Comparing Partial AUCs When Profiles Cross

Why a Total-AUC Ranking Can Hide a Window-Specific Reversal

Two concentration–time profiles can cross: one profile is higher during some part of the observation period, while the other is higher elsewhere. Their total areas summarize exposure across the full period, but that summary can conceal which profile has more area in a particular time window. This distinction matters whenever the timing of exposure is relevant to the question being asked.

The comparison is descriptive. A larger partial AUC in a window means a larger integrated concentration over that window, not necessarily a higher concentration at every time within it. Nor does a ranking of calculated areas, by itself, establish a statistically significant difference, bioequivalence, or a clinical consequence.

Definition: For a concentration–time profile \(C(t)\), the partial AUC over the interval \([a,b]\) is the area under the profile between times \(a\) and \(b\), denoted \(\mathrm{AUC}_{a,b}\). For two profiles A and B observed on the same time scale, their window-specific difference is \(\Delta_{a,b}=\mathrm{AUC}_{A,a,b}-\mathrm{AUC}_{B,a,b}\). A positive difference means Profile A has the larger area in that window; a negative difference means Profile B does.

When concentrations are measured at discrete times, the trapezoidal rule estimates the area between successive observations. If adjacent concentrations are treated as connected by straight lines, the interval area is the width of the interval multiplied by the average of its endpoint concentrations. Partial-AUC boundaries that do not coincide with sample times require concentrations at those boundaries; the calculation must use the specified boundary values, rather than silently moving the window to nearby samples.

\mathrm{AUC}_{a,b}\ \approx\ \sum_{i:\,[t_i,t_{i+1}]\subseteq[a,b]} \frac{C(t_i)+C(t_{i+1})}{2}(t_{i+1}-t_i)

The concentration unit multiplied by the time unit gives the AUC unit. For example, concentrations in mg/L and times in hours produce AUC in mg·h/L. Comparisons require a common window, compatible time and concentration units, and a clear definition of the profiles being compared. The numerical examples below use hypothetical concentrations and apply the trapezoidal rule between the stated times.

How Crossings Affect Area Comparisons

Define the pointwise difference \(D(t)=C_A(t)-C_B(t)\). The difference between the partial AUCs is the integral of that difference over the window. Where Profile A is higher, \(D(t)\) is positive and contributes area in A’s favor; where Profile B is higher, \(D(t)\) is negative and contributes area in B’s favor. A crossing changes the sign of \(D(t)\), but does not by itself determine which profile has the larger area over a whole window.

\Delta_{a,b}=\int_a^b D(t)\,dt =\int_a^b\bigl(C_A(t)-C_B(t)\bigr)\,dt

If a window is divided into nonoverlapping subwindows, its difference is the sum of their differences. This follows by additivity of integration: for \(a<c<b\), the integral from \(a\) to \(b\) equals the integral from \(a\) to \(c\) plus the integral from \(c\) to \(b\). Thus positive and negative contributions can offset. A positive total difference does not require Profile A to be higher throughout the window.

\Delta_{a,b}=\Delta_{a,c}+\Delta_{c,b}

For piecewise-linear profiles, this accounting can be done with trapezoids. If a crossing occurs inside a sampling interval and its timing is useful for interpretation, linear interpolation gives the crossing time on that interval. Splitting the area calculation there makes the positive and negative contributions visible; it does not change the total area obtained by the trapezoidal rule.

Worked Example: Total AUC Favors A, but a Late Window Favors B

Consider these hypothetical profiles, with concentrations in mg/L and time in hours. The observation period is 0 to 3 hours, and straight-line interpolation is used between adjacent observations.

Time (h)	Profile A (mg/L)	Profile B (mg/L)
0	0	0
1	6	3
2	5	7
3	2	1

Total AUC for A. The three trapezoid areas are \(\frac{0+6}{2}(1)=3\), \(\frac{6+5}{2}(1)=5.5\), and \(\frac{5+2}{2}(1)=3.5\) mg·h/L. Therefore, \(\mathrm{AUC}_{A,0,3}=3+5.5+3.5=12\) mg·h/L.

