Why Functional Form Matters
A continuous covariate such as body weight or renal function does not usually divide people into only a few groups. A population PK model therefore needs a relationship that describes how a typical parameter value changes across the covariate’s range. That relationship is its functional form. Choosing a form affects both the fitted pattern within the data and the predictions made for patients whose covariate values differ from the reference.
The forms in this tutorial describe the covariate effect on a typical parameter, denoted \(\theta_{\mathrm{typ}}(x)\), where \(x\) is the covariate value. They do not by themselves describe all individual variability or establish that the covariate causes a change in the parameter. As in covariate model building, the scientific rationale, data support, and predictive consequences matter more than searching for a form that happens to produce a small p-value.
Four Common Forms
The same covariate can be related to a parameter in different ways. The parameter \(\theta\) is often positive, as with clearance or volume. A log-scale relationship can then preserve positivity, while an additive relationship on the original scale needs to be checked for plausible positive predictions.
Power form: A multiplicative change determined by the covariate ratio and exponent \(\beta\), requiring positive \(x\) and \(x_{\mathrm{ref}}\).
Exponential form: A multiplicative change whose logarithm is linear in the difference from the reference value.
Piecewise form: A relationship with different slopes in separate covariate regions; a continuous form can change slope at a knot without a jump in the predicted parameter.
In the linear form, \(\beta\) has units of parameter per unit of \(x\); it is the additive change in \(\theta_{\mathrm{typ}}\) for a one-unit increase in \(x\). In the power form, \(\beta\) is dimensionless and is an elasticity: a 1% increase in \(x\) corresponds approximately to a \(\beta\)% change in the typical parameter locally. In the exponential form, \(\beta\) has units reciprocal to \(x\). An increase of \(\Delta x\) multiplies the parameter by \(\exp(\beta\Delta x)\). These interpretations depend on the stated model form and do not imply a causal effect.
Worked Example: Comparing Additive and Power Predictions for Weight
Suppose, hypothetically, that a model describes typical clearance at reference body weight \(x_{\mathrm{ref}}=70\) kg as \(\theta_{\mathrm{ref}}=5.00\) L/h. Compare an additive linear form with \(\beta=0.025\) L/h per kg and a power form with \(\beta=0.75\) at 100 kg. Assume the observed weights span 50 to 100 kg.
Linear prediction. The difference from reference weight is \(100-70=30\) kg. Thus:
Power prediction. The weight ratio is dimensionless, \(100/70\), so:
Both forms give 5.00 L/h at 70 kg, but their predictions at 100 kg differ by about 0.784 L/h. These values illustrate the consequence of the two assumed shapes; they are not evidence that either form is preferable. The power form assumes a constant elasticity across weight, whereas the linear form assumes a constant absolute increase per kilogram.
Exponential and Piecewise Relationships
For the exponential form, the log of the typical parameter is linear in the covariate difference. It is useful when changes are naturally interpreted proportionally, but the size of the percentage change per covariate unit is fixed on the log scale. In particular, a change of \(\Delta x\) multiplies the reference prediction by \(\exp(\beta\Delta x)\); it does not add a fixed quantity to the parameter.
Worked Example: Exponential Relationship With Renal Function
In a hypothetical model, typical clearance is 4.00 L/h at a renal-function reference value of 60 mL/min. Suppose the exponential slope is \(\beta=0.00400\) per mL/min. Calculate typical clearance at 30 and 90 mL/min.
At 30 mL/min, the difference from reference is \(-30\) mL/min:
At 90 mL/min, the difference is \(+30\) mL/min:
The units in each exponent cancel. A 10 mL/min increase multiplies the prediction by \(\exp(0.0400)\), about 1.0408, or an increase of approximately 4.08%. Equal increases in \(x\) therefore have equal proportional effects under this model, though their absolute changes in L/h differ as the starting prediction changes.
A single slope may be too restrictive when the relationship plausibly changes across the covariate range. A continuous piecewise form uses a knot \(c\), with one slope below the knot and another above it. To keep the prediction anchored correctly at \(x_{\mathrm{ref}}\), including when the reference lies above the knot, use the following hinge form. Here \((z)_+=\max(z,0)\).
For \(x<c\), the hinge term \((x-c)_+\) is zero, so the slope of log typical parameter with respect to \(x\) is \(\beta_L\). For \(x>c\), that hinge increases with slope 1, making the total slope \(\beta_L+(\beta_R-\beta_L)=\beta_R\). At \(x=c\), the model is continuous, but its derivative is undefined unless \(\beta_L=\beta_R\). At \(x=x_{\mathrm{ref}}\), the two hinge terms cancel and the covariate difference is zero, so \(\log\{\theta_{\mathrm{typ}}(x_{\mathrm{ref}})\}=\log(\theta_{\mathrm{ref}})\). The model is therefore anchored at the reference value regardless of which side of the knot it occupies. Because the expression is continuous at \(c\), the prediction does not jump there.
Worked Example: A Continuous Piecewise Relationship
Consider a hypothetical piecewise log-linear relationship for clearance, with \(x_{\mathrm{ref}}=70\) mL/min, \(\theta_{\mathrm{ref}}=4.00\) L/h, knot \(c=60\) mL/min, \(\beta_L=0.00400\) per mL/min, and \(\beta_R=0.0150\) per mL/min. Calculate the predictions at 30, 60, and 90 mL/min.
