Why Residual Spread Matters
A residual plot can reveal more than whether predictions are systematically too high or too low. The width of the residual cloud also matters. If errors are small at low predicted concentrations and much larger at high concentrations, a constant-variance observation model may not describe the data. Conversely, a widening raw-residual funnel is not, by itself, proof that the error model is wrong: the model’s mean predictions, the choice of residual, and the range of observations all affect the appearance of a plot.
The aim is to assess whether residual spread changes systematically with prediction or time, and whether that change is consistent with the assumed observation-error variance. This is a diagnostic question, not a reason to select an error model from a plot alone. A plausible variance model should also make sense for the measurement scale and the scientific setting.
Three Common Variance Patterns
An additive error model assumes that the conditional observation variance is constant over the prediction range. Its residual standard deviation is therefore roughly the same at small and large predictions. A proportional error model instead makes the standard deviation increase in proportion to the magnitude of the prediction. A combined model allows a constant component and a prediction-dependent component.
Here \(\sigma_{\mathrm{add}}\) has the units of the response, while \(\sigma_{\mathrm{prop}}\) is dimensionless when \(f_{ij}\) and \(y_{ij}\) use the same units. In the combined model, the two error components are assumed to have mean zero, to be independent of one another, and to have variances \(\sigma_{\mathrm{add}}^2\) and \(\sigma_{\mathrm{prop}}^2\), respectively. These assumptions give the displayed variance: the variance of a sum of independent, mean-zero terms is the sum of their variances. If a model uses a different error construction, its variance should be derived from that construction rather than assumed to match this formula.
The corresponding residual standard deviations are \(\sigma_{\mathrm{add}}\), \(\sigma_{\mathrm{prop}}|f_{ij}|\), and \(\sqrt{\sigma_{\mathrm{add}}^2+\sigma_{\mathrm{prop}}^2f_{ij}^2}\). Thus a proportional model has roughly constant relative spread: its residual standard deviation divided by \(|f_{ij}|\) is \(\sigma_{\mathrm{prop}}\). A combined model often looks more additive at low predictions and more proportional at high predictions. The transition is not a sharp boundary; it depends on the sizes of the two components.
Worked Examples
Worked Example: Recognizing a Proportional Pattern
Suppose a hypothetical concentration data set has four model predictions of 2, 4, 6, and 8 mg/L, with residuals of \(-0.2\), \(0.4\), \(-0.6\), and \(0.8\) mg/L. At four larger predictions of 20, 40, 60, and 80 mg/L, the residuals are \(-2\), \(4\), \(-6\), and \(8\) mg/L. Each residual is hypothetical and is being used only to illustrate the pattern.
At the lower predictions, the absolute residual-to-prediction ratios are \(0.2/2=0.10\), \(0.4/4=0.10\), \(0.6/6=0.10\), and \(0.8/8=0.10\). At the higher predictions, the ratios are \(2/20=0.10\), \(4/40=0.10\), \(6/60=0.10\), and \(8/80=0.10\). The absolute residuals grow with prediction, but the relative residual size remains 10%.
This pattern is consistent with proportional spread: the residual standard deviation would be expected to scale with the prediction. It is not enough to establish that a proportional error model is correct; the example contains only eight residuals, and real data would be assessed for the overall pattern, including whether residuals are centered near zero.
Worked Example: Calculating Spread Under a Combined Error Model
Suppose a hypothetical combined error model for concentrations measured in mg/L has \(\sigma_{\mathrm{add}}=2\) mg/L and \(\sigma_{\mathrm{prop}}=0.15\). Calculate the expected residual standard deviation at predictions of 10 and 50 mg/L.
At \(f_{ij}=10\) mg/L, the variance is \(2^2+(0.15)^2(10)^2=4+2.25=6.25\;(\mathrm{mg/L})^2\). The standard deviation is \(\sqrt{6.25}=2.50\) mg/L.
