Additive Random Effects on the Parameter Scale

Adding Variability on the Parameter Scale

A population parameter equation must do two things: define the population value and describe how individuals vary around it. In an additive parameter model, the individual value is the population value plus an individual random deviation. This is a direct way to express between-individual variation in the units of the parameter itself. Its simplicity also makes an important question unavoidable: can the resulting individual parameter take values that the model says are impossible?

For example, clearance and volume are positive, while a fraction must lie between zero and one. An additive normal random effect has no such boundaries. Before using the formulation for one of these parameters, calculate or otherwise assess how much probability it assigns outside the parameter's allowable range.

Definition: An additive random-effects model for an individual parameter is \(P_i=\theta+\eta_i\), where \(P_i\) is the parameter for individual \(i\), \(\theta\) is its population value, and \(\eta_i\) is that individual's random deviation. In the usual normal model, \(\eta_i\sim N(0,\omega^2)\), independently across individuals. Thus \(P_i\sim N(\theta,\omega^2)\), with mean \(\theta\) and standard deviation \(\omega\). The random effect and parameter have the same units; \(\omega^2\) has squared parameter units.

The equation \(P_i=\theta+\eta_i\) defines the additive structure. The normality assumption is an additional distributional choice: without specifying a distribution for \(\eta_i\), the equation alone does not determine probabilities of extreme or invalid values. Under the normal model, the random effect is centered at zero, so the average of the individual parameters in the population is \(\theta\). Since a normal distribution has support over the entire real line, it assigns some probability to values below any finite lower bound and above any finite upper bound.

Meaning and Consequences of the Equation

Because the effect is added on the parameter scale, \(\eta_i\) is an absolute difference from \(\theta\), not a percentage difference. If \(\theta\) is measured in liters per hour, then \(\eta_i\) and \(\omega\) are also measured in liters per hour. A one-unit increase in \(\eta_i\) changes \(P_i\) by one unit, regardless of the size of \(\theta\). In contrast, the relative size of the variability depends on both \(\omega\) and \(\theta\).

The equation also makes a simple calculation possible for any specified individual deviation: substitute that value of \(\eta_i\) and add it to \(\theta\). A deviation of zero gives \(P_i=\theta\); a positive deviation puts the individual above \(\theta\), and a negative deviation puts the individual below it. This arithmetic does not guarantee that the result is physically or scientifically admissible.

Worked Example: Constructing Individual Clearance Values

Suppose a hypothetical model uses \(CL_i=\theta+\eta_i\), with \(\theta=5.0\) L/h and \(\eta_i\sim N(0,(2.0\text{ L/h})^2)\). For an individual with \(\eta_i=-1.2\) L/h, calculate clearance and assess the chance of a negative value under the population distribution.

Individual value. Substituting the given deviation gives \(CL_i=5.0\text{ L/h}+(-1.2\text{ L/h})=3.8\text{ L/h}\). This is a positive clearance.

Probability of a negative value. Since \(CL_i\sim N(5.0,(2.0)^2)\), standardizing the zero boundary gives

P(CL_i<0)=P\left(\frac{CL_i-5.0}{2.0}<\frac{0-5.0}{2.0}\right) =\Phi(-2.5)\approx 0.00621.

Here \(\Phi\) is the standard normal cumulative distribution function. The model assigns about 0.621% probability to negative clearance. That probability is small but not zero; whether it is negligible for the intended use requires a scientific judgment, not just the fact that this example's particular individual had a positive value.

For a strictly positive parameter, the probability of invalid values can be calculated directly. If \(\theta>0\), then \(P_i<0\) exactly when \(\eta_i<-\theta\). Under the normal assumption,

P(P_i<0)=\Phi\left(-\frac{\theta}{\omega}\right).

The ratio \(\theta/\omega\) therefore controls the lower-tail risk. A larger population value relative to the random-effect standard deviation means less probability below zero. For a parameter bounded above as well, the upper boundary must also be checked; positivity alone does not establish plausibility.

