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Population PK/PD Model Building · Tutorial 423 of 598

Additive And Proportional Random Effects On PD Parameters

Learn how additive and exponential random effects change individual PD parameters, and how to compare their scale, positivity, and between-subject variability.

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What You'll Learn

  • Derive additive and proportional-linear random-effects parameterizations from their scale assumptions.
  • Explain why an exponential random effect keeps a positive typical parameter positive.
  • Calculate the probability of nonpositive values under additive and proportional-linear models.
  • Interpret the typical value, arithmetic mean, standard deviation, and coefficient of variation under each parameterization.
  • Diagnose when a parameter’s scale or constraints make a random-effects form implausible.

Why the Random-Effects Scale Matters

In a population PD model, a typical parameter describes the population level, while a subject-specific random effect allows an individual parameter to differ from that level. The scale on which that difference is expressed matters. For a positive parameter such as an Emax maximum effect or an EC50 concentration, an additive deviation can produce an impossible negative value. A multiplicative deviation instead makes the size of variation depend on the parameter’s magnitude, and an exponential deviation preserves positivity by construction.

The hierarchical structure of population PD models separates typical parameters, subject-level deviations, and observation-level variability. Here the focus is the form of the subject-level deviation for one parameter. It is not enough to say that a parameter “has between-subject variability”; the model must specify whether that variability is measured in the parameter’s original units or on a relative, transformed scale.

Definition: Let \(\theta\) be the typical value of a PD parameter and \(\theta_i\) its value for subject \(i\). A subject-level random effect \(\eta_i\) represents the deviation of subject \(i\) from the typical value. Under a mean-zero random-effects convention, \(E(\eta_i)=0\); its variance is denoted \(\omega^2\). The parameterization specifies how \(\eta_i\) determines \(\theta_i\).

Three Parameterizations

An additive random effect places the deviation directly on the parameter scale:

$$ \theta_i=\theta+\eta_i. $$

Here \(\eta_i\) and \(\theta_i\) have the same units as \(\theta\). If \(E(\eta_i)=0\) and \(\operatorname{Var}(\eta_i)=\omega^2\), then \(E(\theta_i)=\theta\) and \(\operatorname{Var}(\theta_i)=\omega^2\). The absolute spread is constant across the parameter scale; the relative spread is not. In particular, when \(\theta>0\), the coefficient of variation is \(\omega/\theta\), provided the individual parameter distribution is meaningful for that summary.

A proportional-linear random effect places the deviation on a scale relative to \(\theta\):

$$ \theta_i=\theta(1+\eta_i). $$

The random effect is dimensionless. With \(E(\eta_i)=0\) and \(\operatorname{Var}(\eta_i)=\omega^2\), the mean is \(\theta\), the variance is \(\theta^2\omega^2\), and, for \(\theta>0\), the coefficient of variation is \(\omega\) when the distribution has a positive mean. Thus the standard deviation increases in proportion to the typical parameter. However, this form does not guarantee positivity: any realization with \(\eta_i<-1\) gives \(\theta_i<0\).

An exponential random effect, often called an exponential or log-scale parameterization, is

$$ \theta_i=\theta\exp(\eta_i). $$

For \(\theta>0\), this guarantees \(\theta_i>0\) for every finite real value of \(\eta_i\). Taking logarithms gives \(\log(\theta_i)=\log(\theta)+\eta_i\), so the random effect is an additive deviation on the log scale. This form is multiplicative on the original scale: a given change in \(\eta_i\) corresponds to a ratio of individual to typical parameter values, rather than a fixed difference.

What “Typical” Means Under the Exponential Form

The exponential form requires care when interpreting \(\theta\). If \(E(\eta_i)=0\), then \(\theta\) is the value at \(\eta_i=0\), and the geometric mean of \(\theta_i\) is \(\theta\). It is not generally the arithmetic mean. For a concrete derivation, suppose \(\eta_i\sim N(0,\omega^2)\). The normal moment-generating function gives \(E[\exp(t\eta_i)]=\exp(t^2\omega^2/2)\). Therefore:

$$ E(\theta_i)=\theta E[\exp(\eta_i)] =\theta\exp\left(\frac{\omega^2}{2}\right), $$

and, using \(E(\theta_i^2)=\theta^2\exp(2\omega^2)\),

$$ \operatorname{Var}(\theta_i) =\theta^2\exp(\omega^2)\left[\exp(\omega^2)-1\right], \qquad CV(\theta_i)=\sqrt{\exp(\omega^2)-1}. $$