At \(f_{ij}=50\) mg/L, the variance is \(2^2+(0.15)^2(50)^2=4+56.25=60.25\;(\mathrm{mg/L})^2\). The standard deviation is \(\sqrt{60.25}=7.762\) mg/L.
The additive component contributes a 2 mg/L standard deviation at either prediction. The proportional component contributes \(0.15(10)=1.5\) mg/L at the lower prediction and \(0.15(50)=7.5\) mg/L at the higher prediction. The combined standard deviations are not the sums \(2+1.5\) and \(2+7.5\); independent variances add, so their square root gives the total standard deviation.
Worked Example: Checking Whether Standardization Removes a Funnel
Consider hypothetical predictions of 10, 20, 40, and 80 mg/L with residuals of \(-1.8\), \(1.2\), \(-5.0\), and \(9.0\) mg/L. Suppose a candidate combined error model has \(\sigma_{\mathrm{add}}=1\) mg/L and \(\sigma_{\mathrm{prop}}=0.10\). For each observation, divide the residual by the model’s predicted residual standard deviation.
The standard deviations at the four predictions are, respectively, \(\sqrt{1+0.10^2(10)^2}=\sqrt{2}=1.414\) mg/L, \(\sqrt{1+0.10^2(20)^2}=\sqrt{5}=2.236\) mg/L, \(\sqrt{1+0.10^2(40)^2}=\sqrt{17}=4.123\) mg/L, and \(\sqrt{1+0.10^2(80)^2}=\sqrt{65}=8.062\) mg/L.
The standardized residuals are therefore \(-1.8/1.414=-1.273\), \(1.2/2.236=0.537\), \(-5.0/4.123=-1.213\), and \(9.0/8.062=1.116\). The raw residual magnitudes increase across these predictions, while the standardized residual magnitudes are broadly similar. These four points do not show an obvious remaining increase in spread after scaling by the proposed variance model. They are too few to confirm the model, and the calculation does not establish that the residuals are independent or that the mean model is adequate.
How to Diagnose a Residual Funnel
A residual-versus-prediction plot is a direct way to inspect whether residual width changes across the fitted range. Begin by checking whether the residuals are centered around zero. A consistent upward or downward shift, or a curved centerline, indicates a pattern in the mean as well as—or instead of—a change in spread. Next, compare the vertical width of the residual cloud at low, middle, and high predictions. Look for a systematic change, not simply one large residual or a wider-looking region with very few observations.
The same question can be asked against time since dose if the proposed variance model or measurement process implies that spread should change over time. However, a time-related pattern may also arise because the predicted response changes over time. Compare the observed pattern with the variance predicted by the error model; do not infer a particular error structure from the time axis alone.
Raw residuals are measured in response units, so their width should be compared with the modeled conditional standard deviation. A useful check is to form a standardized residual by dividing \(e_{ij}\) by the square root of the fitted conditional variance. If the variance model is adequate, the standardized residuals should no longer show a systematic change in spread across the prediction range. This is a diagnostic expectation, not a guarantee that every plotted region will have exactly the same width.
The prediction used for the plot also matters. A population-level prediction does not condition on an individual’s estimated random effects, so residuals against it can reflect both between-individual differences and observation error. Conditional residuals based on individual predictions focus more directly on within-individual deviations, but individual predictions are estimated from the same subject’s data and can make some patterns less visible. Compare plots with their targets in mind; do not treat every funnel against a population prediction as evidence of a wrong residual variance function.
A Short Justification for Standardizing
Suppose the conditional residual has mean zero and variance \(V_{ij}\), where \(V_{ij}\) is the variance specified by the observation model at that prediction. Dividing by \(\sqrt{V_{ij}}\) rescales its conditional variance:
This explains why a suitable variance function should remove systematic changes in spread from standardized residuals. It does not prove that the standardized residuals follow a particular distribution, that their means are correctly modeled, or that observations are independent. It also assumes that the variance used for scaling is the relevant conditional variance and has been estimated reasonably well.
