Worked Identification of a Stimulatory Indirect Model

Why Fit Both Stimulatory Mechanisms?

A measured response that changes after dosing does not, by itself, reveal which process the drug altered. A drug can stimulate production of the measured response, or it can stimulate its loss. These mechanisms imply different directions of change from baseline and different rates of approach to the on-drug equilibrium. Fitting both to the same data makes those implications explicit.

The worked dataset below is deliberately simple: the drug effect is treated as constant during exposure, the response begins at baseline, and the response is observed during exposure and after washout. This isolates the identification question. In an applied analysis, drug concentrations, measurement error, baseline variability, and the dosing history also need to be represented appropriately.

Definition: Let \(R(t)\) be a positive response, in response units, and let \(R_0\) be its baseline equilibrium in the absence of drug. Let \(k_{\text{in}}\) be the baseline production rate, in response units per day, and \(k_{\text{out}}\) the baseline fractional loss rate, in day\(^{-1}\), so \(R_0=k_{\text{in}}/k_{\text{out}}\). During a constant drug effect, let \(S\geq 0\) denote the fractional stimulation. Production stimulation multiplies production by \(1+S\); loss stimulation multiplies the fractional loss rate by \(1+S\).

Model Predictions During Constant Exposure

The two candidate models use the same baseline parameters but place the drug effect on different processes. The assumed effect \(S\) is constant during the interval considered. This is an approximation for an exposure that is nearly constant or for an interval in which the effect can reasonably be treated as fixed.

\begin{aligned} \text{Production stimulation:}\qquad \frac{dR}{dt} &= k_{\text{in}}(1+S)-k_{\text{out}}R,\\ R_{\text{ss,prod}} &= R_0(1+S),\qquad \lambda_{\text{prod}}=k_{\text{out}};\\[4pt] \text{Loss stimulation:}\qquad \frac{dR}{dt} &= k_{\text{in}}-k_{\text{out}}(1+S)R,\\ R_{\text{ss,loss}} &= \frac{R_0}{1+S},\qquad \lambda_{\text{loss}}=k_{\text{out}}(1+S). \end{aligned}

Here \(R_{\text{ss}}\) is the equilibrium response under the constant effect, and \(\lambda\) is the rate at which the response approaches that equilibrium. Both rates have units of day\(^{-1}\). If the response at the start of the interval is \(R(0)\), either model has the solution

R(t)=R_{\text{ss}}+\bigl(R(0)-R_{\text{ss}}\bigr)e^{-\lambda t}.

This solution follows by subtracting the equilibrium from the differential equation. For example, in the loss-stimulation model, define \(D(t)=R(t)-R_{\text{ss,loss}}\). Because \(k_{\text{in}}=k_{\text{out}}(1+S)R_{\text{ss,loss}}\), the derivative is \(dD/dt=-k_{\text{out}}(1+S)D\). Separating variables and integrating gives \(D(t)=D(0)e^{-k_{\text{out}}(1+S)t}\), which yields the stated response curve. The production model gives the same exponential form, with its own equilibrium and rate.

When the response starts at baseline, \(R(0)=R_0\), a positive production stimulus predicts an increase toward an equilibrium above baseline. A positive loss stimulus predicts a decrease toward an equilibrium below baseline. The equilibrium and rate are linked: for loss stimulation, a larger \(S\) both lowers the equilibrium and increases the approach rate. They are not independent fitting parameters under this model.

Worked Example: Fitting a Loss-Stimulation Profile

Suppose a hypothetical response is measured in units, with baseline \(R_0=100\) units. During constant drug exposure, assume the observations are generated by loss stimulation with \(k_{\text{out}}=0.1200\) day\(^{-1}\) and \(S=0.5000\). The corresponding baseline production rate is \(k_{\text{in}}=R_0k_{\text{out}}=12.00\) units/day. Calculate the on-drug equilibrium and rate, then predict the responses at the listed times.

Equilibrium and rate. The loss-stimulation model gives

R_{\text{ss,loss}}=\frac{100}{1+0.5000}=66.67\ \text{units}, \qquad \lambda_{\text{loss}}=0.1200(1+0.5000)=0.1800\ \text{day}^{-1}.

Starting from 100 units, the fitted response is \(R(t)=66.67+(100-66.67)e^{-0.1800t}\). Substituting times in days gives the following predictions; the displayed data are rounded to two decimal places.

