From First-Order Absorption to a Stochastic Amount Model
In a deterministic oral model, drug leaves a depot at a rate proportional to the depot amount, enters the central compartment according to bioavailability, and is eliminated from the central compartment. A stochastic model must say more than “add noise”: it must specify which physical or statistical component fluctuates, how the noise enters the amount equations, and whether related changes in the two compartments are linked.
Let \(D_t\) be the amount in the oral depot and \(C_t\) the central amount at time \(t\). Let \(k_a>0\) be the first-order absorption rate, \(k_e>0\) the first-order elimination rate, \(F\in(0,1]\) the bioavailability fraction, and \(V>0\) the central volume of distribution. The measured model concentration is \(C_t/V\). The parameters \(k_a\) and \(k_e\) have units of inverse time; the noise coefficients below have units of inverse square-root time.
The standard existence and uniqueness result for SDEs applies: the drift and diffusion coefficients here are globally Lipschitz and have at most linear growth. Thus, for each finite time horizon, the model has a unique strong solution with finite second moments. The equation is written in the Itô convention. In particular, its drift terms are the Itô drifts; replacing them with a Stratonovich equation would generally change those drift terms.
What the Noise Terms Represent
The term \(-\sigma_aD_t\,dW_{a,t}\) perturbs loss from the depot, while \(F\sigma_aD_t\,dW_{a,t}\) gives the corresponding perturbation to central input. Because the same Brownian motion appears in both equations, a fluctuation that increases depot loss also increases central input, and a fluctuation in the opposite direction does the reverse. This term represents uncertainty in the timing or size of transfer, not two unrelated disturbances.
The terms \(\sigma_DD_t\,dW_{D,t}\) and \(\sigma_CC_t\,dW_{C,t}\) instead represent compartment-specific process variation. They are not absorption or elimination rates. For example, \(\sigma_C\) allows central amount to fluctuate beyond what is implied by the specified absorption and elimination drifts. Its Brownian motion is independent of the other two in this formulation. The concentration trajectory is obtained by scaling the central amount path by \(1/V\); thus, the same amount fluctuation appears in concentration in proportion to \(1/V\).
Worked Example: Expected Concentration Under First-Order Absorption
Take a single dose with \(D_0=100\) mg and \(C_0=0\) mg, \(F=0.8\), \(k_a=0.8\) h\(^{-1}\), \(k_e=0.2\) h\(^{-1}\), and \(V=20\) L. Find the model’s expected concentration at 2 hours and the time of the peak of its expected concentration.
Taking expectations in the depot equation gives \(dE[D_t]/dt=-k_aE[D_t]\), because the Itô integral has mean zero. Hence \(E[D_t]=100e^{-0.8t}\). The central mean satisfies \(dE[C_t]/dt=Fk_aE[D_t]-k_eE[C_t]\), with \(E[C_0]=0\). Solving this linear equation gives $$ E[C_t]=\frac{Fk_aD_0}{k_a-k_e} \left(e^{-k_et}-e^{-k_at}\right). $$
At \(t=2\), \(Fk_aD_0/(k_a-k_e)=0.8(0.8)(100)/0.6=106.6667\) mg. Also, \(e^{-0.4}\approx0.67032\) and \(e^{-1.6}\approx0.20190\). Therefore \(E[C_2]\approx106.6667(0.67032-0.20190)=49.97\) mg and \(E[C_2/V]\approx49.97/20=2.50\) mg/L.
Differentiating the mean amount and setting its derivative to zero gives \(k_ae^{-k_at}=k_ee^{-k_et}\). Thus the peak time is \(\log(k_a/k_e)/(k_a-k_e)=\log(4)/0.6\approx2.31\) hours. This is the peak of the expected concentration curve; it does not assert that each stochastic path peaks at that time.
Two Useful Consequences of the Specification
The following results clarify what the model’s noise does and does not change. In both derivations, the initial amounts are fixed, and the model has finite second moments on finite time intervals as stated above.
Proof. Integrate the depot SDE from \(0\) to \(t\). Its Brownian integrals have expectation zero: the integrands are adapted and square-integrable on every finite interval by the finite-second-moment property. Therefore \(m_D(t)=E[D_t]\) satisfies \(m_D(t)=D_0-k_a\int_0^t m_D(s)\,ds\), or \(m_D'(t)=-k_am_D(t)\), with \(m_D(0)=D_0\). This yields \(m_D(t)=D_0e^{-k_at}\).
