Why Stratify a Virtual Population?
A PBPK virtual population may need enough subjects in every genotype group to characterize subgroup exposures, even when some genotypes are uncommon. Sampling each virtual subject’s genotype according to its population frequency can leave a rare group with very few simulated subjects. A stratified design controls how many subjects are simulated in each group. The key is to keep two quantities distinct: the simulation allocation and the genotype frequency in the target population.
The allocation determines how much computational effort is spent on each genotype. It does not, by itself, determine that genotype’s contribution to population exposure. To recover population summaries after using deliberately chosen subgroup sizes, combine the subgroup results using target-population genotype frequencies. These weights should correspond to the population the analysis is intended to represent.
The exposure \(Y\) must be defined consistently across groups—for example, AUC over the same dosing interval and expressed in the same units. A subgroup simulation should represent the intended distribution of virtual subjects conditional on belonging to that genotype group, including the specified PBPK assumptions and relevant covariates. The earlier material on preserving dependence among virtual-subject covariates is relevant when constructing those within-group subjects. This tutorial focuses on how to allocate the groups and combine their simulated exposure results.
Choosing Controlled Group Sizes
A proportional allocation chooses \(n_g\) approximately in proportion to \(q_g\). It is efficient when the main aim is to represent the target population directly and every group will still have enough subjects. A controlled or disproportionate allocation instead sets group sizes to meet analysis needs, such as a minimum number in each genotype group. This can improve the precision of subgroup-specific simulation summaries, but population-level results must then be reweighted.
For integer counts, a practical allocation begins with a total simulation budget \(N\) and planned counts \(n_g\) satisfying \(\sum_g n_g=N\). If counts are chosen by rounding \(Nq_g\), the rounded counts may not sum exactly to \(N\); assign the remaining subjects in a documented way. Alternatively, specify minimum group sizes and distribute the remaining budget according to a chosen rule. In either case, record \(q_g\) separately from \(n_g\). A group assigned zero subjects cannot contribute a simulated subgroup distribution, regardless of its target frequency.
Worked Example: Planning a Cohort With a Rare Genotype
Suppose three genotype groups have target frequencies \(q_1=0.70\), \(q_2=0.25\), and \(q_3=0.05\). A simulation budget of 1,000 subjects is available. Proportional allocation gives \(n_1=700\), \(n_2=250\), and \(n_3=50\). If the analyst instead requires at least 150 simulated subjects in each group, one possible controlled allocation is \(n_1=600\), \(n_2=250\), and \(n_3=150\), still totaling 1,000.
The controlled allocation deliberately gives group 3 a larger share of simulations than its target frequency: \(150/1000=0.15\), compared with \(q_3=0.05\). This is appropriate for studying that subgroup, but the simulated cohort’s unweighted genotype composition is not the target population composition. Population exposure summaries must use weights \(0.70\), \(0.25\), and \(0.05\), not the allocation proportions \(0.60\), \(0.25\), and \(0.15\).
Frequency-Weighted Means and Variance
For each group, the simulated mean exposure is \(\bar{Y}_g\), the arithmetic mean of its \(n_g\) simulated exposure values. If the subgroup simulation adequately represents the conditional exposure distribution for genotype \(g\), the target-population mean is estimated by weighting these subgroup means by their target frequencies:
This is not generally the same as the unweighted mean across all simulated subjects. That unweighted mean gives each simulated subject equal weight, and therefore weights group \(g\) in proportion to \(n_g\). It is appropriate as a target-population estimate only when the allocation proportions match the target frequencies (or by coincidence when the subgroup means make the results equal).
Worked Example: Reweighting Subgroup Mean Exposures
Consider three hypothetical genotype groups with target frequencies \(0.70\), \(0.25\), and \(0.05\). Their simulated mean AUC values are respectively \(8\), \(12\), and \(20\) mg·h/L. Suppose the groups were simulated in equal numbers. Calculate the frequency-weighted population mean and compare it with the unweighted mean of the three subgroup means.
Frequency-weighted mean. Apply the target frequencies:
Unweighted subgroup mean. Giving each group equal weight produces \((8+12+20)/3=13.33\) mg·h/L. That result describes an equally weighted mixture of the three genotype groups, not the stated target population. The difference occurs because the highest-mean group is rare in the target population but was given one-third of the unweighted contribution.
Population variability reflects both exposure variation within genotype groups and differences between their mean exposures. If \(\sigma_g^2\) is the conditional exposure variance in group \(g\), then the population variance has the decomposition below. In a simulation, replace each \(\sigma_g^2\) by the corresponding subgroup sample variance \(s_g^2\) to estimate the within-group component.
Why the Variance Decomposition Holds
Let \(Y\) be an exposure drawn from the target population and let \(G\) denote its genotype group. Within group \(g\), the conditional mean is \(\mu_g\) and the conditional variance is \(\sigma_g^2\). Write the deviation from the population mean as \(Y-\mu_{\mathrm{pop}}=(Y-\mu_g)+(\mu_g-\mu_{\mathrm{pop}})\). Squaring and taking the conditional expectation within group \(g\), the cross-product has expectation zero because \(E(Y-\mu_g\mid G=g)=0\). Thus:
Averaging this conditional quantity across groups with probabilities \(q_g\) gives the stated result. The decomposition depends on using the same target frequencies in both terms, and on the genotype groups forming a complete partition of the target population. It shows why averaging only the subgroup variances understates total population variability whenever subgroup means differ.
