Why Functional Form Matters
An exposure-response model describes how the expected outcome changes with exposure, after accounting for the assumptions and covariates relevant to the analysis. The chosen functional form determines whether the modeled response changes at a constant rate, approaches a plateau, or can change slope across the exposure range. Those differences matter for interpreting observed data and for predicting responses at exposures that may occur under a proposed dosing regimen.
No curve is established as correct merely because it fits the observed points. In particular, a visually plausible bend does not establish a plateau if most observations are below the exposures needed to reveal one. Functional-form choices should be grounded in the scientific question, the exposure range, and what the data can identify. As in earlier material on exposure measurement error, uncertainty in the individual exposure estimates can also affect the apparent shape; the models here treat exposure as the predictor supplied to the analysis.
For a simple fit, suppose the observations satisfy \(Y_i=m(C_i;\theta)+\varepsilon_i\), where \(C_i\) is exposure, \(\theta\) is the parameter vector, and \(\varepsilon_i\) is a residual. Least squares estimates \(\theta\) by minimizing \(\sum_i [Y_i-m(C_i;\theta)]^2\). This fitting criterion is most naturally interpreted when the conditional mean is correctly specified and residuals have mean zero, are independent across observations, and have constant variance. If residual variance changes with exposure or observations are dependent, ordinary least squares uncertainty calculations may not apply; the curve itself should not be interpreted as fixing those problems.
Linear and Log-Linear Relationships
A linear mean model uses \(m(C)=\beta_0+\beta_1C\). The slope \(\beta_1\) is the expected outcome change per one exposure unit, and \(\beta_0\) is the modeled mean at zero exposure. The units of the slope are outcome units per exposure unit. A linear curve may be useful over a limited range, even when it would be implausible over a much broader range. Its unbounded increase or decrease should be considered before extrapolation.
Here, log-linear means linear in the logarithm of exposure: \(m(C)=\beta_0+\beta_1\log(C/C_{\mathrm{ref}})\), where \(C_{\mathrm{ref}}>0\) is a specified reference exposure and \(C>0\). The ratio inside the logarithm is dimensionless. A doubling of exposure changes the modeled mean by \(\beta_1\log(2)\), regardless of the starting exposure, within the range where the model is used. This form is not the same as modeling the logarithm of the outcome.
Worked Example: Linear Versus Log-Linear Interpretation
Suppose a hypothetical analysis gives \(\hat m(C)=12+0.30C\), where \(C\) is in mg/L and the outcome is a symptom score in points. From 10 to 20 mg/L, the modeled mean changes from \(12+0.30(10)=15\) to \(12+0.30(20)=18\), an increase of 3 points. The slope is 0.30 points per mg/L.
For a separate log-linear fit, suppose \(\hat m(C)=14+2.0\log(C/10)\). At 10 mg/L the fitted mean is 14 points. At 20 mg/L it is \(14+2.0\log(2)=14+1.386=15.39\) points. Thus, doubling exposure from 10 to 20 mg/L corresponds to a modeled increase of 1.386 points. The same model predicts the same increase for any doubling, but not for every fixed increase in mg/L. These are hypothetical fits; neither equation alone establishes a causal effect.
Emax and Sigmoid Emax Models
When a response is expected to approach a plateau, an Emax curve can represent a baseline mean plus an exposure-dependent change. For a nonnegative exposure measure \(C\), one common parameterization is:
Here \(E_0\) is the modeled mean at zero exposure, \(E_{\max}\) is the maximum change from that baseline as exposure tends to infinity, and \(EC_{50}>0\) is the exposure producing half of that maximum change. \(E_{\max}\) has outcome units; \(EC_{50}\) has exposure units. If the response decreases with exposure, \(E_{\max}\) can be negative. The sign should reflect the outcome's direction and scientific interpretation, not be constrained to be positive by habit.
For \(E_{\max}\ne0\), the exposure-dependent component is half its limiting value at \(C=EC_{50}\), since \(EC_{50}/(EC_{50}+EC_{50})=1/2\). The curve approaches \(E_0+E_{\max}\) as \(C\) increases, but only approaches that limit; it does not reach it at any finite exposure. The model presumes a monotonic change toward a single plateau. It cannot represent an interior peak followed by decline.
