Weighting Information for Assay Precision

Why Assay Precision Changes a Sampling Design

A PK sample is useful not only because its expected concentration responds to the parameters of interest, but also because the assay measures that concentration with finite precision. Two samples at equally informative times under an equal-variance error model need not contribute equally when one sample type has greater variance, or when assay variance changes across the concentration range. Sampling design must therefore consider both the model's sensitivities and the observation error.

The key distinction is between a precision weight and the information contributed by a sample. A precision weight is the inverse of its variance. The sample's information also depends on how strongly its expected measurement changes when the model parameters change. An observation with low variance can still contribute little information about a particular parameter if its sensitivity to that parameter is small.

Definition: For independent Gaussian observations with model means $f_i(\theta)$ and known, fixed variances $v_i$, the local sensitivity-based Fisher information matrix is $$ I(\theta)=\sum_{i=1}^{n}\frac{1}{v_i}\,g_i(\theta)g_i(\theta)^\mathsf{T}, \qquad g_i(\theta)=\frac{\partial f_i(\theta)}{\partial\theta}. $$ The factor $1/v_i$ is the precision weight. The sensitivity vector $g_i$ describes how the expected observation changes with the parameters. This expression applies when the observation variances are treated as known and fixed with respect to $\theta$.

This is a local quantity: for a nonlinear PK model, the sensitivities are evaluated at specified nominal parameter values. With a fixed number of observations and a nonsingular information matrix, a local D-optimal design maximizes $\det I(\theta)$. Its determinant measures joint information across the parameters in the specified parameterization; maximizing it reduces the volume of the corresponding local uncertainty region. The earlier material on local D-optimal designs provides the design context for this criterion.

What Changes When Variance Depends on Concentration?

If assay variance is calibrated as a function of concentration, it is important to distinguish two calculations. In a common local weighted-sensitivity calculation, the variance is evaluated at nominal concentrations and then treated as fixed: each observation contributes $g_i g_i^\mathsf{T}/v_i$. This is appropriate when the variances are known design-specific quantities, or when the calculation deliberately freezes them at their nominal values.

If the statistical model instead states that the variance itself changes with the unknown parameters through the predicted concentration, the exact Gaussian Fisher information has an additional term. For independent normal observations $Y_i\sim N(f_i(\theta),v_i(\theta))$, with differentiable means and positive variances, its $(a,b)$ entry is

I_{ab}(\theta)=\sum_i\left[ \frac{1}{v_i}\frac{\partial f_i}{\partial\theta_a} \frac{\partial f_i}{\partial\theta_b} +\frac{1}{2v_i^2}\frac{\partial v_i}{\partial\theta_a} \frac{\partial v_i}{\partial\theta_b} \right].

The second term represents information in changes in the distribution's spread, not just changes in its mean. It should be included only when the variance model is part of the likelihood and its dependence on the parameters is known. A variance estimated imprecisely from assay validation should not automatically be treated as known information.

Write down the observation model.
Specify the predicted concentration, the error distribution, and whether the variance is fixed or depends on model parameters.

Evaluate sensitivities and variances.
At each candidate time, calculate the mean sensitivity vector and the applicable assay variance at the nominal parameter values.

Assemble and compare information.
Sum the weighted sensitivity contributions, then compare candidate designs using the criterion appropriate to the design question.

Justification of the Weighted Information

For a single observation $Y\sim N(f(\theta),v)$ with known, fixed variance $v$, write $g=\partial f/\partial\theta$. Its log-likelihood, omitting a constant, is $\ell=-\tfrac12\log v-\tfrac{(Y-f)^2}{2v}$. Differentiating gives the score vector $U=\partial\ell/\partial\theta=(Y-f)g/v$. Because $E(Y-f)=0$ and $E((Y-f)^2)=v$, the expected outer product of the score is

E(UU^\mathsf{T}) =\frac{E((Y-f)^2)}{v^2}gg^\mathsf{T} =\frac{1}{v}gg^\mathsf{T}.

Independent observations have additive information, so summing this contribution gives the stated matrix. If $v$ depends on $\theta$, differentiating the log-likelihood also produces a score term involving the centered squared residual. For a normal distribution, its expected contribution yields the additional $\tfrac12(\partial v/\partial\theta_a)(\partial v/\partial\theta_b)/v^2$ term. Thus inverse-variance weighting alone is not the full Fisher information in every heteroscedastic likelihood.

Worked Example: Combining Two Assay Precisions in an IV Bolus Model

Consider the hypothetical one-compartment IV bolus model $C(t)=A\exp(-kt)$, with nominal $A=10$ mg/L and $k=0.5$ h$^{-1}$. Suppose independent concentration errors at 1 h and 3 h are normal with known standard deviations 0.5 mg/L and 1.0 mg/L, respectively. Treat these two assay variances as fixed. Calculate the local information for $\theta=(A,k)$.

