Tutorials › Nonlinear Mixed-Effects Models (NLME) › Choosing the Parameter Scale for Covariate Effects

Functional Forms for Covariates · Tutorial 488 of 797

Choosing the Parameter Scale for Covariate Effects

Learn how additive and log-linear covariate relationships differ in parameter changes, interpretation, and constraints, and how to check whether either scale is scientifically plausible.

Intermediate 9 min read

What You'll Learn

  • Write additive and log-linear covariate effects for a typical NLME parameter.
  • Derive the parameter change and ratio associated with a covariate change.
  • Interpret the units of each slope and the percentage change implied by a log-scale slope.
  • Check positivity and identify when an additive model predicts implausible values.
  • Compare the models while holding the covariate range and other model components fixed.
  • Diagnose scale mismatch and recognize the limits of choosing a scale from fit alone.

Why the Parameter Scale Matters

A covariate relationship can be linear on the parameter scale or linear on the logarithm of that parameter. These are different model assumptions, even when they use the same covariate and the same reference value. They imply different changes in the parameter as the covariate changes, and they impose different constraints on possible parameter values.

For example, an increase of one unit in a covariate produces a constant absolute change under an additive relationship. Under a log-linear relationship, it produces a constant multiplicative change. The distinction matters in a nonlinear mixed-effects model because parameters such as clearance, volume, and rate constants are often constrained to be positive. The scale should reflect both the scientific interpretation and the plausible range of the parameter, not just which formula is easier to write.

Definition: Let \(\theta_{\mathrm{typ}}(x)\) be the typical value of a model parameter at covariate value \(x\), with other covariates held fixed. Let \(x_0\) be a reference value and \(\theta_0=\theta_{\mathrm{typ}}(x_0)\). An additive parameter-scale effect is \(\theta_{\mathrm{typ}}(x)=\theta_0+\beta_A(x-x_0)\). A linear effect on the log parameter is \(\log\theta_{\mathrm{typ}}(x)=\log\theta_0+\beta_L(x-x_0)\), equivalently \(\theta_{\mathrm{typ}}(x)=\theta_0\exp\{\beta_L(x-x_0)\}\), where \(\theta_0>0\).

These equations describe the covariate component of the typical parameter. They do not, by themselves, specify the random-effects distribution, residual error model, or the rest of the structural model. In an NLME analysis, those components must still be defined and assessed separately.

Deriving the Different Changes

Write \(\Delta x=x-x_0\). For the additive model, subtracting the parameter at the reference value gives

$$ \theta_{\mathrm{typ}}(x)-\theta_0=\beta_A\Delta x. $$

Thus, for a fixed change \(\Delta x\), the absolute parameter change is constant wherever the model is used. The units of \(\beta_A\) are parameter units per covariate unit. For example, if \(\theta\) is clearance in L/h and \(x\) is weight in kg, then \(\beta_A\) has units L/h/kg.

For the log-linear model, exponentiating both sides of its defining equation gives

$$ \theta_{\mathrm{typ}}(x)=\theta_0\exp(\beta_L\Delta x). $$

Dividing by \(\theta_0\) shows the multiplicative change:

$$ \frac{\theta_{\mathrm{typ}}(x)}{\theta_0}=\exp(\beta_L\Delta x). $$

The corresponding relative change is \(\exp(\beta_L\Delta x)-1\), or \(100\{\exp(\beta_L\Delta x)-1\}\%\). The slope \(\beta_L\) has units reciprocal to the covariate units: for a covariate measured in kg, it is per kg. It is not generally correct to call \(100\beta_L\%\) the exact change for a one-unit increase; that is an approximation when \(\beta_L\) is small. The exact percentage change for one unit is \(100\{\exp(\beta_L)-1\}\%\).

Key distinction: Additive effects imply a constant absolute change per covariate unit. Log-linear effects imply a constant ratio per covariate unit, and therefore a change in absolute amount that depends on the current parameter value.

