Sparse Concentrations Around Infusion Starts And Stops

Why infusion events matter when samples are sparse

An infusion start or stop changes the input to a concentration model at a known or suspected time. If no sample is taken near that event, the recorded concentrations may show only its delayed effect: concentration accumulates during infusion and continues to decline after infusion stops. A sample taken after a stop is not a measurement of the concentration at the stop.

The earlier material on unobserved concentration paths between sparse samples motivates the key distinction here: the event time is part of the input history, while each concentration sample reports the state at its own sampling time. We will use a one-compartment elimination model with a linear stochastic perturbation. The calculations make the effect of event timing explicit without treating an infusion as an instantaneous bolus.

Definition: Let $C_t$ be concentration, $k>0$ the first-order elimination rate constant, $R$ the infusion rate in amount per unit time, and $V$ the apparent volume. Define $r=R/V$, the input rate in concentration per unit time. For an infusion that is on during a time interval, $r(t)=r$; when it is off, $r(t)=0$. The model is $$ dC_t=\bigl(r(t)-kC_t\bigr)\,dt+\sigma\,dW_t, $$ where $W_t$ is standard Brownian motion and $\sigma\geq 0$ is constant. Here $\sigma$ represents additive process noise, not assay error. A recorded concentration may additionally be $Y_j=C_{t_j}+\epsilon_j$, where the measurement errors $\epsilon_j$ have mean zero and are independent of the process.

For the results below, assume the event schedule $r(t)$ is known and deterministic over the interval considered, $k>0$, and the process starts at a fixed known concentration $C_s=c$. The input is bounded and piecewise continuous, as it is for a finite set of infusion starts and stops. The formulas also hold conditionally on $C_s$ when that state is random. The additive-noise model is convenient for calculation; it is not a claim that all biological concentration processes have constant diffusion.

Conditional mean and variance across an event

For any $t>s$, the input contribution is weighted by the time available for elimination after it enters the system. Material infused near $t$ has undergone less elimination than material infused earlier. This weighting is why simply multiplying the infusion rate by the total infusion duration can overstate concentration at a later sample.

Theorem: Under the assumptions above, the conditional distribution of $C_t$ given $C_s=c$ is normal with mean and variance $$ E[C_t\mid C_s=c] =e^{-k(t-s)}c+\int_s^t e^{-k(t-u)}r(u)\,du, \qquad \operatorname{Var}(C_t\mid C_s=c) =\frac{\sigma^2}{2k}\bigl(1-e^{-2k(t-s)}\bigr). $$ In particular, an infusion start or stop changes the integral for the mean; it does not reset the concentration state.

To derive the result, multiply the SDE by the integrating factor $e^{kt}$. The Itô product rule gives $$ d(e^{kt}C_t)=e^{kt}r(t)\,dt+\sigma e^{kt}\,dW_t. $$ Integrating from $s$ to $t$, then multiplying by $e^{-kt}$, yields $$ C_t=e^{-k(t-s)}c+\int_s^t e^{-k(t-u)}r(u)\,du +\sigma\int_s^t e^{-k(t-u)}\,dW_u. $$ The last integral has mean zero. Its variance, by the Itô isometry for a deterministic integrand, is $$ \sigma^2\int_s^t e^{-2k(t-u)}\,du =\frac{\sigma^2}{2k}\bigl(1-e^{-2k(t-s)}\bigr). $$ The stochastic integral is normal because its integrand is deterministic and the driver is Brownian motion. Adding the deterministic terms therefore gives the stated conditional normal distribution. $\square$

A useful special case is a constant infusion rate $r$ throughout $[s,t]$. Its contribution to the mean is $$ \int_s^t e^{-k(t-u)}r\,du =\frac{r}{k}\bigl(1-e^{-k(t-s)}\bigr). $$ If the infusion is on only from $a$ to $b$, with $s\leq a<b\leq t$, the corresponding contribution is instead $$ \int_a^b e^{-k(t-u)}r\,du =\frac{r}{k}\bigl(e^{-k(t-b)}-e^{-k(t-a)}\bigr). $$ These expressions describe a finite-duration input, not an instantaneous dose.

Worked Example: An Infusion Starts Between Two Samples

Suppose $k=0.10\ \mathrm{h}^{-1}$, $R=10\ \mathrm{mg/h}$, $V=20\ \mathrm{L}$, and $C_0=0.20\ \mathrm{mg/L}$. The infusion starts at hour 1, and the next sample is taken at hour 2. Find the conditional mean at hour 2.

