AUC Sensitivity to Sampling Density

Why Sampling Density Matters for AUC

An AUC calculated from observed concentrations is an approximation to the area under the underlying concentration–time curve. Between two observations, the linear trapezoidal rule joins the observed concentrations with a straight line. If the true profile bends away from that line, the area of the trapezoid differs from the area under the curve. The size and direction of this difference depend on both the spacing between observations and the curve’s curvature.

This matters when comparing profiles collected on different schedules. A larger estimated AUC might reflect a genuinely larger exposure, but it might also partly reflect how well the sampling grid captures the curve. The calculations below isolate that numerical issue: they compare estimates of the same hypothetical continuous profile, not estimates from different participants or treatments.

Definition: For observations $(t_i,C_i)$, where $t_i$ is time and $C_i$ is concentration, the linear-trapezoidal estimate over $t_0$ to $t_n$ is $$ AUC_{0-T}^{\mathrm{lin}}=\sum_{i=1}^{n}\frac{C_{i-1}+C_i}{2}(t_i-t_{i-1}). $$ Each interval contributes its width multiplied by the average of its two endpoint concentrations. The estimate depends on the observations and the interpolation rule; it is not, by itself, the exact integral of the unobserved continuous profile.

If concentration is measured in mg/L and time in hours, each interval contribution and the total AUC have units of mg·h/L. A finer grid means shorter gaps between observations, not necessarily more precise concentration measurements. This tutorial considers how those gaps affect the interpolation component of the estimate.

Curvature Determines the Direction of Local Error

Consider one interval $[a,b]$, with width $h=b-a$, and suppose the underlying concentration function $C(t)$ has a continuous second derivative there. Let $T$ be the trapezoid area and $I$ the exact integral on that interval. With this sign convention, a positive error means the trapezoid overestimates the exact area.

Key result: For a twice continuously differentiable function on one interval, the trapezoidal error can be written $$ T-I=\frac{h^3}{12}C''(\xi) $$ for some $\xi$ between $a$ and $b$. In particular, convex curvature ($C''>0$ throughout the interval) makes the trapezoid overestimate the area, while concave curvature ($C''<0$ throughout) makes it underestimate the area.

The result follows directly for a quadratic function, and Taylor expansion explains the same leading error for a smooth curve. Write the expansion about $a$ as $C(a+x)=C(a)+C'(a)x+\tfrac{1}{2}C''(a)x^2+\cdots$, for $0\leq x\leq h$. Integrating the displayed terms gives an area of $C(a)h+C'(a)h^2/2+C''(a)h^3/6$. Applying the trapezoid rule to those terms gives $C(a)h+C'(a)h^2/2+C''(a)h^3/4$. Subtracting the integral from the trapezoid gives $C''(a)h^3/12$ to this order. The exact error formula above follows from the trapezoidal remainder theorem; when curvature varies, the remainder evaluates the second derivative at some point within the interval.

The $h^3$ factor is important: for a similar amount of curvature, doubling a single interval’s width multiplies its leading local error by about eight. For a profile observed over a fixed duration, subdividing into shorter intervals generally reduces the accumulated error. But the direction of the error can change across the profile. A rising limb, a rounded peak, and a declining limb need not have the same curvature, so interval errors can partly cancel.

Worked Example: A Concave Profile on a Sparse and Dense Grid

Suppose a hypothetical concentration profile on $0$ to $4$ hours is $C(t)=2+6t-1.5t^2$, in mg/L. Its curvature is $C''(t)=-3$ mg/L per hour squared throughout, so each trapezoid should underestimate the exact area. Compare observations every 4 hours with observations every hour.

Sparse grid. The endpoint concentrations are $C(0)=2$ and $C(4)=2$ mg/L. The trapezoidal estimate is $$ AUC_{\mathrm{sparse}}=\frac{2+2}{2}(4-0)=8\ \text{mg·h/L}. $$

Dense grid. The concentrations at $t=0,1,2,3,4$ hours are $2,6.5,8,6.5,2$ mg/L. The four contributions are $$ \begin{aligned} AUC_{\mathrm{dense}}&=\frac{2+6.5}{2}(1)+\frac{6.5+8}{2}(1)\\ &\quad+\frac{8+6.5}{2}(1)+\frac{6.5+2}{2}(1)\\ &=4.25+7.25+7.25+4.25=23\ \text{mg·h/L}. \end{aligned} $$

Exact area for comparison. Integrating the stated curve gives $[2t+3t^2-0.5t^3]_0^4=8+48-32=24$ mg·h/L. Thus the sparse estimate is 16 mg·h/L below the exact area, whereas the dense estimate is 1 mg·h/L below it. Both are below the exact value, as the concavity predicts, and the dense-grid estimate is closer.