Time after exposure begins (days)	Substitution	Response (units)
0	\(66.67+33.33e^0\)	100.00
2	\(66.67+33.33e^{-0.3600}\)	89.92
4	\(66.67+33.33e^{-0.7200}\)	82.89
8	\(66.67+33.33e^{-1.440}\)	74.56
12	\(66.67+33.33e^{-2.160}\)	70.51

For instance, at day 2, \(e^{-0.3600}\approx0.6977\), so \(R(2)\approx66.67+33.33(0.6977)=89.92\) units. These hypothetical observations are exactly consistent with the specified model apart from rounding. With real data, one would estimate parameters by fitting the model while accounting for measurement error rather than expect exact agreement.

Fitting the Competing Production-Stimulation Model

For a response that falls from baseline during exposure, the direction of change is already a strong constraint. With \(R(0)=R_0\) and \(S\geq0\), the production-stimulation equilibrium is \(R_0(1+S)\), which is at least \(R_0\). Its trajectory therefore stays at baseline when \(S=0\), or rises above baseline when \(S>0\). It cannot generate the observed decline under the stated assumptions.

Worked Example: Comparing the Two Mechanisms on the Same Data

Use the five hypothetical responses from the preceding example, at 0, 2, 4, 8, and 12 days. Treat the baseline as known to be 100 units, and compare the models by the unweighted sum of squared residuals (SSE), the sum of squared differences between observations and predictions.

Loss stimulation. With \(S=0.5000\) and \(k_{\text{out}}=0.1200\) day\(^{-1}\), the predicted values are 100.00, 89.92, 82.89, 74.56, and 70.51 units. Since these values generated the hypothetical data, the unrounded SSE is zero; the small discrepancy from rounded displayed values is only rounding.

Production stimulation. For any allowed \(S\geq0\), the model cannot decline from the known baseline. Its best directionally compatible fit is at the boundary \(S=0\), which predicts 100 units at every time. Using the rounded observations, its SSE is

\begin{aligned} \text{SSE} &=(100.00-100.00)^2+(89.92-100.00)^2+(82.89-100.00)^2\\ &\quad +(74.56-100.00)^2+(70.51-100.00)^2\\ &=0+101.61+292.75+647.19+869.66\\ &=1911.21\ \text{units}^2. \end{aligned}

The fitted production-stimulation effect is at the no-effect boundary, and the response remains flat. At that boundary, the production model’s response curve does not identify its turnover rate from these observations: when \(S=0\) and the response begins at equilibrium, every value of \(k_{\text{out}}\) predicts the same flat curve. The contrast is thus not simply between two estimated turnover rates. One model explains the observed decline with a positive effect and a defined on-drug rate; the other cannot reproduce the direction of change with a positive production stimulus.

This comparison assumes the response truly begins at baseline, baseline is stable, and the drug effect is stimulatory and nonnegative. If baseline is uncertain or drifting, or if the effect changes substantially during exposure, the apparent direction and the model comparison may change. The SSE calculation illustrates fit to this constructed dataset; for noisy study data, a likelihood model should reflect the measurement-error distribution and any unequal variances.

What Washout Adds

On-drug measurements distinguish the direction of the two proposed effects in this example, but they do not automatically establish the baseline turnover rate precisely. Once the drug effect has disappeared, both mechanisms reduce to the baseline equation \(dR/dt=k_{\text{in}}-k_{\text{out}}R\). A washout profile returning toward baseline can therefore inform \(k_{\text{out}}\), but the return alone does not reveal whether the preceding on-drug change arose from production or loss stimulation.

Worked Example: Estimating Baseline Turnover From Washout

The loss-stimulation fit predicts 70.51 units at day 12. Suppose drug effect has ended at that time and the response is 81.75 units at day 16. During washout, the baseline equilibrium is 100 units and the model is \(R(t)=100+(R(12)-100)e^{-k_{\text{out}}(t-12)}\).

Rearranging the equation gives an estimate from the day-16 observation:

\begin{aligned} k_{\text{out}} &=-\frac{1}{4}\ln\left(\frac{81.75-100}{70.51-100}\right)\\ &=-\frac{1}{4}\ln\left(\frac{-18.25}{-29.49}\right) \approx 0.1200\ \text{day}^{-1}. \end{aligned}

The numerator and denominator inside the logarithm are both negative deviations from baseline, so their ratio is positive, as required. The estimate agrees with the value used to construct the hypothetical data. The calculation assumes that the drug effect has ended, baseline is stable, and the exponential return model applies over the four-day interval. A second washout observation at day 20 would provide a useful check: the model predicts \(R(20)=100+(70.51-100)e^{-0.9600}\approx88.71\) units.

