Why Residual-Error Starting Values Need Their Own Check
A pharmacokinetic prediction can be plausible while its residual-error starting values are not. If the residual error is initialized too small, the likelihood can treat ordinary assay variation as strong evidence against the model. If it is initialized too large, meaningful concentration differences may initially carry little weight. Either problem can make optimization less stable or encourage the fit to compensate for an unsuitable error structure.
The purpose of an initial value is not to declare the final assay error known. It is to give estimation a reasonable scale, in the units and parameterization of the model being fit. A useful first check compares observed concentrations with predictions from a preliminary structural model. Those predictions need not be final, but they should be calculated consistently for the records used in the check.
Residual-Error Models and Their Parameters
Let \(Y_{ij}\) denote an observed concentration for subject \(i\) at observation \(j\), and let \(F_{ij}\) be the corresponding model prediction. Define the prediction residual as \(r_{ij}=Y_{ij}-F_{ij}\). The residual model describes the variation of \(Y_{ij}\) around \(F_{ij}\), conditional on that prediction. The examples below use untransformed concentrations.
Under these assumptions, the conditional variances are \(\sigma_{\mathrm{add}}^2\), \(F_{ij}^2\sigma_{\mathrm{prop}}^2\), and \(\sigma_{\mathrm{add}}^2+F_{ij}^2\sigma_{\mathrm{prop}}^2\), respectively. If the errors are also normal, these variance expressions define normal observation models conditional on \(F_{ij}\). Normality is an additional assumption, not a consequence of the variance equations.
Some software uses a different combined-error parameterization: for example, it may define the conditional standard deviation as a sum, rather than define the variance as a sum of independent components. Other implementations use variance parameters where this tutorial uses standard deviations. Check the model equation and units in the estimation code before transferring a numerical starting value. A value for a variance cannot be substituted directly for a value for its standard deviation.
Estimating a Starting Scale From Residuals
For a proposed prediction \(F_{ij}\), a residual gives a signed discrepancy in concentration units. The squared residual gives a rough contribution to residual variance. Starting values can be obtained by summarizing these discrepancies in a way that matches the model: roughly constant absolute scatter suggests an additive scale; roughly constant relative scatter suggests a proportional scale. These summaries are only preliminary because residuals also reflect prediction error, sampling, and possible model misspecification.
Worked Example: A Proportional Starting Value
Suppose four hypothetical observed concentrations, in mg/L, are paired with preliminary predictions as shown: \(F=(2,5,10,20)\) mg/L and \(Y=(2.2,4.5,11,18)\) mg/L. The residuals are \(r=Y-F=(0.2,-0.5,1,-2)\) mg/L. Their relative values are \(r/F=(0.10,-0.10,0.10,-0.10)\).
For a proportional model, \(r/F\) is an approximate realization of the relative error. Assuming the errors are centered at zero, use their root mean square as a simple moment-based starting scale:
Thus a reasonable preliminary value is \(\sigma_{\mathrm{prop},0}=0.10\), or about 10% relative standard deviation under this parameterization. This is a starting value based on four residuals, not a precise estimate of assay variability. The predictions are all positive; ratios would be unstable or undefined at zero or near-zero predictions, which is one reason a purely proportional model needs particular scrutiny at low concentrations.
For an additive model, the corresponding rough scale is the root mean square of the residuals in concentration units. The calculation assumes residuals are centered around zero. A persistent nonzero residual mean is a warning about prediction bias, not a reason to inflate the error scale until the bias is hidden.
Worked Example: An Additive Starting Value
Consider four hypothetical records with predictions \(F=(4,8,12,16)\) mg/L and observations \(Y=(4.3,7.7,12.4,15.6)\) mg/L. The residuals are \(r=(0.3,-0.3,0.4,-0.4)\) mg/L. Their mean is zero, so this small set does not show an overall signed offset.
For constant additive variance, the root mean square residual gives a simple initial scale:
A candidate starting value is therefore \(0.354\) mg/L. Its units are important: entering \(0.354\) as a relative error would not describe the same model. With a larger data set, inspect residual spread across the prediction range before treating a single additive scale as plausible.
Deriving a Combined-Error Starting Value
In the combined model, the additive and proportional components are independent and have mean zero. Expanding the variance of their sum shows how residual spread should change with the prediction:
The last equality uses both independence, which makes the covariance zero, and the specified component variances. Consequently, if residual variance can be summarized at different prediction levels, it should be approximately linear in \(F^2\): the intercept corresponds to \(\sigma_{\mathrm{add}}^2\), and the slope corresponds to \(\sigma_{\mathrm{prop}}^2\). This is a diagnostic and a source of rough starting values, not a guarantee that an ordinary regression of individual squared residuals will estimate those quantities reliably. Squared residuals are noisy, and a small sample can produce an unstable or even negative fitted intercept or slope. A negative component variance is not a valid starting value; reconsider the summaries, the data, or the model rather than interpreting it as a negative error scale.
Worked Example: Separating Additive and Proportional Components
Suppose four hypothetical observations have predictions \(F=2\) mg/L, with residuals \(+0.5\) and \(-0.5\) mg/L, and four have predictions \(F=10\) mg/L, with residuals \(+1.3\) and \(-1.3\) mg/L. Within each prediction level, the residual mean is zero. The mean squared residuals are \(0.25\) (mg/L)\(^2\) at \(F=2\) and \(1.69\) (mg/L)\(^2\) at \(F=10\).