Worked Examples

Worked Example: Comparing Absolute and Relative Changes

Suppose, hypothetically, that the typical clearance at a weight of 70 kg is 5.00 L/h. Compare two candidate relationships for weight: an additive slope of \(0.0300\) L/h/kg and a log-linear slope of \(0.0150\) per kg. Calculate the typical clearance at 90 kg under each relationship.

The covariate difference is \(\Delta x=90-70=20\) kg. Under the additive model,

$$ \theta_{\mathrm{typ}}(90)=5.00+(0.0300)(20)=5.00+0.600=5.60\ \text{L/h}. $$

Under the log-linear model,

$$ \theta_{\mathrm{typ}}(90)=5.00\exp\{(0.0150)(20)\} =5.00\exp(0.300) =5.00(1.3499) =6.75\ \text{L/h}. $$

The additive model predicts an increase of 0.600 L/h, or 12.0% of the reference value. The log-linear model predicts a ratio of 1.3499, corresponding to an increase of about 35.0%. Both models agree on the reference value but not on how clearance changes away from it.

Worked Example: Checking Positivity Over a Covariate Range

Suppose a hypothetical positive parameter is 4.00 units at \(x_0=50\) kg. An additive candidate has slope \(-0.0800\) units/kg. A log-linear candidate has slope \(-0.0200\) per kg. Calculate both predictions at 100 kg and determine whether the additive model remains positive over the full interval from 50 to 110 kg.

At 100 kg, \(\Delta x=50\) kg. The additive prediction is

$$ \theta_{\mathrm{typ}}(100)=4.00+(-0.0800)(50)=4.00-4.00=0. $$

It reaches zero at \(x=50+4.00/0.0800=100\) kg and is negative above 100 kg. In particular, at 110 kg it predicts \(4.00+(-0.0800)(60)=-0.800\) units. If the parameter must be positive, this additive relationship is not admissible across the stated interval.

The log-linear prediction at 100 kg is

$$ \theta_{\mathrm{typ}}(100)=4.00\exp\{(-0.0200)(50)\} =4.00\exp(-1) =1.47\ \text{units}. $$

At every finite \(x\), the exponential is positive, so this model stays positive, including at 110 kg: \(4.00\exp(-1.20)=1.20\) units. Positivity alone does not establish that the predictions are scientifically reasonable; the decrease might still be too large.

Worked Example: Interpreting a Log-Scale Slope

In a hypothetical model, the typical value of a positive parameter is 2.00 units at age 40 years, and the log-linear age slope is \(0.0250\) per year. Find the typical value at age 50 and the exact percentage change over that ten-year increase.

Here \(\Delta x=10\) years. The predicted ratio is

$$ \exp\{(0.0250)(10)\}=\exp(0.250)=1.2840. $$

Therefore, the predicted parameter value is \(2.00(1.2840)=2.57\) units. The exact percentage change is \(100(1.2840-1)=28.4\%\). The slope times ten is \(0.250\) on the log scale; it is not an exact 25.0% increase on the parameter scale.

For a ten-year decrease, the ratio is \(\exp(-0.250)=0.7788\), a decrease of \(100(0.7788-1)=-22.1\%\). The percentage increase and decrease are not symmetric because they are multiplicative changes in opposite directions.

When the Two Scales Can Look Similar

Near the reference value, the models may make similar predictions if their slopes are chosen appropriately. The local rate of change for the additive model is \(\beta_A\). Differentiating the log-linear expression gives

$$ \frac{d\theta_{\mathrm{typ}}(x)}{dx} =\beta_L\theta_0\exp\{\beta_L(x-x_0)\} =\beta_L\theta_{\mathrm{typ}}(x). $$

At the reference value this derivative is \(\beta_L\theta_0\). Thus, matching the local slopes at \(x_0\) requires \(\beta_A=\beta_L\theta_0\). Away from \(x_0\), however, the log-linear slope on the parameter scale changes with \(\theta_{\mathrm{typ}}(x)\), while the additive slope remains constant. Local agreement does not imply agreement across a broad covariate range.

