particular, residuals should not be summarized as showing a single “too-rapid” or “too-slow” response when the onset and washout implications differ.
Mechanism Mismatch: Production Versus Loss Inhibition
The initial direction of change provides a basic mechanism check. Suppose concentration is approximately constant over a short interval, the response begins at \(R_0\), and the effect is \(I>0\). Using \(k_{\mathrm{in}}=k_{\mathrm{out}}R_0\), the initial slopes are:
Thus, when the observed response rises after dosing but a production-inhibition fit predicts a fall, early residuals are likely to be positive as the observation pulls above the prediction. If the fitted production-inhibition model is adjusted to produce a rise through other parameter choices, the residual sequence can be more complicated; the initial-slope argument is a local comparison under the stated baseline and constant-effect conditions, not a guarantee about every fitted profile.
Worked Example: Comparing Initial Slopes Under the Two Mechanisms
Consider a hypothetical baseline response \(R_0=100\) response units, \(k_{\mathrm{in}}=20\) response units per day, and \(k_{\mathrm{out}}=0.20\) per day. These satisfy the baseline condition because \(0.20\times100=20\). Let the inhibition be \(I=0.50\) during a short interval.
For production inhibition, the initial slope is \(20(1-0.50)-0.20(100)=10-20=-10\) response units per day. For loss inhibition, it is \(20-0.20(1-0.50)(100)=20-10=+10\) response units per day.
The sign difference follows directly from which rate is inhibited. For an additional check, assume the concentration and inhibition remain constant for 0.1 day. The production-inhibition equation has equilibrium \(10/0.20=50\) response units, so its solution gives \(R(0.1)=50+(100-50)e^{-0.20(0.1)}=99.01\) response units. The loss-inhibition equation has equilibrium \(20/0.10=200\) response units, giving \(R(0.1)=200+(100-200)e^{-0.10(0.1)}=101.00\) response units.
The two model predictions move in opposite directions from baseline over this interval. If hypothetical observations instead rise, fitting production inhibition without adequate flexibility may leave positive early residuals. The particular residual pattern after fitting depends on the observations, sampling times, error model, and estimated parameters; the calculation establishes the local model contrast, not a universal residual signature.
Turnover-Rate Error and Washout Residuals
When drug effect has ended, a first-order turnover response returns toward baseline exponentially. If the response at washout start is \(R_s\), then \(R(t)=R_0+(R_s-R_0)e^{-k_{\mathrm{out}}t}\), where \(t\) is time since washout. This relation makes the direction of turnover-rate error explicit. A larger \(k_{\mathrm{out}}\) predicts faster approach to baseline; a smaller value predicts slower approach.
For a response below baseline, an assumed rate that is too large predicts values closer to baseline than the true values. Its residuals \(Y-\hat{Y}\) tend to be negative during washout. For a response above baseline, the same too-large rate predicts a faster decline toward baseline, so the residual direction is reversed. Therefore, “positive residual means turnover is too fast” is not a general rule; the sign depends on whether the response is above or below baseline.
Worked Example: An Assumed Turnover Rate That Is Too Large
Suppose a hypothetical response is 60 response units at washout start and has baseline 100 response units. The true turnover rate is \(0.20\) per hour, but a fitted model uses \(0.40\) per hour. Assume the drug effect is absent after washout and compare predictions at 1 and 3 hours.
At 1 hour, the true expected response is \(100+(60-100)e^{-0.20(1)}=100-40(0.8187)=67.25\) response units. The fitted prediction is \(100-40e^{-0.40}=73.19\) response units. Thus the residual is \(67.25-73.19=-5.94\) response units.
At 3 hours, the true expected response is \(100-40e^{-0.60}=78.05\) response units. The fitted prediction is \(100-40e^{-1.20}=87.95\) response units. The residual is \(78.05-87.95=-9.90\) response units. In this below-baseline example, the fitted response returns toward baseline too quickly, leaving negative residuals that grow in magnitude over these times. The calculations compare expected profiles without random measurement error; actual observations would also vary around their expectations.