The same expectation argument in the central equation gives \(m_C'(t)=Fk_aD_0e^{-k_at}-k_em_C(t)\), with \(m_C(0)=C_0\). Multiplication by the integrating factor \(e^{k_et}\) and integration yield $$ m_C(t)=C_0e^{-k_et} +Fk_aD_0e^{-k_et}\int_0^t e^{(k_e-k_a)s}\,ds. $$ For \(k_a\ne k_e\), evaluating the integral gives the stated formula. For \(k_a=k_e=k\), the integral is \(t\), giving \(m_C(t)=e^{-kt}(C_0+FkD_0t)\). The result concerns means only: it does not imply that individual paths equal the deterministic solution. \(\square\)
Derivation. Since \(F\) is constant, \(dM_t=dD_t+dC_t/F\). Substitution gives the depot drift \(-k_aD_t\,dt\) and central input drift \(+k_aD_t\,dt\), which sum to zero. The paired Brownian terms are \(-\sigma_aD_t\,dW_{a,t}\) and \(+\sigma_aD_t\,dW_{a,t}\), which also sum to zero. The remaining central drift is \(-k_eC_t/F\,dt\), and the separate process-noise terms remain as shown. Thus the specified absorption fluctuation redistributes adjusted amount between compartments, whereas elimination reduces it and the separate process noise can change it. This is an algebraic balance for the model, not a claim that each noisy path is physically bounded below by zero. \(\square\)
Local Variability and Trajectory Interpretation
Quadratic variation gives a precise way to separate instantaneous noise contributions. For a short interval \(dt\), the diffusion coefficients determine the leading-order conditional variance and covariance increments. With the three Brownian motions independent, the results are $$ \begin{aligned} \frac{d[D]_t}{dt} &=D_t^2(\sigma_a^2+\sigma_D^2),\\ \frac{d[C]_t}{dt} &=F^2\sigma_a^2D_t^2+\sigma_C^2C_t^2,\\ \frac{d[D,C]_t}{dt} &=-F\sigma_a^2D_t^2. \end{aligned} $$
The negative cross-variation is the local signature of the paired absorption flux. A depot-only process noise term increases depot variation without directly adding central variation; central process noise increases central variation without directly adding depot variation. If the volume is fixed, concentration’s quadratic variation is \(1/V^2\) times that of \(C_t\). These rates describe local path variability, not the full variance at a later observation time, which also depends on how the drift propagates earlier fluctuations.
Worked Example: Separating Instantaneous Noise Contributions
Suppose \(D_t=60\) mg, \(C_t=30\) mg, \(F=0.8\), \(\sigma_a=0.15\), \(\sigma_D=0.10\), and \(\sigma_C=0.20\), with time measured in hours. Calculate the three local quadratic-variation rates.
For depot amount, \(D_t^2(\sigma_a^2+\sigma_D^2)=60^2(0.15^2+0.10^2)=3600(0.0325)=117\) mg\(^2\)/h.
For central amount, the absorption contribution is \((F\sigma_aD_t)^2=(0.8(0.15)(60))^2=7.2^2=51.84\) mg\(^2\)/h. The central process-noise contribution is \((\sigma_CC_t)^2=(0.20(30))^2=6^2=36\) mg\(^2\)/h. The total rate is therefore \(51.84+36=87.84\) mg\(^2\)/h.
The cross-variation rate is \(-F\sigma_a^2D_t^2=-0.8(0.15^2)(3600)=-64.8\) mg\(^2\)/h. Its negative sign confirms the opposing directions of the paired flux. These are instantaneous rates, not claims that a one-hour variance equals the listed rate.
Worked Example: One Euler–Maruyama Step
For illustration, take \(D_t=50\) mg, \(C_t=20\) mg, \(k_a=0.4\) h\(^{-1}\), \(k_e=0.1\) h\(^{-1}\), \(F=0.8\), and noise coefficients \(\sigma_a=0.2\), \(\sigma_D=0.1\), and \(\sigma_C=0.15\). Let \(\Delta t=0.25\) h and use the Brownian increments \(\Delta W_a=0.10\), \(\Delta W_D=-0.05\), and \(\Delta W_C=0.08\), each in units of \(\sqrt{\text{h}}\). Calculate one Euler–Maruyama update.
