Simultaneous Bands for Entire Concentration-Time Curves

Why a Whole-Curve Band Is Different

A concentration-time profile is a curve, not a collection of unrelated predictions. If a PBPK model is simulated under specified input distributions and a fixed dosing scenario, each simulation produces concentrations at many times. A prediction interval at one time addresses the concentration there; a simultaneous band addresses whether one complete simulated curve remains inside its limits throughout a stated time domain.

The distinction matters whenever the question concerns a profile as a whole: for example, whether a plausible exposure trajectory stays within a range over a dosing interval. The band is tied to the predictive distribution specified by the simulation, including the uncertainties and variability represented in that simulation. It does not automatically account for model limitations or uncertainty omitted from the inputs.

Definition: Let $C(t)$ denote a random concentration-time curve over a domain $t\in\mathcal{T}$. A band with lower limit $L(t)$ and upper limit $U(t)$ has simultaneous predictive coverage $1-\alpha$ if $$ P\{L(t)\leq C(t)\leq U(t)\text{ for every }t\in\mathcal{T}\}\geq 1-\alpha. $$ A pointwise band instead targets $P\{L(t)\leq C(t)\leq U(t)\}\geq 1-\alpha$ separately at each specified time. The latter does not, by itself, guarantee the former.

In practice, curves are usually stored on a finite grid $t_1,\ldots,t_J$. A simultaneous guarantee on that grid concerns all $J$ recorded concentrations. It is not automatically a guarantee at every time between grid points. That distinction should be stated when the reported band is based on discrete outputs.

Build a Band From Curve-Level Deviations

A direct simulation-based approach first chooses a center $m(t)$ and a positive scale $s(t)$ for the predictive distribution at each time. For example, these can be the simulated mean and standard deviation, or the median and a suitable scale. The choices affect the shape and interpretation of the band; neither is required to make the calibration procedure work. On a time grid, score each complete calibration curve by its largest standardized departure from the center:

D_b=\max_{1\leq j\leq J} \left|\frac{C_b(t_j)-m(t_j)}{s(t_j)}\right|.

Here $C_b(t_j)$ is the concentration from calibration simulation $b$ at time $t_j$. A large $D_b$ means that at least one point on that curve is far from the center relative to the local scale. Let $q$ be a high quantile of these curve-level scores. The band is

L(t_j)=m(t_j)-q\,s(t_j),\qquad U(t_j)=m(t_j)+q\,s(t_j).

The same $q$ is used across the grid. Consequently, the curve-level maximum—not a separate quantile at each time—determines how wide the band must be. The resulting limits need not be symmetric around a pointwise median in the original concentration scale if a transformed scale is used instead.

Calibration and Its Justification

For a transparent finite-simulation calibration, divide independent simulations into a training set and a calibration set. Use the training set to fix $m(t)$ and $s(t)$. Then calculate a score $D_b$ for every calibration curve using those fixed functions. With $B$ calibration scores, sort them as $D_{(1)}\leq\cdots\leq D_{(B)}$, and choose

k=\left\lceil (B+1)(1-\alpha)\right\rceil,\qquad q=D_{(k)}.

This finite-rank rule applies when $k\leq B$; if $k=B+1$, the rule provides no finite cutoff. It is useful to distinguish this rank-based calibration from simply taking an interpolated empirical quantile, whose finite-sample coverage need not follow the same rank argument.

Assumptions for the rank guarantee: Conditional on the independent training simulations, the calibration curves and a new target curve must be exchangeable draws from the same predictive distribution, and the score must be defined in the same way for each. The rank rule then gives coverage of at least $1-\alpha$ on the stated grid, with ties making coverage conservative. This is a guarantee about draws from the specified predictive distribution, not a guarantee that the PBPK model is scientifically correct.

The justification is an order-statistic argument. Conditional on the training set, the $B$ calibration scores and the new curve's score are exchangeable. With no ties, the rank of the new score among these $B+1$ scores is equally likely to be any rank from 1 to $B+1$. The new score is no greater than the $k$th calibration score when its rank is at most $k$, an event with probability $k/(B+1)$. Because $k=\lceil(B+1)(1-\alpha)\rceil$, this probability is at least $1-\alpha$. If the new score is at most $q$, then every standardized deviation on the grid is at most $q$, which is exactly the event that its full grid curve lies inside the band. Ties cannot reduce the coverage.

