Why Window Placement Matters
A partial AUC summarizes exposure over a specified time interval. That interval is part of the metric: moving one or both boundaries can change the reported value even when the underlying concentration profile is unchanged. The effect can be particularly noticeable when concentrations are changing rapidly near a boundary, as may occur around an early or late peak.
The introductory material on partial-AUC window boundaries establishes how a window is specified, and the earlier material on interpolating concentrations and calculating partial AUC with trapezoids establishes how to estimate its area from sampled data. Here the focus is the sensitivity of that area to a shift in window placement. The key comparison is between the exposure entering the window at its later end and the exposure leaving it at its earlier end.
For interpretation, assume the profile is nonnegative and integrable over the times considered. For the derivative result below, assume additionally that \(C(t)\) is continuous near the moving boundaries. In practice, concentrations are observed only at scheduled sample times. A trapezoidal calculation treats the profile between adjacent observations as linear; its result is therefore an estimate based on that interpolation, not a guarantee about the unobserved true curve.
Finite Shifts: Exposure Added and Lost
A shift moves both boundaries. The segment at the end of the original window is no longer necessarily the segment added: rather, the shifted window gains the strip from \(b\) to \(b+\delta\) and loses the strip from \(a\) to \(a+\delta\), for a positive shift \(\delta\). This description holds whether or not the two windows overlap.
The antiderivative gives a direct justification that does not require the windows to overlap:
For \(0<\delta<b-a\), the windows overlap, and the common area cancels in a geometric picture. For \(\delta\ge b-a\), they may just touch or be disjoint, so that picture must not be used as the general proof. The antiderivative argument remains valid in either case.
The identity also gives a useful condition for the direction of change. For \(\delta>0\), the partial AUC increases exactly when the average concentration over \([b,b+\delta]\) exceeds the average concentration over \([a,a+\delta]\), since the two strips have equal duration. A concentration value at one instant does not, on its own, determine the result of a finite shift.
Local Sensitivity at the Moving Boundaries
When \(C(t)\) is continuous at the boundaries, the derivative of the shifted partial AUC with respect to the shift \(s\) is:
This follows from writing the area as \(F(b+s)-F(a+s)\) and differentiating each term. For a small positive shift \(\delta\), the first-order change is approximately \(\delta\{C(b)-C(a)\}\). This approximation is local: for a larger shift, concentrations can vary substantially across each strip, so the finite-strip integrals, rather than only the concentrations at the initial boundaries, determine the change.
The derivative has units of concentration: AUC divided by time has the units of concentration. If the concentration at the end boundary is much higher than at the start boundary, a small rightward shift tends to increase the partial AUC. If it is lower, a small rightward shift tends to decrease it. Rapid changes near either boundary can make the difference vary quickly with the shift, so the approximate effect may not persist over a wide range of placements.
Worked Example: Shifting a Window After an Early Peak
Suppose a hypothetical concentration profile has measured values \(C(0)=0\), \(C(2)=12\), \(C(4)=8\), \(C(6)=3\), and \(C(8)=1\), in mg/L. Use linear interpolation between samples and trapezoids to compare windows \([1,4]\) and \([2,5]\) hours.
The interpolation assumption is applied only between the listed sample times. It gives \(C(1)=6\) between 0 and 2 hours, and \(C(5)=5.5\) between 4 and 6 hours. The lost strip is \([1,2]\), with area \((6+12)(2-1)/2=9\) mg·h/L. The added strip is \([4,5]\), with area \((8+5.5)(5-4)/2=6.75\) mg·h/L.
Therefore the shift changes the partial AUC by \(6.75-9=-2.25\) mg·h/L. The original window area is \(9+(12+8)(4-2)/2=29\) mg·h/L. The shifted window area is \((12+8)(4-2)/2+6.75=26.75\) mg·h/L, consistent with the strip calculation. The early peak leaves relatively high concentrations in the lost strip, so this rightward shift reduces the captured area.
Worked Example: Shifting a Window Toward a Later Peak
Consider a second hypothetical profile with \(C(0)=0\), \(C(2)=2\), \(C(4)=5\), \(C(6)=12\), and \(C(8)=8\) mg/L. Under the same piecewise-linear interpolation, compare \([1,4]\) with \([2,5]\) hours.
