Why Compare the Designs Under the Same Budget?
A fixed robust design chooses all sampling times before data collection, aiming to perform acceptably across plausible pharmacokinetic parameters. An adaptive design uses interim data to select at least some later sampling times. Adaptation can improve alignment between sampling and the unknown PK profile, but it also uses part of the budget to learn which profile is plausible. A fair comparison therefore holds the total sampling budget and the uncertainty assumptions constant, and evaluates the full adaptive rule—not just its best-case schedule.
The comparison below uses a deliberately simple one-compartment model to make the tradeoffs calculable. The numerical results are hypothetical and illustrate the design comparison; they are not results from a clinical study. The model omits several features of real PK data, so its conclusions should be read as conditional on the stated assumptions.
For either design, specify the performance criterion before comparing results. For example, one may compare expected Fisher information under a probability distribution for the unknown parameters, or compare the worst-case information over a defined set of parameter values. These criteria answer different questions: average performance depends on the uncertainty weights, whereas worst-case performance emphasizes protection against an unfavorable scenario.
A One-Compartment PK Model and Its Information
Suppose concentration after an intravenous bolus is modeled as \(C(t)=C_0\exp(-kt)+\epsilon\), where \(t\) is time in hours, \(C_0=1.00\) mg/L is known, \(k\) is the elimination rate constant in h\(^{-1}\), and independent measurement errors satisfy \(\epsilon\sim N(0,\sigma^2)\) with \(\sigma=0.100\) mg/L. The target is \(k\). This example treats \(C_0\) as known and omits between-subject variability; these assumptions make the comparison one-dimensional.
It is convenient to express information about \(\eta=\log k\). For one observation at time \(t\), the model mean is \(\mu(t)=C_0\exp(-kt)\). By the chain rule, its sensitivity to \(\eta\) is \(\partial\mu(t)/\partial\eta=-ktC_0\exp(-kt)\), since \(\partial k/\partial\eta=k\). With independent normal errors of known variance, the Fisher information for \(\eta\) from that observation is the squared sensitivity divided by \(\sigma^2\).
Here \(I_{\eta}\) is information about the dimensionless log parameter. For independent observations, information adds. The formula describes expected information under the assumed model; it is not a guarantee about the precision obtained in any one data set.
Worked Example: Information at One Sampling Time
For the hypothetical value \(k=0.100\) h\(^{-1}\), calculate the information in one observation at \(t=8.00\) h.
The dimensionless product is \(kt=(0.100\text{ h}^{-1})(8.00\text{ h})=0.800\). Substituting in the information formula gives
The units of \(k\) and \(t\) cancel, as required for information about \(\eta=\log k\). This calculation assumes the stated known \(C_0\), independent errors, and constant error variance. Under this model, information is greatest when \(kt=1\), so this observation is near the most informative time for \(k=0.100\) h\(^{-1}\).
The time that maximizes information depends on \(k\). To see this, write the time-dependent part as \(f(x)=x^2\exp(-2x)\), where \(x=kt\). Its derivative is \(f'(x)=2x\exp(-2x)(1-x)\). For \(x>0\), the derivative is positive below 1 and negative above 1, so the maximum occurs at \(x=1\), or \(t=1/k\). This is a justification for why parameter uncertainty matters: a time chosen for one elimination rate need not be optimal for another.
Worked Comparison: A Fixed Schedule and an Adaptive Rule
Suppose the plausible parameter values are \(k=0.100\) h\(^{-1}\) and \(k=0.400\) h\(^{-1}\), each with probability 0.50 for the design comparison. Both designs have a budget of four independent concentration observations at nominal sampling times. Replicate observations at a nominal time are allowed, as if assigned across otherwise identical individuals. A fixed robust candidate uses times 2, 3, 8, and 9 h. An adaptive rule takes two pilot observations at 2 and 8 h, classifies the profile as low or high elimination from the pilot data, then assigns its two remaining observations at 10 h if classified low or 2.5 h if classified high.
The candidate fixed schedule is a prespecified compromise: it spreads observations across earlier and later times. The adaptive design rule includes both its pilot times and its data-dependent choice. Neither schedule is asserted to be globally optimal among all possible designs; the point is to compare a concrete robust schedule with a concrete adaptive rule at the same total budget.
Worked Example: Information From the Fixed Robust Schedule
For \(k=0.100\) h\(^{-1}\), the information at 2, 3, 8, and 9 h is approximately 2.681, 4.939, 12.92, and 13.39. Thus the total is
For \(k=0.400\) h\(^{-1}\), the corresponding values are approximately 12.92, 13.06, 1.701, and 0.9676, giving
With equal scenario probabilities, average information is \((33.93+28.65)/2=31.29\). The worst-case information across these two scenarios is 28.65. These two summaries are not interchangeable: the first averages over the stated probabilities; the second reports the less informative of the two scenarios.
For the adaptive rule, information from the two pilot observations is included regardless of the classification. At the low value of \(k\), these pilot times provide \(2.681+12.92=15.60\) units of information. At the high value, they provide \(12.92+1.701=14.62\). The remaining two observations can add substantial information when the rule chooses times suited to the true profile. However, the choice is based on noisy interim data, so it need not always be suited to that profile.
Worked Example: Adaptive Performance With Imperfect Classification
First consider an idealized upper-bound calculation in which the interim classification is always correct. When \(k=0.100\) h\(^{-1}\), each remaining observation at 10 h has \(kt=1\) and contributes \(100\exp(-2)=13.53\), so total information is \(15.60+2(13.53)=42.67\). When \(k=0.400\) h\(^{-1}\), each remaining observation at 2.5 h also has \(kt=1\), giving total information \(14.62+2(13.53)=41.69\). These values assume perfect knowledge from the pilot and should not be treated as the expected performance of a noisy adaptive rule.
