Why the Scale of a Random Effect Matters
A population pharmacokinetic model distinguishes a typical parameter value from variation among individuals. The scale on which that variation is added determines what an individual random effect means. An additive effect on clearance in L/h describes an absolute difference in L/h; an additive effect on log clearance describes a multiplicative difference in clearance. These are different models, with different implied distributions and practical consequences.
The same choice applies to volume and to rate constants such as \(k=CL/V\). A model may estimate an effect on a transformed parameter, but predictions and interpretation generally concern the original parameter scale. The mapping between the two scales should therefore be made explicit.
Identity and Log Scales
With the identity transformation, \(g(\theta)=\theta\), the model is additive on the parameter's original scale:
If \(\eta_i\sim N(0,\omega^2)\), the model assigns a normal distribution to \(\theta_i\), centered at \(\theta_{\mathrm{typ}}\), with standard deviation \(\omega\) in the parameter's units. For clearance, \(\omega\) is measured in L/h; for volume, it is measured in L. An effect of \(+1\) L/h means the same absolute increment regardless of the starting clearance.
For a positive parameter, a common alternative is the natural-log transformation:
Here, \(\eta_i\) is dimensionless. If \(\eta_i\sim N(0,\omega^2)\), individual values are positive and their distribution is lognormal. A particular effect \(\eta_i\) multiplies the typical value by \(\exp(\eta_i)\); it does not add a fixed number of units. Equal positive and negative effects have equal and opposite magnitudes on the log-ratio scale, but they do not yield equal absolute changes on the original scale.
The normality, zero-centered effects, and common variance in these examples are modeling assumptions, not consequences of choosing a transformation. In a population model, effects are usually treated as draws from a common distribution across individuals. Whether this distribution and its implied predictions are reasonable must be assessed for the application.
Worked Example: Equal Log-Scale Effects, Unequal Clearance Changes
Suppose typical clearance is \(5.00\) L/h and the log-clearance model is \(CL_i=5.00\exp(\eta_i)\). Compare individuals with \(\eta_i=+0.200\) and \(\eta_i=-0.200\). These values are hypothetical.
For the positive effect:
The change from typical clearance is \(6.1070-5.00=+1.1070\) L/h. For the negative effect:
The change is \(4.0937-5.00=-0.9063\) L/h. The two effects correspond to multiplication by reciprocal factors, \(\exp(0.200)\) and \(\exp(-0.200)\), but the absolute changes differ because each is measured from the same typical value in a multiplicative model. The positive effect is a \(22.14\%\) increase; the negative effect is an \(18.13\%\) decrease.
How the Choice of Scale Changes Predictions
An identity-scale normal model permits every real value. For a parameter that must be positive, this means it can assign some probability to impossible negative clearances or volumes. The risk depends on the typical value relative to the random-effect standard deviation. A log-scale normal effect instead maps every finite \(\eta_i\) to a positive parameter value, although positivity alone does not guarantee clinically plausible predictions.
The two parameterizations also differ in how variability changes with the typical value. Under an identity-scale model, the standard deviation is constant in the parameter's units. Under a log-scale model, the same effect produces a proportional change, so the absolute difference grows with the typical value. For small \(\eta_i\), the approximation \(\exp(\eta_i)\approx 1+\eta_i\) makes a log-scale effect resemble a proportional effect, but this is an approximation near zero, not an identity for large effects.
Worked Example: Checking Negative Predictions Under an Additive Model
Suppose a hypothetical clearance model is \(CL_i=5.00+\eta_i\) L/h, with \(\eta_i\sim N(0,1.00^2)\) in \((\mathrm{L/h})^2\). The additive standard deviation is \(1.00\) L/h. The model predicts negative clearance when \(\eta_i<-5.00\) L/h.
Standardizing gives \(Z=\eta_i/1.00\), where \(Z\sim N(0,1)\), so:
This probability is very small for these hypothetical values, but it is not zero. If the additive standard deviation were large relative to typical clearance, negative predictions could become much more frequent and call the model's suitability into question. By contrast, with \(CL_i=5.00\exp(\eta_i)\), every finite effect gives positive clearance. A log-scale model can still predict extremely small or very large values, so the distribution of predictions should be examined rather than judged by positivity alone.
Mapping Clearance and Volume Effects to a Rate Constant
In a linear one-compartment model, the elimination rate constant is \(k_i=CL_i/V_i\). If clearance and volume each have log-scale effects, then the rate constant also has a log-scale expression. This follows by substituting the individual parameter models and applying the log rule for a ratio:
Thus the individual log-rate-constant deviation is the clearance effect minus the volume effect. A positive clearance effect raises \(k_i\), while a positive volume effect lowers it, all else equal. If both effects vary, the distribution of their difference depends on their joint distribution; it cannot be fully characterized from the two marginal variances alone. No independence assumption is implied by the algebra.
