First-Order Absorption With Multiplicative Noise

Why the Convention Matters for First-Order Absorption

Let $A_t$ denote the amount of drug in an absorption depot. Without process noise, first-order absorption is described by $dA_t=-k_aA_t\,dt$, where $k_a$ is the deterministic absorption rate. A state-proportional noise term makes fluctuations larger when more drug is present, but its effect on the drift depends on whether the stochastic differential equation is interpreted in the Ito or Stratonovich sense.

The conversion is especially easy to misread here: the Ito drift includes a positive correction, but the deterministic absorption coefficient in the Stratonovich equation remains $k_a$. The correction is not an extra physical absorption rate. It is the adjustment needed to express the same stochastic model using a different integration convention.

Definition: Consider a scalar depot amount $A_t$ satisfying the Stratonovich equation $$ dA_t=-k_aA_t\,dt+\sigma A_t\circ dW_t,\qquad A_0=a_0>0. $$ Here $W_t$ is standard Brownian motion, $k_a\geq 0$ is constant, and $\sigma$ is a constant noise coefficient. The symbol $\circ\,dW_t$ denotes Stratonovich integration. The corresponding Ito equation has drift $\left(-k_a+\tfrac12\sigma^2\right)A_t$.

Assume the Brownian motion is defined on a filtered probability space satisfying the usual conditions, and that $a_0$, $k_a$, and $\sigma$ are deterministic. The coefficients are linear and globally Lipschitz, so the equation has a unique global solution. If time is measured in days, $k_a$ has units day$^{-1}$, while $\sigma$ has units day$^{-1/2}$; thus $\sigma^2/2$ has the same units as $k_a$.

Deriving the Conversion for This Model

For a continuous semimartingale integrand $H_t$, the Stratonovich-to-Ito integral identity is $\int H_t\circ dW_t=\int H_t\,dW_t+\tfrac12[H,W]_t$, where brackets denote quadratic covariation. Apply it with $H_t=\sigma A_t$. In an Ito equation with diffusion coefficient $\sigma A_t$, the state and Brownian motion have covariation $d[A,W]_t=\sigma A_t\,dt$: the finite-variation drift contributes no quadratic covariation. Therefore $d[\sigma A,W]_t=\sigma^2A_t\,dt$.

Theorem: Under the assumptions above, the Stratonovich first-order absorption equation and its Ito form describe the same process, and are related by $$ dA_t=\left(-k_a+\frac{\sigma^2}{2}\right)A_t\,dt+\sigma A_t\,dW_t. $$ More generally, for scalar coefficients $b$ and $g$ with sufficient differentiability and regularity for the conversion formula, $dX_t=b(X_t)\,dt+g(X_t)\circ dW_t$ converts to Ito drift $b(x)+\tfrac12g(x)g'(x)$.

For completeness, in this model the conversion follows directly from the bracket identity. The Stratonovich diffusion integral contributes its Ito integral plus half the covariation just calculated. Adding that finite-variation term to the original drift gives

-k_aA_t\,dt+\sigma A_t\circ dW_t = \left(-k_aA_t+\frac12\sigma^2A_t\right)dt +\sigma A_t\,dW_t.

The same correction follows from the general formula with $b(x)=-k_ax$, $g(x)=\sigma x$, and $g'(x)=\sigma$. Hence $\tfrac12g(x)g'(x)=\tfrac12\sigma^2x$, proving the stated conversion. $\square$

Worked Example: Converting and Summarizing a Stratonovich Absorption Model

Suppose $A_0=50$ mg, $k_a=0.36$ day$^{-1}$, and $\sigma=0.40$ day$^{-1/2}$. Write the Ito equation and find the mean and median amount after one day under the Stratonovich model.

The correction is $$ \frac12\sigma^2=\frac12(0.40)^2=0.08\ \text{day}^{-1}. $$ Thus the Ito drift coefficient is $-0.36+0.08=-0.28$ day$^{-1}$, and the equivalent Ito equation is $dA_t=-0.28A_t\,dt+0.40A_t\,dW_t$.

