Interpreting Weighted Residuals At Low And High Concentrations

Why the Same Concentration Difference Can Mean Different Things

A residual is the difference between an observed concentration and a model prediction. But an absolute difference alone does not say whether the observation is surprising. A difference of 1 concentration unit may be substantial when the predicted concentration is 2, yet minor when the prediction is 50. The residual-error model supplies the scale needed to judge that discrepancy.

This matters when reviewing population pharmacokinetic diagnostics: weighting changes the size of a residual relative to the expected observation error. It does not change the observed concentration, the prediction, or the raw difference between them. A small weighted residual can indicate a discrepancy that is modest relative to the assumed error, even if its absolute size looks large.

Definition: For observation \(Y_{ij}\) from subject \(i\) at occasion \(j\), let \(f_{ij}\) be the model-predicted concentration and define the raw residual \(e_{ij}=Y_{ij}-f_{ij}\). If the model assigns this observation an error variance \(V_{ij}\), its model-based standardized residual is \(r_{ij}=e_{ij}/\sqrt{V_{ij}}\). The denominator is the modeled residual standard deviation, in concentration units. Thus \(r_{ij}\) is dimensionless and measures the discrepancy in units of its expected error.

The standardized residual defined here captures the central weighting idea. Particular diagnostic residuals, including conditional weighted residuals, can involve additional adjustments. Those adjustments do not change the key point of this tutorial: the assumed error variance determines how strongly a given concentration discrepancy is scaled.

Proportional Error: Weighting by Relative Size

Under a proportional residual-error model, the residual standard deviation increases in proportion to the predicted concentration. One common form is \(Y_{ij}=f_{ij}(1+\epsilon_{ij})\), where \(\epsilon_{ij}\) has mean zero and standard deviation \(\sigma_{\mathrm{prop}}\). Assuming this error is normal, the conditional mean of \(Y_{ij}\) is \(f_{ij}\), and its conditional variance is \(\sigma_{\mathrm{prop}}^2f_{ij}^2\). Here \(\sigma_{\mathrm{prop}}\) is dimensionless.

V_{ij}=\sigma_{\mathrm{prop}}^2f_{ij}^2, \qquad r_{ij}=\frac{Y_{ij}-f_{ij}}{\sigma_{\mathrm{prop}}f_{ij}}.

For a fixed absolute discrepancy, the denominator grows with the prediction, so the absolute standardized residual decreases as the predicted concentration rises. Conversely, for a fixed relative discrepancy—such as a 20% deviation—the standardized residual is the same at low and high predictions, provided the proportional error model applies throughout that range. This is the central interpretation: proportional weighting judges discrepancies relative to predicted concentration, not simply by their absolute size.

Worked Example: Equal Relative Errors Under Proportional Error

Suppose a hypothetical model uses proportional error with \(\sigma_{\mathrm{prop}}=0.20\). Compare an observation of 2.6 concentration units when the prediction is 2, with an observation of 26 units when the prediction is 20. Assume the stated proportional error model is appropriate at both concentrations.

For the first observation, the raw residual is \(2.6-2=0.6\) units and the modeled standard deviation is \(0.20(2)=0.4\) units. Therefore,

r=\frac{2.6-2}{0.20(2)} =\frac{0.6}{0.4} =1.50.

For the second, the raw residual is \(26-20=6\) units and the modeled standard deviation is \(0.20(20)=4\) units:

r=\frac{26-20}{0.20(20)} =\frac{6}{4} =1.50.

The absolute differences are 0.6 and 6 units, but each observation is 30% above its prediction. Under this model, both discrepancies are 1.50 modeled standard deviations above the prediction. This numerical equivalence is a consequence of proportional weighting, not evidence that the observations or their absolute errors are identical.

Combined Error: Additive at Low Concentrations, Proportional at High

A combined additive–proportional error model includes both a baseline error scale and an error scale that rises with concentration. A common formulation is \(Y_{ij}=f_{ij}+\epsilon_{a,ij}+f_{ij}\epsilon_{p,ij}\), where the additive and proportional error terms are independent, have mean zero, and have standard deviations \(\sigma_{\mathrm{add}}\) and \(\sigma_{\mathrm{prop}}\). The additive term and \(\sigma_{\mathrm{add}}\) are measured in concentration units; \(\sigma_{\mathrm{prop}}\) is dimensionless. Under these assumptions, variances add:

V_{ij}=\sigma_{\mathrm{add}}^2+\sigma_{\mathrm{prop}}^2f_{ij}^2, \qquad r_{ij}= \frac{Y_{ij}-f_{ij}} {\sqrt{\sigma_{\mathrm{add}}^2+\sigma_{\mathrm{prop}}^2f_{ij}^2}}.