Worked Example: Combining Individual Clearance and Volume Effects
For a hypothetical individual, let \(CL_{\mathrm{typ}}=6.00\) L/h, \(V_{\mathrm{typ}}=20.0\) L, \(\eta_{CL,i}=0.150\), and \(\eta_{V,i}=-0.100\). Both parameters have log-scale effects. First calculate the individual clearance and volume:
The typical rate constant is \(6.00/20.0=0.3000\ \mathrm{h}^{-1}\). The individual rate constant can be calculated directly from the individual values, or from the combined effect:
The units reduce to \(\mathrm{h}^{-1}\), as required for a rate constant. The clearance effect raises clearance and the negative volume effect lowers volume; both changes raise \(k_i\). The rate constant is approximately \(28.4\%\) above its typical value, since \(0.3852/0.3000\approx1.284\).
Assumptions, Diagnostics, and Alternatives
A random-effects parameterization should be checked against the quantity it is meant to describe. Useful checks include examining estimated individual effects and model-predicted distributions on the original parameter scale, checking for implausible clearance, volume, or rate-constant values, and assessing whether the model systematically misses observations across individuals. If an identity-scale model assigns substantial probability to negative values, or a log-scale model generates extreme values, the chosen distribution or parameterization may be unsuitable. Numerical estimation problems can also arise when the data do not adequately identify the typical value or the between-individual variability.
The main alternatives are not merely different notations; they encode different assumptions:
| Parameterization | Individual parameter | Effect interpretation | Main consideration |
|---|---|---|---|
| Identity scale | \(\theta_{\mathrm{typ}}+\eta_i\) | Absolute change in the parameter's units | May predict negative values for parameters that must be positive |
| Log scale | \(\theta_{\mathrm{typ}}\exp(\eta_i)\) | Multiplicative change; \(\exp(\eta_i)\) is the individual-to-typical ratio | Preserves positivity but can imply highly skewed or extreme predictions |
| Fixed value without an individual effect | \(\theta_i=\theta_{\mathrm{typ}}\) | No between-individual variation in that parameter | May be too restrictive if meaningful individual variation exists |
A model with no random effect is a useful conceptual comparison, but it is not automatically an adequate alternative: it assumes that the parameter does not vary across individuals in the modeled population. Conversely, adding a random effect is not a remedy for every mismatch. A structural-model error, an unsuitable observation-error model, or an omitted predictor can also produce systematic discrepancies.
Common Mistakes and Limitations
- Confusing additive and multiplicative effects. On the log scale, \(\eta_i=0.20\) does not mean an increase of 0.20 units of clearance. It means multiplication by \(\exp(0.20)\), about 1.221.
- Assuming equal log effects give equal absolute changes. Effects \(+a\) and \(-a\) give reciprocal ratios, not equal-and-opposite absolute differences on the original scale.
- Ignoring units. An identity-scale effect on clearance has units of clearance, whereas a log-scale effect is dimensionless. The resulting rate constant must have units of inverse time.
- Assuming positive values guarantee a good model. A log transformation prevents negative predictions, but does not rule out implausibly small or large values, poor fit, or weak parameter identification.
- Treating the typical value as a universal individual value. The typical parameter is the model value at zero random effect. It does not describe every individual when variability is included.
The main limitation of the transformation approach is that its convenient mathematical properties do not establish that its distribution matches the population. The implied original-scale distribution should be considered alongside clinical knowledge and the information available in the data. Interpretation of typical values, medians, and means for lognormal parameters requires care; the key point here is that the transformation sets the scale on which the random effect is additive.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions.
- In an identity-scale clearance model, what are the units of \(\eta_i\) if clearance is measured in L/h?
- For a log-clearance model, what original-scale ratio corresponds to \(\eta_i=0.10\)?
- If clearance and volume have log effects of \(0.20\) and \(0.05\), respectively, what is the individual deviation in \(\log(k)\)?
- Why does a log-scale effect prevent negative clearance but not guarantee plausible clearance predictions?
Solutions. (1) \(\eta_i\) is measured in L/h because it is added directly to clearance. (2) The ratio is \(\exp(0.10)\approx1.105\), so clearance is about \(10.5\%\) above its typical value. (3) The deviation is \(0.20-0.05=0.15\), because \(\log(k_i)\) contains the clearance effect minus the volume effect. (4) Exponentiation maps every finite real effect to a positive multiplier, but very large positive or negative effects can still produce extreme or near-zero clearances.