The depot drift contribution is \(-k_aD_t\Delta t=-0.4(50)(0.25)=-5\) mg. Its absorption-noise contribution is \(-\sigma_aD_t\Delta W_a=-0.2(50)(0.10)=-1\) mg; its separate process-noise contribution is \(\sigma_DD_t\Delta W_D=0.1(50)(-0.05)=-0.25\) mg. Thus \(D_{t+\Delta t}\approx50-5-1-0.25=43.75\) mg.
The central drift is \((Fk_aD_t-k_eC_t)\Delta t=(0.8(0.4)(50)-0.1(20))(0.25)=(16-2)(0.25)=3.5\) mg. The absorption-noise contribution is \(F\sigma_aD_t\Delta W_a=0.8(0.2)(50)(0.10)=0.8\) mg, and the central process-noise contribution is \(\sigma_CC_t\Delta W_C=0.15(20)(0.08)=0.24\) mg. Therefore \(C_{t+\Delta t}\approx20+3.5+0.8+0.24=24.54\) mg. With \(V=10\) L, this gives a one-step concentration of about \(2.454\) mg/L. This numerical update illustrates the terms; it is an approximation, not an exact finite-time transition.
Diagnostics, Alternatives, and Limitations
A useful diagnostic begins by checking whether simulated trajectories reflect the intended noise structure. The shared absorption increment should produce opposing adjusted-amount changes in \(D_t\) and \(C_t/F\); treating those increments as independent would remove the model’s paired-flux property. Check units, the signs in both equations, and whether the simulated quadratic-variation patterns agree with the coefficients. When comparing observed data with model predictions, distinguish process variability along the latent path from observation or residual error: changing the observation-error model does not repair a misspecified process model.
This formulation also has important limitations. Although \(D_t\) remains nonnegative from a positive initial value under this multiplicative depot equation, the central equation contains a Brownian absorption-flux increment that can be negative over a time step. Consequently, \(C_t\) is not guaranteed to remain nonnegative. A simulation with negative central amounts or concentrations is a warning that this unconstrained diffusion may be unsuitable for the intended application; silently truncating negative values changes the model. The paired Brownian flux is a local stochastic approximation, not a literal nonnegative random absorption rate at every instant.
The deterministic ODE is an alternative when unexplained path variation is not supported; it sets the process-noise coefficients to zero. Random effects on \(k_a\) or \(k_e\) are another alternative when variability is primarily between subjects or persists as a subject-specific parameter difference. They do not, by themselves, represent continuously fluctuating within-subject process noise. The present model is useful when the specific within-path variability and paired transfer structure are scientifically defensible. It is limited by its diffusion approximation, its fixed parameters and volume in this formulation, and its lack of guaranteed central-compartment positivity.
Check Your Understanding
Answer each question, then compare with the brief solutions.
- If \(\sigma_a=0\) but \(\sigma_D\) and \(\sigma_C\) are positive, which noise remains, and does the absorption-flux cancellation still apply?
- For \(F=0.5\) and \(\sigma_aD_t=2\) mg/\(\sqrt{\text{h}}\), what are the paired absorption diffusion coefficients in \(D_t\) and \(C_t\)?
- Why does the formula for the mean concentration not establish that a typical stochastic concentration path follows the deterministic curve?
- If \(k_a=k_e=k\), what is \(E[C_t]\) when \(C_0=0\)?
- What feature of the model is contradicted if the absorption-noise contribution increases \(D_t\) while decreasing \(C_t\) for a shared positive \(dW_a\) increment?
Solutions. (1) The depot and central process-noise terms remain. The paired absorption terms are both zero, so their cancellation is trivially retained. (2) The depot coefficient is \(-2\) mg/\(\sqrt{\text{h}}\), and the central coefficient is \(F(2)=1\) mg/\(\sqrt{\text{h}}\); a positive shared increment decreases depot amount and increases central amount. (3) An expectation describes an average over paths, while individual paths include Brownian fluctuations and can differ from that average. (4) The equal-rate limit is \(E[C_t]=F kD_0t e^{-kt}\). (5) It reverses the specified paired-flux signs: for a positive shared increment, the depot absorption term is negative and the central absorption term is positive.