Peak, Trough, and Mid-Interval Samples

Why Sample at Different Parts of a Dosing Interval?

At repeated-dose steady state, samples taken just after a dose, midway through an interval, and just before the next dose do not provide interchangeable information. A peak and a trough are separated in time, so their concentrations can distinguish the overall exposure scale from the rate of decline. An intermediate sample can add information, but its value depends on the other sampling times and on the error model. The design question is not simply whether a time is called “peak,” “trough,” or “mid-interval”; it is how the model’s sensitivity changes across the planned samples.

We will compare these times for a one-compartment model with instantaneous IV boluses and first-order elimination. This deliberately simple setting makes the information calculation explicit. The conclusions depend on the model and assumptions; in particular, an oral concentration peak need not occur immediately after a dose.

Model and assumptions: A dose \(D\) is administered by instantaneous IV bolus every \(\tau\) time units. At steady state, for \(0 \leq t < \tau\), concentration is \(C(t)=\frac{D}{V}\frac{e^{-kt}}{1-e^{-k\tau}}\), where \(V\) is volume and \(k>0\) is the first-order elimination rate, in inverse time units. Dose, dosing interval, and sample times are known. For the calculations below, observed log concentrations equal predicted log concentrations plus independent, normally distributed errors with mean zero and known standard deviation \(\sigma\), the same at every time.

The error assumption means that \(\log C_{\mathrm{obs}}(t)=\log C(t)+\epsilon\), with \(\epsilon\sim N(0,\sigma^2)\). Thus the observations are lognormal on the original concentration scale. This is not an additive, constant-variance error model for concentrations themselves. We use \(\alpha=\log(D/V)\) and estimate \(\theta=(\alpha,k)\). Treating \(\alpha\) as a parameter lets us separate the log concentration scale from the elimination rate; it does not require estimating \(D\) or \(V\) separately.

Information from One Sample

Taking logs gives \(\log C(t)=\alpha-kt-\log(1-e^{-k\tau})\). For a sample at time \(t\), its sensitivity vector is the derivative of this predicted log concentration with respect to the parameters. Under the stated normal error model, one observation contributes the outer product of that vector, divided by \(\sigma^2\), to the Fisher information matrix.

\mathbf{g}(t)= \begin{pmatrix} \frac{\partial \log C(t)}{\partial \alpha}\\[2pt] \frac{\partial \log C(t)}{\partial k} \end{pmatrix} = \begin{pmatrix} 1\\[2pt] -t-\frac{\tau}{e^{k\tau}-1} \end{pmatrix}, \qquad \mathbf{I}(t)=\frac{1}{\sigma^2}\mathbf{g}(t)\mathbf{g}(t)^\mathsf{T}.

Every sample has sensitivity 1 to \(\alpha\), but sensitivity to \(k\) changes with time. A sample alone contributes a rank-one matrix, so it cannot identify both parameters with finite uncertainty. With \(n\) independent samples at times \(t_1,\ldots,t_n\), define \(s_i=-t_i-\frac{\tau}{e^{k\tau}-1}\), the sensitivity to \(k\). The total information is

\mathbf{I}(\theta)=\frac{1}{\sigma^2} \begin{pmatrix} n & \sum_i s_i\\ \sum_i s_i & \sum_i s_i^2 \end{pmatrix}.

This expression shows why sample count alone is not enough. Information about the slope-like parameter \(k\), after allowing for an unknown intercept \(\alpha\), depends on how much the sensitivities \(s_i\) vary across sample times. If they are all identical, the information matrix is singular: the observations cannot separate a change in scale from a change in elimination rate.

Key result: Let \(\bar{s}=\frac{1}{n}\sum_i s_i\) and \(SS_s=\sum_i(s_i-\bar{s})^2\). For this two-parameter log-concentration model, the information determinant is \(nSS_s/\sigma^4\), and the \(k\)-entry of the inverse information matrix is \(\sigma^2/SS_s\), giving a local/asymptotic variance approximation and, for unbiased estimators under regularity conditions, a Cramér–Rao lower bound. Thus, at fixed sample count and error variance, greater spread in sensitivities gives greater information for \(k\). Since \(s_i\) changes linearly with \(t_i\), that spread is the same as the spread in sampling times.

Worked Comparisons at Steady State

Worked Example: Peak, Mid-Interval, and Trough Sensitivities

Consider a hypothetical IV bolus regimen with \(\tau=12\) h and \(k=0.2000\) h\(^{-1}\). Compare one sample at the peak, \(t=0\) just after the bolus, one at the midpoint \(t=6\) h, and one at the trough \(t=12\) h just before the next dose. The trough is represented by the left-hand limit as \(t\to\tau^-\); it is not the post-dose concentration at \(t=0\), which is larger by the factor \(e^{k\tau}=e^{2.4}\).