To obtain the distribution, apply the Stratonovich chain rule to $\log A_t$, which is valid because the solution stays positive: $$ d\log A_t=-0.36\,dt+0.40\circ dW_t. $$ Consequently, $\log A_1=\log(50)-0.36+0.40W_1$, and $W_1\sim N(0,1)$. The median is the exponential of the median of this normal log-amount: $$ \operatorname{Median}(A_1)=50e^{-0.36}\approx 34.88\ \text{mg}. $$

For a normal random variable $Z\sim N(0,1)$, $E[e^{cZ}]=e^{c^2/2}$. Therefore $$ E[A_1]=50e^{-0.36}E[e^{0.40W_1}] =50e^{-0.36}e^{(0.40)^2/2} =50e^{-0.28}\approx 37.79\ \text{mg}. $$ The mean exceeds the median because the lognormal distribution is right-skewed. The two values are checked using $e^{-0.36}\approx0.697676$ and $e^{-0.28}\approx0.755784$.

Solution, Positivity, and What the Drift Means

The Stratonovich chain rule also gives the solution at any $t\geq0$: $$ A_t=a_0\exp\left(-k_at+\sigma W_t\right). $$ It is positive for every finite $t$, almost surely. This formula separates the deterministic log-scale absorption term, $-k_at$, from the random Brownian term. It also permits direct calculation of the distribution: $\log A_t\sim N(\log a_0-k_at,\sigma^2t)$.

Result: For $t>0$, the Stratonovich model has $$ \operatorname{Median}(A_t)=a_0e^{-k_at}, \qquad E[A_t]=a_0e^{(-k_a+\sigma^2/2)t}. $$ The median follows from the median of a normal log-amount. The mean follows from $E[e^{\sigma W_t}]=e^{\sigma^2t/2}$, since $W_t\sim N(0,t)$.

This result shows why the Ito drift coefficient should not be read in isolation as the deterministic absorption rate. In the equivalent Ito equation, the coefficient is $-k_a+\sigma^2/2$, and it can even be positive. Yet the Stratonovich model still has a negative deterministic drift $-k_aA_t$; the positive term in its Ito representation compensates for the difference between the conventions. The mean can increase when $\sigma^2/2>k_a$, even while the median declines for $k_a>0$.

Worked Example: A Positive Ito Drift With a Declining Median

Let $A_0=100$ mg, $k_a=0.05$ day$^{-1}$, and $\sigma=0.40$ day$^{-1/2}$. In the Ito representation, the drift coefficient is $$ -k_a+\frac12\sigma^2=-0.05+\frac12(0.40)^2=-0.05+0.08=0.03\ \text{day}^{-1}. $$ It is positive, but the equivalent Stratonovich model still has deterministic absorption coefficient $k_a=0.05$ day$^{-1}$.

After one day, the Stratonovich model’s median is $100e^{-0.05}\approx95.12$ mg, which is below the initial amount. Its mean is $100e^{(-0.05+0.08)}=100e^{0.03}\approx103.05$ mg, which is above the initial amount. These calculations are consistent: a declining median does not imply a declining mean when the distribution is sufficiently right-skewed.

Converting in the Other Direction and Comparing Specifications

A model written in Ito form must be converted in the opposite direction if it is to be expressed as a Stratonovich equation. For diffusion coefficient $g(x)=\sigma x$, subtract $\tfrac12g(x)g'(x)=\tfrac12\sigma^2x$ from the Ito drift. For example, the Ito equation $dA_t=-0.20A_t\,dt+0.30A_t\,dW_t$ is equivalent to the Stratonovich equation with drift coefficient $$ -0.20-\frac12(0.30)^2=-0.20-0.045=-0.245\ \text{day}^{-1}. $$ It is not equivalent to a Stratonovich drift coefficient of $-0.20$.

Specification	Drift coefficient	Meaning in this comparison
Stratonovich absorption model	$-k_a$	The stated deterministic log-scale absorption coefficient
Equivalent Ito equation	$-k_a+\sigma^2/2$	The drift after accounting for the convention conversion
Deterministic model, $\sigma=0$	$-k_a$	Ordinary first-order absorption; no correction is present

These are alternative representations, not three models to combine. If a model is specified as Ito with drift $-k_aA_t$, that is a different stochastic model from a Stratonovich equation with drift $-k_aA_t$ whenever $\sigma\ne0$. In the Ito-specified model, the mean is $a_0e^{-k_at}$; in the Stratonovich-specified model, it is $a_0e^{(-k_a+\sigma^2/2)t}$. A deterministic model is a useful reference but omits random path variation entirely.