This also clarifies the interpretation. An additive slope describes an absolute change, while a log-linear slope describes a constant proportional change. Which is more appropriate depends on the scientific question and on how the parameter is expected to behave. A parameter that changes by roughly the same amount per unit of a covariate may be described naturally on its original scale. A parameter expected to change by roughly the same proportion may be described naturally on the log scale.

Assumptions, Diagnostics, and Alternatives

Either formula assumes the specified relationship is a reasonable approximation over the covariate range being modeled, with other covariates held fixed. The additive form assumes a constant slope on the parameter scale. The log-linear form assumes a constant slope on the log-parameter scale and requires a positive reference value. Neither assumption should be extended automatically beyond the observed range.

Useful checks include plotting model predictions against the covariate over the observed range, inspecting residual diagnostics for systematic patterns, and checking whether predicted parameter values remain scientifically plausible. A pattern in residuals may signal that the covariate relationship is not adequately represented, but does not by itself identify which alternative scale is correct. Sparse observations at the extremes make assessments there less reliable.

When comparing candidate scales, keep the data, covariate range, other model components, and fitting conditions comparable. Assess the resulting fit and parameter behavior together; a small improvement in fit does not alone establish a scientifically meaningful relationship. Consider whether slopes have interpretable units, whether predictions respect known bounds, and whether the model behaves sensibly at the covariate values relevant to use.

FeatureAdditive parameter scaleLinear log-parameter scale
Model form\(\theta_0+\beta_A(x-x_0)\)\(\theta_0\exp\{\beta_L(x-x_0)\}\)
Change for a fixed covariate incrementConstant absolute changeConstant ratio; absolute change depends on the current value
Slope interpretationParameter units per covariate unitLog units, or reciprocal covariate units
PositivityNot guaranteedGuaranteed for finite covariate values if \(\theta_0>0\)

The log-linear form is not a universal solution to positivity: it cannot represent a parameter that is zero or negative, and it may predict values that are implausibly small or large. Conversely, the additive form may be reasonable over a restricted range even for a parameter that must remain positive, provided its predictions stay positive throughout that range. If neither form captures the scientific pattern, a more flexible relationship may be considered, but flexibility does not remove the need to check interpretability and extrapolation.

Common Pitfall

A frequent error is to say, “The log-scale slope is 0.02, so the parameter rises by exactly 2% per covariate unit.” The coefficient is a change in the log parameter. For one unit, the exact ratio is \(\exp(0.02)\), so the exact percentage increase is \(100\{\exp(0.02)-1\}=2.02\%\), not exactly 2%. The approximation is close for a small coefficient, but the formula matters, especially across several units. Another error is to choose the log-linear form simply because the parameter is positive: positivity is a useful constraint, not proof that proportional change is the right scientific assumption.

Key takeaway: Choose the scale by asking what should remain constant as the covariate changes: the parameter’s absolute increment or its ratio. Then verify units, positivity, plausibility, and behavior over the covariate range of interest.

Check Your Understanding

Answer the questions, then compare with the brief solutions below.

  1. For an additive relationship, what are the units and interpretation of the slope?
  2. A log-linear slope is \(0.0100\) per unit of \(x\). What is the exact percentage change for a one-unit increase?
  3. Why can an additive model be unsuitable for a positive parameter even if its predictions are positive near the reference value?
  4. State one limitation of the log-linear form that is not solved merely by its positivity guarantee.

Solutions. (1) The slope has units of parameter units per covariate unit and gives the constant absolute change per unit. (2) \(100\{\exp(0.0100)-1\}=1.01\%\). (3) Its straight-line prediction can reach zero and become negative elsewhere in the modeled range. (4) It requires a positive parameter and can still predict scientifically implausible values or extrapolate poorly.