Maximin Design for Worst-Case Information

Why Optimize the Worst Case?

A design that performs well at a single best guess for the model parameters may perform poorly if that guess is wrong. One response is to evaluate performance across a specified range of plausible parameter values and protect against the least favorable value in that range. This is the central idea of maximin design: choose the feasible design whose worst-case information is as large as possible.

The parameter range is part of the design problem, not a detail to add afterward. A worst case is always a worst case within the region that has been specified. If that region is too narrow, the design may offer little protection against realistic uncertainty; if it includes implausible extremes, the criterion may be dominated by them.

Definition: Let $d$ denote a feasible design, let $\theta$ denote the unknown model parameter vector, and let $\Theta$ be the specified parameter region. Let $U(d,\theta)$ be a scalar information criterion, with larger values defined to be better. The maximin design is $$ d^* \in \arg\max_{d\in\mathcal{D}}\ \min_{\theta\in\Theta} U(d,\theta), $$ where $\mathcal{D}$ is the feasible design set. The inner minimum is the design's least favorable, or worst-case, criterion value over $\Theta$; the outer maximum chooses the design with the largest such value.

The criterion $U$ must be chosen before comparing designs. It could be scalar Fisher information for a target parameter, or a scalar criterion derived from an information matrix. Maximin design does not itself prescribe which scientific target or information summary to use. The results below assume that higher $U$ consistently means better performance for the stated goal.

Conditions and a Useful Justification

For a finite list of scenarios and a finite list of candidate designs, each minimum and maximum exists. For continuous choices, a common sufficient set of conditions is that $\mathcal{D}$ and $\Theta$ are nonempty compact sets and $U(d,\theta)$ is continuous jointly in $d$ and $\theta$. Under these conditions the inner minimum is attained, and the resulting worst-case value is continuous in $d$; consequently a maximizing design exists. In practice, constraints such as allowed sampling times, sample counts, and operational limits must be included in $\mathcal{D}$.

A simple derivation explains why worst-case optimization can create a compromise. Suppose there are two scenarios and a design allocation $q$ determines their information values $U_1(q)$ and $U_2(q)$. The objective is not their average but the lower envelope, $\min\{U_1(q),U_2(q)\}$. If one value is increasing in $q$ and the other is decreasing, an interior optimum can occur where they are equal: moving either way then worsens the smaller value. This is a useful guide, but equality is not a universal rule; constraints, unequal slopes, or additional scenarios can put the optimum at a boundary.

Worked Example: Choosing Among Scenario-Based Designs

Suppose three feasible sampling designs have been evaluated using the same scalar information criterion, in information units. The rows represent three prespecified parameter scenarios. All numbers below are hypothetical.

Design	Scenario 1	Scenario 2	Scenario 3	Worst-case information
A	80	20	65	20
B	55	55	55	55
C	35	95	45	35

For each design, take the smallest entry across the scenarios. Thus $U_{\min}(A)=\min(80,20,65)=20$, $U_{\min}(B)=\min(55,55,55)=55$, and $U_{\min}(C)=\min(35,95,45)=35$, all in information units.

The largest of these worst-case values is 55, so design B is the maximin choice. Design C is excellent in Scenario 2, but that does not compensate under this criterion for its weaker value of 35 in Scenario 1. This conclusion is conditional on treating the three scenarios as the complete specified region and on using the same criterion for every comparison.

Worked Sampling-Time Example

Consider a hypothetical one-compartment elimination model with known initial concentration $C_0$, concentration measured once at time $t$, and independent normally distributed measurement error with known standard deviation $\sigma$. The mean concentration is $C_0e^{-kt}$, where $k$ is the elimination rate constant. Assume the only uncertain model parameter is $k$, which lies in $\Theta=[0.2,1.0]\ \mathrm{h}^{-1}$. A feasible sampling time must lie in $0\leq t\leq 3$ h.

For a normal observation with known variance $\sigma^2$, the Fisher information for scalar parameter $k$ is the squared derivative of the mean with respect to $k$, divided by the variance. This model therefore gives

U(t,k)=I_k(t,k) =\frac{1}{\sigma^2}\left(\frac{\partial (C_0e^{-kt})}{\partial k}\right)^2 =\frac{C_0^2}{\sigma^2}t^2e^{-2kt}.

This expression assumes that $C_0$ and $\sigma$ are known, the mean model is correct, and the measurement error has the stated variance. It is information for $k$ under this model, not a guarantee about the accuracy of an estimate if those assumptions fail.

Worked Example: Selecting a Sampling Time Against the Fastest Elimination

Take hypothetical values $C_0=100\ \mathrm{mg/L}$ and $\sigma=5\ \mathrm{mg/L}$. Compare feasible times $t=0.5,\ 1.0,\ 2.0$ h. For any fixed positive $t$, $t^2e^{-2kt}$ decreases as $k$ increases, since

\frac{\partial U(t,k)}{\partial k} =-2t\,U(t,k)<0 \qquad (t>0).

Thus the least favorable value in $k\in[0.2,1.0]\ \mathrm{h}^{-1}$ is $k=1.0\ \mathrm{h}^{-1}$. Since $C_0^2/\sigma^2=10000/25=400$, the worst-case information at each candidate time is

\begin{aligned} U(0.5,1.0)&=400(0.5)^2e^{-2(1.0)(0.5)} =100e^{-1}=36.79\ \mathrm{h}^2,\\ U(1.0,1.0)&=400(1.0)^2e^{-2(1.0)(1.0)} =400e^{-2}=54.13\ \mathrm{h}^2,\\ U(2.0,1.0)&=400(2.0)^2e^{-2(1.0)(2.0)} =1600e^{-4}=29.30\ \mathrm{h}^2. \end{aligned}

Among these three candidates, $t=1.0$ h has the largest worst-case information, 54.13 $\mathrm{h}^2$, and is the maximin choice on this grid. The units follow from differentiating concentration with respect to $k$, which has units $\mathrm{h}^{-1}$.