A Short Justification for the Washout Pattern
The turnover diagnostic follows from the residual expected when the true and fitted washout profiles share \(R_0\) and \(R_s\). Write \(k_T\) for the true rate and \(k_F\) for the fitted rate. Then:
If the response starts below baseline, \(R_s-R_0<0\). If \(k_F>k_T\) and \(t>0\), then \(e^{-k_Tt}>e^{-k_Ft}\), so the parenthesized term is positive and \(e(t)<0\). This proves the negative expected residual direction in the preceding example. If the response starts above baseline, \(R_s-R_0>0\), and the same rate ordering yields \(e(t)>0\). If the fitted rate is too small, these signs reverse. The result assumes drug effect is absent, both profiles share the same starting response and baseline, and the only difference is the turnover rate.
Diagnostics, Comparisons, and Limitations
A useful diagnostic begins with residuals plotted against actual time, with dosing, concentration, and response phases marked. Also inspect residuals against predictions and concentration. Use raw residuals to understand discrepancies in response units; use standardized residuals when the assumed error model gives different variances across observations. For example, under proportional error, a raw deviation of 5 units may be relatively small at a prediction of 200 and large at a prediction of 10.
Look for runs of residuals with the same sign, systematic curvature, or a change in pattern around peak concentration or washout. These can signal temporal misspecification, but sparse sampling may miss a phase entirely. Repeated observations from the same individual may also be correlated, so a sequence of residuals should not automatically be treated as independent evidence. Residuals calculated after parameter estimation are constrained by the fitting process; they are not independent tests of each model feature.
Residual inspection complements, rather than replaces, likelihood-based estimation and model assessment. A likelihood comparison evaluates how well candidate models account for the data under their specified error distributions and parameter constraints. Residual plots show where and when discrepancies occur, which can guide model checking. Neither a visually appealing residual plot nor a better fit alone proves a biological mechanism. A wrong PK input, baseline drift, incorrect observation variance, delayed effect, or omitted response process can mimic some patterns attributed to turnover or mechanism.
When production inhibition is fitted to loss-inhibition data, a systematic pattern may arise because the models act on different terms and can initially predict opposite directions. When turnover is wrong, washout residuals may show a phase-specific trend. But if parameters such as baseline, effect magnitude, and turnover are estimated together, they can redistribute error across time. Residual patterns are therefore most informative when interpreted alongside the model equations, concentration-time data, sampling schedule, and plausible sources of variability.
The practical objective is not to diagnose a mechanism from one residual. It is to identify reproducible, time-structured discrepancies that motivate checking assumptions or estimating a scientifically plausible alternative. Conclusions should remain tentative unless supported by the complete model comparison and relevant design information.
Check Your Understanding
For each item, identify the residual direction or calculation and give a phase-specific interpretation.
- An observation is 42 response units and its prediction is 47. What is the residual, and what does its sign mean?
- A response is below baseline during washout. The fitted turnover rate is larger than the true rate. What sign of expected residual follows, assuming the profiles share a starting value and baseline?
- At baseline, \(k_{\mathrm{in}}=15\) response units per hour, \(R_0=75\) response units, and \(I=0.40\). Find the initial slopes for production inhibition and loss inhibition.
- Why does a run of positive residuals not uniquely establish that production inhibition is the wrong mechanism?
Solutions. (1) \(e=42-47=-5\) response units; the observation is 5 units below the prediction. (2) The expected residual is negative: with a below-baseline response, the too-large fitted rate predicts faster recovery and values above the true profile. (3) The baseline condition gives \(k_{\mathrm{out}}=15/75=0.20\) per hour. Production inhibition has initial slope \(15(1-0.40)-0.20(75)=9-15=-6\) response units per hour; loss inhibition has slope \(15-0.20(1-0.40)(75)=15-9=+6\) response units per hour. (4) A run is a diagnostic clue, not a unique cause: measurement variation, misspecified concentration input, baseline drift, turnover error, or an unsuitable error model can also produce structured residuals.