Independent Noise For Multiple Compartments

Independent Drivers and Compartment Noise

A multi-compartment model can represent random fluctuations in each compartment with its own Wiener driver. This is a modeling choice about the sources of continuous process noise: it says that the driving Brownian motions have independent increments and zero cross-variation. It does not, by itself, say that compartment amounts are independent. Their deterministic drift may couple them, and their diffusion coefficients may depend on the whole state.

Let $X_t=(X_{1,t},\ldots,X_{m,t})^\mathsf{T}$ be the vector of compartment states. Let $W_t=(W_{1,t},\ldots,W_{m,t})^\mathsf{T}$ consist of independent standard Wiener processes, Brownian relative to the filtration used for the model. A convenient specification assigns the $i$th state its own driver $W_i$:

dX_{i,t}=b_i(t,X_t)\,dt+g_i(t,X_t)\,dW_{i,t},\qquad i=1,\ldots,m.

In vector notation, the diffusion matrix is diagonal:

dX_t=b(t,X_t)\,dt+G(t,X_t)\,dW_t,\qquad G(t,x)=\operatorname{diag}\bigl(g_1(t,x),\ldots,g_m(t,x)\bigr).

The coefficient $g_i(t,x)$ is allowed to depend on every component of $x$. “Diagonal” specifies which Wiener increment directly enters each equation; it does not require each equation's coefficient to depend only on its own state. For a well-defined strong solution, a standard sufficient set of assumptions is that $b$ and $G$ are measurable in time, globally Lipschitz in the state uniformly over the time interval, and satisfy a uniform linear-growth bound. The initial state is measurable at time zero, and the Wiener process is Brownian relative to the filtration. Local versions of these conditions can be used up to an explosion time.

What Diagonal Diffusion Encodes

For a diffusion driven by a Brownian vector with independent components, the instantaneous covariance matrix is $G G^\mathsf{T}$. With diagonal $G$, this matrix is diagonal: each state has its own instantaneous variance rate, and distinct states have zero instantaneous covariance rate. The precise pathwise statement uses quadratic covariation.

Theorem: Suppose $W_1,\ldots,W_m$ are independent standard Wiener processes and $g_i(t,X_t)$ are predictable, locally square-integrable integrands. Define the continuous local martingale $M_i(t)=\int_0^t g_i(s,X_s)\,dW_i(s)$. Then $$ [M_i,M_j]_t= \begin{cases} \displaystyle\int_0^t g_i(s,X_s)^2\,ds,&i=j,\\[4pt] 0,&i\ne j. \end{cases} $$ Here $[M_i,M_j]$ denotes quadratic covariation. Thus a diagonal diffusion gives zero cross-quadratic variation between the direct noise terms in different state equations.

Proof. Independent Wiener components satisfy $[W_i,W_j]_t=0$ for $i\ne j$, while $[W_i,W_i]_t=t$. For simple predictable integrands, the stochastic integrals are sums of coefficients known at the start of each time interval multiplied by the corresponding Brownian increments. Their quadratic covariation is therefore the sum of the products of those coefficients times the Brownian cross-variation. This gives $[\,\int g_i\,dW_i,\int g_j\,dW_j\,]_t=\int_0^t g_i(s,X_s)g_j(s,X_s)\,d[W_i,W_j]_s$. General locally square-integrable predictable integrands follow by approximation by simple integrands in the Itô-integral norm, first after stopping to make the integrands square-integrable and then removing the stopping. Substituting the Brownian cross-variations gives the displayed result. $\square$

For a short time interval of length $h$, the Euler increment at a fixed state $x$ is $\Delta X^{E}=b(t,x)h+G(t,x)\Delta W$, where $\Delta W\sim N(0,hI)$. Consequently its conditional covariance is exactly $hG(t,x)G(t,x)^\mathsf{T}$. This is the local covariance encoded by the SDE; it is not a claim that the exact finite-time transition distribution is Gaussian.

Worked Example: Two Compartments With Separate Noise

Consider central and peripheral amounts $A_c$ and $A_p$, with elimination from the central compartment and exchange between compartments. Specify independent drivers $W_c$ and $W_p$:

\begin{aligned} dA_c&=\bigl(u(t)-(k_{10}+k_{12})A_c+k_{21}A_p\bigr)\,dt +\sigma_c A_c\,dW_c,\\ dA_p&=\bigl(k_{12}A_c-k_{21}A_p\bigr)\,dt +\sigma_p A_p\,dW_p. \end{aligned}

The drift represents input, elimination, and intercompartmental transfer. The diagonal diffusion matrix is $G=\operatorname{diag}(\sigma_c A_c,\sigma_p A_p)$. Take $u=0$, $k_{10}=0.2$, $k_{12}=0.3$, $k_{21}=0.1$ per hour, $\sigma_c=0.1$ and $\sigma_p=0.2$ per square-root hour, and evaluate at $A_c=10$, $A_p=4$. The drift rates are

b_c=-(0.2+0.3)(10)+(0.1)(4)=-5+0.4=-4.6, \qquad b_p=(0.3)(10)-(0.1)(4)=3-0.4=2.6.

The diffusion coefficients at this state are $g_c=(0.1)(10)=1$ and $g_p=(0.2)(4)=0.8$. Thus the instantaneous covariance rate is

GG^\mathsf{T} = \begin{pmatrix}1&0\\0&0.8\end{pmatrix} \begin{pmatrix}1&0\\0&0.8\end{pmatrix} = \begin{pmatrix}1&0\\0&0.64\end{pmatrix}.

