Stochastic Absorption Rates Versus Depot Noise

Two Different Places to Put the Noise

A first-order absorption model removes drug from a depot in proportion to the amount currently there. A stochastic extension can perturb the absorption flux, or it can add process noise directly to the depot amount. Those choices can produce similar-looking depot equations, but they describe different mechanisms: one fluctuates the transfer between compartments, while the other adds or removes depot amount without necessarily transferring it to the central compartment.

Let $D_t$ be the depot amount and $C_t$ the central amount at time $t$. Let $k_a>0$ be the first-order absorption constant and $k_e\geq 0$ the first-order elimination constant. Assume an Itô formulation, deterministic initial amounts $D_0,C_0\geq 0$, and Brownian drivers adapted to the model filtration. The stochastic integrands below are assumed square-integrable on each finite time interval. A dose can be represented by setting $D_0$ to the dose amount.

Definition: In the stochastic absorption-flux model, the absorption flux has a deterministic first-order component and a diffusion component: $$ dJ_t=k_aD_t\,dt+\sigma_aD_t\,dW_t. $$ The compartment equations are $$ dD_t=-dJ_t,\qquad dC_t=dJ_t-k_eC_t\,dt. $$ Here $\sigma_a\geq 0$ controls the size of the flux perturbation, and the same Brownian motion $W_t$ appears with opposite signs in the two compartments.

The shared Brownian term is essential to this flux interpretation: a random amount leaving the depot through the stochastic part of $dJ_t$ enters the central compartment through that same part. The resulting equations are

\begin{aligned} dD_t&=-k_aD_t\,dt-\sigma_aD_t\,dW_t,\\ dC_t&=(k_aD_t-k_eC_t)\,dt+\sigma_aD_t\,dW_t. \end{aligned}

The sign of the Brownian term in the depot equation is a convention: replacing $W_t$ by $-W_t$ does not change its distribution. What matters is that the central equation uses the opposite sign to the depot equation for the same driver.

Definition: In the additive depot-noise model, diffusion is added directly to the depot amount: $$ \begin{aligned} dD_t&=-k_aD_t\,dt+\eta\,dW_t^{D},\\ dC_t&=(k_aD_t-k_eC_t)\,dt. \end{aligned} $$ The coefficient $\eta\geq 0$ has units of amount divided by the square root of time. This specification does not, by itself, transfer the depot noise to the central compartment.

What the Depot Equations Imply

The first-order drift is the same in both models, but the diffusion coefficients differ: the flux model has a coefficient proportional to the current depot amount, whereas additive depot noise has a constant coefficient. The following calculations make the consequences explicit. They use the standard Itô integral mean and isometry properties.

Result: For the stochastic absorption-flux model, $$ E[D_t]=D_0e^{-k_at},\qquad \operatorname{Var}(D_t)=D_0^2e^{-2k_at}\bigl(e^{\sigma_a^2t}-1\bigr). $$ Moreover, $D_t>0$ almost surely for all finite $t$ when $D_0>0$.

Derivation. The depot equation is $dD_t=-k_aD_t\,dt-\sigma_aD_t\,dW_t$. Applying Itô's formula to $\log D_t$, up to any time at which $D_t$ might reach zero, gives

d\log D_t =\left(-k_a-\frac{\sigma_a^2}{2}\right)dt-\sigma_a\,dW_t.

Integrating yields the geometric solution

D_t=D_0\exp\left[-\left(k_a+\frac{\sigma_a^2}{2}\right)t-\sigma_aW_t\right].

This expression is finite and strictly positive at every finite $t$ when $D_0>0$, so it also confirms that zero is not reached. Since $W_t\sim N(0,t)$, the normal moment-generating function gives

\begin{aligned} E[D_t] &=D_0e^{-(k_a+\sigma_a^2/2)t}E[e^{-\sigma_aW_t}] =D_0e^{-(k_a+\sigma_a^2/2)t}e^{\sigma_a^2t/2} =D_0e^{-k_at},\\ E[D_t^2] &=D_0^2e^{-2(k_a+\sigma_a^2/2)t}E[e^{-2\sigma_aW_t}] =D_0^2e^{-2k_at+\sigma_a^2t}. \end{aligned}

Subtracting the squared mean from the second moment gives the stated variance. The random amount remains positive, but its path need not decrease at every instant: the Brownian component can temporarily outweigh the deterministic absorption drift.