First calculate the interval-dependent term:

\frac{\tau}{e^{k\tau}-1} =\frac{12}{e^{(0.2000)(12)}-1} =\frac{12}{e^{2.4}-1} \approx 1.1972\text{ h}.

The sensitivities to \(k\) are therefore \(s(0)=-1.1972\) h, \(s(6)=-7.1972\) h, and \(s(12)=-13.1972\) h. Each sample’s information contribution is \(\sigma^{-2}(1,s_i)^\mathsf{T}(1,s_i)\). The peak contributes information about both parameters, but its \(k\)-sensitivity is relatively small in magnitude; the trough has a much larger-magnitude \(k\)-sensitivity. The midpoint supplies a sensitivity between those two. None of the three contributions alone identifies both parameters.

For all three samples, the mean sensitivity is \((-1.1972-7.1972-13.1972)/3=-7.1972\) h. Their squared deviations from that mean are \((6)^2\), \(0^2\), and \((-6)^2\), so \(SS_s=72.00\) h\(^2\). This centered spread, not a sum of uncentered differences from an unrelated reference value, is the quantity used to calculate information about \(k\).

Worked Example: Information and Parameter Uncertainty

Use the same three times, and suppose the hypothetical log-error standard deviation is \(\sigma=0.1000\). The sum of sensitivities is \(-21.5916\) h. The total information matrix is

\mathbf{I}= \frac{1}{0.1000^2} \begin{pmatrix} 3 & -21.5916\\ -21.5916 & 1.1972^2+7.1972^2+13.1972^2 \end{pmatrix} \approx \begin{pmatrix} 300 & -2159.16\\ -2159.16 & 22739.9 \end{pmatrix}.

The variance of the estimated elimination rate, accounting for the unknown \(\alpha\), is \(\sigma^2/SS_s\). Substituting gives \(\operatorname{Var}(\hat{k})=0.01000/72.00=0.0001389\) h\(^{-2}\), and hence \(SE(\hat{k})=\sqrt{0.0001389}=0.01179\) h\(^{-1}\). The units are inverse hours, the same as \(k\).

For \(\alpha\), the inverse information matrix gives \(\operatorname{Var}(\hat{\alpha})=\sigma^2(1/n+\bar{s}^2/SS_s)\). Here this is \(0.01000(1/3+7.1972^2/72.00)=0.01053\), so \(SE(\hat{\alpha})=0.1026\). This is uncertainty on the log scale. These standard errors describe the local information-based approximation under the assumed model and error structure; they are not guarantees of performance if the model or assumptions fail.

Worked Example: A Four-Sample Design and Predicted Uncertainty

Now compare a hypothetical four-sample plan at \(t=(0,0,7.742,12)\) h: two peak replicates, one intermediate sample, and one trough. Replicates are allowed mathematically; whether two peak draws are operationally feasible is a separate design constraint. The mean time is \((0+0+7.742+12)/4=4.9355\) h. The centered time sum of squares is

SS_t=(0-4.9355)^2+(0-4.9355)^2+(7.742-4.9355)^2+(12-4.9355)^2 \approx 106.5\text{ h}^2.

Because \(s_i=-t_i-1.1972\), the centered sensitivity sum of squares is also \(SS_s=106.5\) h\(^2\). This is the value used in the information calculation. With \(\sigma=0.1000\), \(SE(\hat{k})=0.1000/\sqrt{106.5}=0.009690\) h\(^{-1}\). The smaller standard error than in the three-sample example reflects both the additional observation and the sampling-time spread; it is not a same-budget comparison.

The design also predicts log concentration at an unsampled time. For a sample time \(t_*=6\) h, the sensitivity is \(s_*=-7.1972\) h. With unknown \(\alpha\) and \(k\), the estimated mean log concentration has approximate variance \(\sigma^2[1/n+(s_*-\bar{s})^2/SS_s]\). Here \(\bar{s}=-(4.9355+1.1972)=-6.1327\) h, so

SE\{\widehat{\log C(6)}\} =0.1000\sqrt{\frac14+\frac{(-7.1972+6.1327)^2}{106.5}} \approx 0.05105.