Choosing and Comparing Error Structures
| Error structure | Expected raw-residual spread | Useful interpretation |
|---|---|---|
| Additive | Approximately constant in response units | Absolute measurement error is similar across prediction levels. |
| Proportional | Increases in proportion to the absolute prediction | Relative error is approximately stable across prediction levels. |
| Combined | Has a constant component and a prediction-dependent component | Allows absolute and relative contributions to coexist. |
| Transformed-response model | Depends on the transformation and its implied original-scale model | Changes the scale on which the response and error are modeled. |
A transformed-response model can sometimes express multiplicative variability conveniently, but it is not automatically interchangeable with a proportional error model on the original scale. The transformation changes the scale of the modeled response and may change how predictions and residuals are interpreted. The chosen model must be appropriate for the response scale and for the data, including whether the transformation is defined for all observations.
When comparing candidate error structures, ask whether the fitted variance function is scientifically plausible and whether it removes the systematic spread pattern without introducing a new one. A more flexible variance function may fit the observed residuals better simply because it has more parameters. The plot is evidence to consider alongside model fit and the adequacy of the mean predictions, not a stand-alone selection rule.
Common Mistakes and Limitations
- Calling every funnel heteroscedasticity in the error model. A funnel in raw residuals is evidence of changing spread, but it may also reflect predictions that are systematically wrong, the use of population rather than conditional predictions, or limited data in parts of the prediction range. Check the residual center and the prediction target.
- Confusing a larger residual with a larger relative error. A residual of 5 mg/L may be large at a prediction of 10 mg/L and modest at a prediction of 100 mg/L. Compare residuals with the modeled standard deviation or with the prediction when assessing proportional spread.
- Adding standard deviations instead of variances. In the independent combined-error construction, variances add. The total standard deviation is the square root of the sum of the component variances.
- Assuming a good spread pattern proves the whole model is adequate. Standardizing can make residual widths look stable even when the mean model is biased or another dependence remains. Spread diagnostics address conditional variance, not every model assumption.
- Overinterpreting a small or unevenly populated plot. A few observations at high predictions can create an apparent funnel or conceal one. Sparse regions provide weak visual evidence, and subject-level clustering can make pooled plots misleading.
A residual funnel matters because an inadequate variance model can misrepresent the precision assigned to observations and affect parameter estimation and uncertainty calculations. Still, no single plot identifies the cause with certainty. The most defensible conclusion is proportional to the evidence: describe the pattern, say whether the candidate error model accounts for it, and note remaining uncertainty.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions below.
- In an additive error model, what should happen to the conditional residual standard deviation as the prediction increases?
- A proportional error model has \(\sigma_{\mathrm{prop}}=0.12\). What is its residual standard deviation at a prediction of 25 mg/L?
- For a combined model with \(\sigma_{\mathrm{add}}=3\) mg/L and \(\sigma_{\mathrm{prop}}=0.20\), calculate the residual standard deviation at a prediction of 10 mg/L.
- A plot has a widening raw-residual funnel, but standardized residuals have similar spread across predictions. What does this suggest, and what does it not prove?
- Why can a funnel against population predictions arise even if the observation-error variance model is reasonable?
Solutions. (1) It should remain approximately constant in response units: it equals \(\sigma_{\mathrm{add}}\). (2) The standard deviation is \(0.12(25)=3.00\) mg/L. (3) The variance is \(3^2+0.20^2(10)^2=9+4=13\;(\mathrm{mg/L})^2\), so the standard deviation is \(\sqrt{13}=3.606\) mg/L. (4) The proposed variance function may account for the changing raw spread, but this does not prove that the mean model is correct, that the residual distribution is otherwise appropriate, or that dependence is absent. (5) Population predictions do not condition on individual random effects, so between-individual differences can contribute to the residual spread as well as observation error.