These moment formulas depend on the normal assumption for \(\eta_i\). Positivity does not: for any finite real \(\eta_i\), \(\exp(\eta_i)>0\). If a model specifically needs \(\theta\) to equal the arithmetic population mean, one possible normal-error form is \(\theta_i=\theta\exp(\eta_i-\omega^2/2)\). That adjustment changes the interpretation of \(\theta\) and should be stated explicitly rather than silently assumed.

Worked Example: Additive Variability Can Produce Negative Emax Values

Suppose the typical Emax parameter is \(\theta=4.00\) effect units and the additive random effect is normally distributed with mean zero and standard deviation \(3.00\) effect units. Thus \(\theta_i=4.00+\eta_i\), with \(\eta_i\sim N(0,9.00)\). Find the probability that an individual Emax is nonpositive.

The condition \(4.00+\eta_i\leq 0\) is equivalent to \(\eta_i\leq-4.00\). Standardizing by the random-effect standard deviation gives

$$ P(\theta_i\leq0) =P\left(\frac{\eta_i}{3.00}\leq\frac{-4.00}{3.00}\right) =\Phi(-1.3333) \approx0.09121. $$

The model therefore assigns about \(9.121\%\) probability to a nonpositive individual Emax. The additive model’s mean is still \(4.00\) effect units and its standard deviation is \(3.00\) effect units, but the nonpositive tail makes this form questionable if Emax must be positive. The probability calculation uses the stated normal assumption; the possibility of negative values follows from the additive form itself.

Worked Example: Proportional-Linear Effects Have Relative Scale but No Positivity Guarantee

Let the typical value of a positive PD parameter be \(\theta=4.00\) units, and let \(\theta_i=4.00(1+\eta_i)\), where \(\eta_i\sim N(0,0.250^2)\). Calculate the mean, standard deviation, coefficient of variation, and probability of a nonpositive individual value.

The mean is \(E(\theta_i)=4.00[1+E(\eta_i)]=4.00\) units. The standard deviation is \(4.00(0.250)=1.00\) unit, so the coefficient of variation is \(1.00/4.00=0.250\), or \(25.0\%\). For positivity, \(4.00(1+\eta_i)\leq0\) exactly when \(\eta_i\leq-1\). Hence

$$ P(\theta_i\leq0) =P\left(\frac{\eta_i}{0.250}\leq\frac{-1}{0.250}\right) =\Phi(-4.000) \approx0.00003167. $$

This is about \(0.003167\%\), a small but nonzero probability. The proportional-linear model expresses variability relative to the typical parameter, but unlike the exponential form, it cannot rule out nonpositive values. Whether that tail is acceptable depends on the parameter’s scientific meaning and the range of variability supported by the data.

Worked Example: Interpreting an Exponential Random Effect

Suppose \(\theta=4.00\) units and \(\theta_i=4.00\exp(\eta_i)\), with \(\eta_i\sim N(0,0.300^2)\). Calculate the arithmetic mean, coefficient of variation, and central 95% range of individual values. The range uses the central 95% interval for \(\eta_i\), approximately \(-1.96(0.300)\) to \(1.96(0.300)\).

The arithmetic mean is

$$ E(\theta_i)=4.00\exp(0.300^2/2) =4.00\exp(0.0450) \approx4.184\text{ units}. $$

The coefficient of variation is

$$ CV(\theta_i)=\sqrt{\exp(0.300^2)-1} =\sqrt{\exp(0.0900)-1} \approx0.3069, $$

or approximately \(30.69\%\). The central 95% range obtained by transforming the random-effect interval is

$$ \left[4.00\exp(-0.588),\;4.00\exp(0.588)\right] \approx[2.222,\;7.202]\text{ units}. $$

Every value in this range is positive. The central value corresponding to \(\eta_i=0\) is \(4.00\) units, while the arithmetic mean is approximately \(4.184\) units. The difference is a consequence of the nonlinear exponential transformation, not an arithmetic error.

