Support Points in Fixed-Size D-Optimal Schedules

Why the Number of Sampling Times Is Not the Whole Story

A schedule with many distinct sampling times may sound more informative than one that repeats a few times. But the number of times alone does not determine how well a model’s parameters can be estimated. Each time contributes a local sensitivity vector: its direction indicates which parameter combinations the measurement informs, while its magnitude indicates the scale of that information. Replicates add information in an existing direction; a new time can add a different direction, or nearly repeat one already present.

For a fixed number of individual concentration measurements, the design question is therefore about the balance between where measurements occur and how often each time is used. The determinant of the information matrix provides a precise way to examine that balance. The discussion below assumes a local design: sensitivities are evaluated at a specified nominal parameter value.

Definition: A support point is a distinct sampling time used in a schedule. Its replicate count is the number of measurements taken at that time. If the schedule has \(n\) measurements, \(m\) distinct support points, and counts \(r_1,\ldots,r_m\), then \(\sum_{j=1}^{m} r_j=n\).

Information From a Fixed-Size Schedule

Suppose the concentration measured at time \(t\) has mean \(\mu(t,\theta)\), where \(\theta\) is a vector of \(p\) parameters. Assume the measurements satisfy \(Y_i=\mu(t_i,\theta)+\epsilon_i\), with independent errors \(\epsilon_i\sim N(0,\sigma^2)\), a common known variance \(\sigma^2\), and fixed sampling times. Assume also that \(\mu\) is differentiable with respect to each parameter. Under these conditions, the Fisher information for \(\theta\), evaluated at a nominal value \(\theta_0\), is

I(\theta_0)=\frac{1}{\sigma^2}\sum_{i=1}^{n} f(t_i)f(t_i)^\mathsf{T}, \qquad f(t)=\left.\frac{\partial \mu(t,\theta)}{\partial\theta}\right|_{\theta=\theta_0}.

Here \(f(t)\) is a column vector of \(p\) sensitivities. A repeated measurement at time \(t_j\) contributes the same outer product \(f(t_j)f(t_j)^\mathsf{T}\) again. Grouping measurements by support point gives

I(\theta_0)=\frac{1}{\sigma^2}\sum_{j=1}^{m} r_j f(t_j)f(t_j)^\mathsf{T}.

For a full-rank information matrix, D-optimality compares schedules by the determinant of \(I\): a larger determinant corresponds to a smaller volume for the local joint uncertainty ellipsoid, under the usual information-based approximation. The fixed number \(n\) matters because allocating a measurement to one time means it cannot also be allocated to another.

Key result: A schedule can have a positive determinant only if its sensitivity vectors span the \(p\)-dimensional parameter space. Thus \(m\geq p\) is necessary for full rank, but not sufficient: \(p\) or more support points can still have sensitivities that are linearly dependent.

How Support Points Enter the Determinant

Write \(X\) for the \(n\times p\) matrix whose \(i\)th row is \(f(t_i)^\mathsf{T}\). Then \(I=\sigma^{-2}X^\mathsf{T}X\). The Cauchy–Binet identity expresses the determinant of this product as a sum over all subsets of \(p\) measurement rows:

\det I = \sigma^{-2p} \sum_{\substack{S\subseteq\{1,\ldots,n\}\\|S|=p}} \det(X_S)^2.

In this expression, \(X_S\) is the square matrix formed from the \(p\) rows selected by \(S\). Each squared determinant measures how much independent sensitivity those \(p\) measurements provide. If two selected measurements are replicates at the same time, their rows are identical, so that term is zero. Grouping the nonzero terms by distinct support points yields an equivalent formula:

\det I = \sigma^{-2p} \sum_{\substack{J\subseteq\{1,\ldots,m\}\\|J|=p}} \left(\prod_{j\in J}r_j\right) \det(F_J)^2,

where \(F_J\) is the \(p\times p\) matrix with rows \(f(t_j)^\mathsf{T}\) for the support points in \(J\). To see the multiplicity factor, select one measurement from each of the \(p\) chosen support points. There are \(\prod_{j\in J}r_j\) such selections, and each gives the same squared determinant. Selecting two measurements at the same support point gives zero because those rows coincide.