Checking a Bounded Parameter

Fractions such as bioavailability or a pathway fraction cannot be below zero or above one. An additive normal effect does not preserve either bound. The following calculation checks both tails rather than treating a positive value as automatically valid.

Worked Example: Additive Variation in a Fraction

Consider a hypothetical fraction \(F_i=0.70+\eta_i\), where \(\eta_i\sim N(0,0.20^2)\). Calculate the probability that the model assigns \(F_i\) outside the allowable interval from zero to one.

Values below zero require \(\eta_i<-0.70\). Values above one require \(\eta_i>0.30\). Standardizing each threshold,

P(F_i<0\text{ or }F_i>1) =\Phi\left(\frac{-0.70}{0.20}\right) +1-\Phi\left(\frac{1.00-0.70}{0.20}\right).

The standardized cutoffs are \(-3.5\) and \(1.5\). Using \(\Phi(-3.5)\approx0.000233\) and \(1-\Phi(1.5)\approx0.066807\),

P(F_i<0\text{ or }F_i>1) \approx0.000233+0.066807=0.067040.

Thus about 6.70% of the model's population distribution lies outside the permitted range: about 0.0233% below zero and 6.68% above one. The upper-bound violation dominates. This is not fixed by noting that \(0.70\) itself is valid, nor by checking only the lower tail.

A Boundary-Risk Calculation

A useful design-stage check is to ask how small \(\omega\) must be, relative to \(\theta\), to keep the negative-value probability below a chosen tolerance. This is a mathematical screen, not a universal rule for selecting or accepting a model.

State the model and bound.
Assume \(P_i=\theta+\eta_i\), \(\eta_i\sim N(0,\omega^2)\), with \(\theta>0\). The invalid event for a positive parameter is \(P_i<0\).

Express the probability.
Standardization gives \(P(P_i<0)=\Phi(-\theta/\omega)\).

Apply the chosen tolerance.
For a tolerance \(q\), require \(\Phi(-\theta/\omega)\le q\). Since \(\Phi\) is increasing, this is equivalent to \(-\theta/\omega\le\Phi^{-1}(q)\).

Interpret the condition.
For \(q<0.5\), the condition becomes \(\omega\le\theta/|\Phi^{-1}(q)|\). It quantifies a distributional constraint; it does not establish that the model is otherwise appropriate.

Worked Example: Limiting the Negative Tail to Five Percent

Suppose a hypothetical positive parameter has \(\theta=10\) units. Find the largest \(\omega\) for which the normal additive model assigns at most 5% probability to negative values.

Set \(q=0.05\), for which \(\Phi^{-1}(0.05)\approx-1.645\). The condition is

\Phi\left(-\frac{10}{\omega}\right)\le0.05 \quad\Longleftrightarrow\quad -\frac{10}{\omega}\le-1.645 \quad\Longleftrightarrow\quad \omega\le\frac{10}{1.645}\approx6.079\text{ units}.

At \(\omega=6.0\) units, the negative-tail probability is \(\Phi(-10/6)\approx\Phi(-1.667)\approx0.0478\), or 4.78%. At \(\omega=7.0\) units, it is \(\Phi(-10/7)\approx\Phi(-1.429)\approx0.0766\), or 7.66%. The first value meets the specified screen and the second does not. The 5% tolerance is a hypothetical decision criterion, not a general standard.

What the Calculation Does and Does Not Establish

The boundary calculation follows directly from the distributional assumption. Its steps are to identify the invalid region, translate that region into a condition on \(\eta_i\), standardize using its mean zero and standard deviation \(\omega\), and evaluate the corresponding normal tail. For a bounded interval \(L\le P_i\le U\), the out-of-range probability is

P(P_i<L\text{ or }P_i>U) =\Phi\left(\frac{L-\theta}{\omega}\right) +1-\Phi\left(\frac{U-\theta}{\omega}\right).

