Why compare curves rather than simply choose the best fit?
An exposure-response model translates exposure into an expected outcome. Several curves may describe the observed data similarly while implying different responses at clinically important exposures. A flexible curve can follow random variation and improve the likelihood, yet make less reliable predictions where data are sparse. A simple curve may be stable but miss meaningful curvature. The aim is therefore not to select the curve with the smallest residual sum or the most appealing shape in isolation. It is to compare fit, complexity, predictive discrepancies, and the exposure range supported by observations.
The earlier material on nonlinear exposure-response functional forms introduced plausible curve families. Here the task is to compare such candidates. For a fair comparison, they should address the same outcome, analysis population, observation times, covariates, and observations, and use likelihoods defined on the same basis. Model comparisons do not repair confounding, exposure measurement error, or misspecified outcome distributions; those features must be handled in the analysis before the curve comparison is interpreted.
Likelihood comparisons and their conditions
For observations \(y_1,\ldots,y_n\) and model parameters \(\theta\), the likelihood \(L(\theta)\) is the probability density or probability of the observed data under that model. The maximized log-likelihood is \(\ell=\log L(\hat{\theta})\), where \(\hat{\theta}\) denotes the parameter estimates that maximize the likelihood. A larger maximized likelihood indicates better in-sample fit, but it cannot by itself penalize added flexibility: a model that contains another as a special case cannot fit worse at its maximum.
A likelihood-ratio test is available for nested models: the restricted model must be obtained by imposing constraints on parameters of the unrestricted model. Under the null hypothesis that those constraints hold, and under regularity conditions such as identifiable parameters, an interior null value, and an appropriate likelihood model, the statistic is approximately chi-square distributed with degrees of freedom equal to the number of independent constraints. It is not a general test for comparing any two curves.
The statistic is nonnegative because the unrestricted model includes all parameter values allowed by the restricted model, so \(L_{\mathrm{unrestricted}}\geq L_{\mathrm{restricted}}\). Under the null, the likelihood-ratio result follows asymptotically from the fact that the log-likelihood difference for a regular set of \(df\) constraints has the chi-square limit described above. Boundary parameters, non-identifiability, or other failures of regularity can invalidate the usual reference distribution; for example, the standard chi-square result should not be assumed automatically when testing a parameter constrained to a boundary.
Worked Example: Testing an Added Curvature Term
Suppose a hypothetical continuous efficacy outcome is modeled as normally distributed around an exposure-dependent mean, with independent observations and a common residual variance. The restricted mean model is linear in exposure, while the unrestricted model adds a quadratic term. Thus, setting the quadratic coefficient to zero recovers the linear model. Assume both models use the same observations and likelihood, and the added coefficient is an identifiable interior parameter.
Suppose the fitted linear model has \(-2\ell=214.0\), and the fitted quadratic model has \(-2\ell=208.0\). The likelihood-ratio statistic is $$ \Lambda=214.0-208.0=6.000. $$
There is one added parameter, so \(df=1\). The upper-tail probability for a \(\chi^2_1\) statistic of 6.000 is approximately \(p=0.0143\). At \(\alpha=0.05\), this is evidence against the restricted linear mean shape under the stated model and regularity assumptions. It does not establish that the quadratic curve is mechanistically correct, that the association is causal, or that its predictions are reliable outside the observed exposure range.
Information criteria for non-nested shapes
When candidates are not nested, such as a linear curve and an Emax curve, a likelihood-ratio test generally does not apply. Information criteria provide a different comparison by trading off maximized likelihood against parameter count. For \(k\) estimated parameters and \(n\) observations, the common forms are
For either criterion, lower values indicate a preferred balance among the candidates being compared; neither value is an absolute goodness-of-fit measure. AIC and BIC use different penalties, and their differences are relative to the candidate set, data, and likelihood. The parameter count must include estimated quantities contributing to the likelihood, such as a residual variance when it is estimated. The observations and likelihood basis must be comparable. For example, criteria from different outcomes, different subsets of observations, or incompatible likelihood definitions are not a sound direct ranking.