The sensitivities are $\partial C/\partial A=\exp(-kt)$ and $\partial C/\partial k=-tC(t)$. The nominal concentrations are $C(1)=10e^{-0.5}=6.065$ mg/L and $C(3)=10e^{-1.5}=2.231$ mg/L. Thus the sensitivity vectors, with entries in the parameter order $(A,k)$, are approximately $g_1=(0.6065,-6.065)$ and $g_3=(0.2231,-6.694)$. The weights are $1/0.5^2=4.000$ L$^2$/mg$^2$ and $1/1.0^2=1.000$ L$^2$/mg$^2$.

Using $I=4g_1g_1^\mathsf{T}+g_3g_3^\mathsf{T}$, the contributions are approximately

4g_1g_1^\mathsf{T}\approx \begin{pmatrix} 1.472&-14.72\\ -14.72&147.2 \end{pmatrix}, \qquad g_3g_3^\mathsf{T}\approx \begin{pmatrix} 0.04979&-1.494\\ -1.494&44.81 \end{pmatrix}.

Adding them gives $I\approx\begin{pmatrix}1.521&-16.21\\-16.21&192.0\end{pmatrix}$. For a check on the determinant, the determinant of the two-row sensitivity matrix is approximately $-2.707$, so $\det(I)=4(1)(-2.707)^2\approx29.31$. The 1 h observation receives four times the precision weight of the 3 h observation, but it does not provide four times the information about every parameter: its contribution also depends on both sensitivity components.

Precision Weighting Can Shift Preferred PK Times

The following example isolates the effect of assay precision on a two-parameter design. Consider log concentrations from the hypothetical one-compartment model, with mean $\log C(t)=\eta-kt$, where $\eta=\log A$. Assume independent normal errors on the log scale, with variance known for each candidate time. For $\theta=(\eta,k)$, the mean sensitivity is $g(t)=(1,-t)^\mathsf{T}$. With two observations at $t_i,t_j$ and weights $w_i=1/v_i$, direct expansion of the information determinant gives

\det\{w_i g(t_i)g(t_i)^\mathsf{T} +w_j g(t_j)g(t_j)^\mathsf{T}\} =w_iw_j(t_j-t_i)^2.

For example, the matrix has diagonal entries $w_i+w_j$ and $w_it_i^2+w_jt_j^2$, and off-diagonal entry $-w_it_i-w_jt_j$. Expanding its determinant cancels the $w_i^2$ and $w_j^2$ terms, leaving $w_iw_j(t_j-t_i)^2$. This shows why separation matters and why the weights can change the preferred pair.

Worked Example: A Noisy Late Sample Changes the Best Pair

Suppose two samples are selected from 1, 2, and 4 h. First assume every log concentration has variance 0.04, giving weight 25 at each time. The determinant scores for pairs (1,2), (1,4), and (2,4) are, respectively, $25(25)(2-1)^2=625$, $25(25)(4-1)^2=5625$, and $25(25)(4-2)^2=2500$. The equal-precision choice is therefore the pair at 1 and 4 h.

Now suppose assay validation gives log-error variances 0.01, 0.04, and 1.00 at 1, 2, and 4 h. The corresponding weights are 100, 25, and 1. The candidate scores become

\begin{aligned} (1,2):\;&100(25)(2-1)^2=2500,\\ (1,4):\;&100(1)(4-1)^2=900,\\ (2,4):\;&25(1)(4-2)^2=100. \end{aligned}

The highest score is now 2500 for the 1 h and 2 h pair. For confirmation, its information matrix is $\begin{pmatrix}125&-150\\-150&200\end{pmatrix}$, with determinant $125(200)-150^2=2500$. The 1 h and 4 h pair gives $\begin{pmatrix}101&-104\\-104&116\end{pmatrix}$, with determinant $101(116)-104^2=900$. Thus the late sample's greater time separation does not compensate for its very low precision in this design. The result is conditional on the specified nominal model and the assumed log-scale variances.

Concentration-Dependent Variance: Weight Is Not the Whole Story

A variance that increases with concentration does not imply that high-concentration samples are automatically uninformative. For example, sensitivity to log-amplitude is proportional to concentration. Both the sensitivity and variance must be considered. In addition, when the variance function depends on the model parameters, one must decide whether to use the fixed-variance local approximation or the full Gaussian information.