This decomposition is a direct justification for considering both placement and replication. A support point matters in combination with other points: its sensitivity must contribute to a linearly independent set. Its replicate count increases the number of measurement subsets involving it, but does not create a new sensitivity direction by itself. If \(m<p\), the sum has no terms and \(\det I=0\). If \(m\geq p\), the determinant is positive only if at least one \(p\)-point subset has nonzero determinant.

Worked Example: A Third Support Point Adds a New Direction

Consider a hypothetical two-parameter model with measurement-error variance \(\sigma^2=1\). At three candidate times, suppose the local sensitivity vectors are \(f_1=(1,0)^\mathsf{T}\), \(f_2=(0,1)^\mathsf{T}\), and \(f_3=(1,1)^\mathsf{T}\). Compare two schedules with \(n=4\): schedule A takes two measurements at each of the first two times; schedule B takes two at the first time and one at each of the other times.

For schedule A, the information matrix is \(I_A=2f_1f_1^\mathsf{T}+2f_2f_2^\mathsf{T}\), or

I_A= \begin{pmatrix}2&0\\0&2\end{pmatrix}, \qquad \det(I_A)=2(2)-0^2=4.

For schedule B, \(I_B=2f_1f_1^\mathsf{T}+f_2f_2^\mathsf{T}+f_3f_3^\mathsf{T}\), giving

I_B= \begin{pmatrix}3&1\\1&2\end{pmatrix}, \qquad \det(I_B)=3(2)-1(1)=5.

The determinant decomposition gives the same result: the pair contributions are \(2(1)^2\) for points 1 and 2, \(2(1)^2\) for points 1 and 3, and \(1(1)^2\) for points 2 and 3, for a total of \(5\). In this hypothetical comparison, reallocating one replicate to a third support point increases the determinant from 4 to 5 because that point contributes useful sensitivity combinations.

A PK Illustration: New Times and Replicates Compete

For a one-compartment IV bolus concentration model, write \(\mu(t,\theta)=\exp(a-kt)\), where \(a=\log A\), \(A\) is the initial concentration, and \(k\) is the elimination rate constant. For parameters \(\theta=(a,k)^\mathsf{T}\), the sensitivity vector is

f(t)=\mu(t,\theta) \begin{pmatrix}1\\-t\end{pmatrix}.

For two times \(t_i\) and \(t_j\), the determinant of the two sensitivity rows is \(\mu(t_i)\mu(t_j)(t_i-t_j)\). Therefore, with two parameters and common variance \(\sigma^2\), their contribution to the information determinant is \(\sigma^{-4}\mu(t_i)^2\mu(t_j)^2(t_i-t_j)^2\), multiplied by the product of their replicate counts. Separation in time helps create distinct sensitivity directions, but declining concentrations also reduce the magnitude of later sensitivities.

Worked Example: Reallocating a Replicate in a Hypothetical PK Schedule

Suppose the nominal values are \(a=0\) and \(k=0.5\) per hour, and candidate times are 0, 1, and 2 hours. The corresponding sensitivity vectors are approximately \(f_0=(1,0)^\mathsf{T}\), \(f_1=(0.6065,-0.6065)^\mathsf{T}\), and \(f_2=(0.3679,-0.7358)^\mathsf{T}\). Compare schedule A, with two measurements at 0 hours and two at 1 hour, with schedule B, with two at 0 hours, one at 1 hour, and one at 2 hours. Both have four measurements.

For two parameters, each pair of distinct support points contributes its squared row determinant times the product of its replicate counts. In schedule A, the determinant for the 0- and 1-hour sensitivities is \(-0.6065\). Thus

\det(I_A)=\sigma^{-4}\,4(0.6065^2) \approx 1.4715\,\sigma^{-4}.

For schedule B, the pairwise determinants are approximately \(-0.6065\) for 0 and 1 hours, \(-0.7358\) for 0 and 2 hours, and \(-0.2231\) for 1 and 2 hours. Their squared values are approximately 0.3679, 0.5413, and 0.04979. The multiplicity products are 2, 2, and 1, respectively, so

\det(I_B) = \sigma^{-4}\left[2(0.3679)+2(0.5413)+1(0.04979)\right] \approx 1.8682\,\sigma^{-4}.

In this hypothetical local comparison, schedule B has the larger determinant, despite replacing one 1-hour replicate with a 2-hour measurement. The improvement is not a general rule that later or more numerous support points are preferable. It results from the particular sensitivities at these nominal values, times, and error variance assumption.