Worked Example: Comparing Linear and Emax Curves
Consider 80 hypothetical independent efficacy observations. Compare a linear mean curve with an Emax mean curve, both with an estimated normal residual variance. The linear model estimates an intercept, slope, and variance, so \(k=3\); the Emax model estimates a baseline, maximum change, half-maximal exposure, and variance, so \(k=4\). Suppose the maximized values are \(-2\ell=202.4\) for the linear model and \(-2\ell=198.0\) for Emax.
The AIC values are $$ \mathrm{AIC}_{\mathrm{linear}}=202.4+2(3)=208.4,\qquad \mathrm{AIC}_{\mathrm{Emax}}=198.0+2(4)=206.0. $$
Since \(\log(80)\approx4.382\), the BIC values are $$ \mathrm{BIC}_{\mathrm{linear}}=202.4+3(4.382)=215.5,\qquad \mathrm{BIC}_{\mathrm{Emax}}=198.0+4(4.382)=215.5. $$
The displayed BIC values round to the same tenth; before rounding, the linear value is about 215.546 and the Emax value about 215.528, a negligible difference of roughly 0.018. AIC favors Emax by 2.4 points, while BIC provides essentially no separation. The appropriate conclusion is modest relative support for Emax by AIC, not decisive evidence that Emax is the true shape. The slight likelihood improvement may not justify the extra flexibility for the intended use.
Check predictions where the data lie
A compact predictive check asks whether fitted curves reproduce the observed outcome pattern across exposure. For a set of exposure groups or representative exposure values, compare observed summaries with model predictions and their uncertainty. A standardized discrepancy can be computed as the observed summary minus its prediction, divided by the prediction-scale standard error. It is a descriptive diagnostic, not automatically a hypothesis test: its interpretation depends on how the standard error was obtained, whether the summaries are independent, and whether uncertainty in the fitted curve is included.
Worked Example: Comparing In-Range Predictions
Suppose hypothetical observed mean efficacy values are available at exposures \(C=0.2, 0.5, 1.0,\) and \(1.5\) mg/L. For illustration, assign each observed mean a prediction-scale standard error of 1.5 outcome units. Compare the fitted Emax curve \(f(C)=10+30C/(2+C)\) with the fitted linear curve \(g(C)=10+9C\). The values below are hypothetical and are used only to illustrate the comparison.
| Exposure (mg/L) | Observed mean (outcome units) | Emax prediction | Linear prediction | Emax standardized discrepancy | Linear standardized discrepancy |
|---|---|---|---|---|---|
| 0.2 | 12.8 | 12.73 | 11.8 | 0.05 | 0.67 |
| 0.5 | 16.2 | 16.00 | 14.5 | 0.13 | 1.13 |
| 1.0 | 20.4 | 20.00 | 19.0 | 0.27 | 0.93 |
| 1.5 | 23.1 | 22.86 | 23.5 | 0.16 | -0.27 |
For example, at \(C=1.5\) mg/L, the Emax prediction is \(10+30(1.5)/(2+1.5)=10+45/3.5=22.857\), or 22.86 outcome units. Its standardized discrepancy is \((23.1-22.857)/1.5=0.162\), approximately 0.16. At \(C=0.5\), the linear prediction is \(10+9(0.5)=14.5\), giving discrepancy \((16.2-14.5)/1.5=1.13\).
These summaries show that the Emax curve tracks the four means more closely in this illustration, particularly at moderate exposures. They do not establish that its predictions are calibrated for individual patients, nor do four group means demonstrate that the assumed standard errors or outcome model are correct. Inspect the pattern of discrepancies, not just an aggregate score: a systematic run of positive or negative discrepancies can reveal a shape mismatch that a single fit statistic obscures.