At low predicted concentrations, the additive component may dominate the variance, giving an approximately constant error scale. At high concentrations, the proportional component may dominate, making the standard deviation rise approximately in proportion to the prediction. The changeover is not a universal concentration threshold: it depends on both error parameters. The two variance contributions are equal when \(\sigma_{\mathrm{add}}^2=\sigma_{\mathrm{prop}}^2f_{ij}^2\), or \(f_{ij}=\sigma_{\mathrm{add}}/\sigma_{\mathrm{prop}}\), for positive parameter values.

Worked Example: The Same Absolute Error at Low and High Concentrations

Consider a hypothetical combined-error model with \(\sigma_{\mathrm{add}}=0.50\) concentration units and \(\sigma_{\mathrm{prop}}=0.20\). Compare a positive raw residual of 0.60 units at predictions of 1 and 10 units. The model assigns the same parameters at both concentrations.

At a prediction of 1, the modeled variance is \(0.50^2+0.20^2(1)^2=0.25+0.04=0.29\) square concentration units. Its standard deviation is \(\sqrt{0.29}=0.5385\) units, so:

r=\frac{0.60}{\sqrt{0.50^2+0.20^2(1)^2}} =\frac{0.60}{0.5385} \approx 1.114.

At a prediction of 10, the variance is \(0.50^2+0.20^2(10)^2=0.25+4.00=4.25\) square concentration units. The standard deviation is \(\sqrt{4.25}=2.062\) units:

r=\frac{0.60}{\sqrt{0.50^2+0.20^2(10)^2}} =\frac{0.60}{2.062} \approx 0.291.

The 0.60-unit discrepancy is about 1.114 modeled standard deviations at the lower prediction but only 0.291 at the higher prediction. The proportional component contributes little to the low-concentration variance and dominates the high-concentration variance. These values describe discrepancy relative to the specified model; they do not establish whether the error parameters or the model are appropriate.

Why the Weighting Formula Has This Interpretation

The interpretation follows directly from dividing the residual by its standard deviation. If an error model says that an observation typically varies more at a particular concentration, then a given absolute departure at that concentration is less exceptional on the model's scale. For proportional error, substitute \(V_{ij}=\sigma_{\mathrm{prop}}^2f_{ij}^2\) into the standardized-residual definition and take the positive square root of the variance:

r_{ij} =\frac{Y_{ij}-f_{ij}}{\sqrt{\sigma_{\mathrm{prop}}^2f_{ij}^2}} =\frac{Y_{ij}-f_{ij}}{\sigma_{\mathrm{prop}}f_{ij}} =\frac{(Y_{ij}-f_{ij})/f_{ij}}{\sigma_{\mathrm{prop}}}.

The last expression shows that the residual is the relative discrepancy divided by its modeled relative-error standard deviation. This derivation assumes \(f_{ij}>0\) and \(\sigma_{\mathrm{prop}}>0\). If \(f_{ij}=0\), a purely proportional model assigns zero variance and the standardized residual is undefined; this is one reason a proportional-only model may be unsuitable near zero.

In a normal observation model, the squared standardized residual \(r_{ij}^2\) is the squared-error part of the observation's contribution to the negative log-likelihood. If the variance depends on the prediction, the likelihood also contains a variance term, proportional to \(\log(V_{ij})\). Thus interpreting the size of a weighted residual is not the same as claiming that \(r_{ij}^2\) alone is the full likelihood contribution or a complete model comparison.

Worked Example: Distinguishing a Relative Pattern From a Weighted Pattern

Suppose a hypothetical proportional-error model has \(\sigma_{\mathrm{prop}}=0.25\). At a prediction of 4 concentration units, an observation is 5 units; at a prediction of 16 units, an observation is 20 units. Both are above prediction, and the relative deviation is 25% in each case.

At 4 units, the raw residual is \(5-4=1\) unit and the modeled standard deviation is \(0.25(4)=1\) unit. Thus \(r=1/1=1.00\). At 16 units, the raw residual is \(20-16=4\) units and the standard deviation is \(0.25(16)=4\) units, so \(r=4/4=1.00\). Both observations have the same positive standardized residual, although their raw residuals differ by a factor of four.

Now consider instead an observation of 17 units at the prediction of 16. Its raw residual is \(17-16=1\) unit, but its standardized residual is \(1/4=0.25\). The absolute residual matches the low-concentration example's 1 unit, yet the relative discrepancy is only 6.25% and the weighted residual is smaller. A raw-residual display and a standardized-residual display therefore answer different questions: absolute error versus discrepancy relative to the assumed error scale.