Choosing and Checking the Scale

The parameter’s meaning and units provide the first check. An additive effect is natural when differences in the original units are scientifically meaningful and values may legitimately cross zero. A proportional or exponential effect is more natural when differences are better described as percentages or ratios. For a strictly positive parameter, the exponential form enforces the constraint; the proportional-linear form may be adequate only if its probability of invalid values is negligible and such values are not otherwise consequential.

These forms also differ in how their variability behaves as the typical parameter changes. Under an additive model, the absolute standard deviation remains \(\omega\), so the relative variability \(\omega/\theta\) grows as \(\theta\) becomes smaller. Under a proportional-linear model, the standard deviation is \(\theta\omega\), and the coefficient of variation is constant when \(\theta>0\). Under the normal exponential model, the coefficient of variation is determined by \(\omega\), while the arithmetic mean is shifted above the typical value. These are different modeling assumptions, not merely different notations for the same model.

Useful diagnostics begin with consequences of the fitted parameterization. For an additive or proportional-linear model on a parameter that must be positive, calculate or simulate the implied fraction of nonpositive values under the fitted random-effects model. Also inspect the scale of the estimated variability relative to the parameter: a modest absolute standard deviation can imply extreme relative variability when the typical value is small. For an exponential model, check whether the implied arithmetic mean and range are scientifically plausible, rather than treating the typical value as the arithmetic mean. In all cases, assess whether estimates are stable and whether the resulting individual parameters give sensible PD predictions over the observed concentration and response ranges.

A parameterization cannot by itself repair weak information in the data. If the observations do not inform between-subject variation separately from other sources of variation, changing additive to exponential form may change the interpretation without resolving that limitation. Nor does positivity guarantee biological plausibility: an exponential distribution can produce very large positive values. The assumed scale and range should therefore be justified by scientific meaning and checked against predictions, not selected solely because one form is mathematically convenient.

Common Mistake and Comparison

Common mistake: “The exponential model has mean-zero \(\eta_i\), so the population mean of \(\theta_i\) is \(\theta\).” This is generally false. With normal \(\eta_i\), \(E(\theta_i)=\theta\exp(\omega^2/2)\), which exceeds \(\theta\) whenever \(\omega^2>0\). The correction is to distinguish the parameter value at \(\eta_i=0\) from the arithmetic mean, or to use and explicitly report a mean-adjusted parameterization if the arithmetic mean is intended.

FormScale of random effectPositive values guaranteed?Variability interpretation
\(\theta_i=\theta+\eta_i\)Original parameter unitsNoConstant absolute spread
\(\theta_i=\theta(1+\eta_i)\)Relative, dimensionlessNoSpread proportional to \(\theta\)
\(\theta_i=\theta\exp(\eta_i)\)Log scale; multiplicative on original scaleYes, for \(\theta>0\)Relative spread; normal effects give a lognormal parameter distribution

The central trade-off is therefore between an interpretable absolute deviation, an interpretable relative deviation, and a positivity constraint. No one form is universally best. State the parameterization, the random-effect scale, and the meaning assigned to \(\theta\) whenever reporting or interpreting a population PD model.

Check Your Understanding

Answer each question, then compare with the brief solutions below.

  1. For an additive model with \(\theta=6.00\) units and random-effect standard deviation \(1.50\) units, what are the mean and standard deviation of \(\theta_i\), assuming a mean-zero random effect?
  2. In \(\theta_i=\theta(1+\eta_i)\), with \(\theta>0\), what condition on \(\eta_i\) produces a negative individual parameter?
  3. For a normal exponential random effect with \(\omega=0.400\), is \(\theta\) the arithmetic mean of \(\theta_i\)? If not, give the mean as a multiple of \(\theta\).
  4. Which form guarantees a positive individual value for a positive \(\theta\), and what limitation remains even then?

Solutions. (1) The mean is \(6.00\) units and the standard deviation is \(1.50\) units, since \(\theta_i=\theta+\eta_i\) and \(E(\eta_i)=0\). (2) A negative value occurs when \(\eta_i<-1\); equality gives zero. (3) No. The mean is \(\theta\exp(0.400^2/2)=\theta\exp(0.0800)\approx1.083\theta\). (4) The exponential form guarantees positivity for \(\theta>0\) and finite \(\eta_i\), but it can still assign implausibly large positive values or imply an arithmetic mean different from \(\theta\).

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