Result: For the additive depot-noise model, $$ E[D_t]=D_0e^{-k_at},\qquad \operatorname{Var}(D_t)=\frac{\eta^2}{2k_a}\left(1-e^{-2k_at}\right). $$ For $\eta>0$ and $t>0$, $D_t$ has a normal distribution and therefore has positive probability of being negative.

Derivation. Multiply the depot equation by the integrating factor $e^{k_at}$. The Itô product rule gives $d(e^{k_at}D_t)=\eta e^{k_at}dW_t^D$, and hence

D_t=D_0e^{-k_at}+\eta\int_0^t e^{-k_a(t-s)}\,dW_s^D.

The stochastic integral has mean zero. By the Itô isometry its variance is

\eta^2\int_0^t e^{-2k_a(t-s)}\,ds =\frac{\eta^2}{2k_a}\left(1-e^{-2k_at}\right).

The integrand is deterministic, so the integral is normally distributed. Its variance is strictly positive for $t>0$ and $\eta>0$; a nondegenerate normal distribution assigns positive probability to every interval, including intervals below zero. This proves the positivity warning, not just a possibility suggested by a simulation. $\square$

Worked Examples

Worked Example: Variability Under a Noisy Absorption Flux

Take $D_0=100$ mg, $k_a=0.8$ h$^{-1}$, and $\sigma_a=0.3$ h$^{-1/2}$. Find the depot mean and variance after one hour under the flux model.

The derived formulas give

E[D_1]=100e^{-0.8}=44.93\text{ mg}.

For the variance,

\begin{aligned} \operatorname{Var}(D_1) &=100^2e^{-1.6}(e^{0.09}-1)\\ &\approx 10000(0.20190)(0.09417) \approx 190.1\text{ mg}^2. \end{aligned}

The standard deviation is approximately $\sqrt{190.1}=13.79$ mg. The mean is unchanged from deterministic first-order absorption, but the stochastic rate perturbation creates variability that is proportional to the initial depot amount.

Worked Example: Negative Depot Amounts Under Additive Noise

Let $D_0=1$ mg, $k_a=1$ h$^{-1}$, and $\eta=0.5$ mg h$^{-1/2}$. At $t=1$ h, the depot mean is

E[D_1]=e^{-1}=0.3679\text{ mg}.

The variance and standard deviation are

\begin{aligned} \operatorname{Var}(D_1) &=\frac{0.5^2}{2}(1-e^{-2}) \approx 0.1081\text{ mg}^2,\\ \operatorname{SD}(D_1)&\approx \sqrt{0.1081}=0.3288\text{ mg}. \end{aligned}

Because $D_1$ is normal, its probability of being negative is

P(D_1<0) =\Phi\left(\frac{0-0.3679}{0.3288}\right) =\Phi(-1.119)\approx 0.132,

where $\Phi$ is the standard normal distribution function. Thus this specification assigns about a 13.2% probability to a negative depot amount at one hour. That may be unacceptable when $D_t$ represents a physical drug amount.

Worked Example: Checking Whether Noise Represents Transfer

Suppose the current state is $D_t=80$ mg and $C_t=30$ mg, with $k_a=0.5$ h$^{-1}$, $k_e=0.2$ h$^{-1}$, and $\sigma_a=0.1$ h$^{-1/2}$. In the flux model, the deterministic depot drift is $-0.5(80)=-40$ mg/h, and its diffusion coefficient is $-0.1(80)=-8$ mg h$^{-1/2}$. The central drift is $0.5(80)-0.2(30)=34$ mg/h, and its diffusion coefficient is $+8$ mg h$^{-1/2}$.

Adding the two compartment equations cancels both absorption terms:

d(D_t+C_t)=-k_eC_t\,dt.