Worked Example: Additive and Proportional Assay Error

For illustration, let a single concentration measurement have mean $C$ and Gaussian assay variance $v(C)=0.2^2+(0.10C)^2$, with units of (mg/L)$^2$. Suppose $C=\exp(\eta)$, so the parameter is log-concentration $\eta$. Compare a nominal measurement at $C=8$ mg/L with one at $C=2$ mg/L.

At $C=8$, $v=0.04+0.64=0.68$ (mg/L)$^2$, and $\partial C/\partial\eta=C=8$ mg/L. The fixed-variance mean-sensitivity information is $C^2/v=64/0.68=94.12$. At $C=2$, $v=0.04+0.04=0.08$ (mg/L)$^2$, giving $C^2/v=4/0.08=50.00$. Although the high-concentration assay has larger absolute variance, it contributes more mean information about $\eta$ in this comparison.

If the stated variance function is treated as an exact, known part of the Gaussian model, then $\partial v/\partial\eta=0.02C^2$, in (mg/L)$^2$. The full information is $C^2/v+\tfrac12[(\partial v/\partial\eta)/v]^2$. At $C=8$, the variance term is $\tfrac12(1.28/0.68)^2=1.772$, giving total information $95.89$. At $C=2$, it is $\tfrac12(0.08/0.08)^2=0.5000$, giving total information $50.50$. These totals assume the variance model and its parameters are known; if that assumption is not credible, the extra variance-based term should not be relied upon in design.

Diagnostics, Alternatives, and Limitations

Before trusting a weighted design, check whether its weights represent the measurement process that will actually be used. A variance curve estimated from limited validation data may be uncertain, and a small change in that curve can produce large changes in a design when candidate scores are close. Examine the design across plausible assay variances and nominal PK parameters rather than reporting only one optimized schedule.

Check residual scale by concentration and sample type. If residual spread systematically differs from the assumed variance, the weights may be miscalibrated. Standardized residuals, formed by dividing residuals by their assumed standard deviations, should not show a clear remaining trend in spread.
Inspect the weighted information contributions. An extremely large weight can make the design depend almost entirely on one sample. Confirm that this observation is operationally reliable and that the complete matrix remains nonsingular and numerically stable.
Check the variance model's role. If variance is parameter-dependent, state whether the calculation uses frozen nominal weights or the full likelihood information. Mixing the two approaches can lead to incorrect information scores.
Compare designs under plausible alternatives. Recompute scores under equal variance, alternative variance curves, and realistic sample-type variances. If the selected times change substantially, assay characterization is a design priority.

Unweighted sensitivity information is a useful comparison when precision is effectively constant across times. Inverse-variance weighted sensitivity information is appropriate for known, fixed heteroscedastic variances. A full likelihood calculation is needed when a specified probability model makes variance parameter-dependent. Replication can improve information by adding contributions at a time, but it may not replace sampling at times with distinct parameter sensitivities. Each approach relies on assumptions about the error distribution, variance calibration, and nominal model; none removes uncertainty about whether those assumptions hold.

A common mistake is to select the most widely separated times because the unweighted sensitivity vectors seem most distinct, then assume the same pair remains optimal under any assay conditions. The correction is to calculate each candidate's weighted information, using the appropriate variance model, and compare the design criterion. Another mistake is to call $1/v_i$ the information from observation $i$. It is only the precision weight; the sensitivity vector, and in a full heteroscedastic likelihood the variance derivatives, determine the complete contribution.

Key takeaway: Assay precision enters a local sampling design through variance-weighted sensitivities. When the variance changes across concentrations or sample types, recompute the candidate information matrices: the best sampling times can change, and a parameter-dependent variance model may require more than simple inverse-variance weighting.

Check Your Understanding

Answer each question using the stated assumptions. Brief solutions follow the questions.

Two independent observations have the same sensitivity vector, but one has variance 0.25 and the other variance 1.00. What is the ratio of their sensitivity-based information contributions?
For the log-concentration model in the schedule example, calculate the determinant score for the pair at 1 and 4 h if their weights are 10 and 2.
Why can a high-concentration observation contribute more information about log-amplitude even when its assay variance is larger?
When should the variance-derivative term be included in Gaussian Fisher information?

Solutions. (1) The weights are 4 and 1, so the first contribution is four times the second because their sensitivity vectors are identical. (2) The score is $10(2)(4-1)^2=180$. (3) Information depends on both precision and sensitivity; for $C=\exp(\eta)$, the mean sensitivity to $\eta$ is $C$, so the contribution is $C^2/v$, not $1/v$ alone. (4) Include the term when the specified Gaussian likelihood has a known, parameter-dependent variance and its derivatives with respect to the parameters are part of the model; do not add it merely because observed variances differ across samples.