Exposure support, alternatives, and failure modes
The observed exposure distribution limits what shape can be learned. If exposure values cluster in a narrow interval, very different curves can be nearly indistinguishable there. Sparse observations at the extremes weaken evidence about the low-exposure baseline or high-exposure plateau. Predictions outside the observed range are extrapolations, even if a fitted equation produces a precise-looking number.
Worked Example: A Plateau That the Data Do Not Establish
Suppose the exposure observations in the preceding hypothetical example range only from 0.2 to 1.5 mg/L. At \(C=3.0\) mg/L, the Emax curve predicts \(10+30(3)/(2+3)=28.0\) outcome units, while the linear curve predicts \(10+9(3)=37.0\) outcome units. The difference is \(37.0-28.0=9.0\) outcome units, much larger than their differences within the observed range.
There are no observations near 3.0 mg/L in this example to resolve that difference. The Emax model's eventual plateau is a property of its functional form, not evidence that the observed data establish saturation. The comparison may support describing the in-range pattern, but it does not warrant claiming that either curve predicts the outcome accurately at 3.0 mg/L.
The main alternatives answer different questions. A likelihood-ratio test assesses whether a specified nested restriction is compatible with the data under its assumptions. AIC and BIC rank relative fit-complexity trade-offs, including between non-nested candidates. Predictive checks reveal where predictions miss and whether residual patterns are systematic. More flexible options, such as splines, can represent complex curvature but need enough exposure support and can be unstable where observations are sparse. A mechanistically plausible function can be more interpretable and parsimonious, but biological plausibility does not excuse poor fit. No single method replaces the others.
Diagnostics for exposure-response model fit provide the broader context for examining residual patterns and influential observations. In a shape comparison, warning signs include a flexible model tracking isolated fluctuations, parameter estimates that change sharply when a few observations are removed, large uncertainty in curvature or plateau parameters, and competing curves that diverge mainly outside the data range. Reconsider the outcome likelihood, covariate structure, and exposure support when these occur; do not respond automatically by adding terms until the fit improves.
Use the same observations, outcome model, covariates, and likelihood basis; state how the curves differ.
Use a likelihood-ratio test only for nested models with suitable regularity conditions; use information criteria for relative comparison of non-nested candidates.
Compare discrepancies across the observed exposure range and identify where predictions are extrapolations.
Describe what the data support for the intended exposure range, and retain uncertainty when candidates are similar or support is weak.
A frequent mistake is to choose the curve with the largest likelihood and then treat its shape as established. Because added parameters can improve in-sample fit, that reasoning favors complexity without considering overfitting. The correction is to compare suitable fit-complexity criteria, inspect predictive discrepancies, and check how much exposure support exists for the features being interpreted. A related error is to report a likelihood-ratio p-value for non-nested curves; the ordinary nested-model reference distribution does not justify that comparison.
Check Your Understanding
Answer each question, then compare with the brief solutions.
- A restricted model has \(-2\ell=150\), and a nested unrestricted model has \(-2\ell=146\), with two added parameters. What is the likelihood-ratio statistic and its degrees of freedom?
- Two models fitted to the same data have AIC values 120 and 124. What can be concluded, and what cannot be concluded?
- At an exposure of 2 mg/L, the observed mean is 18 outcome units, the predicted mean is 15, and the prediction-scale standard error is 2 units. Calculate the standardized discrepancy.
- Why is a predicted plateau at an exposure far above all observed values weak evidence for saturation?
Solutions. (1) The statistic is \(150-146=4\), with \(df=2\); its use as an approximate chi-square test requires the nested-model assumptions and regularity conditions. (2) The model with AIC 120 has the lower AIC and is preferred on this relative criterion among these candidates. This does not prove it is true, establish adequate fit, or guarantee better predictions in every setting. (3) The discrepancy is \((18-15)/2=1.50\), or 1.50 prediction-scale standard errors above the prediction. It is descriptive unless additional assumptions justify inferential use. (4) With no observations near or beyond that exposure, the data do not distinguish a plateau from other shapes there; the plateau follows from the chosen function and is an extrapolation.