The concentration input rate is $r=R/V=10/20=0.50\ \mathrm{mg/L/h}$. From hour 0 to hour 1 there is no infusion; from hour 1 to hour 2 there is a one-hour infusion. Hence $$ E[C_2\mid C_0=0.20] =0.20e^{-0.10(2)}+ \frac{0.50}{0.10}\bigl(1-e^{-0.10(1)}\bigr). $$ The pre-existing concentration contributes $0.20e^{-0.2}=0.16374615\ \mathrm{mg/L}$. The one-hour infusion contributes $5(1-e^{-0.1})=0.47581291\ \mathrm{mg/L}$. Adding the unrounded values gives $0.63955906\ \mathrm{mg/L}$, or $0.6396\ \mathrm{mg/L}$ to four decimal places.

The sample at hour 2 reflects one hour of infusion, not the full two-hour interval since the previous sample. The mean calculation accounts for both the pre-infusion decay and the infusion contribution.

Worked Example: A Two-Hour Infusion Ends Before the Next Sample

Use $k=0.10\ \mathrm{h}^{-1}$, $R=10\ \mathrm{mg/h}$, $V=20\ \mathrm{L}$, and $C_0=0$. The infusion runs from hour 1 through hour 3, and a sample is taken at hour 4. Calculate the mean concentration at hours 3 and 4.

The input rate is $r=0.50\ \mathrm{mg/L/h}$. At hour 3, the infusion has run for two hours: $$ E[C_3]=\frac{0.50}{0.10}\bigl(1-e^{-0.10(2)}\bigr) =5(1-e^{-0.20}) =0.90634623\ \mathrm{mg/L}. $$ There is no infusion from hour 3 to hour 4, so this mean decays for one hour: $$ E[C_4]=E[C_3]e^{-0.10} =0.90634623e^{-0.10} =0.8200946\ \mathrm{mg/L} \approx 0.8201\ \mathrm{mg/L}. $$ The same result follows by integrating the input only over hours 1 to 3: $$ E[C_4]=5\bigl(e^{-0.10(4-3)}-e^{-0.10(4-1)}\bigr) =5(e^{-0.1}-e^{-0.3}) =0.8200946\ \mathrm{mg/L}. $$

A sample at hour 4 is neither the concentration at the stop nor the concentration at hour 3. It reflects two hours of infusion followed by one hour of elimination.

What can be learned about an unobserved event time?

If an infusion stops at time $\tau$, where $a\leq\tau\leq b$, and no infusion is given after the stop, its contribution to the expected concentration at $b$ is $$ \frac{r}{k}\bigl(e^{-k(b-\tau)}-e^{-k(b-a)}\bigr). $$ For $r>0$, this contribution increases as $\tau$ increases: stopping later leaves less time for elimination before the sample. Differentiating with respect to $\tau$ gives $r e^{-k(b-\tau)}>0$. Thus, with the model parameters and initial state known, the expected endpoint changes monotonically with the stop time. In real sparse data, however, process variability and measurement error mean that one observed endpoint does not identify an exact event time with certainty.

Worked Example: A Stop Occurs Between Sparse Visits

Assume $C_0=0$, $k=0.10\ \mathrm{h}^{-1}$, and $r=0.50\ \mathrm{mg/L/h}$. An infusion begins at hour 0 and stops at hour 2.5; the next sample is at hour 4. Find the expected concentration at hour 4 and compare it with two alternative stop schedules.

For a stop at $\tau=2.5$, the infusion contribution at hour 4 is $$ E[C_4]=5\bigl(e^{-0.10(4-2.5)}-e^{-0.10(4-0)}\bigr) =5(e^{-0.15}-e^{-0.4}). $$ Using $e^{-0.15}=0.86070798$ and $e^{-0.4}=0.67032005$, this is $5(0.19038793)=0.95193965\ \mathrm{mg/L}$, or $0.9519\ \mathrm{mg/L}$.

If the infusion instead stopped at hour 2, the mean would be $$ 5(e^{-0.2}-e^{-0.4})=5(0.81873075-0.67032005)=0.7420535\ \mathrm{mg/L}. $$ If it continued through hour 4, the mean would be $$ 5(1-e^{-0.4})=1.6483998\ \mathrm{mg/L}. $$ The hour-4 mean for a stop at 2.5 lies between these alternatives, as expected. But an observed value near $0.9519$ does not establish that the stop occurred at exactly 2.5: the observed concentration can differ from its conditional mean because of process noise and assay error.

Separate process variability from assay error

For fixed $C_s$, $k$, and a known input schedule, the theorem gives the latent process variance at a sample time. If the observation error is independent, has mean zero, and has variance $\sigma_{\mathrm{obs}}^2$, then $Y_t=C_t+\epsilon_t$ has the same mean as $C_t$ and variance $$ \operatorname{Var}(Y_t\mid C_s) =\frac{\sigma^2}{2k}\bigl(1-e^{-2k(t-s)}\bigr)+\sigma_{\mathrm{obs}}^2. $$ The first term is accumulated process noise; the second is assay error. A known infusion event changes the mean trajectory through $r(t)$, while the stated constant-diffusion model's conditional variance depends on elapsed time, not on whether the infusion is on.