If the feasible time is continuous, the same monotonicity result reduces the inner minimization to $k=1.0\ \mathrm{h}^{-1}$. The remaining objective is proportional to $t^2e^{-2t}$. Its derivative is $2te^{-2t}(1-t)$, so it increases for $0<t<1$ h and decreases for $t>1$ h. The continuous maximin time is therefore $t=1.0$ h, which is feasible. This conclusion depends on the stated single-observation model and the interval for $k$.

Maximin Allocation Across Sampling Times

A design may allocate a fixed number of subjects across candidate sampling times rather than choose only one time. The next example illustrates why a balanced allocation can protect against parameter uncertainty. Assume subjects contribute independent information, and suppose the information values per subject have been computed under the same model and criterion for two plausible parameter scenarios. The values are hypothetical and are supplied as inputs to the allocation problem.

Worked Example: Balancing Early and Late Samples

There are 10 subjects to allocate between an early and a late sampling time. Let $q$ be the proportion assigned to the early time, so $q\in\{0,0.1,\ldots,1\}$. Suppose the information per subject is as follows:

Parameter scenario	Early time	Late time
Fast elimination	8 information units	2 information units
Slow elimination	2 information units	8 information units

The total information in the fast-elimination scenario is $10[8q+2(1-q)]=10(2+6q)$ units. In the slow-elimination scenario it is $10[2q+8(1-q)]=10(8-6q)$ units. The worst-case total is therefore

U_{\min}(q)=10\min\{2+6q,\ 8-6q\}.

The two scenario values are equal when $2+6q=8-6q$, giving $q=0.5$. For $q<0.5$, $2+6q<8-6q$, so $U_{\min}(q)=10(2+6q)$, which increases with $q$. For $q>0.5$, $8-6q<2+6q$, so $U_{\min}(q)=10(8-6q)$, which decreases with $q$. Hence $q=0.5$ maximizes the minimum. It assigns five subjects to each time and yields 50 information units in either scenario.

The equality point is feasible with 10 subjects. If the total were odd, the allocation would be discrete and the two nearest feasible proportions would need to be compared directly; an exact equalization might not be possible.

Diagnostics, Alternatives, and Limitations

Maximin results are sensitive to how the region and criterion are constructed. Check the following before treating an optimizer as robust:

Check the least favorable point. Report which parameter value or scenario determines the minimum for the selected design. If it is an unexpected boundary or an isolated numerical artifact, inspect the model and the region.
Check coverage of the region. A coarse grid can miss a lower value between grid points. Refine the grid near apparent minima, or use a justified analytic argument, as in the sampling-time example.
Check feasibility and numerical stability. Confirm that the chosen design satisfies sampling, cost, and operational constraints. Near-singular information calculations or unstable optimization can make apparent worst-case values unreliable.
Check sensitivity to the region. Recompute the design after reasonable changes to parameter bounds. A design that changes sharply when a boundary moves slightly should be reported as region-sensitive.

Maximin design differs from a nominal design, which optimizes the criterion at one selected parameter value, and from Bayesian expected-utility design, which averages performance according to a probability distribution. Scenario-based robust design supplies a set of cases to evaluate; maximin specifies that the smallest criterion value over those cases drives the choice. These approaches answer different questions: average performance can tolerate a poor scenario if other scenarios compensate, whereas maximin does not permit that compensation. Minimax-regret design is another alternative, but it uses a different comparison against scenario-specific best performance.

The maximin criterion is deliberately conservative. A very low value at one plausible but rare or poorly justified corner of $\Theta$ can dominate the decision. Conversely, excluding a genuinely plausible parameter value can create a false impression of protection. Maximin also depends on the chosen scalar information criterion: robust performance for one target does not imply robust performance for every parameter or scientific objective. State the region, criterion, constraints, and worst-case point alongside the selected design.

Common mistake: “Design A has higher information in most scenarios, so it must be the maximin choice.” This reasoning averages implicitly and ignores the defining operation. For each candidate, first compute its minimum criterion value over the entire specified region; only then select the candidate with the largest minimum. In the scenario table above, design B wins despite design A's larger values in two scenarios.

Check Your Understanding

Answer each question using the maximin objective and the assumptions stated in the examples.

Two designs have scenario-wise information values $(12,4)$ and $(7,7)$. Which is maximin, and what is its worst-case value?
In the one-sample elimination model, why is the worst-case $k$ the upper endpoint of the interval for every fixed $t>0$?
For the sampling-time model, what continuous time maximizes worst-case information when $k\in[0.2,0.8]\ \mathrm{h}^{-1}$ and $0\leq t\leq 3$ h?
In the allocation example, explain why the lower of the two scenario information lines determines the objective.
Name one way a numerical maximin result can appear robust when it is not.

Solutions. (1) The minima are 4 and 7, so the second design is maximin, with worst-case value 7. (2) For $t>0$, $\partial U/\partial k=-2tU<0$, so information decreases throughout the interval as $k$ increases; its minimum occurs at $k=0.8\ \mathrm{h}^{-1}$. (3) At the worst-case $k=0.8$, the time-dependent factor is $t^2e^{-1.6t}$. Its derivative is $2te^{-1.6t}(1-0.8t)$, which changes from positive to negative at $t=1/0.8=1.25$ h. This feasible time maximizes the worst-case information. (4) Maximin evaluates the smaller scenario value for each allocation; the larger value cannot offset a smaller worst-case value. (5) A coarse parameter grid may omit a lower information value between grid points, or the specified parameter region may omit plausible values.