What Refinement Can—and Cannot—Guarantee

If the profile has a bounded second derivative, say $|C''(t)|\leq M$, the local error expression gives a useful composite bound. For intervals with widths $h_i=t_i-t_{i-1}$, the absolute error in the total trapezoidal estimate is no greater than the sum of the absolute local errors:

\left|AUC_{0-T}^{\mathrm{lin}}-\int_{t_0}^{t_n}C(t)\,dt\right| \leq \frac{M}{12}\sum_{i=1}^{n}h_i^3.

With equal intervals of width $h$, the sum is $nh^3$. Since the total duration is $nh$, this is proportional to $h^2$ for fixed duration. This gives a precise sense in which reducing the largest gaps improves the worst-case error bound for a sufficiently smooth profile.

The bound does not say that every newly calculated AUC must be closer to the exact AUC than the previous estimate. It bounds the error using a curvature limit, which is usually unknown from sparse observations alone. Also, the errors from different intervals have signs: positive errors on convex sections may offset negative errors on concave sections. Refinement changes those cancellations as well as the size of each interval’s error.

Worked Example: A Convex Profile and the Effect of Halving the Interval

Consider the hypothetical profile $C(t)=1+t^2$ mg/L from $0$ to $2$ hours. This curve is convex because $C''(t)=2$. Compare one interval of width 2 hours with two intervals of width 1 hour.

One interval. The endpoint concentrations are 1 and 5 mg/L, giving $$ AUC_{\mathrm{coarse}}=\frac{1+5}{2}(2)=6\ \text{mg·h/L}. $$

Two intervals. The midpoint concentration is $C(1)=2$ mg/L, so $$ AUC_{\mathrm{refined}}=\frac{1+2}{2}(1)+\frac{2+5}{2}(1)=1.5+3.5=5\ \text{mg·h/L}. $$

The exact integral is $[t+t^3/3]_0^2=2+8/3=4.667$ mg·h/L. The coarse estimate exceeds it by 1.333 mg·h/L; the refined estimate exceeds it by 0.333 mg·h/L. The direction is consistent with convexity, and halving the interval reduces the error by a factor of four here. For this quadratic, the second derivative is constant, so the local error formula is exact and halving the width reduces each local error by $2^3=8$; there are two refined intervals, making the total error one quarter of the coarse error.

A more subtle point is that a refined estimate need not always move closer to the exact integral. The following example uses a smooth, nonnegative curve to show how opposite curvature can produce cancellation on one grid.

Worked Example: Refinement Can Break Error Cancellation

Let $C(t)=2+\sin(2\pi t)$ mg/L over $0$ to $1$ hour. The curve remains nonnegative, and its exact area is $2$ mg·h/L because the sine term integrates to zero over this interval.

Coarse grid. At both endpoints, the concentration is 2 mg/L. A single trapezoid gives $$ AUC_{\mathrm{coarse}}=\frac{2+2}{2}(1)=2\ \text{mg·h/L}, $$ which happens to equal the exact area.

Refined grid. At $t=0.25$, the concentration is 3 mg/L; at $t=0.5$, it is 2 mg/L; and at $t=0.75$, it is 1 mg/L. Four intervals of width 0.25 hours give $$ \begin{aligned} AUC_{\mathrm{refined}}&=0.25\left(\frac{2+3}{2}+\frac{3+2}{2}+\frac{2+1}{2}+\frac{1+2}{2}\right)\\ &=0.25(2.5+2.5+1.5+1.5)=2.00\ \text{mg·h/L}. \end{aligned} $$

This particular refinement still gives the exact area, so use the midpoint-free nested refinement at $t=0.25$ and $0.75$ with the central observed point omitted only for the original grid: the coarse grid $\{0,0.5,1\}$ has concentrations $2,2,2$ and estimate 2, while the refined four-interval grid also gives 2. The example therefore shows exact cancellation, not a worsening estimate. For a clear case of refinement worsening the estimate, use $C(t)=2+\sin(2\pi t)+0.2\sin(4\pi t)$. Its exact area remains 2; the coarse endpoint trapezoid is 2, while at quarter points the added term is zero and the same refined estimate is also 2. These symmetric grids preserve cancellation as well.

Those symmetric examples illustrate why a carefully selected numerical counterexample matters. Here is a non-symmetric, fully calculated case. Take $C(t)=2+\sin(2\pi t)+0.5\sin(\pi t)$ mg/L on $0$ to $1$ hour. Its exact area is $2+1/\pi=2.318$ mg·h/L. A single endpoint trapezoid is 2 mg·h/L, an underestimate. On the two-interval grid at $t=0.5$, the midpoint concentration is $2.5$ mg/L, so the estimate is $\tfrac{1}{2}\{(2+2.5)/2+(2.5+2)/2\}=2.25$ mg·h/L, still an underestimate and closer. This example does not worsen either. In general, a refinement can worsen accuracy when the coarse-grid error benefits from cancellation between curved sections; whether that happens depends on the locations of the new observations and the curve. Consequently, “more samples always move the estimate closer” is not a general theorem. The error bound and convergence as the maximum interval width shrinks are more reliable statements.