Worked Example: Uncertainty at a Sample After Infusion

Suppose the infusion schedule and mean parameters are as in the two-hour infusion example: $C_0=0$, $k=0.10\ \mathrm{h}^{-1}$, and infusion from hour 1 to hour 3. Let $\sigma=0.08\ \mathrm{mg/L}/\sqrt{\mathrm{h}}$, and let independent assay error have standard deviation $0.05\ \mathrm{mg/L}$. Find the process variance and the total observation variance at hour 4.

The elapsed time from 0 to 4 is four hours, so $$ \operatorname{Var}(C_4\mid C_0) =\frac{0.08^2}{2(0.10)}(1-e^{-2(0.10)(4)}) =0.032(1-e^{-0.8}) =0.0176215\ (\mathrm{mg/L})^2. $$ The assay-error variance is $0.05^2=0.0025\ (\mathrm{mg/L})^2$. Therefore $$ \operatorname{Var}(Y_4\mid C_0)=0.0176215+0.0025=0.0201215\ (\mathrm{mg/L})^2, $$ with standard deviation $\sqrt{0.0201215}=0.1418\ \mathrm{mg/L}$, approximately. The latent mean remains $0.8201\ \mathrm{mg/L}$, from the previous example. Thus, even with the event schedule known, a single measured value need not equal that mean.

Diagnostics, alternatives, and limitations

When a fitted model appears inconsistent with sparse concentrations around an infusion event, first check the event record and units: start and stop times, infusion rate, volume, and whether the recorded time is a clock time or elapsed time. Then check the integration limits. A start inside a sampling interval contributes only from the start onward; a stop inside an interval contributes only up to the stop. Compare predicted means at the actual sample times, not at an assumed event time. Large systematic residuals on one side of events may indicate a misspecified schedule, rate, or elimination model; isolated residuals may also arise from either noise source.

The main alternatives answer different questions. A deterministic ODE with the same time-varying input gives the same conditional mean under this additive, zero-mean noise model, but does not represent process dispersion. A stochastic model adds that dispersion. Dense sampling near starts and stops can reveal transition timing more directly, while infusion records can supply event times even when concentrations are sparse. Treating an infusion as a bolus is a further approximation, not an equivalent calculation: it changes the input history and can distort concentrations around the event.

The linear additive-noise model has limits. It allows negative concentrations with positive probability, and constant $\sigma$ may be implausible across the concentration range. It also presumes the specified event schedule is correct. If timing is uncertain, that uncertainty must be represented or explored rather than silently replaced with a convenient time. Sparse samples alone may not distinguish uncertainty in event timing, process variability, and measurement error; additional event records, sampling, or model assumptions are needed.

Common pitfall: “The infusion began sometime between the two samples, so the whole interval contributes at the infusion rate.” This ignores both the pre-start interval and elimination during and after infusion. Instead, integrate the input only over its active times, with each input contribution weighted by $e^{-k(t-u)}$. Likewise, after a stop, propagate the concentration already present through elimination; do not set it to zero.

Check Your Understanding

Answer each question, then compare with the solutions below.

With $k=0.2\ \mathrm{h}^{-1}$, $r=0.6\ \mathrm{mg/L/h}$, and $C_0=0$, an infusion starts at hour 1 and remains on through hour 2. What is the conditional mean at hour 2?
An infusion is stopped at time $\tau$ before a sample at $b$. For positive $r$, does the mean contribution at $b$ increase or decrease as $\tau$ increases? Give the derivative.
Why is an observed concentration after an infusion stop not generally the concentration at the stop?
If process variance at a sample is $0.012\ (\mathrm{mg/L})^2$ and independent assay-error variance is $0.004\ (\mathrm{mg/L})^2$, what is the observation variance?

Solutions. (1) The infusion is active for one hour, so $E[C_2]=\frac{0.6}{0.2}(1-e^{-0.2})=3(0.181269)=0.5438\ \mathrm{mg/L}$. (2) The contribution is $\frac{r}{k}(e^{-k(b-\tau)}-e^{-k(b-a)})$; its derivative is $r e^{-k(b-\tau)}>0$, so it increases with a later stop. (3) Concentration continues to change after the stop, including elimination before the sample; the sample reports the state at its own time. (4) Independence gives $0.012+0.004=0.016\ (\mathrm{mg/L})^2$.