Thus the stochastic absorption flux changes where the amount is, but contributes no net random change to the combined amount. Now compare an additive depot model with $\eta=8$ mg h$^{-1/2}$ at the same state. Its depot diffusion coefficient is $+8$, and its central diffusion coefficient is zero. The combined amount then has diffusion coefficient $+8$, so the noise is not a transfer from depot to central compartment. Matching the instantaneous depot noise size does not make the two formulations equivalent.

Mass Balance, Assumptions, and Diagnostics

The flux formulation preserves absorption as an internal transfer: adding its compartment equations gives $d(D_t+C_t)=-k_eC_t\,dt$, with no Brownian term. This cancellation is exact because the same driver and equal-and-opposite diffusion coefficients were specified. In the additive formulation, adding the equations instead gives

d(D_t+C_t)=-k_eC_t\,dt+\eta\,dW_t^D.

That can be a reasonable model if the depot is subject to an external, unobserved input or loss, but it should not be labelled a fluctuating absorption flux unless the corresponding transfer into the central compartment is also specified. The flux model's diffusion is proportional to $D_t$; additive depot diffusion has a constant size even as the depot approaches zero.

A white-noise perturbation of a rate also needs careful interpretation. In the flux model, $k_a\,dt+\sigma_a\,dW_t$ is a diffusion approximation for fluctuating flux, not a nonnegative instantaneous rate: Brownian increments can have either sign. If a model instead requires a genuinely nonnegative random rate $K_t$ integrated over time, then for a specified nonnegative, integrable rate path the depot obeys $dD_t=-K_tD_t\,dt$ and $D_t=D_0\exp(-\int_0^tK_s\,ds)$. That random-rate model is not the same as the white-noise SDE, and it has no Itô correction of the kind appearing in the geometric solution.

Useful checks when choosing or fitting a formulation include whether the depot can become negative, whether total amount changes randomly in the absence of elimination, and whether the fitted diffusion magnitude is expected to scale with the amount present. Simulated paths that cross below zero, or a model that implies random total drug creation when only absorption is intended, are signs to revisit the specification. Constraining or transforming a state to enforce positivity changes the model and should not be treated as a harmless numerical repair.

The alternatives therefore answer different questions. Use a stochastic absorption flux when uncertainty represents fluctuating transfer and the same random transfer should enter both compartments. Use additive depot diffusion when the hypothesized uncertainty acts directly on the depot amount. If a nonnegative random absorption rate is scientifically essential, specify that rate process explicitly rather than interpreting Brownian flux noise as a positive rate.

Common pitfall: Writing $dD_t=-k_aD_t\,dt+\eta\,dW_t$ and calling $\eta\,dW_t$ “noise in the absorption rate” is generally incorrect. It adds amount to or removes amount from the depot, independently of the current depot amount; it does not specify a transfer to the central compartment. To model stochastic transfer, put the diffusion in a shared flux with opposite signs in the depot and central equations.

Check Your Understanding

Try each question, then compare with the brief solutions.

In the flux model, what is the diffusion coefficient in the equation for $C_t$ if the depot diffusion coefficient is $-\sigma_aD_t$?
For the additive model with $k_a=2$ h$^{-1}$, $\eta=0.4$ mg h$^{-1/2}$, and $t=0.5$ h, calculate $\operatorname{Var}(D_t)$.
Why does the stochastic flux cancel from the equation for $D_t+C_t$, while additive depot noise does not?
Does a Brownian perturbation $\sigma_a\,dW_t$ define a nonnegative instantaneous absorption rate? Explain.

Solutions. (1) It is $+\sigma_aD_t$, using the same Brownian driver, so the stochastic transfer has opposite signs in the two compartments. (2) The variance is $\frac{0.4^2}{2(2)}(1-e^{-2(2)(0.5)})=0.04(1-e^{-2})\approx0.0346$ mg$^2$. (3) The flux model uses equal-and-opposite noise terms for one internal transfer; additive depot noise has no matching central term. (4) No. Brownian increments can be positive or negative, so this is a diffusion approximation for flux fluctuations, not a nonnegative instantaneous rate.

Stochastic Absorption Rates Versus Depot Noise

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