Worked Example: Separating Within- and Between-Group Variability
Use the same target frequencies \(0.70\), \(0.25\), and \(0.05\), and hypothetical subgroup mean AUC values \(8\), \(12\), and \(20\) mg·h/L. Suppose their subgroup variances are \(4\), \(9\), and \(25\) \((\mathrm{mg\cdot h/L})^2\). The weighted mean is \(9.6\) mg·h/L, as calculated above.
Within-group contribution:
Between-group contribution: subtract the weighted mean from each subgroup mean, square the differences, and weight them:
The total population variance is \(6.30+8.640=14.94\ (\mathrm{mg\cdot h/L})^2\), and its standard deviation is \(\sqrt{14.94}=3.865\) mg·h/L. The within-group term alone would miss the additional variability associated with differences among genotype-specific means.
Combining Exposure Distributions
A population exposure distribution is a mixture of the conditional distributions in the genotype groups. For any threshold \(y\), its cumulative distribution function is the weighted sum of the subgroup cumulative distribution functions:
where \(F_g(y)=P(Y\leq y\mid G=g)\). The same principle gives a population probability of exceeding a threshold by weighting the corresponding subgroup probabilities. To estimate a population quantile, first form the weighted mixture distribution and then find the value at which its cumulative probability reaches the desired level. In general, a population quantile is not the frequency-weighted average of subgroup quantiles.
Worked Example: Estimating an Exceedance Probability and a Quantile
Suppose each genotype group has four hypothetical simulated AUC values in mg·h/L. Group 1 has \(6,8,10,12\); group 2 has \(10,14,16,18\); and group 3 has \(15,20,25,30\). Their target frequencies are \(0.70\), \(0.25\), and \(0.05\). Estimate the population probability that AUC exceeds \(15\) mg·h/L.
In group 1, none of four values exceeds 15, so the subgroup proportion is \(0/4=0\). In group 2, two of four exceed 15, giving \(2/4=0.50\). In group 3, three of four exceed 15, giving \(3/4=0.75\). Therefore:
The estimated population exceedance probability is 16.25%. To obtain the empirical population 90th percentile, assign each simulated subject its group’s target frequency divided by four. The resulting mass at each group 1 value is \(0.70/4=0.175\), at each group 2 value is \(0.25/4=0.0625\), and at each group 3 value is \(0.05/4=0.0125\). In increasing order, the cumulative weight reaches \(0.8375\) at 15 mg·h/L and \(0.9000\) at 16 mg·h/L. Using the quantile convention “smallest value whose cumulative weight is at least 0.90,” the empirical population 90th percentile is 16 mg·h/L.
Diagnostics, Alternatives, and Limitations
Several checks help identify a stratified analysis that is not answering its intended question. Confirm that the genotype groups are exhaustive and non-overlapping, that the target frequencies sum to one, and that each group’s simulated exposures use the same outcome definition and units. Check that every group with positive target frequency has simulated subjects. Compare allocation proportions \(n_g/N\) with target frequencies \(q_g\); a difference is not automatically an error, but it signals that unweighted pooled summaries will generally target the wrong mixture. Also inspect subgroup exposure summaries and empirical distributions for implausible values, unexpectedly sparse tails, or computational failures.
Proportional sampling is a direct alternative when a cohort is intended to mirror target frequencies. It simplifies unweighted summaries but may provide little information about rare groups. Controlled stratification improves subgroup coverage and allows deliberate group comparisons, while requiring explicit weighting for population summaries. A third approach is to simulate within groups and then generate a frequency-proportional resample for a particular analysis. That can make the target mixture visually or computationally convenient, but it does not add information to the original subgroup simulations.
Frequency weighting cannot correct a poorly represented conditional distribution: a subgroup with too few or unrepresentative virtual subjects can still yield unstable estimates, especially for tail probabilities and quantiles. Nor can weighting repair incorrect target frequencies, omitted genotype groups, or a PBPK model that does not adequately represent the intended population. Results describe the specified target mixture under the model and assumptions; they are not, by themselves, evidence that genotype causes an exposure difference. Further evaluation of genotype–physiology dependence is a separate question.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions below.
- Two groups have target frequencies 0.80 and 0.20, but are simulated in equal numbers. Their mean exposures are 5 and 15 mg/L. What is the target-weighted mean?
- Why can an unweighted mean over all virtual subjects be misleading after disproportionate stratification?
- A genotype group has target frequency 0.10, within-group variance 4 \((\mathrm{mg/L})^2\), and mean exposure 12 mg/L. The population mean is 10 mg/L. What is this group’s weighted within-group variance contribution and weighted between-group contribution?
- Two genotype groups have target frequencies 0.75 and 0.25. Their estimated probabilities of exceeding a threshold are 0.10 and 0.50. What is the population exceedance probability?
- Is the population 90th percentile generally the frequency-weighted average of the genotype-specific 90th percentiles? Explain briefly.
Solutions. (1) \(0.80(5)+0.20(15)=7.0\) mg/L. (2) It weights each group by its simulation fraction, which may differ from the target frequency; it therefore describes the simulated allocation mixture rather than the target population mixture. (3) The within-group contribution is \(0.10(4)=0.40\ (\mathrm{mg/L})^2\). The between-group contribution is \(0.10(12-10)^2=0.40\ (\mathrm{mg/L})^2\). (4) \(0.75(0.10)+0.25(0.50)=0.075+0.125=0.20\), or 20%. (5) No. Form the frequency-weighted mixture distribution and locate its 90th percentile; mixture quantiles are generally not averages of subgroup quantiles.