Set each observed mean squared residual equal to the combined variance at its prediction level. Write \(a=\sigma_{\mathrm{add}}^2\) and \(b=\sigma_{\mathrm{prop}}^2\). Then:
Subtracting the first equation from the second gives \(96b=1.44\), so \(b=0.0150\). Substituting into the first equation gives \(a=0.25-4(0.0150)=0.190\). Taking square roots yields:
These values reproduce the two summarized variances: at \(F=2\), \(0.190+4(0.0150)=0.250\) (mg/L)\(^2\); at \(F=10\), \(0.190+100(0.0150)=1.69\) (mg/L)\(^2\). The result is a moment-based initialization from deliberately simple grouped data. With real data, residuals from the same subject or nearby times may not be independent, and the two-level calculation does not resolve prediction bias or other sources of variation.
Assay Noise Near the Quantification Limit
A purely proportional model predicts that residual standard deviation shrinks toward zero as \(F\) approaches zero. Real assays may retain an absolute noise floor, so relative error can become large at low concentrations even when the absolute error remains modest. An additive component can represent such a floor over a suitable range; a combined model allows the absolute component to dominate at low predictions and the proportional component to matter more as predictions rise.
For instance, with \(\sigma_{\mathrm{add}}=0.30\) mg/L and \(\sigma_{\mathrm{prop}}=0.10\), the combined standard deviation at \(F=1\) mg/L is \(\sqrt{0.30^2+(0.10\times1)^2}=0.316\) mg/L. At \(F=10\) mg/L it is \(\sqrt{0.30^2+(0.10\times10)^2}=1.044\) mg/L. The relative size of the additive contribution is much greater at the lower prediction. This illustrates why a relative-error summary calculated near zero can be misleading.
Use assay validation or quality-control information at relevant concentration levels when available. It can help distinguish known analytical imprecision from discrepancies caused by the structural model or other unmodeled variability. Observations below a quantification limit are not ordinary exact concentration measurements. Replacing them by the limit or by a fixed fraction of it and treating that substituted value as exact can distort residual summaries; handle such records consistently with the model and data-analysis plan.
Diagnostics, Alternatives, and Limits
Before using a residual summary as a starting value, check whether its assumptions are at least plausible. Residuals should not show a strong systematic shift from zero across prediction, time, or other relevant design features. A trend can signal a poor preliminary prediction, not merely an incorrect residual scale. Compare absolute residuals and relative residuals against predictions: roughly constant absolute spread supports an additive component, while roughly constant relative spread supports a proportional component. A spread that changes with prediction may support a combined model, but does not by itself prove it is the right model.
Where there are enough observations, summarize residual spread across prediction ranges rather than relying on a few individual squared residuals. Check for extreme observations, negative predictions, data-unit mismatches, and inconsistent handling of records near the quantification limit. Also check that the residual definition matches the prediction used: mixing prediction types or including records with substantially different model conditions can obscure the error pattern. Residuals within a subject may be related, so treating every record as independent evidence for a precise scale can overstate how informative a small data set is.
The main practical alternatives are to start from assay precision data, use a simple residual summary from model predictions, or choose a plausible scale from prior model-development experience and assess whether estimation is sensitive to it. Assay data can be more directly informative about analytical noise, but may not describe every source of unexplained observation-level variation in the pharmacometric data. Prediction residuals reflect both assay noise and any shortcomings in the predictions. A plausible informed value avoids an unstable calculation but should be checked rather than assumed correct. These approaches inform initialization; none makes a misspecified error model adequate.
Residual-based calculations are approximate because the preliminary predictions are themselves uncertain and can be wrong. They do not establish the final parameter estimates, prove normality, or identify the cause of unexplained variation. Their value is practical: they give the optimizer an appropriate order of magnitude and reveal obvious mismatches before a more complex estimation step. If residual spread changes sharply, the mean pattern is systematic, or candidate scales depend heavily on a few records, revisit the model and data rather than trusting one numerical summary.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions.
- A proportional model has \(\sigma_{\mathrm{prop}}=0.08\) and prediction \(F=5\) mg/L. What is its conditional residual standard deviation?
- For a combined model, \(\sigma_{\mathrm{add}}=0.20\) mg/L, \(\sigma_{\mathrm{prop}}=0.10\), and \(F=3\) mg/L. Calculate its conditional residual variance and standard deviation.
- Residuals are approximately constant in absolute size across the prediction range, but their relative size increases as predictions fall. Which error component is suggested by this pattern, and what caution applies near the quantification limit?
- A squared-residual regression against \(F^2\) gives a negative intercept. Can the square root of that intercept be used as an additive standard deviation? What should be done instead?
- Why is an observation substituted at the quantification limit not automatically suitable as an exact value in a residual-scale calculation?
Solutions. 1. The standard deviation is \(F\sigma_{\mathrm{prop}}=5(0.08)=0.40\) mg/L. 2. The variance is \(0.20^2+(3\times0.10)^2=0.04+0.09=0.130\) (mg/L)\(^2\), and the standard deviation is \(\sqrt{0.130}=0.361\) mg/L. 3. The pattern suggests an additive component because absolute scatter is roughly stable; near the limit, relative error can become large, and censored or poorly quantified observations need appropriate treatment rather than naive ratios. 4. No. A standard deviation cannot be the square root of a negative variance. Check the summaries, sample size, predictions, and model; use defensible nonnegative starting values or another source such as assay precision information. 5. The substituted value is not the exact measured concentration, so treating it as exact can misrepresent its residual and the uncertainty associated with censoring.