Unobserved Concentration Paths Between Sparse Samples

What Sparse Samples Leave Unseen

A concentration measured at two visits constrains the path between them, but does not reveal it. The concentration may decline smoothly, change slope as distribution gives way to elimination, or fluctuate around a model-based trend. Those distinct latent paths can agree at the observed times. Sparse data therefore support comparisons among paths under stated assumptions, not a direct reconstruction of what happened between visits.

The distinction matters in pharmacokinetics because a path is often clinically relevant: it may determine the timing of a peak, a trough, or a period above a target concentration. A deterministic reference model supplies one path for specified parameters and dosing conditions. A stochastic model instead allows a distribution of paths. Neither makes unobserved values known simply because the endpoints were measured.

Definition: A latent concentration path is the time-indexed sequence of underlying concentrations \(C(t)\) in a subject, whether or not concentrations were measured at every time. A conditional path distribution describes uncertainty about unobserved states after conditioning on specified observations and model assumptions. It is not the same as measurement error in the observed assays.

For numerical work below, we sometimes treat endpoint concentrations as known exactly to isolate path uncertainty. In practice, an assay is an imperfect observation of the latent state. When measurement error matters, uncertainty in the endpoints must also be included; treating noisy assays as exact can make a bridge seem more certain than it is.

Different Paths Can Fit the Same Samples

A useful first comparison is between a one-compartment exponential decline and a two-component decline. The latter can represent a rapid distribution phase followed by slower elimination. Both can be evaluated at the observed times, but sparse sampling may miss the change in slope.

Worked Example: A Rapid Distribution Phase Between Visits

Suppose a concentration follows the deterministic two-component reference curve \(C(t)=8e^{-2t}+2e^{-0.1t}\), with time in hours and concentration in consistent units. The first exponential declines rapidly; the second declines slowly. Calculate concentrations at 0, 1, and 2 hours, then compare the midpoint with log-linear interpolation between the endpoints.

At time 0, \(C(0)=8+2=10\). At 1 hour, \(C(1)=8e^{-2}+2e^{-0.1}\). Using \(e^{-2}\approx0.135335\) and \(e^{-0.1}\approx0.904837\),

C(1)\approx 8(0.135335)+2(0.904837) =1.08268+1.80967 =2.89235\approx 2.892.

At 2 hours, \(C(2)=8e^{-4}+2e^{-0.2}\). Using \(e^{-4}\approx0.0183156\) and \(e^{-0.2}\approx0.818731\),

C(2)\approx 8(0.0183156)+2(0.818731) =0.146525+1.637462 =1.783987\approx 1.7840.

If only \(C(0)=10\) and \(C(2)\approx1.7840\) were known, log-linear interpolation would put the midpoint at the geometric mean: \(\sqrt{10(1.7840)}=\sqrt{17.84}\approx4.2237\). The two-component model instead gives about 2.892 at 1 hour. Both paths agree with the two endpoints, but they imply very different midpoint concentrations. The difference arises because rapid distribution makes the early decline steep. The samples alone do not establish that this particular two-component curve is the true path.

This example compares a specified deterministic curve with a simple interpolation; it does not estimate a stochastic bridge. To quantify uncertainty in an unobserved state, we need a process model and a conditioning calculation.

A Conditional Path Model on the Log Scale

Concentration is positive, so it is often useful to model its logarithm \(X(t)=\log C(t)\). Consider the additive-noise model \(dX(t)=a(t)\,dt+\sigma\,dW(t)\), where \(a(t)\) is a specified, integrable drift, \(\sigma>0\) is constant, and \(W(t)\) has independent Gaussian increments with variance equal to elapsed time. For elimination on the log scale, a natural drift is \(a(t)=-k(t)\), with \(k(t)\) an elimination-rate function. A changing \(k(t)\) changes the expected log-concentration slope.

Assume observations at times \(s<t\), exact latent endpoints \(X(s)=x_s\) and \(X(t)=x_t\), and the stated additive-noise model on the whole interval. The interior distribution below is conditional on both endpoints. The drift is assumed deterministic, and the process innovations over disjoint intervals are independent.