Worked Example: Pointwise Coverage Does Not Ensure Whole-Curve Coverage

Suppose a three-time-point band has 95% coverage at each time. To isolate the distinction, assume the three events of being inside the band are independent. What is the probability that a simulated curve lies inside at all three times?

Let $A_j$ be the event that the curve is inside the band at time $t_j$. Under the stated assumptions, $P(A_j)=0.95$ for each $j$, and the events are independent. Therefore, $$ P(A_1\cap A_2\cap A_3)=P(A_1)P(A_2)P(A_3) =0.95^3=0.857375. $$

Thus, the curve is covered at all three time points with probability about 85.7%, not 95%. Independence is only an illustrative assumption: concentrations at nearby times are often dependent, and their dependence changes the joint coverage. The key point is that three marginal 95% statements do not establish 95% simultaneous coverage.

Worked Example: Calibrating a Grid-Based Band

Suppose a training set has fixed the center and scale at three times as follows. Nine independent calibration curves produce the sorted maximum standardized scores shown below. Construct the band for a target grid coverage of $1-\alpha=0.80$.

The center concentrations are $m(t_1)=10$, $m(t_2)=4$, and $m(t_3)=1$ mg/L. The corresponding scales are $s(t_1)=2$, $s(t_2)=1$, and $s(t_3)=0.25$ mg/L. The sorted calibration scores are $0.42, 0.55, 0.61, 0.73, 0.81, 0.94, 1.02, 1.17, 1.40$.

There are $B=9$ calibration curves, so $$ k=\left\lceil(9+1)(0.80)\right\rceil =\lceil8\rceil=8. $$ The eighth sorted score is $q=D_{(8)}=1.17$. Apply $m(t_j)\pm q\,s(t_j)$ at each time:

At $t_1$, the half-width is $1.17(2)=2.34$ mg/L, giving limits $10-2.34=7.66$ and $10+2.34=12.34$ mg/L. At $t_2$, the half-width is $1.17(1)=1.17$ mg/L, giving limits $2.83$ and $5.17$ mg/L. At $t_3$, the half-width is $1.17(0.25)=0.2925$ mg/L, giving limits $0.7075$ and $1.2925$ mg/L.

The common multiplier is calibrated from the largest standardized deviation across each entire calibration curve. Under the exchangeability assumptions, the rank argument targets at least 80% coverage over these three grid points for a new curve from the same predictive distribution. It does not assert coverage between the grid points.

Positive Concentrations and Transformations

An additive band can extend below zero even though concentrations cannot. One option is to build the band on the log-concentration scale when concentrations are positive and the predictive spread is more naturally represented multiplicatively. Define $Z(t)=\log C(t)$, choose a center $m_Z(t)$ and positive scale $s_Z(t)$ on that scale, and calibrate the maximum standardized absolute log deviation. Transform the limits back with the exponential function:

L(t)=\exp\{m_Z(t)-q\,s_Z(t)\},\qquad U(t)=\exp\{m_Z(t)+q\,s_Z(t)\}.

Worked Example: A Positive Band on the Log Scale

Suppose an independently trained log-scale center and scale are fixed at three times. The centers correspond to concentrations of 8, 2, and 0.5 mg/L; the log-scale scales are $0.20$, $0.30$, and $0.40$. Ten independent calibration curves have sorted maximum absolute standardized log deviations $0.45, 0.58, 0.66, 0.79, 0.83, 0.96, 1.04, 1.21, 1.35, 1.62$. For target grid coverage 80%, calculate the band.

Here $B=10$, so $k=\lceil(10+1)(0.80)\rceil=9$, and $q=1.35$. At the first time, the log-scale half-width is $1.35(0.20)=0.27$. The concentration limits are $8\exp(-0.27)=6.11$ and $8\exp(0.27)=10.48$ mg/L.

At the second time, the half-width is $1.35(0.30)=0.405$, giving $2\exp(-0.405)=1.33$ and $2\exp(0.405)=3.00$ mg/L. At the third time, the half-width is $1.35(0.40)=0.54$, giving $0.5\exp(-0.54)=0.291$ and $0.5\exp(0.54)=0.858$ mg/L. These limits are positive and reflect multiplicative rather than additive departures. The rank-based interpretation applies on the log scale and therefore also to the corresponding concentration band.