At 30 mL/min, \((30-60)_+=0\) and \((70-60)_+=10\). The change in log clearance from reference is:
Therefore, \(\theta_{\mathrm{typ}}(30)=4.00\exp(-0.270)\approx3.054\) L/h. At the knot, the change is \((0.00400)(60-70)+(0.0110)(0-10)=-0.150\), giving \(4.00\exp(-0.150)\approx3.443\) L/h.
At 90 mL/min, \((90-60)_+=30\), so the change in log clearance is:
Thus, \(\theta_{\mathrm{typ}}(90)=4.00\exp(0.300)\approx5.399\) L/h. The reference prediction is still exactly 4.00 L/h at 70 mL/min, even though the reference is above the knot. The slope changes at 60 mL/min, but the predicted clearance remains continuous there.
Choosing and Checking a Form
A functional form should be considered in relation to the scientific question and the available covariate range. The table summarizes the assumptions that most directly distinguish the forms.
| Form | What is constant across \(x\)? | Key checks |
|---|---|---|
| Linear | Additive change per covariate unit | Are predictions plausible and positive where required? |
| Power | Elasticity of the parameter with respect to \(x\) | Are \(x\) and \(x_{\mathrm{ref}}\) positive, and is a ratio-based effect plausible? |
| Exponential | Change in log parameter per covariate unit | Is a constant proportional change over the range plausible? |
| Piecewise | Log-scale slope within each region | Is there a rationale and adequate data around the knot and on both sides? |
The linear form is simple to interpret, but can predict a zero or negative value when extrapolated far enough in the direction of a negative slope. The power and exponential forms preserve positivity when the reference parameter is positive, but positivity alone does not ensure plausible predictions: either can grow or shrink rapidly beyond the observed range. A piecewise form can represent a change in trend, but adds flexibility and requires data support for estimating both slopes and the knot’s location. A knot chosen after repeatedly examining outcomes can make the apparent pattern overly optimistic.
Diagnostics should include plotting the covariate against relevant individual or model-based parameter information, checking fitted trends across the observed range, and examining whether residual patterns suggest systematic misfit. Sparse observations at one end, a few influential covariate values, implausible predictions, or large changes in the estimated slope under modest modeling choices are warnings. No diagnostic plot proves a functional form correct; it can reveal when a proposed form is not describing the data adequately. Inference and predictive checks should account for the population model’s uncertainty rather than treating estimated individual parameters as error-free observations.
Interpret predictions outside the observed covariate range as extrapolations. For instance, a power relationship supported by weights from 50 to 100 kg does not establish that the same elasticity applies at 25 or 200 kg. A plausible trend within the data can become implausible outside it. State the range used to inform the relationship, and distinguish model-based extrapolation from evidence directly supported by observations.
Common Pitfall: Treating Reference Values as Intercepts
A frequent error is to assume that a parameter written as \(\theta_{\mathrm{ref}}\) will automatically be the prediction at \(x_{\mathrm{ref}}\). That is true only if the formula is actually anchored there. In a piecewise expression, using \(\log(\theta_{\mathrm{ref}})+\beta_L(c-x_{\mathrm{ref}})+\beta_R(x-c)\) for \(x>c\) does not generally preserve the reference prediction when \(x_{\mathrm{ref}}>c\). Substituting \(x=x_{\mathrm{ref}}\) yields an additional slope contribution, not necessarily zero. Use the anchored hinge expression above, or explicitly constrain the reference to the left of the knot and verify the formula at the reference.
Another common mistake is to compare slope values across forms as if they had the same meaning. A linear slope is in parameter units per covariate unit; a power exponent is dimensionless; an exponential slope is in inverse covariate units. These coefficients cannot be compared directly. Compare the predicted parameter values, proportional changes, and scientific implications over a shared covariate range.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions.
- A linear clearance model has a reference prediction of 3.00 L/h at 60 kg and slope 0.020 L/h per kg. What is its prediction at 80 kg, and what does the slope mean?
- In a power model with exponent 0.75, what approximate percent change in the parameter corresponds to a 1% increase in the covariate?
- An exponential model has \(\beta=0.010\) per unit of \(x\). What multiplicative change does it predict for an increase of 5 units?
- Why must a piecewise model be checked at its reference value when the reference lies above the knot?
- A linear model fits observed weights from 50 to 100 kg. Is its prediction at 140 kg direct evidence about clearance at that weight? Explain.
Solutions. (1) The difference is 20 kg, so \(3.00+(0.020)(20)=3.40\) L/h. The slope represents an additive increase of 0.020 L/h per kilogram under this model. (2) Approximately 0.75%, since the power exponent is the elasticity. (3) The multiplier is \(\exp(0.010\times5)=\exp(0.050)\approx1.051\), an increase of about 5.13%. (4) A poorly anchored piecewise formula can add a nonzero slope contribution at the reference, so substitution must confirm that \(\theta_{\mathrm{typ}}(x_{\mathrm{ref}})=\theta_{\mathrm{ref}}\). (5) No. That prediction is an extrapolation beyond the observed weight range; the data do not directly support the assumed relationship at 140 kg.