Observations, Diagnostics, and Limits

A convincing mechanism assessment uses the full time course, not only the first and last response values. The following checks are particularly useful:

Direction from baseline. Under the constant positive-effect assumptions, production stimulation predicts an upward departure from baseline, whereas loss stimulation predicts a downward departure.
Approach to equilibrium. The response should approach its on-drug equilibrium smoothly and monotonically in these first-order models. Systematic curvature or reversals during constant exposure suggest changing drug effect, baseline drift, measurement problems, or a model that is too simple.
Linked equilibrium and rate. For loss stimulation, \(R_{\text{ss,loss}}/R_0=1/(1+S)\) and \(\lambda_{\text{loss}}/k_{\text{out}}=1+S\). A lower fitted equilibrium should correspond to a faster on-drug approach relative to baseline under this model. If the data require these quantities to vary independently, the model’s effect structure is inadequate.
Exposure and sampling. Drug concentration or a justified exposure surrogate should be recorded alongside response. Samples near onset, during the transition, near the apparent plateau, and after washout provide different information. Sparse late samples alone may show a changed level without identifying the rate that produced it.
Residuals and uncertainty. Residuals that systematically change sign over time indicate a time-course mismatch. Wide parameter intervals, strong parameter correlations, or a flat likelihood profile warn that a numerical fit is not a precise identification.

The main alternative to selecting a mechanism from response direction is to fit both candidates to the observed data with an explicit error model and compare their likelihoods or predictive performance. Such a comparison is more informative when both mechanisms can plausibly describe the data. A positive loss-stimulation model and a positive production-stimulation model make opposite predictions from a stable baseline, so a clear decline can rule against the latter under the assumptions here. If the data are noisy, small, or affected by baseline drift, neither visual direction nor a lower unweighted SSE is sufficient by itself; uncertainty and model assumptions matter.

These simple models also have limitations. The assumed constant \(S\) may not hold when concentration changes quickly, and a delayed drug effect may require an effect compartment or another time-varying effect description. Interindividual variation, assay error, measurement limits, and changes in baseline production or loss can blur the predicted trajectories. A response profile can support a mechanism without proving that the drug acts directly on that process. Additional concentrations, doses, washout observations, or independent information about production or loss may be needed.

Common mistake: A falling response is sometimes treated as evidence that production stimulation is merely “weak.” That is not correct for a positive production-stimulation model starting at baseline: any \(S>0\) raises its equilibrium and predicts an increase, while \(S=0\) predicts no change. The correction is to check the direction implied by the model equations before interpreting fitted parameter values. A fit at \(S=0\) is a no-effect boundary result, not evidence that a positive production stimulus explains a decline.

Key takeaway: Under constant positive stimulation, production stimulation raises the response equilibrium and leaves the approach rate at \(k_{\text{out}}\); loss stimulation lowers the equilibrium and increases the approach rate to \(k_{\text{out}}(1+S)\). On-drug observations reveal the predicted direction, while washout observations help estimate baseline turnover but do not, by themselves, identify the on-drug mechanism.

Check Your Understanding

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

If \(R_0=80\) units and \(S=0.25\), what are the production-stimulation and loss-stimulation equilibria?
For \(k_{\text{out}}=0.10\) day\(^{-1}\) and \(S=0.25\), what is the loss-stimulation approach rate during exposure?
Why can a positive production-stimulation model starting at baseline not explain a sustained decline during constant exposure?
What does a flat fitted production-stimulation curve at \(S=0\) imply about estimating \(k_{\text{out}}\) from a response that starts at baseline?
Why are washout observations useful, and why do they not alone distinguish production stimulation from loss stimulation?

Solutions. 1. Production stimulation gives \(80(1.25)=100\) units; loss stimulation gives \(80/1.25=64.00\) units. 2. The rate is \(0.10(1.25)=0.1250\) day\(^{-1}\). 3. Its equilibrium is \(R_0(1+S)\), at least \(R_0\) for \(S\geq0\), so from baseline it stays flat at \(S=0\) or rises for \(S>0\). 4. When \(S=0\) and the response starts at equilibrium, the predicted curve is flat for any \(k_{\text{out}}\), so that profile does not identify the rate. 5. After drug effect ends, the response returns toward baseline at the baseline turnover rate, helping estimate that rate; both mechanisms share the same baseline equation during washout, so the return does not identify which process the drug stimulated during exposure.