Alternatives, Diagnostics, and Limitations

Pointwise prediction bands are useful when the scientific question concerns selected times individually. They may be narrower than simultaneous bands, but they do not control the probability of at least one excursion over a grid. A grid-based Bonferroni construction offers another option: use pointwise error probability $\alpha/J$ at each of $J$ times. The union bound then ensures that the probability of any grid-point failure is at most $\alpha$, without requiring independence. This can be conservative, especially when concentrations at neighboring times are strongly dependent. Direct maximum-score calibration uses the simulated dependence among times and can be less conservative for the specified predictive distribution.

A simultaneous band is not a confidence band for the mean concentration curve. It is intended to contain a new random curve from a predictive distribution. Nor does a nominal calibration level remove uncertainty from a finite number of simulations: a small calibration set gives coarse ranks and unstable estimates of the band. The output also depends on the predictive distribution, center, scale, time domain, and time grid chosen in advance.

Several checks help identify a band that is not serving its intended purpose:

Check the target and simulation design. Training and calibration curves must represent the same scenario and predictive target. Changing the dosing regimen or input distribution only for calibration breaks the exchangeability argument.
Check the scale. Each $s(t_j)$ must be positive and meaningful. A zero or extremely small scale makes standardized scores undefined or unstable; a log transformation also requires positive concentrations at the times being transformed.
Check calibration size and Monte Carlo stability. The selected order statistic can change substantially with few calibration curves. Repeated independent runs or larger simulation sets can reveal unstable widths; Monte Carlo precision should not be confused with predictive coverage.
Check the grid. A curve can leave the band between recorded times, particularly where concentrations change rapidly or have a sharp peak. Use a sufficiently fine grid, inspect regions of rapid change, or assess interpolation error before making a continuous-time claim.
Check the curve-level failures. On an independent validation set, count curves that leave the band at least once. Also record where excursions occur. Repeated failures in a time region can reveal inadequate resolution or an inappropriate center and scale, although a finite validation set has sampling variability.

A band only represents sources of variability and uncertainty included in the PBPK simulation. A narrow band can therefore be misleading if important input uncertainty, dependence, or variability was omitted. Conversely, an unusually wide band can result from an extreme simulated curve or from poor numerical behavior. Neither width alone nor a successful calibration count verifies the scientific adequacy of the model.

Common pitfall: “The band is 95% pointwise, so it contains 95% of complete curves.” This confuses separate marginal coverage statements with a joint event. Whole-curve coverage requires calibration of a curve-level quantity, such as the maximum deviation over the specified time grid, or another method that explicitly controls joint coverage.

Check Your Understanding

Use the stated assumptions in each question. Solutions follow each prompt.

A band has 90% pointwise coverage at each of four times. Does this establish 90% coverage of the entire four-point curve? Solution: No. Pointwise coverage describes four marginal events. Their joint probability depends on their dependence and is not determined to be 90% by those statements.
There are $B=19$ calibration curves and the target coverage is 95%. Which calibration rank is used? Solution: $k=\lceil(19+1)(0.95)\rceil=\lceil19\rceil=19$, so use the largest calibration score, $D_{(19)}$. Under exchangeability, the rank argument gives at least $19/20=95\%$ grid coverage.
Why must the center and scale be fixed independently of the calibration curves for the stated rank argument? Solution: The argument treats the calibration scores and a new score, conditional on training, as exchangeable. If calibration curves also determine the center or scale, the calibration scores are used both to fit the band and to assess its deviations; the stated independent-calibration rank argument no longer applies as given.
A positive concentration band is constructed on the log scale with $m_Z(t)=\log(4)$, $s_Z(t)=0.25$, and $q=2$. Find the limits. Solution: The limits are $\exp\{\log(4)-0.5\}=4\exp(-0.5)=2.43$ mg/L and $\exp\{\log(4)+0.5\}=4\exp(0.5)=6.59$ mg/L.
A reported band contains every simulated validation curve at its recorded grid points. What does this establish about times between the grid points? Solution: Nothing by itself. Grid coverage does not establish continuous-time coverage; between-grid behavior must be assessed, for example by refining the grid in regions of rapid change and evaluating interpolation or numerical error.