PK State, Drift, Diffusion, And Wiener Notation

Writing the State Equation Clearly

A pharmacokinetic SDE is easier to interpret when its components answer four separate questions: what quantities form the state, what is their deterministic rate of change, how does random variation enter, and how many Wiener drivers are used? These questions are connected, but their answers need not have the same dimension. A model can track several amounts while using just one Wiener driver, or use several drivers to affect a single state.

We use the Ito convention selected in the earlier material on Ito versus Stratonovich interpretations. The equations below describe continuous evolution on an interval without a dosing jump. The state may contain amounts, concentrations, or a mixture of explicitly defined PK quantities.

Definition: Let $X(t)\in\mathbb{R}^d$ be a column vector of $d$ PK states, and let $W(t)\in\mathbb{R}^m$ be a standard $m$-dimensional Wiener process. An Ito SDE for the state is $$ dX(t)=b(t,X(t))\,dt+G(t,X(t))\,dW(t). $$ Here $b(t,x)\in\mathbb{R}^d$ is the drift vector, and $G(t,x)\in\mathbb{R}^{d\times m}$ is the diffusion matrix. The state dimension is $d$; the number of Wiener drivers is $m$.

The dimensions follow from matrix multiplication. The increment $dW(t)$ is an $m\times1$ column, so $G\,dW$ is $d\times1$, matching $dX$. The drift $b\,dt$ must also be $d\times1$. If $W_j$ is the $j$-th Wiener driver, then column $j$ of $G$ gives the contribution of that driver to each state. In components,

dX_i(t)=b_i(t,X(t))\,dt+\sum_{j=1}^{m}G_{ij}(t,X(t))\,dW_j(t), \qquad i=1,\ldots,d.

When time is measured in hours, $dt$ has units of hours and $dW_j$ has units of the square root of hours. Thus, if state $X_i$ is measured in mg, $b_i$ has units mg/hour and $G_{ij}$ has units mg/$\sqrt{\text{hour}}$. A Wiener driver is not a dose, an amount, or a random rate: it is the stochastic input whose increments are scaled by the diffusion coefficients.

From Coefficients to Local State Changes

The drift describes the leading short-time mean change, while the diffusion matrix determines the leading short-time covariance. This distinction is especially useful when checking whether a proposed SDE has been written consistently.

Theorem: Suppose $b(t,x)$ and $G(t,x)$ are continuous in $(t,x)$, globally Lipschitz in $x$, and have at most linear growth. Let the solution start from the nonrandom state $X(t)=x$, with finite second moments. As $h$ decreases to zero, $$ E[X(t+h)-x]=b(t,x)h+o(h), \qquad \operatorname{Cov}(X(t+h)-x)=G(t,x)G(t,x)^{T}h+o(h). $$

Proof. Write the increment as the sum of a drift integral and a stochastic integral:

X(t+h)-x=D_h+M_h, \quad D_h=\int_t^{t+h}b(s,X(s))\,ds, \quad M_h=\int_t^{t+h}G(s,X(s))\,dW(s).

The Ito integral has conditional mean zero given the information at time $t$, so $E[M_h]=0$. Continuity of the coefficients and of the solution in mean square at the initial time gives $E[b(s,X(s))]=b(t,x)+o(1)$ as $s$ decreases to $t$. Integrating over an interval of length $h$ yields $E[D_h]=b(t,x)h+o(h)$, proving the mean statement.

For the covariance, the Ito isometry, applied componentwise, gives

E[M_hM_h^T]=\int_t^{t+h}E[G(s,X(s))G(s,X(s))^T]\,ds =G(t,x)G(t,x)^T h+o(h).

The drift integral has second moment of order $h^2$, because its integrand has bounded second moments on a sufficiently short interval. Its covariance is therefore $O(h^2)$. By the Cauchy–Schwarz inequality, the cross-covariance between $D_h$ and $M_h$ is $O(h^{3/2})$: the respective second moments are $O(h^2)$ and $O(h)$. Both terms are $o(h)$. Subtracting the mean affects the covariance only through terms of order $h^2$. The leading covariance is therefore the one displayed above. $\square$

The matrix $GG^T$, not $G$ itself, is the instantaneous covariance rate of the state increments. Its dimensions are $d\times d$. The diagonal entries give the leading variance rates of individual states; the off-diagonal entries give their leading covariance rates.