Total AUC for B. The corresponding areas are \(\frac{0+3}{2}(1)=1.5\), \(\frac{3+7}{2}(1)=5\), and \(\frac{7+1}{2}(1)=4\) mg·h/L. Therefore, \(\mathrm{AUC}_{B,0,3}=1.5+5+4=10.5\) mg·h/L. Across the full period, A has 1.5 mg·h/L more area.

Compare two windows. Over 0–2 hours, A has \(3+5.5=8.5\) mg·h/L and B has \(1.5+5=6.5\) mg·h/L. A therefore has 2 mg·h/L more area in this early window. Over 2–3 hours, A has \(\frac{5+2}{2}(1)=3.5\) mg·h/L, while B has \(\frac{7+1}{2}(1)=4\) mg·h/L. B has 0.5 mg·h/L more area in this late window.

The total ranking favors A, but the late-window ranking favors B. The profiles cross between the observations at 1 and 2 hours, and again between 2 and 3 hours. The area arithmetic—not merely the existence of a crossing—shows how much each window contributes to the total difference.

Locating a Crossing Does Not Settle the Window Ranking

A crossing identifies where the instantaneous concentrations are equal under the chosen interpolation. It does not identify which profile has the larger partial AUC on either side of the crossing without considering the widths and sizes of the differences. The next calculation shows why a window that contains a crossing can still have a nonzero area difference.

Worked Example: A Window Contains Both Positive and Negative Contributions

Use the hypothetical concentrations from 1 to 2 hours in the preceding example. Under straight-line interpolation, Profile A falls from 6 to 5 mg/L, while Profile B rises from 3 to 7 mg/L. Let \(u\) be time in hours after 1 hour. Then \(C_A=6-u\) and \(C_B=3+4u\). Equating the profiles gives \(6-u=3+4u\), so \(u=0.6\) and the crossing occurs at 1.6 hours. Both concentrations there are 5.4 mg/L.

From 1 to 1.6 hours, A is higher. Its area is \(\frac{6+5.4}{2}(0.6)=3.42\) mg·h/L, and B’s is \(\frac{3+5.4}{2}(0.6)=2.52\) mg·h/L. This portion contributes \(3.42-2.52=0.90\) mg·h/L in A’s favor.

From 1.6 to 2 hours, B is higher. A’s area is \(\frac{5.4+5}{2}(0.4)=2.08\) mg·h/L, while B’s is \(\frac{5.4+7}{2}(0.4)=2.48\) mg·h/L. This portion contributes \(2.08-2.48=-0.40\) mg·h/L to the difference.

Adding the signed contributions gives \(0.90+(-0.40)=0.50\) mg·h/L. Directly applying the trapezoidal rule over the entire 1–2 hour interval gives A \(=\frac{6+5}{2}(1)=5.5\) mg·h/L and B \(=\frac{3+7}{2}(1)=5\) mg·h/L, the same difference. Thus A has the larger area in the whole window, even though B is higher during its final 0.4 hours.

Window Choice and Interpretation

A partial-AUC comparison answers a question tied to a specified interval. Moving a boundary can change how much of a crossing is included, how positive and negative contributions offset, and even which profile has the larger window area. The boundaries should therefore be selected to match the scientific question and stated before interpreting the comparison. Comparing different windows for A and B does not isolate a profile difference because the calculations no longer cover the same time period.

Worked Example: A Late-Window Reversal With a Different Sampling Grid

Suppose two new hypothetical profiles are measured at 0, 2, and 4 hours. Concentrations are in mg/L, with linear interpolation between observations.

Time (h)	Profile A (mg/L)	Profile B (mg/L)
0	0	0
2	8	4
4	2	8

Over 0–2 hours, the areas are \(\frac{0+8}{2}(2)=8\) mg·h/L for A and \(\frac{0+4}{2}(2)=4\) mg·h/L for B. Over 2–4 hours, they are \(\frac{8+2}{2}(2)=10\) mg·h/L for A and \(\frac{4+8}{2}(2)=12\) mg·h/L for B. Thus total AUCs are 18 mg·h/L for A and 16 mg·h/L for B, while B has 2 mg·h/L more area in the late window.