Bridge formula: Write \(M(s,t)=\int_s^t a(v)\,dv\). For \(s<u<t\), under the additive-noise log-state model, \(X(u)\mid X(s)=x_s,X(t)=x_t\) is normal with mean \(x_s+M(s,u)+\frac{u-s}{t-s}\{x_t-x_s-M(s,t)\}\) and variance \(\sigma^2\frac{(u-s)(t-u)}{t-s}\). The corresponding concentration is lognormal, since \(C(u)=e^{X(u)}\).

The mean starts from the drift-predicted value at \(u\), then adjusts for the difference between the observed endpoint and the endpoint predicted from \(x_s\). The adjustment is proportional to the fraction of the interval elapsed. The variance describes process uncertainty remaining after both endpoints are fixed; it is not the variance of an assay.

Why the Bridge Formula Holds

Here is a derivation. Let \(h_1=u-s\), \(h_2=t-u\), and \(T=t-s=h_1+h_2\). The Gaussian increment model gives \(X(u)=x_s+M(s,u)+\epsilon_1\) and \(X(t)=x_s+M(s,t)+\epsilon_1+\epsilon_2\), where \(\epsilon_1\) and \(\epsilon_2\) are independent, with distributions \(N(0,\sigma^2h_1)\) and \(N(0,\sigma^2h_2)\). Define the endpoint residual \(r=x_t-x_s-M(s,t)\). Conditional on the endpoints, \(\epsilon_1+\epsilon_2=r\).

For independent centered normal variables, the conditional distribution of the first given their sum is normal. This can be checked directly by writing its density as proportional to \(\exp[-z^2/(2\sigma^2h_1)-(r-z)^2/(2\sigma^2h_2)]\) and completing the square in \(z\). The coefficient of the squared centered term gives conditional variance \(\sigma^2h_1h_2/T\), and the center is \(h_1r/T\). Substituting this conditional mean and variance into \(X(u)=x_s+M(s,u)+\epsilon_1\) proves the formula.

For a constant drift, the bridge mean simplifies to straight-line interpolation on the log scale: the deterministic drift cancels from the conditional mean. The bridge variance remains positive at interior times, even though the endpoints are fixed. It tends to zero as \(u\) approaches either endpoint.

Variance result: The bridge variance is largest at the midpoint \(u=(s+t)/2\), where it equals \(\sigma^2(t-s)/4\). To verify this, write \(h=u-s\); the variance is \(\sigma^2h(T-h)/T\). Since \(h(T-h)=T^2/4-(h-T/2)^2\), it is maximized at \(h=T/2\). Thus, with this model and fixed endpoints, the interior uncertainty grows with the sampling gap.

Worked Conditional Comparisons

Worked Example: Elimination Changes Within the Sampling Gap

Let the interval be 0 to 2 days, with endpoint concentrations \(C(0)=10\) and \(C(2)=2\). Use \(X=\log C\), \(\sigma=0.4\) log-concentration units per square-root day, and \(a(t)=-k(t)\), where \(k(t)=0.1\) per day on days 0 to 1 and \(k(t)=1\) per day on days 1 to 2. Find the conditional distribution at day 1.

The starting log concentration is \(x_0=\log 10\approx2.302585\), and the endpoint is \(x_2=\log2\approx0.693147\). The accumulated drift to day 1 is \(-0.1\); over the full interval it is \(-1.1\). Therefore the endpoint residual relative to the drift prediction is \(r=x_2-x_0-(-1.1)\approx0.693147-2.302585+1.1=-0.509438\).

The bridge mean at day 1 is the drift-predicted value \(x_0-0.1\) plus half the endpoint residual: \(2.302585-0.1-0.254719=1.947866\). The conditional variance is \(0.4^2(1)(1)/2=0.08\), so the standard deviation is \(\sqrt{0.08}\approx0.28284\). The conditional median concentration is \(e^{1.947866}\approx7.01\). A central 95% interval on the log scale is \(1.947866\pm1.96(0.28284)\), or approximately \([1.3935,2.5022]\); exponentiating gives a concentration interval of approximately \([4.03,12.21]\).

The wide interval warns against treating the conditional median as an observed fact. The change in elimination affects the drift-predicted path, while the stochastic model still allows substantial uncertainty at the unobserved visit. Different drift or diffusion assumptions would produce different conditional distributions.