Worked PK Notation

Worked Example: One Amount State With Multiplicative Diffusion

Let $A(t)$ be an amount in mg, and suppose, between doses, that

dA(t)=-kA(t)\,dt+\sigma A(t)\,dW(t), \qquad k=0.15\ \text{hour}^{-1}, \quad \sigma=0.20\ \text{hour}^{-1/2}.

There is one state, so $d=1$, and one Wiener driver, so $m=1$. The state vector is $X=A$, the drift vector is the scalar $b(A)=-0.15A$, and the diffusion matrix is the $1\times1$ matrix $G(A)=0.20A$.

At $A=40$ mg, the drift is $-0.15(40)=-6$ mg/hour. The diffusion coefficient is $0.20(40)=8$ mg/$\sqrt{\text{hour}}$. Therefore, for a short interval of length $h$ hours, the leading mean change is $-6h$ mg, and the leading variance is $G^2h=64h$ mg$^2$. The units check: $(\text{mg}/\sqrt{\text{hour}})^2\text{hour}=\text{mg}^2$.

The negative drift represents the model’s mean elimination trend at this state. The diffusion coefficient scales random fluctuations with the current amount. It does not mean the amount changes by 8 mg every hour; the corresponding short-time standard deviation is approximately $8\sqrt{h}$ mg.

Worked Example: Converting an Amount State to Concentration

Suppose a one-compartment amount follows $dA=-0.15A\,dt+0.20A\,dW$, with $A$ in mg and time in hours. Let the fixed volume be $V=20$ L and define concentration $C=A/V$, measured in mg/L.

Because $V$ is constant, $C=A/20$ and therefore $dC=dA/20$. Substitution gives

dC=-0.15C\,dt+0.20C\,dW.

For example, $A=40$ mg corresponds to $C=40/20=2$ mg/L. The concentration drift is $-0.15(2)=-0.30$ mg/(L·hour), and the diffusion coefficient is $0.20(2)=0.40$ mg/(L·$\sqrt{\text{hour}}$). Equivalently, dividing the amount coefficients $-6$ mg/hour and $8$ mg/$\sqrt{\text{hour}}$ by 20 L gives those same values. A constant change of units rescales the state, drift, and diffusion together.

Worked Example: Two States and One Wiener Driver

Consider two amount states $X=(A_1,A_2)^T$, each in mg, and one Wiener driver $W$. For illustration, let

b(A_1,A_2)= \begin{pmatrix} -0.2A_1+0.1A_2\\ 0.2A_1-0.3A_2 \end{pmatrix}, \qquad G(A_1,A_2)= \begin{pmatrix} 0.1A_1\\ 0.08A_2 \end{pmatrix}.

Here $d=2$ and $m=1$: $b$ is $2\times1$, $G$ is $2\times1$, and $dW$ is $1\times1$. At $A_1=10$ mg and $A_2=5$ mg, the drift is

b= \begin{pmatrix} -0.2(10)+0.1(5)\\ 0.2(10)-0.3(5) \end{pmatrix} = \begin{pmatrix} -1.5\\ 0.5 \end{pmatrix} \text{ mg/hour}, \qquad G= \begin{pmatrix} 1\\ 0.4 \end{pmatrix} \text{ mg}/\sqrt{\text{hour}}.

The state covariance rate is

GG^T= \begin{pmatrix} 1\\ 0.4 \end{pmatrix} \begin{pmatrix} 1&0.4 \end{pmatrix} = \begin{pmatrix} 1&0.4\\ 0.4&0.16 \end{pmatrix} \text{ mg}^2/\text{hour}.