Comparing Grids and Diagnosing Sensitivity

A practical sensitivity check uses the same profile and common start and end times, then calculates the trapezoidal AUC on the available grid and on a denser grid. Where possible, make the denser grid nested: retain the sparse-grid observations and add intermediate observations. Report the estimates, their difference, and the intervals contributing most to that difference. If the grids come from separate profiles, observed differences also include profile variation and measurement error, so they cannot be attributed to sampling density alone.

A large difference between sparse- and dense-grid estimates is evidence that the estimate is sensitive to the observation schedule over that profile. A small difference is reassuring for that comparison, but it does not prove that both estimates are close to the unknown true integral: both grids can miss curvature in the same region. Sparse endpoint concentrations alone cannot reveal every bend between observations.

Inspect the gaps. Long gaps across a rapidly changing portion of the profile are a reason for caution. A large gap on a nearly straight portion may matter less, but straightness cannot be established from just its endpoints.
Compare interval contributions. Identify intervals where adding an intermediate observation changes the estimated area substantially. Concentration changes and interval width together determine the trapezoid contribution.
Look at changes in slope. For three consecutive observations, compare the slopes $(C_i-C_{i-1})/(t_i-t_{i-1})$ and $(C_{i+1}-C_i)/(t_{i+1}-t_i)$. A change suggests curvature near those observations; it does not fully identify curvature inside a wide gap.
Check robustness to a plausible grid change. A material shift after removing or adding observations should be reported as sensitivity, not hidden by choosing whichever estimate is preferred.

These diagnostics assess numerical sensitivity, not all sources of uncertainty. They cannot establish the exact area without information about the curve between samples. A model-based estimate can integrate a specified continuous concentration function, but its result depends on the model and its assumptions. The linear trapezoidal rule instead uses the observed concentrations and a piecewise-linear interpolation assumption. Neither approach makes the unobserved profile directly known.

Alternatives and Limitations

Approach	How it represents the interval	What to keep in mind
Linear trapezoidal rule	Joins consecutive concentrations by a straight line.	Simple and transparent; error depends on interval width and curvature.
Log-trapezoidal rule	Uses a log-linear, or exponential, change between positive endpoint concentrations.	Can be suitable for an interval following exponential decline; it requires positive concentrations and a defensible interval shape.
Model-based integration	Integrates a fitted continuous concentration–time function.	Can use a mechanistic or statistical description of the curve, but adds dependence on model form and fit.

The log-trapezoidal rule is not automatically better simply because a profile declines. It is most defensible when the assumed exponential change is reasonable across the interval. Conversely, linear interpolation does not claim that every concentration changes linearly in reality; it defines a reproducible numerical approximation. A good comparison names the interpolation method and the sampling grid, rather than treating an AUC value as independent of how it was obtained.

Common Mistakes

Assuming a denser grid must give a larger AUC. The direction depends on curvature. Convex intervals tend to be overestimated and concave intervals underestimated by linear trapezoids.
Assuming every refinement improves accuracy. A refined estimate often improves as gaps shrink, but curvature changes and error cancellation mean that a particular refinement can move away from the exact integral.
Treating sparse-versus-dense agreement as proof of accuracy. Two grids may share a bias or miss the same feature between observations. Agreement is a useful sensitivity check, not a guarantee.
Calling a difference caused by sampling when profiles differ. To isolate grid sensitivity, compare estimates from the same underlying profile or carefully account for profile variation.

The central conclusion is that AUC sensitivity is governed not simply by the number of samples but by the placement of samples relative to the profile’s curvature. Shorter gaps reduce the error bound for a smooth profile, while paired-grid comparisons reveal whether a particular estimate changes materially under denser observation.

Check Your Understanding

Work through the questions, then compare your reasoning with the solutions below.

On an interval where $C''(t)>0$ throughout, does the linear trapezoidal estimate tend to lie above or below the exact area? State the condition behind your answer.
A single interval of width 3 hours is split into three intervals of width 1 hour. If the absolute curvature is bounded by the same constant $M$, how does the composite error bound compare before and after splitting?
Two nested grids give nearly identical AUC estimates. What does this support, and what does it not establish?
Why can a denser grid sometimes produce a less accurate estimate than a sparse grid?

Solutions. (1) Above: for a twice continuously differentiable profile with positive second derivative throughout, $T-I=h^3C''(\xi)/12>0$. (2) The coarse bound is $M(3^3)/12=27M/12$. The three-interval bound is $3M(1^3)/12=3M/12$, one ninth as large. This compares bounds, not necessarily the realized errors. (3) It supports low sensitivity to that particular grid change for that profile; it does not prove that either estimate equals the true integral or that both grids capture all curvature. (4) Refinement can alter cancellation between positive errors on convex portions and negative errors on concave portions. There is no general guarantee that each refinement moves the estimate closer, even though the error bound shrinks as the maximum interval width is reduced for a smooth profile.