Worked Example: Uncertainty Across a Long Gap

Suppose the endpoints of a long intervisit interval are both 4 concentration units, and a fitted bridge model gives a midpoint log-concentration mean of \(\log4\) and standard deviation 0.25. Calculate a central 95% interval for the midpoint concentration.

On the log scale, the interval is \(\log4\pm1.96(0.25)=\log4\pm0.49\). Since exponentiation is increasing, the corresponding concentration interval is \(4e^{-0.49}\) to \(4e^{0.49}\). Using \(e^{-0.49}\approx0.612626\) and \(e^{0.49}\approx1.632316\), the endpoints are \(4(0.612626)\approx2.4505\) and \(4(1.632316)\approx6.5293\). Thus the interval is approximately \([2.45,6.53]\), not a point concentration of 4.

As another log-scale calculation, a conditional mean of 1.4513 and standard deviation 0.283 give \(1.4513\pm1.96(0.283)\). Since \(1.96(0.283)=0.55468\), the interval is \([0.89662,2.00598]\), or approximately \([0.897,2.006]\). These are log concentrations; exponentiating is necessary to report an interval on the concentration scale.

A long interval leaves more opportunity for process variation to accumulate. The precise amount depends on the diffusion model and the endpoint conditioning, so elapsed time alone is not enough to calculate a numerical interval.

Diagnostics, Alternatives, and Limitations

A conditional path is a model-based comparison, not a recovered history. Check whether the assumed drift is pharmacologically credible over the interval, whether the diffusion magnitude is supported by the available data, and whether the observation model has been kept separate from process variation. In particular, inspect residual patterns at observed times, compare predictions across plausible drift and diffusion specifications, and report whether endpoint assays were treated as exact or noisy.

Common warning signs include a bridge that implies implausible concentrations, intervals that change dramatically under small modeling choices, and apparent certainty produced by fixing noisy endpoints. A single diffusion value may also be inadequate when variability changes over time or depends on concentration. If a different model better reflects those features, the bridge distribution must be recalculated under that model rather than retaining the formula above.

Common pitfall: “The measured concentrations are 10 at the first visit and 2 at the second, so the concentration halfway between them was 6.” This assumes a particular linear path on the concentration scale. Log-linear interpolation instead gives a geometric midpoint, and a stochastic process model gives a distribution of possible midpoint values. Neither interpolation is established by the endpoints alone.

The main alternatives answer different questions. A deterministic ODE reference model gives a single concentration curve conditional on its structure, parameters, and dosing history. Interpolation gives a convenient curve between observations but does not represent process uncertainty. A conditional stochastic bridge gives a distribution over unobserved states under a specified process model. With assay error, a full analysis conditions on the observations through both the process model and the observation model; it should not simply equate a measured value with the latent endpoint.

Limitations remain even when the calculation is correct. The additive-noise log model implies positive concentrations after exponentiation, but its Gaussian log fluctuations may allow implausibly extreme values. The constant-\(\sigma\) assumption may not fit concentration-dependent or time-varying variability. A path distribution can be weakly identified with sparse observations, and different plausible models may agree at visits while disagreeing between them. Report the assumptions and uncertainty rather than presenting one latent path as measured data.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions.

Two exact endpoint concentrations are 9 and 1. What does a log-linear interpolation give halfway between them, and why is that not proof of the true midpoint?
For a constant-drift additive-noise log model, what is the bridge variance at the midpoint of an interval of length \(T\)?
If \(k(t)\) increases during a sampling gap, which part of the log-state model changes directly: the drift, the diffusion variance, or both?
Why can treating assay measurements as exact endpoints understate uncertainty about an interior concentration?

Solutions. (1) The log-linear midpoint is \(\sqrt{9\cdot1}=3\), the geometric mean. It is one interpolation choice, not a value identified by the endpoints. (2) The midpoint variance is \(\sigma^2T/4\), from \(\sigma^2(u-s)(t-u)/(t-s)\) with both half-intervals equal to \(T/2\). (3) Increasing \(k(t)\) changes the drift \(a(t)=-k(t)\); under the stated constant-\(\sigma\) model it does not directly change the diffusion variance. (4) Noisy assays leave uncertainty about the latent endpoint states themselves. Fixing them at their measured values removes that uncertainty and can make the conditional interior distribution too narrow.