Assumptions, Diagnostics, and Failure Modes

The formulas are useful only to the extent that the error model represents the observation process. The simple standardized-residual interpretation assumes that the prediction is the appropriate conditional mean, the variance formula is appropriate, and the error terms have the stated structure. Under a normal model, errors are also assumed to be conditionally normal. Independence across observations is often assumed in the basic residual model; serial dependence or other correlation changes the joint interpretation even if each marginal variance looks plausible.

In fitted population models, parameters and predictions are estimated, and residuals may also depend on subject-level information. Consequently, diagnostic residuals are not automatically independent draws from a standard normal distribution. Their behavior can be affected by estimation, the construction of the particular residual, and the information available for each subject. The earlier material on conditional weighted residuals and their approximate standard-normal behavior provides context for that qualification.

When reviewing low- and high-concentration residuals, ask whether their spread changes in a way consistent with the assumed error model, and whether systematic positive or negative residuals persist across a range. A weighted residual pattern can highlight bias or an incorrect variance model, but it cannot by itself identify the cause. For example, a concentration-dependent pattern could reflect a misspecified structural prediction, an unsuitable error model, influential observations, or dependence not represented by the model.

Proportional-only error near zero. As \(f_{ij}\) approaches zero, the modeled standard deviation approaches zero. A small absolute residual can then produce a very large standardized residual. Check whether an additive component is scientifically and empirically needed rather than treating every large low-end residual as a structural-model failure.
Combined error dominated by one component. If the additive term is negligible over the entire observed range, or the proportional term is negligible, the nominal combined model may behave much like a single-component model. Interpret the effective weighting over the actual prediction range.
Wrong variance scale. If residual spread is systematically larger or smaller than the model implies, standardized residuals may be too small or too large. A smooth-looking weighted pattern is not proof that the variance model is correct.
Confusing sign with magnitude. A positive residual means observation above prediction; a negative residual means below. Weighting changes magnitude, not sign, when the modeled standard deviation is positive.

Choosing a Residual Scale Without Overclaiming

Raw residuals, relative residuals, and model-standardized residuals are complementary rather than interchangeable. Raw residuals retain concentration units and show absolute discrepancies. Relative residuals express departures as a fraction of a positive prediction and are natural for proportional error, but become unstable or undefined near zero. Model-standardized residuals account for the specified variance model and permit comparison on a common nominal scale, but their interpretation depends on that model being adequate.

A common mistake is to say, “The high-concentration residual is six times larger, so that observation is six times more poorly predicted.” That conclusion uses raw differences without accounting for their expected variability. If the variance rises with concentration, the larger raw residual may be entirely consistent with proportional error. The correction is to compare both the raw difference and its standardized value, stating which scale supports the conclusion.

The opposite mistake is to treat a modest weighted residual as proof of a good prediction. Weighting can make a large absolute discrepancy modest relative to a large modeled standard deviation. If that standard deviation is overstated, the weighted residual can conceal a practically important absolute deviation. Review the assumed error scale alongside residuals, and retain concentration units when absolute accuracy matters.

Key takeaway: Proportional error scales residuals by predicted concentration, while combined error uses both a baseline additive scale and a concentration-dependent scale. A weighted residual is a discrepancy relative to the assumed error variance—not an error-free verdict on model fit.

Check Your Understanding

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

Under proportional error with \(\sigma_{\mathrm{prop}}=0.10\), what is the standardized residual for an observation 2 concentration units above a prediction of 10 units?
Why can an absolute residual of 0.5 units be more concerning at a low prediction than at a high prediction under proportional error?
In a combined-error model, which component is more likely to dominate at low concentrations, and why is this not guaranteed at every concentration?
A positive raw residual has a negative standardized residual. Is this possible under the definition in this tutorial? Explain.
Why is a weighted residual alone insufficient to establish that the model fits well?

Solutions. (1) The standard deviation is \(0.10(10)=1\) unit, so \(r=2/1=2.00\). (2) The proportional standard deviation is \(\sigma_{\mathrm{prop}}f\), which is smaller at a lower positive prediction; the same absolute discrepancy is therefore larger relative to that scale. (3) The additive component may dominate because it does not shrink with the prediction, but dominance depends on whether \(\sigma_{\mathrm{add}}^2\) exceeds \(\sigma_{\mathrm{prop}}^2f^2\) over the concentration considered. (4) No. The denominator is positive when the variance is positive, so standardization preserves the residual's sign. (5) It depends on the assumed variance and prediction being appropriate, and fitted residuals can have additional dependence or estimation effects; a modest value cannot rule out bias or a misspecified error model.