Two Different Questions About Population Exposure
When subjects differ in characteristics that affect exposure, a population summary can be formed in at least two ways. One approach predicts exposure for each subject and then averages those predictions. Another first averages the subject characteristics and predicts exposure for a subject with those average characteristics. If the exposure function is nonlinear, these procedures generally give different answers.
The distinction matters because the two summaries answer different questions. The average of subject-specific predictions describes the average predicted exposure across subjects. The prediction at average characteristics describes a constructed subject whose characteristics equal the population averages. That constructed subject need not have the population’s average predicted exposure.
For clarity, suppose exposure is represented by a subject-specific prediction \(g(X_i)\). In the examples below, use the familiar relationship \(AUC=D/CL\) for a hypothetical single dose with complete exposure observation under linear elimination. Here \(D\) is dose and \(CL\) is clearance; this relationship serves only as a simple nonlinear illustration. The key feature is that exposure varies inversely with clearance, so averaging clearance before predicting exposure is not the same as averaging the individual exposure predictions.
Worked Calculations
Worked Example: Two Subjects With Different Clearances
Suppose two hypothetical subjects each receive a dose of \(100\) mg. Their clearances are \(2\) L/h and \(6\) L/h. Calculate their mean individual AUC and the typical-subject AUC at the mean clearance.
For each subject, \(AUC_i=D/CL_i\). The individual predictions are \(100/2=50\) mg·h/L and \(100/6=50/3\) mg·h/L. Therefore, the mean individual AUC is calculated without rounding the component predictions:
The mean clearance is \((2+6)/2=4\) L/h. The typical-subject AUC at that clearance is \(100/4=25.00\) mg·h/L. Thus, for these two subjects, mean individual AUC exceeds AUC at mean clearance by \(100/3-25=25/3=8.333\) mg·h/L. This is a difference between two summaries, not evidence that either calculation is erroneous.
Worked Example: The Same Mean Clearance, Different Variability
Compare two hypothetical two-subject populations, with dose \(100\) mg for every subject. Population A has clearances of \(2\) and \(6\) L/h. Population B has clearances of \(3\) and \(5\) L/h. Both have mean clearance \(4\) L/h.
For Population A, the mean individual AUC is \((100/2+100/6)/2=100/3=33.33\) mg·h/L, using the unrounded values in the calculation. For Population B, it is \((100/3+100/5)/2=(100/3+20)/2=80/3\approx26.67\) mg·h/L.
At the common mean clearance of \(4\) L/h, both typical-subject predictions are \(100/4=25.00\) mg·h/L. The different mean individual AUCs show that the mean clearance alone does not determine the mean exposure when the exposure function is nonlinear. Here the more dispersed clearances in Population A produce the larger average inverse clearance and hence the larger mean AUC.
Worked Example: Averaging Predictions Across Four Subjects
For four hypothetical subjects given \(100\) mg, let clearances be \(2, 2, 4,\) and \(8\) L/h. Compute both summaries and state what each represents.
The individual AUC predictions are \(50, 50, 25,\) and \(12.5\) mg·h/L. Their mean is \((50+50+25+12.5)/4=137.5/4=34.375\) mg·h/L. The mean clearance is \((2+2+4+8)/4=4\) L/h, so the AUC at mean clearance is \(100/4=25.00\) mg·h/L.
The first value is the average predicted AUC across these four subjects, with each subject counted once. The second is the prediction for a constructed subject with clearance \(4\) L/h. It is not the average of the four predictions. Reporting either value should make clear which population summary is intended.
Why Nonlinearity Creates a Difference
For the inverse-clearance example, define \(f(c)=D/c\) for positive clearance \(c\) and fixed positive dose \(D\). Its second derivative is \(f''(c)=2D/c^3\), which is positive for every \(c>0\). Thus \(f\) is strictly convex. Jensen’s inequality states that, for a convex function and a random variable with a finite expectation for which the function’s expectation exists, \(E[f(X)]\geq f(E[X])\). Consequently, if clearance is positive and the relevant expectations are finite,
For a finite set of subjects, the same statement applies to their empirical distribution: the average of \(D/CL_i\) is at least \(D\) divided by the arithmetic mean clearance. With strict convexity, the inequality is strict when the clearances are not all equal. This direction of inequality is specific to a convex exposure function such as \(D/CL\); it is not a universal rule for every nonlinear function.