The late-window profiles cross under linear interpolation. On that interval, A is \(14-3t\) and B is \(2t\), so equality occurs when \(14-3t=2t\), or \(t=2.8\) hours. The crossing helps describe the timing of the reversal, while the trapezoid calculations establish the window ranking. The coarser sampling grid also means the computed areas rely on the straight-line approximation over two-hour intervals.

Assumptions, Diagnostics, and Alternatives

These calculations assume both profiles refer to the same time origin and are expressed in compatible concentration units. The compared windows must have the same boundaries. With discrete observations, the trapezoidal calculation treats the concentration between each pair of points as linear; it is exact for a truly linear segment and an approximation otherwise. If a boundary falls between observations, its concentration must be determined using the prespecified interpolation method before calculating the partial AUC.

Several checks help reveal when a comparison may be misleading:

Check the window labels and endpoints. Confirm that both areas cover precisely the same start and end times.
Check units and arithmetic. Each trapezoid uses average concentration multiplied by elapsed time, and all component areas must have the same AUC units.
Inspect the profiles, not just the totals. A crossing or a rapid change between sparse samples may make the chosen linear interpolation consequential.
Reconcile component areas with the total. If nonoverlapping windows partition the observation period, their partial AUCs should sum to the corresponding total AUC, up to rounding.
Separate area from pointwise statements. A larger partial AUC does not imply a higher concentration at every time in that window.

The principal alternative to a window-specific comparison is to report only the total AUC over the observation period. That is a useful summary when whole-period exposure is the target, but it can hide timing differences. Reporting the time profiles alongside prespecified partial AUCs makes those differences visible. A different interpolation or integration approach may be considered if the profile shape or sampling schedule makes the straight-line assumption unsuitable; whichever method is used must be stated and applied consistently.

Partial-AUC rankings are descriptive summaries of the profiles and the chosen windows. A ranking alone does not quantify uncertainty, account for subject-to-subject variability, or justify a regulatory equivalence conclusion. Those questions require methods beyond this area comparison. The interpretation is also limited to the time interval actually measured: a comparison on a finite observation window does not establish what happens outside it.

Common Pitfall

Common mistake: “A has the larger total AUC, so A must have the larger partial AUC in every window.” This is incorrect because total AUC adds the contributions from all parts of the period; a sufficiently large advantage in one window can outweigh a disadvantage in another. In the first worked example, A’s total is 12 mg·h/L versus B’s 10.5 mg·h/L, yet B’s late-window area is 4 mg·h/L versus A’s 3.5 mg·h/L. Compare the areas over the window of interest rather than inferring them from the total ranking.

Key takeaway: When profiles cross, compare partial AUCs on explicitly matched windows. The total-AUC ranking summarizes the full period, while each window-specific ranking reflects the net area of positive and negative concentration differences within that window.

Check Your Understanding

Answer the questions using the trapezoidal rule and the stated linear-interpolation assumption. Solutions are included below.

In the first worked example, what are the partial AUCs for A and B over 2–3 hours, and which profile has the larger area in that window?
For two profiles with differences of \(+1.2\) mg·h/L in an early window and \(-0.7\) mg·h/L in a later window, what is their combined difference across those windows?
Two profiles have concentrations 4 and 2 mg/L at 1 hour, and 2 and 6 mg/L at 3 hours, respectively. Under linear interpolation, when do they cross between these times?
Why does a larger partial AUC not prove that a profile is higher at every time in that window?

Solutions. (1) A has \(\frac{5+2}{2}(1)=3.5\) mg·h/L; B has \(\frac{7+1}{2}(1)=4\) mg·h/L, so B is larger by 0.5 mg·h/L. (2) The combined difference is \(1.2+(-0.7)=0.5\) mg·h/L, so A has the larger combined area by 0.5 mg·h/L. (3) Let \(u\) be hours after 1 hour. The first profile is \(4-u\), and the second is \(2+2u\). Solving \(4-u=2+2u\) gives \(u=2/3\), so the crossing is at about 1.67 hours. (4) A partial AUC is the net integrated difference: positive and negative contributions can offset, so the area ranking does not require one profile to be higher throughout.