Why Dose Normalization Needs Care
Dividing a concentration by dose can be useful for describing exposure, especially when concentrations are expected to scale proportionally with dose. But this transformation also changes the size and variance of residuals. A pattern that looks like evidence for a particular residual error model after dose normalization may be a mathematical consequence of dividing by dose, rather than evidence about the error process that generated the observations.
The central distinction is between describing normalized concentrations and choosing a residual model for the concentrations the analysis actually models. Dose-normalized residuals can help reveal patterns, but their variance must be interpreted on the normalized scale. A residual model should be assessed against the observation scale and structural predictions used in the likelihood.
For clarity, consider first a residual model on the original concentration scale. An additive model has \(C_{ij}=m_{ij}+\epsilon_{ij}\), where the errors have mean zero and variance \(\sigma_a^2\). A proportional model can be written \(C_{ij}=m_{ij}(1+\epsilon_{ij})\), where the errors have mean zero and variance \(\sigma_p^2\); equivalently, the residual variance is \(\sigma_p^2m_{ij}^2\). A combined model includes both additive and proportional contributions, often giving variance \(\sigma_a^2+\sigma_p^2m_{ij}^2\) when the two error terms are independent. These models describe variance as a function of the unnormalized prediction \(m_{ij}\), not as a function of dose-normalized residuals.
What Division by Dose Does to Variance
Suppose dose is fixed for the observations being compared. Dividing a residual by that fixed dose divides its variance by the square of the dose. This result follows directly from the variance rule for a constant multiplier.
Thus, if the original residual variance is additive and constant, the variance of the dose-normalized residual decreases as \(1/D_i^2\). Conversely, if the original error is proportional and the model prediction scales exactly with dose, dose-normalized residual variance can be constant across doses at a fixed sampling time. That constancy does not establish that the original error was additive. It is also the expected result of proportional error under the stated scaling condition.
The proportional-error result depends on the normalized prediction \(m_{ij}/D_i\). If the prediction is proportional to dose at the same time and under otherwise comparable conditions, this ratio is constant across doses. If the dose–concentration relationship is not proportional, or if observations are taken at different times or under different conditions, it need not be constant. The formula therefore does not imply that dose-normalized residual variance must be constant in every data set.
Worked Example: Additive Error Looks Dose-Dependent After Normalization
At one matched sampling time, suppose the model predicts \(0.20\) mg/L after a 100 mg dose and \(0.40\) mg/L after a 200 mg dose. The predictions are proportional to dose. For illustration, suppose the observed concentrations are \(0.25\) and \(0.15\) mg/L at 100 mg, and \(0.45\) and \(0.35\) mg/L at 200 mg. Each observation has an original-scale residual of either \(+0.05\) or \(-0.05\) mg/L.
After dividing by dose, the residuals at 100 mg are \(+0.05/100=+0.000500\) and \(-0.05/100=-0.000500\) (mg/L)/mg. At 200 mg, they are \(+0.05/200=+0.000250\) and \(-0.05/200=-0.000250\) (mg/L)/mg.
The squared residual magnitude on the normalized scale is \(2.50\times10^{-7}\) (mg/L per mg)\(^2\) at 100 mg, and \(6.25\times10^{-8}\) (mg/L per mg)\(^2\) at 200 mg. The latter is one-quarter of the former. Yet the original residual magnitudes are identical. This pattern is exactly what division by dose produces for equal additive errors; it is not, by itself, evidence that the original-scale residual variance declines with dose.
These few observations illustrate the arithmetic, not a reliable variance estimate. In an analysis, a variance pattern must be assessed with sufficient data and in relation to the model's predictions and design.
Proportional Error Can Look Different After Normalization
Under proportional error, the size of a typical residual increases with the predicted concentration. If predictions double with dose, residuals also double, on average in scale. Dividing those residuals by dose can remove that increase when predictions per unit dose are constant. The normalized residuals may then have similar magnitudes at each dose, even though the original-scale residual variance increases with concentration.
Worked Example: Proportional Error Has Equal Normalized Residuals
Use the predictions \(m=0.20\) mg/L at 100 mg and \(m=0.40\) mg/L at 200 mg. Suppose the observed residuals are \(+0.04\) and \(-0.04\) mg/L at the lower dose, and \(+0.08\) and \(-0.08\) mg/L at the higher dose. Each residual is 20% of its prediction in magnitude.
At 100 mg, the dose-normalized residuals are \(+0.04/100=+0.000400\) and \(-0.000400\) (mg/L)/mg. At 200 mg, they are \(+0.08/200=+0.000400\) and \(-0.000400\) (mg/L)/mg. The normalized residual magnitudes are identical.
On the original scale, however, the residual magnitude doubles as the prediction doubles. The proportional model describes this pattern through a constant relative error of \(0.04/0.20=0.20\) and \(0.08/0.40=0.20\). Equal dose-normalized residual magnitudes are compatible with this proportional error model because \(m/D\) is the same at both doses. They do not establish that an additive model on the original scale is appropriate.
When Dose Normalization Does Not Remove Dose Patterns
Even proportional error need not yield equal dose-normalized residual variance. The result depends on how the structural prediction changes with dose. For example, if the concentration response is nonlinear, \(m/D\) may differ between doses. Then proportional error gives normalized residual variance \(\sigma_p^2(m/D)^2\), which can vary with dose. Comparing normalized residual spreads without considering the predictions can therefore suggest a changing error scale when the proportional error model itself has a constant relative error.