Thus, over a short interval $h$, the leading variances are $h$ and $0.16h$ mg$^2$, and the leading covariance is $0.4h$ mg$^2$. The example emphasizes dimensions: two states do not require two Wiener drivers.

Fixed Changes of State Coordinates

A linear change of coordinates offers a systematic way to check how the notation changes when amounts are replaced by concentrations. Let $Y=RX$, where $R$ is a fixed matrix with $q$ rows and $d$ columns. Then $Y$ has $q$ states, while the number of Wiener drivers remains $m$.

Theorem: If $dX=b\,dt+G\,dW$ and $Y=RX$ for a fixed matrix $R$, then $$ dY=Rb\,dt+RG\,dW. $$ If $R$ is invertible, this is a change of coordinates without loss of state information.

Proof. Since $R$ is constant, the differential of the linear map $Y=RX$ is $dY=R\,dX$. Substituting the SDE gives $dY=R(b\,dt+G\,dW)=(Rb)\,dt+(RG)\,dW$. The new drift has dimension $q\times1$, and the new diffusion has dimension $q\times m$, as required. No additional term appears because the transformation is linear and time-independent. $\square$

For $d$ amount states and a fixed diagonal volume matrix $D$, the concentration vector is $C=D^{-1}A$. Taking $R=D^{-1}$ gives $b_C=D^{-1}b_A$ and $G_C=D^{-1}G_A$. This statement assumes fixed volumes. If the transformation depends nonlinearly on the state or on time, the differential calculation requires additional care; the linear rule above should not be applied automatically.

Reading and Checking an SDE

A scalar equation can be a compact representation when there is one state and one driver. A vector equation is usually clearer when several states or drivers are present, because it makes the dimensions visible. Neither notation changes the model by itself. Whichever form is used, each state equation must identify its drift and every driver that contributes to it.

Use these checks before interpreting or implementing a model:

Check dimensions. If there are $d$ states and $m$ drivers, confirm that $b$ is $d\times1$, $G$ is $d\times m$, and $dW$ is $m\times1$.
Check units. For a state measured in mg and time in hours, drift entries must be mg/hour and diffusion entries mg/$\sqrt{\text{hour}}$.
Check the covariance scale. Compute $GG^T$ when interpreting local variances and covariances. Do not treat entries of $G$ as variances.
Check the stated state. An equation for amount is not automatically an equation for concentration; conversion requires the volume and its assumptions to be specified.

A common mistake is to give $G$ the same dimensions as $GG^T$, or to label a diffusion entry as a variance. For example, in the one-state example, $G=8$ mg/$\sqrt{\text{hour}}$, while the variance rate is $G^2=64$ mg$^2$/hour. The correction is to identify first whether a quantity multiplies $dW$ or describes the covariance of the resulting state increment.

The coefficient matrix $G$ is not always uniquely determined by a specified covariance rate: different matrix factors can produce the same $GG^T$. Therefore, a written model should state its Wiener drivers and its chosen diffusion matrix, rather than only reporting a covariance rate. The notation also does not guarantee that amounts or concentrations remain nonnegative, that coefficients are biologically plausible, or that a numerical approximation preserves those properties. Those are separate modeling and computational checks.

Check Your Understanding

Answer each question, then compare with the brief solutions.

A model has three PK states and two Wiener drivers. What are the dimensions of $b$, $G$, and $dW$?
A diffusion coefficient is $0.5$ mg/$\sqrt{\text{hour}}$. What is its contribution to the local variance over an interval of length $h$ hours?
If $C=A/V$ for fixed $V$, how do the drift and diffusion coefficients transform?
Why does a two-state model not necessarily require two Wiener drivers?

Solutions. (1) $b$ is $3\times1$, $G$ is $3\times2$, and $dW$ is $2\times1$. (2) Squaring the coefficient and multiplying by $h$ gives $0.25h$ mg$^2$. (3) Divide both coefficients by $V$, so the concentration drift and diffusion are the amount drift and diffusion scaled by $1/V$. (4) State dimension and driver count are distinct; the diffusion matrix can have two rows and one column, for example.