A sigmoid Emax curve adds a positive shape parameter \(\gamma\):
The parameter \(\gamma\) controls steepness around \(EC_{50}\), while \(EC_{50}\) still gives half of the limiting change. When \(\gamma=1\), this is the Emax model. Larger \(\gamma\) makes the transition more abrupt around \(EC_{50}\); it does not change the limiting values. The expression at \(C=0\) is defined by its continuous limit as \(E_0\).
Worked Example: Interpreting an Emax Curve
Consider a hypothetical fitted curve for a continuous biomarker: \(m(C)=30+20C/(5+C)\), with \(C\) in mg/L and the outcome in units. At \(C=5\) mg/L, \(m(5)=30+20(5)/(5+5)=30+10=40\) units. This is halfway between the baseline of 30 units and the limiting mean of 50 units.
At \(C=15\) mg/L, \(m(15)=30+20(15)/(5+15)=30+15=45\) units. As \(C\) becomes very large, the mean approaches \(30+20=50\) units. The curve implies diminishing increments: increasing exposure from 5 to 15 mg/L raises the modeled mean by 5 units, less than the 10-unit increase from zero to 5 mg/L. These calculations interpret the fitted functional form; they do not prove that the high-exposure plateau is supported by observed data.
Splines: Local Changes in Slope
A spline represents a relationship in connected pieces, allowing the slope to change at selected knot values. A simple one-knot linear spline can be written \(m(C)=\beta_0+\beta_1C+\beta_2(C-k)_+\), where \(k\) is the knot and \((C-k)_+=\max(0,C-k)\). Below \(k\), the slope is \(\beta_1\); above \(k\), it is \(\beta_1+\beta_2\). The model is continuous at \(k\), although its slope may change there. This is not the same as fitting two unrelated lines, which can produce a jump at the boundary.
Worked Example: Calculating a One-Knot Spline
Suppose a hypothetical model for a continuous outcome is \(m(C)=8+0.5C+0.4(C-10)_+\), with exposure in mg/L and outcome in points. At \(C=6\), the positive-part term is zero, so \(m(6)=8+0.5(6)=11\) points. At the knot \(C=10\), \(m(10)=8+0.5(10)=13\) points. At \(C=15\), \(m(15)=8+0.5(15)+0.4(5)=17.5\) points.
Below 10 mg/L, the slope is 0.5 points per mg/L. Above 10 mg/L, it is \(0.5+0.4=0.9\) points per mg/L. The two expressions meet at the knot, so the fitted mean is continuous there. A spline does not automatically represent a plateau: this example's slope increases above the knot.
More knots allow more changes in slope, but each adds flexibility and requires data to estimate it. Knot placement should be specified or justified, and there must be adequate observations across the regions used to estimate local trends. A flexible spline can track random variation when data are sparse, especially near the ends of the exposure range.
What the Exposure Range Can Support
Nonlinear parameter estimates depend on where observations fall relative to the curve's changing regions. For an Emax model, if all observations have \(C\ll EC_{50}\), then \(C/(EC_{50}+C)\) is approximately \(C/EC_{50}\), and the mean is approximately \(E_0+(E_{\max}/EC_{50})C\). In that region, the data may inform the ratio \(E_{\max}/EC_{50}\) more readily than the two parameters separately. If observations are also far below the bend, apparent curvature may be weak. If nearly all observations are at exposures far above \(EC_{50}\), outcomes may be close to the plateau, giving little information about the location of the bend.
This approximation explains a key data-range limitation:
Thus, a fitted \(EC_{50}\) far beyond the observed range can yield a curve that looks nearly linear over the data, with a plateau inferred mainly from model assumptions. Similarly, a sigmoid steepness estimate can be unstable when few observations lie around the transition. Parameter constraints can prevent nonsensical values but cannot create information that was not collected.