A local expansion gives another way to see why variability matters. Let \(X\) be a scalar subject characteristic with mean \(\mu\), and suppose \(f\) is three times differentiable on the convex hull of the support of \(X\). Assume \(|f'''(x)|\leq M\) throughout that convex hull, and \(E[|X-\mu|^3]\) is finite. Taylor’s theorem around \(\mu\) gives
Taking expectations eliminates the first-order term because \(E[X-\mu]=0\), leaving
This is a local approximation, not a replacement for the exact calculation or Jensen’s inequality. It highlights that curvature and variability both contribute to the gap. The derivative bound must hold over the intervals joining \(\mu\) to possible values of \(X\); bounding it only at values in the support is not enough if those intervals extend beyond the support.
Assumptions, Interpretation, and Diagnostics
The calculations require the exposure function to be defined for the subject characteristics being used. In the inverse-clearance illustration, clearance must be positive, dose is held fixed, and the stated AUC relationship is being used within its stated setting. For population expectations, the quantities being compared must exist and be finite. To apply the convexity inequality, the function must be convex over the range of relevant clearances; positivity of the second derivative verifies this for \(D/CL\). If dose varies too, it must be included among the subject characteristics and in the exposure function rather than silently treated as fixed.
The word “typical” requires care. A subject set to the arithmetic mean of measured characteristics is not necessarily an actual subject, a median subject, or a subject with median exposure. If characteristics form a vector, average each component and then apply the function to that vector. The resulting prediction still need not equal the mean of predictions. In addition, the distribution and dependence among characteristics may matter: knowing each characteristic’s mean does not generally suffice to calculate the mean of a nonlinear function of them.
A practical check is to calculate both summaries from the same subjects and the same exposure function. Confirm that each individual prediction uses the intended dose, units, and subject characteristics; then verify the mean characteristic values and the averaging weights. If the two answers are unexpectedly identical, check whether the function is effectively linear over the observed range or whether the inputs have little variation. If their difference is unexpected in direction or size, inspect the function’s curvature over the actual input range and check for calculation or unit errors.
These summaries are alternatives only when they answer the question of interest. Averaging individual predictions is appropriate for the average predicted exposure across the represented subjects when each subject is intended to contribute equally. Predicting at average characteristics can be useful as a compact reference prediction, but it does not by itself describe the average exposure or the spread of exposures. Neither quantity alone describes the full distribution of individual predictions.
Common Pitfall
More generally, for a linear function \(g\), averaging inputs and averaging predictions agree. For a nonlinear function they need not. That distinction is about the mathematical operation and the target summary; it does not, by itself, establish a causal interpretation or quantify uncertainty in either estimate.
Check Your Understanding
Try each question, then compare your reasoning with the short solutions.
- Two subjects have clearances of \(2\) and \(6\) L/h and each receives \(100\) mg. What are the mean individual AUC and AUC at mean clearance?
- Why does \(E[D/CL]\geq D/E[CL]\) hold when dose \(D\) is fixed and positive and clearance is positive?
- What does the prediction at mean characteristics represent, and why need it not equal mean individual exposure?
- If the exposure function is linear over the relevant range, how do the two summaries compare?
Solutions. (1) The mean individual AUC is \((100/2+100/6)/2=100/3=33.33\) mg·h/L, calculated from unrounded values. The mean clearance is \(4\) L/h, giving \(100/4=25.00\) mg·h/L at mean clearance. (2) The function \(f(c)=D/c\) has \(f''(c)=2D/c^3>0\), so it is convex on positive clearances; Jensen’s inequality gives the stated direction when the expectations exist. (3) It is the prediction for a constructed subject whose characteristics equal their averages. Nonlinearity means applying the function after averaging inputs generally differs from averaging the function’s subject-specific outputs. (4) For a linear function, applying the function to the mean input equals the mean of its outputs.