Interpolation gives \(C(1)=1\) and \(C(5)=8.5\) mg/L. The lost strip \([1,2]\) has area \((1+2)(2-1)/2=1.5\) mg·h/L. The added strip \([4,5]\) has area \((5+8.5)(5-4)/2=6.75\) mg·h/L. The net change is \(6.75-1.5=5.25\) mg·h/L.
The original area is \(1.5+(2+5)(4-2)/2=8.5\) mg·h/L, while the shifted area is \((2+5)(4-2)/2+6.75=13.75\) mg·h/L. The difference is \(13.75-8.5=5.25\) mg·h/L. In this case the window moves toward rising concentrations and a later peak, so the added strip carries more exposure than the lost strip.
Worked Example: A Shift With No Window Overlap
To check the finite-shift identity when the windows do not overlap, consider a hypothetical increasing profile \(C(t)=t+1\) mg/L over 0 to 3 hours. Compare \([0,1]\) with \([2,3]\). These equal-width windows are separated, so an explanation based only on cancelling overlapping areas would not apply.
The first area is \((1+2)(1)/2=1.5\) mg·h/L, and the second is \((3+4)(1)/2=3.5\) mg·h/L. The difference is \(2.0\) mg·h/L. In the strip identity, \(\delta=2\): the added strip \([1,3]\) has area \((2+4)(2)/2=6\) mg·h/L, and the lost strip \([0,2]\) has area \((1+3)(2)/2=4\) mg·h/L. Their difference is \(6-4=2.0\) mg·h/L, as required.
Diagnostics, Alternatives, and Limitations
A sensitivity check should make the window definition explicit and examine the profile near both boundaries. A plot with the window marked can reveal whether a boundary falls on a steep rise, steep decline, or sparsely sampled region. Calculating the two strip areas separately shows what drives the change, while recomputing the partial AUC over a few scientifically plausible placements indicates whether the conclusion depends heavily on a particular boundary.
- Check interpolation near each boundary. If a boundary lies between samples, identify the observations used to interpolate its concentration. A long gap or sharp change in that interval makes the trapezoidal estimate less secure.
- Check both edges, not just the peak. A shift adds one strip and removes another. A high concentration near the lost start boundary can outweigh a moderate concentration near the added end boundary.
- Check that the comparison keeps the intended feature fixed. Translating a window keeps its duration fixed; widening a window changes duration as well as placement and is a different comparison.
- Check the time and concentration units. Strip areas and partial AUCs must use the same units and time scale before their difference is interpreted.
Other exposure summaries answer different questions. Total AUC summarizes exposure over a broader stated interval, whereas a partial AUC isolates exposure within a selected time window. \(C_{\max}\) describes a sampled or otherwise defined peak concentration, not the accumulated area. Neither alternative makes a boundary-sensitive partial AUC incorrect; each represents a different feature of the profile. If a scientific question depends on exposure during a particular phase, the window should be justified and specified consistently rather than moved after inspecting which placement gives a preferred result.
A partial AUC can change substantially when boundaries move even though the overall profile and its total exposure are unchanged. Its sensitivity is not necessarily a defect: it can be the intended focus on exposure timing. But it limits comparisons made with differently placed windows, and it can make a summary unstable if the boundary is arbitrary or the sampling grid poorly resolves the local profile. Trapezoidal estimates also inherit the piecewise-linear approximation, which can smooth over rapid changes between observations.
Check Your Understanding
Answer each question using the window-shift results above. Solutions are provided below the questions.
- For a positive shift, what two strip areas determine the change in a fixed-width partial AUC?
- If \(C(a)=4\) mg/L and \(C(b)=10\) mg/L for a continuous profile, what is the local derivative of the shifted partial AUC at shift zero, and what does its sign indicate?
- A window shifts right by 1 hour. The added strip has area 7 mg·h/L and the lost strip has area 9 mg·h/L. What is the change, and does the partial AUC rise or fall?
- Why is a common-area cancellation argument not a sufficient general justification when the shift can be at least as large as the window width?
Solutions. (1) Subtract the area over the lost strip \([a,a+\delta]\) from the area over the added strip \([b,b+\delta]\). (2) The derivative is \(C(b)-C(a)=10-4=6\) mg/L. Its positive sign means that an infinitesimal rightward shift initially increases the partial AUC. (3) The change is \(7-9=-2\) mg·h/L, so the partial AUC falls by 2 mg·h/L. (4) The windows may not overlap, leaving no common area to cancel. The antiderivative derivation establishes the strip identity for every positive shift for which the integrals exist.