Worked Example: Nonlinear Dose Response Changes the Normalized Spread
Suppose a structural model predicts \(0.20\) mg/L after 100 mg and \(0.30\) mg/L after 200 mg. The predictions do not double when dose doubles. Consider residuals of \(+0.02\) and \(-0.02\) mg/L at the 100 mg dose, and \(+0.03\) and \(-0.03\) mg/L at the 200 mg dose. In both groups, the residual magnitude is 10% of the prediction.
The relative residual magnitudes are \(0.02/0.20=0.10\) and \(0.03/0.30=0.10\), consistent with a constant proportional error scale. But the dose-normalized residual magnitudes are \(0.02/100=0.000200\) and \(0.03/200=0.000150\) (mg/L)/mg. Their squared magnitudes are \(4.00\times10^{-8}\) and \(2.25\times10^{-8}\) (mg/L per mg)\(^2\), respectively.
The normalized squared magnitude at 200 mg is \(2.25/4.00=0.5625\), or 56.25% of the value at 100 mg. This difference follows from the changed ratios \(m/D\): they are \(0.00200\) and \(0.00150\) (mg/L)/mg. It would be misleading to interpret the smaller normalized spread as proof that the residual model should change with dose. The proportional model still has the same 10% relative error at both predictions.
How to Assess Residual Error
Error-model choice should follow the model and observation scale, not a transformed plot alone. First specify the concentration prediction and the candidate residual variance on the scale used for the likelihood. Then examine whether residual magnitudes behave as that model predicts. For example, additive error implies roughly constant residual spread across predicted concentrations; proportional error implies spread increasing with prediction and roughly constant relative residuals; a combined model allows an additive component at low predictions and a proportional component as predictions increase.
Dose-normalized plots can still be useful, but their purpose must be clear. Plotting \(C/D\) against time may help describe exposure patterns. Plotting \(e/D\) can show how the residuals look after that same transformation. Neither plot automatically identifies the original-scale residual model. When doses differ, inspect residuals against the original predictions and dose as well as any normalized display. If comparisons are made across sampling times, account for the fact that predicted concentrations—and therefore proportional-error variance—can change with time.
A useful check is to compare residuals with the scale implied by each candidate model. For additive error, use residuals in concentration units. For proportional error, examine relative residuals \(e_{ij}/m_{ij}\), where predictions are positive and not near zero. For a combined error model, consider the model-implied variance \(\sigma_a^2+\sigma_p^2m_{ij}^2\); a residual divided by the square root of this variance is a variance-scaled residual. Such displays are diagnostics, not proof that a model is correct.
| Display or model | What it can show | Important limitation |
|---|---|---|
| Original-scale residuals against prediction | Whether absolute spread is roughly constant or grows with concentration | Patterns can also reflect structural misspecification or other sources of variation |
| Relative residuals \(e/m\) | Whether residual size is roughly proportional to prediction | Unstable when predictions are near zero; does not alone establish a full error model |
| Dose-normalized residuals \(e/D\) | Residual patterns after scaling by dose | Variance is divided by \(D^2\); the pattern can change solely because of normalization |
| Candidate model's variance-scaled residuals | Whether residual spread is broadly compatible with the assumed variance function | Good-looking residuals do not prove the structural or error model is correct |
Look for systematic changes in residual spread with prediction, dose, time, or other relevant design features. Also check for trends in residual means: a nonzero pattern can indicate that the structural predictions are systematically missing the observations, rather than that the variance function is wrong. Sparse data, a narrow range of predictions, and a small number of observations per dose limit what plots can reveal. A single striking normalized pattern should not outweigh the model's scale and the study design.
Common Pitfall and Practical Limits
Wrong reasoning: “The dose-normalized residuals are smaller at higher doses, so the residual error must be lower at high doses.” This conclusion compares variances after rescaling and ignores that dividing by dose itself reduces residual magnitudes. Under constant additive error, the normalized variance falls as \(1/D^2\), even when original-scale residual variance is unchanged.
Correction: State which scale is being modeled, write the implied variance on that scale, and then account for the transformation. For proportional error, check residuals relative to the prediction; for additive error, check absolute residuals; and for dose-normalized residuals, interpret their variance with the \(1/D^2\) factor in mind. A transformation is not a substitute for specifying the error model.
Dose normalization also relies on a meaningful dose scale and does not guarantee that observations are comparable. If predictions are not proportional to dose, or observations differ in sampling time or other conditions, \(m/D\) can vary. Normalized values can then retain systematic differences for structural reasons. Conversely, a normalized plot can conceal original-scale differences that matter for the likelihood. These limitations make normalization a descriptive tool rather than an automatic error-model selection rule.
Check Your Understanding
Answer each question, then compare with the brief solutions.
- If original-scale residuals have constant variance \(\sigma^2\), what is the variance of residuals divided by a fixed dose \(D\)?
- Predictions double with dose, and residual magnitudes also double while relative residuals remain 15%. Which error pattern is more consistent with these facts: additive or proportional?
- Why can proportional error produce different dose-normalized residual spreads at two doses?
- What is one reason not to choose an error model from a dose-normalized residual plot alone?
Solutions. (1) The variance is \(\sigma^2/D^2\), by the variance rule for multiplication by a constant. (2) Proportional error is more consistent: the residual-to-prediction ratio remains 0.15 while absolute residual size increases with the prediction. (3) Under proportional error, normalized variance is \(\sigma_p^2(m/D)^2\); if \(m/D\) differs across doses, the normalized spreads can differ. (4) Division by dose changes residual variance, and the resulting pattern may reflect the transformation rather than the original-scale error process.