Diagnostics, Pitfalls, and Limitations

When checking a conversion, verify the diffusion derivative and the sign of the adjustment. Here $g(x)=\sigma x$, so $g'(x)=\sigma$ and the Stratonovich-to-Ito correction is positive, $+\sigma^2x/2$. Check units as well: a correction proportional to $\sigma^2$, not $\sigma$, has units of a drift rate. Finally, verify that $\sigma=0$ removes the correction and returns the deterministic first-order model.

Common mistake: treating the correction as a changed absorption rate. The reasoning “the Ito drift is $-k_a+\sigma^2/2$, so the physical absorption rate is $k_a-\sigma^2/2$” confuses a convention-dependent drift with the parameter in the Stratonovich specification. Keep the stated $k_a$ distinct from the Ito drift coefficient.
Common mistake: forgetting which direction is being converted. Stratonovich to Ito adds $\sigma^2x/2$; Ito to Stratonovich subtracts it. Applying the same sign in both directions will not preserve the model.
Diagnostic: compare distributions, not just drift labels. For equivalent specifications, the explicit solution and resulting mean and median must agree. A mismatch usually indicates an incorrect sign or a comparison of different models.

This model is deliberately narrow: it assumes constant $k_a$, constant proportional noise, a single Brownian driver, and no dosing input or transfer to a central compartment. Its positivity follows from the explicit exponential solution, but its random paths can increase, and the model does not enforce a deterministic mass-balance interpretation of every fluctuation as absorption. Whether the noise represents the intended biological variability is a modeling decision, not a consequence of the conversion formula.

Worked Example: Checking Whether Two Equations Are Equivalent

Compare the Stratonovich equation $dA_t=-0.20A_t\,dt+0.30A_t\circ dW_t$ with the Ito equation $dA_t=-0.20A_t\,dt+0.30A_t\,dW_t$. They have identical written drift coefficients. Are they equivalent?

For the Stratonovich equation, conversion gives Ito drift $$ -0.20+\frac12(0.30)^2=-0.20+0.045=-0.155\ \text{day}^{-1}. $$ This is not the drift coefficient $-0.20$ in the proposed Ito equation, so the equations are not equivalent.

Their means after one day, starting from $A_0=a_0$, confirm the difference. The Stratonovich model has mean $a_0e^{-0.155}$; the Ito model with drift $-0.20A_t$ has mean $a_0e^{-0.20}$. Since these are unequal for $a_0>0$, equal-looking drift labels under different conventions do not define the same model.

Check Your Understanding

Try each question, then check the worked solutions below.

For $dA_t=-0.30A_t\,dt+0.20A_t\circ dW_t$, what is the equivalent Ito drift coefficient?
For a Stratonovich model with $A_0=20$ mg, $k_a=0.10$ day$^{-1}$, and $\sigma=0.20$ day$^{-1/2}$, find the median and mean after one day.
An Ito model has drift coefficient $-0.12$ day$^{-1}$ and diffusion coefficient $0.50A_t$. What is its equivalent Stratonovich drift coefficient?
In a Stratonovich model with $k_a>0$, can the one-time mean increase while the median decreases? State the condition and explain.

Solutions. (1) The correction is $\tfrac12(0.20)^2=0.02$ day$^{-1}$, so the Ito coefficient is $-0.30+0.02=-0.28$ day$^{-1}$. (2) The median is $20e^{-0.10}\approx18.10$ mg. The mean is $20e^{-0.10+0.02}=20e^{-0.08}\approx18.46$ mg. (3) Subtract the correction $\tfrac12(0.50)^2=0.125$ day$^{-1}$ from the Ito coefficient: the Stratonovich coefficient is $-0.12-0.125=-0.245$ day$^{-1}$. (4) Yes. For $k_a>0$, the median $a_0e^{-k_at}$ decreases; the mean $a_0e^{(-k_a+\sigma^2/2)t}$ increases when $\sigma^2/2>k_a$. This is due to the right-skewed distribution, not a reversal of the median trend.

Specification	Drift coefficient	Meaning in this comparison
Stratonovich absorption model	\(-k_a\)	The stated deterministic log-scale absorption coefficient
Equivalent Ito equation	\(-k_a+\sigma^2/2\)	The drift after accounting for the convention conversion
Deterministic model, \(\sigma=0\)	\(-k_a\)	Ordinary first-order absorption; no correction is present