If \(D/V=10.00\) mg/L, the model predicts \(C(6)=10.00e^{-1.2}/(1-e^{-2.4})\approx3.312\) mg/L. A rough 95% interval for the latent model concentration (the median observed concentration), using the normal approximation on the log scale, is obtained by exponentiating \(\log(3.312)\pm1.96(0.05105)\). This gives approximately 2.997 to 3.661 mg/L. It reflects parameter uncertainty in the latent model concentration under this model; it is not a prediction interval for a future measured concentration, which would also include observation error.

How to Compare Candidate Times

The calculations above suggest a practical sequence for comparing peak, trough, and intermediate samples. First specify the structural model, the parameters of interest, and the residual error model. Then compute the sensitivities at each proposed time and combine their information contributions. Finally, inspect the resulting parameter uncertainty and any prediction uncertainty relevant to the decision. The information matrix is local: its values depend on the parameter values used to calculate the sensitivities.

Define the interval and sample meaning.
Distinguish a post-dose peak from a pre-dose trough, and record actual times relative to the dose.

Calculate weighted sensitivities.
For the stated log-error model, use \(\mathbf{g}(t)/\sigma\). If assay precision differs by time or concentration, the weights must reflect that error model.

Combine information and check rank.
Sum the sample contributions, then verify that the matrix is nonsingular and not nearly singular before interpreting its inverse.

Report uncertainty in context.
Give parameter standard errors in their units and, where useful, quantify uncertainty in a predicted concentration at a stated time.

Every sample has the same direct sensitivity, 1, to the concentration-scale parameter \(\alpha\), while later samples have greater sensitivity magnitude to \(k\). The information about both parameters depends on the full design. But the peak’s information about \(k\) is not zero, because the steady-state accumulation term depends on \(k\). A mid-interval sample can help distinguish curvature or model departures in richer models, but this two-parameter log-linear model has a sensitivity that changes linearly with time. Under this model, an intermediate time does not automatically add more information about \(k\) than an endpoint; its value is determined by its contribution to the overall sensitivity spread and by the other samples.

Diagnostics, Alternatives, and Limitations

Check that the planned samples can actually be collected at the intended times and that the assumed error model is plausible over the expected concentration range. After data collection, inspect residuals against time and predicted concentration, verify dose and sample timestamps, and look for values below the assay’s quantification capability. A poor fit, systematic residual pattern, or very large parameter correlation can indicate that the assumed model does not describe the data adequately. A nearly singular information matrix means the design does not separate the parameters reliably, even if the number of samples appears adequate.

The comparison changes if measurement errors are additive with constant variance on the concentration scale rather than constant variance on the log scale. The sensitivities of concentration are then multiplied by \(C(t)\), and observations at different concentrations receive different effective contributions. Unequal assay precision likewise changes the weights. With oral dosing, absorption, lag time, or a non-instantaneous input can shift the peak and alter the time sensitivities; the immediate post-dose peak convention used here does not apply. These limitations are reasons to calculate information under the intended model and error assumptions, rather than universal rules that “peak is for exposure” and “trough is for elimination.”

Common pitfall: Treating the peak, midpoint, and trough as three equally informative labels—or assuming that a trough always estimates elimination best—ignores the other samples, parameter sensitivities, and residual error. The correction is to calculate each weighted sensitivity and evaluate the combined information matrix for the full proposed schedule.

Check Your Understanding

Answer each question, then compare with the solutions below.

In the stated model, what is the sensitivity of log concentration to \(\alpha\), and why can one sample not estimate both \(\alpha\) and \(k\) precisely?
For \(\tau=12\) h and \(k=0.2000\) h\(^{-1}\), calculate the sensitivity to \(k\) at \(t=6\) h.
For samples at \(t=0,6,12\) h, what is the centered sum of squared sensitivities to \(k\)?
With those three sample times and \(\sigma=0.1000\), calculate the information-based standard error for \(k\).
Why can a predicted mean concentration interval be narrower than a prediction interval for a future observation?

Solutions. (1) The sensitivity to \(\alpha\) is 1. A single sample contributes a rank-one information matrix, so its information has rank one and cannot distinguish two freely varying parameters. (2) The interval term is \(12/(e^{2.4}-1)\approx1.1972\) h, so \(s(6)=-6-1.1972=-7.1972\) h. (3) The three sensitivities are \(-1.1972,-7.1972,-13.1972\) h, with mean \(-7.1972\) h; the centered squared deviations sum to \(36+0+36=72.00\) h\(^2\). (4) \(SE(\hat{k})=\sigma/\sqrt{SS_s}=0.1000/\sqrt{72.00}=0.01179\) h\(^{-1}\). (5) A mean-concentration interval includes uncertainty in estimated parameters, whereas a future measurement also varies because of residual observation error; that added source of variability generally makes its prediction interval wider.