This expression assumes \(L<U\), \(\omega>0\), and the stated normal distribution. The two tail events are disjoint, so their probabilities add. If \(\omega=0\), there is no random variation and the probability is determined by whether \(\theta\) is inside the bounds. For a different distribution of \(\eta_i\), the normal tail formula does not apply; use the specified distribution instead.

A calculated tail probability is a diagnostic of the model's implications, not a test that decides whether the formulation is correct. Consider whether invalid parameter values would make the individual-level model undefined, conflict with the scientific meaning of the parameter, or create unacceptable predictions. Also examine relevant ranges of \(\theta\) and \(\omega\), not just one convenient pair of values. If the parameters are estimated, uncertainty in their estimates means a single plug-in calculation may not capture all uncertainty about the tail probability.

Alternatives and Limitations

The main alternative for a positive parameter is the exponential random-effects formulation covered in the earlier material on exponential random effects: \(P_i=\theta\exp(\eta_i)\), with a centered normal \(\eta_i\). It preserves positivity for every finite \(\eta_i\) and expresses individual differences multiplicatively rather than as fixed-unit additions. These formulations answer different variability questions; one should not be substituted for the other solely because it is computationally convenient.

For a parameter constrained to a finite interval, a formulation that maps an unconstrained random quantity into that interval can preserve the bounds. A truncated distribution is another possible choice, but truncation changes the distribution and generally changes the mean of the random effect unless the model is adjusted accordingly. Each alternative therefore carries assumptions of its own. The additive model remains useful when additive deviations are scientifically meaningful and its support is compatible with the parameter's allowable values.

Diagnostics and Common Pitfalls

Check support, not just the central value. A valid \(\theta\) does not prevent negative or over-bound individual values. Calculate the relevant tail probabilities under the assumed distribution.
Do not treat a small tail as exactly zero. A normal distribution is unbounded. A small calculated probability may be acceptable for a purpose, but it does not make the parameter mathematically constrained.
Keep scale and units consistent. In an additive equation, \(\eta_i\) and \(\omega\) have the same units as \(P_i\). A variance is in squared units.
Do not interpret \(\eta_i\) as a percentage. The equation adds an absolute deviation. A deviation of 2 units has the same absolute effect whether \(\theta\) is 10 units or 100 units.
Do not silently repair invalid values after simulation. Replacing negative draws with zero, for example, creates a different distribution from the stated additive normal model.

In practice, inspect the estimated \(\theta\) and \(\omega\), calculate boundary probabilities, and simulate individual parameter draws when useful. Check whether the resulting values and downstream concentration predictions are scientifically plausible. Frequent boundary violations, numerical failures when invalid values enter the structural model, or strong sensitivity to small changes in \(\omega\) are reasons to revisit the formulation rather than ignore the diagnostic.

Key takeaway: \(P_i=\theta+\eta_i\) places random variation directly on the parameter scale. It is appropriate only when additive deviations are meaningful and the distribution it implies is compatible with the parameter's allowable range. Under normal random effects, calculate the probability of crossing each relevant boundary before relying on the model.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions below.

A parameter is modeled as \(P_i=8+\eta_i\), with \(\eta_i\sim N(0,2^2)\). What is the mean and standard deviation of \(P_i\), and what units does \(\omega\) have relative to \(P_i\)?
For a positive parameter with \(\theta=4\) and \(\omega=1\), express the probability of a negative value using \(\Phi\). Is it zero?
A fraction is modeled additively around \(0.40\). Why is checking only the probability below zero insufficient?
In one sentence, distinguish additive from exponential random effects for a positive parameter.

Solutions. (1) \(P_i\sim N(8,4)\), so its mean is 8 parameter units and its standard deviation is 2 parameter units; \(\omega\) has the same units as \(P_i\). (2) \(P(P_i<0)=\Phi(-4/1)=\Phi(-4)\approx0.0000317\), which is small but not zero. (3) A fraction also has an upper bound of one, so values above one must be included in the invalid probability. (4) An additive effect changes the parameter by an absolute amount in its original units, whereas an exponential effect changes it multiplicatively and keeps it positive.