Now suppose the pilot classification is correct with probability 0.80 under either parameter scenario. This probability is a hypothetical operating characteristic of the rule, supplied for illustration rather than derived from the concentration model. If the true value is low but the rule incorrectly chooses 2.5 h, each added observation contributes \(100(0.25)^2\exp(-0.50)=3.791\). The wrong-choice total is therefore \(15.60+2(3.791)=23.18\). Expected information at the low value is
If the true value is high but the rule incorrectly chooses 10 h, each added observation contributes \(100(4)^2\exp(-8)=0.5367\). The wrong-choice total is \(14.62+2(0.5367)=15.70\). Expected information at the high value is
Averaging over the two equally likely scenarios gives \( (38.77+36.49)/2=37.63\), compared with 31.29 for the fixed schedule. Under these illustrative assumptions, the adaptive rule has higher average information. Its expected information is also higher than the fixed schedule's information in each scenario, but this does not establish that adaptive designs are generally superior: the result depends on the candidate schedules, error model, parameter scenarios, and interim classification accuracy.
How to Make the Comparison and Interpret It
The essential comparison is between two complete design strategies evaluated under the same assumptions. A robust schedule is evaluated at each plausible parameter value and summarized using a declared criterion. An adaptive design is evaluated over both parameter uncertainty and possible interim data, because different pilot outcomes can lead to different later times. In practice, simulation is often needed to estimate this joint performance when the interim rule is more complex than the illustrative classification rule.
Use the same total number of observations, model assumptions, plausible parameter distribution or range, and performance criterion for both strategies.
Calculate its information or other prespecified performance measure at each scenario, then report the chosen average and/or worst-case summary.
Include pilot observations and average over the interim outcomes that determine later sampling times. Do not substitute an oracle schedule that knows the true parameter.
Use simulation or other suitable calculations to assess classification accuracy, precision, bias, interval coverage, and whether the rule respects the sampling budget.
Information is useful for comparing expected local precision under the stated model, but it is not the only relevant measure. A design with higher expected information may still have poor performance in a low-probability region, be sensitive to model misspecification, or provide misleading uncertainty intervals if analysis ignores the adaptation. In this simple normal model with known variance, information adds across observations even when times are chosen adaptively. In more complex analyses, the sampling rule and analysis procedure should be assessed together; simulation can check whether the intended estimator and interval retain their desired operating properties.
| Design feature | Fixed robust schedule | Adaptive rule |
|---|---|---|
| Sampling times | All set before data collection | Later times depend on interim data |
| Response to parameter uncertainty | Balances performance across scenarios in advance | Can tailor later times to interim evidence |
| Evaluation requirement | Evaluate schedule at each plausible parameter value | Evaluate pilot outcomes and resulting schedules jointly |
| Main vulnerability | May be a compromise that is optimal for no scenario | Interim noise may trigger a poorly matched schedule |
Diagnostics, Limitations, and a Common Mistake
A comparison can be misleading if its assumptions are not checked. Examine whether the plausible \(k\) values and their weights represent the intended population of uncertainty. Check that the chosen times are operationally feasible and that the assumed error variance remains reasonable at low concentrations. In actual PK settings, assay limits, residual error that changes with concentration, uncertain \(C_0\), absorption phases, or between-subject variability can alter the information profile. When such features matter, evaluate designs with the intended PK model rather than applying this simplified calculation unchanged.
For an adaptive strategy, also inspect how often the interim data lead to each branch and how performance changes when the classification is wrong. Compare not only average information but also scenario-specific precision and, where relevant, bias and interval coverage. If the result depends strongly on one assumed classification probability or one parameter scenario, that sensitivity is part of the design conclusion, not a detail to omit.
- Common mistake: comparing an adaptive design with an unequal budget. It is incorrect to compare a four-observation fixed schedule with four pilot observations plus additional adaptive observations. The correction is to count the pilot observations within the same total budget.
- Common mistake: giving the adaptive rule perfect information. Choosing the best remaining time using the true \(k\) calculates an oracle benchmark, not a data-driven design. The correction is to average over interim outcomes, including incorrect classifications.
- Common mistake: reporting only average information. A favorable average can hide a weak scenario. The correction is to state the uncertainty weights and, when protection across scenarios matters, report a worst-case or scenario-specific result as well.
Check Your Understanding
Answer each question using the model and hypothetical comparison above.
- Why is the time \(t=1/k\) informative in the example, and what does this imply when \(k\) is uncertain?
- What is the fixed schedule's average information under the equal scenario probabilities, and what is its worst-case information?
- Why is the perfect-classification adaptive result not the appropriate main estimate of data-driven performance?
- Suppose the interim classification is wrong more often than 20%. What aspect of the comparison should be recalculated?
Solutions. (1) The information depends on \(x=kt\) through \(x^2\exp(-2x)\), whose derivative is positive for \(0<x<1\) and negative for \(x>1\); hence its maximum is at \(x=1\), or \(t=1/k\). If \(k\) is uncertain, the maximizing time differs across plausible values. (2) The average is \((33.93+28.65)/2=31.29\) information units; the worst case is \(\min(33.93,28.65)=28.65\). (3) Perfect classification assumes the pilot reveals the true scenario, which noisy data do not guarantee; it is an oracle benchmark. (4) Recalculate adaptive performance by averaging the information from correct and incorrect branches using the revised classification probabilities, then compare it with the fixed schedule under the same parameter weights and budget.