Before interpreting curvature, inspect the exposure distribution and the outcome observations across that distribution. Ask whether observations cover the region where the proposed model bends, where a plateau is supposed to occur, or on both sides of each spline knot. Sparse tails, gaps, clusters caused by dose levels, and outlying observations can make a detailed curve appear more certain than it is. Residual plots and uncertainty intervals for fitted parameters or predictions are useful checks, but a formal model-fit comparison belongs to a separate question; a favorable fit statistic does not by itself establish reliable extrapolation.
Choosing Among the Forms
| Form | Useful feature | Main caution |
|---|---|---|
| Linear | Constant change per exposure unit; straightforward interpretation. | Unbounded extrapolation and no diminishing returns. |
| Log-linear in exposure | Constant modeled change per exposure ratio, such as a doubling. | Requires positive exposure and is not constant per exposure unit. |
| Emax | Monotonic approach toward a plateau with interpretable \(EC_{50}\). | Can be weakly identified if data do not cover the bend or plateau. |
| Sigmoid Emax | Allows a steeper or shallower transition than standard Emax. | Additional shape flexibility requires information around the transition. |
| Linear spline | Allows slope changes at specified knots without requiring a plateau. | Many knots or sparse regions can produce unsupported local shapes. |
These forms answer different scientific descriptions of the relationship; they are not interchangeable parameterizations with identical assumptions. A linear or log-linear form can be adequate across a narrow observed range even if a saturating process is plausible more broadly. An Emax form encodes a particular monotone plateau behavior. A spline can describe local slope changes but usually does not provide a single mechanistic plateau parameter. The practical choice should therefore consider scientific rationale and data coverage together, rather than selecting the most flexible curve by default.
Common Mistakes and Limitations
- Calling \(EC_{50}\) the exposure at the plateau. It is the exposure at half the modeled maximum change, not the exposure at which the curve is effectively saturated.
- Interpreting an estimated plateau as observed fact. If the data do not include sufficiently high exposures, the plateau may be an extrapolation imposed by the Emax form.
- Using a negative or zero \(EC_{50}\) in the standard Emax model. The parameter is defined as positive; a nonpositive estimate makes its usual exposure interpretation invalid.
- Assuming a spline is automatically safer because it is flexible. Flexibility can fit noise, and unsupported knots can create unstable local slopes.
- Reading a fitted association as a causal effect. A curve describes the modeled relationship under its assumptions. Causal interpretation requires the relevant causal assumptions, not simply a nonlinear equation.
All these mean curves also have limitations: they do not, on their own, describe outcome variability, account for repeated measures, or solve confounding and exposure-estimation problems. Their predictions are most defensible where the data provide relevant support. Outside that region, they should be labeled as extrapolations and treated cautiously.
Check Your Understanding
Answer each question, then compare with the solutions.
- A log-linear model uses \(\beta_1=3\) outcome units and natural logarithms. What modeled mean change corresponds to doubling exposure?
- For \(m(C)=4+12C/(6+C)\), calculate the mean at \(C=6\) and state the limiting mean as exposure increases.
- In a sigmoid Emax model, what exposure gives half of the maximum modeled change, and what does \(\gamma\) control?
- For \(m(C)=2+0.2C+0.5(C-8)_+\), give the slopes below and above the knot and calculate \(m(10)\).
- Why might an Emax model be unable to estimate \(E_{\max}\) and \(EC_{50}\) separately when all observed exposures are much smaller than \(EC_{50}\)?
Solutions. (1) The change is \(3\log(2)=2.079\) outcome units. (2) At \(C=6\), \(m(6)=4+12(6)/(6+6)=4+6=10\) units; the limiting mean is \(4+12=16\) units. (3) Half of the maximum change occurs at \(C=EC_{50}\); \(\gamma\) controls steepness around the transition. (4) The slopes are 0.2 and \(0.2+0.5=0.7\) outcome units per exposure unit. At \(C=10\), \(m(10)=2+0.2(10)+0.5(2)=5\) units. (5) When \(C\ll EC_{50}\), the curve is approximately \(E_0+(E_{\max}/EC_{50})C\), so the data may identify the ratio more strongly than either parameter alone.