Support Count, Rank, and Replication

The determinant expansion clarifies three cases that can otherwise be confused. First, if there are fewer than \(p\) support points, the schedule cannot identify all \(p\) local parameter directions, regardless of how many replicates are taken at those points. Second, having at least \(p\) support points does not guarantee full rank: the sensitivities may still be dependent. Third, once the sensitivity vectors span the parameter space, additional replicates can increase the determinant by increasing the number of informative subsets.

Worked Example: Three Parameters Require Independent Sensitivities

Consider three parameters and hypothetical sensitivity vectors \(f_1=(1,0,0)^\mathsf{T}\), \(f_2=(0,1,0)^\mathsf{T}\), and \(f_3=(0,0,1)^\mathsf{T}\). With replicate counts 2, 2, and 1, respectively, and \(\sigma^2=1\), the information matrix is diagonal:

I= \begin{pmatrix}2&0&0\\0&2&0\\0&0&1\end{pmatrix}, \qquad \det(I)=2(2)(1)=4.

There are five measurements but only three support points, and the determinant is positive because the three sensitivity vectors are linearly independent. By contrast, if all five measurements were taken at the first point, the information matrix would be \(\operatorname{diag}(5,0,0)\), with determinant zero. Replication cannot supply the missing second and third directions. Likewise, three support points with sensitivities \((1,0,0)^\mathsf{T}\), \((0,1,0)^\mathsf{T}\), and \((1,1,0)^\mathsf{T}\) would still give a zero determinant: all three vectors lie in a two-dimensional subspace.

Diagnostics, Comparisons, and Limitations

A useful diagnostic is to examine the sensitivity matrix or information matrix, not merely to count distinct times. A zero or nearly zero determinant signals rank deficiency or weak information in at least one parameter direction. A very small singular value of the sensitivity matrix is another warning: the schedule may technically have full rank while remaining close to a collinear configuration. As in the earlier material on sensitivity collinearity, local conclusions depend on the nominal parameter values; a different plausible value can change the sensitivity directions and magnitudes.

D-optimality rewards information volume across the parameter vector as a whole. This differs from judging a schedule by its weakest information direction or by the precision of one particular parameter combination. Those priorities can lead to different designs. Similarly, simply maximizing the number of distinct times ignores whether the new sensitivities are large and independent. The determinant decomposition is useful for understanding a candidate schedule, but it does not by itself establish which allocation is best over a continuous set of possible times.

The formulas above rely on independent measurements with common known variance and on the local sensitivity matrix for a differentiable mean model. If measurement variances differ by time, errors are correlated, or observations are censored, the information contributions must reflect those features. Further, determinant comparisons are tied to a parameterization and scale unless comparisons are made consistently in the same parameterization. A high determinant does not guarantee good precision for every individual parameter, nor does it remove model uncertainty or sensitivity to incorrect nominal values.

Common mistake: “There are two parameters, so two distinct times are sufficient and any additional replicate is wasted.” Two independent sensitivity directions are needed for positive determinant, but the directions must actually be independent. Once they are, replicates can still increase the determinant; they are not wasted. Conversely, adding a time helps only when its sensitivity contributes useful independent information relative to the existing schedule.

Check Your Understanding

Answer each question, then compare with the solutions below.

For \(p=4\) parameters, can a schedule with three support points have a positive information determinant? Explain.
A two-parameter schedule has support sensitivities \(f_1=(1,0)^\mathsf{T}\), \(f_2=(0,1)^\mathsf{T}\), and \(f_3=(1,1)^\mathsf{T}\), with one measurement at each. What is the determinant when \(\sigma^2=1\)?
Why does adding a replicate at a support point not necessarily have zero value, even though two replicate sensitivity rows are identical?
Does having \(m\geq p\) support points guarantee a positive determinant? State the additional condition.

Solutions. (1) No. With fewer than four distinct sensitivity vectors, their span has dimension at most three, so the information matrix has rank at most three. (2) The three pair contributions are \(1^2+1^2+(-1)^2=3\), so \(\det(I)=3\). (3) The replicate can form informative subsets with measurements at other support points; the multiplicity factors increase the number of such subsets. Only a subset containing two identical rows contributes zero. (4) No. At least one set of \(p\) support sensitivities must be linearly independent, so that its squared determinant is positive.