For an Euler step of $h=0.01$ hour at this state, the noise standard deviations are $\sqrt{0.01}(1)=0.1$ amount units and $\sqrt{0.01}(0.8)=0.08$ amount units. Their cross-covariance is zero because the Brownian increments are independent. The drift contributions for the step are $-4.6(0.01)=-0.046$ and $2.6(0.01)=0.026$. These calculations describe a local step, not a guarantee that the model's amounts remain nonnegative under every path.

Separate Drivers Are Not Independent States

The drift in the example transfers amount between compartments. Therefore each state can affect the future trajectory of the other even though their direct Wiener drivers are independent. In addition, a diffusion coefficient may depend on other states. The zero cross-quadratic variation result concerns the direct martingale terms; it does not imply independent state processes or independent finite-time state increments.

Worked Example: A Shared Driver Creates Cross-Covariance

Suppose instead that two compartments are both perturbed by the same Wiener process: $dA_1=\cdots+0.1A_1\,dW$ and $dA_2=\cdots+0.2A_2\,dW$. At $A_1=10$, $A_2=4$, their noise coefficients are $1$ and $0.8$. The diffusion matrix has one column, $G=(1,0.8)^\mathsf{T}$, so the instantaneous covariance matrix is

GG^\mathsf{T} = \begin{pmatrix}1\\0.8\end{pmatrix} \begin{pmatrix}1&0.8\end{pmatrix} = \begin{pmatrix}1&0.8\\0.8&0.64\end{pmatrix}.

The cross-covariance rate is $0.8$, not zero: over a frozen-coefficient Euler step of length $h$, the covariance of the two noise increments is $0.8h$. Using two separate independent Wiener drivers with the same two coefficients would instead give the diagonal matrix $\operatorname{diag}(1,0.64)$. The choice between these specifications changes the model's joint innovations; it is not merely a change in notation.

Worked Example: Zero Cross-Variation Does Not Guarantee Independence

Let $W_1,W_2$ be independent Wiener processes and define $M_1(t)=W_1(t)$ and $M_2(t)=\int_0^t W_1(s)\,dW_2(s)$. This is a diagonal-noise construction for the state $(M_1,M_2)$: its diffusion matrix is $\operatorname{diag}(1,M_1)$. The theorem gives $[M_1,M_2]_t=0$ because the drivers are independent.

Nevertheless, conditional on the path of $W_1$, the Itô integral $M_2(t)$ is normally distributed with mean zero and variance $V=\int_0^t W_1(s)^2\,ds$. This variance is random and depends on $M_1$'s driving path. To verify that it is nonconstant for $t>0$, use $E[W_1(s)^2]=s$ and the Gaussian identity $\operatorname{Cov}(W_1(s)^2,W_1(r)^2)=2\min(s,r)^2$. Then

E[V]=\int_0^t s\,ds=\frac{t^2}{2}, \qquad \operatorname{Var}(V) =2\int_0^t\int_0^t\min(s,r)^2\,ds\,dr =\frac{t^4}{3}>0.

If $M_2(t)$ were independent of the path of $W_1$, its conditional second moment would be constant. Instead $E[M_2(t)^2\mid W_1]=V$, which is nonconstant. Thus the two martingales are dependent despite having zero cross-quadratic variation. Orthogonal direct noise terms should not be described as independent processes without additional justification.

Checking and Interpreting a Specification

A reliable check is to calculate the diffusion covariance, rather than infer it from the way the equations are verbally described. If the Brownian vector has covariance matrix $R$, so that $d[W]_t=R\,dt$, the instantaneous state covariance is $G R G^\mathsf{T}$. The diagonal formula applies when $R=I$. A diagonal $G$ with correlated drivers need not produce a diagonal state covariance.

For model implementation and diagnostics, check the following:

Count the drivers and inspect the columns. With one independent driver per compartment and a diagonal $G$, each row uses its corresponding Wiener component only.
Calculate $G R G^\mathsf{T}$. Its off-diagonal entries are the instantaneous cross-covariance rates. This check catches accidental reuse of a noise source and unintended driver correlation.
Use independent normal draws in an Euler simulation. For a step of length $h$, generate independent $N(0,h)$ increments for independent drivers. Reusing one draw creates a shared-driver model.
Do not diagnose independence from one finite step's sample correlation alone. Drift coupling, state-dependent coefficients, and finite-step effects can induce dependence even when direct cross-quadratic variation is zero. Compare local increments after accounting for drift and the predicted covariance rate.

An alternative is to model correlated innovations directly, using a nonidentity $R$, or to factor a desired covariance matrix into a diffusion matrix and independent standard drivers. These approaches can represent common environmental or physiological perturbations, but they should encode a deliberate covariance structure. Independent diagonal noise is simpler to interpret, but it may be scientifically inadequate if compartments share important sources of process variability.

Finally, additive noise in an amount equation can produce negative amounts, and even multiplicative noise does not automatically settle every positivity question when drift and transfer terms interact. A diagonal diffusion matrix is a statement about the direct innovation structure, not a positivity constraint or a complete biological justification. Inspect simulated paths and consider whether the chosen state scale and noise model are appropriate.

Common pitfall: “There is one noise term in each equation, so the compartment states are independent.” This conclusion is wrong. Separate independent Wiener drivers imply zero instantaneous cross-variation of the direct noise terms when the diffusion matrix is diagonal. They do not remove drift coupling, state-dependent dependence, or dependence over finite time intervals.

Check Your Understanding

Answer each question, then compare with the solutions.

For $G=\operatorname{diag}(2,3)$ and independent drivers, what is the instantaneous covariance matrix?
If two equations use coefficients $a$ and $b$ multiplying the same Wiener process, what is their instantaneous cross-covariance rate?
Does zero cross-quadratic variation prove that two state processes are independent? Give the relevant distinction.
With two independent drivers and diagonal $G$, can the coefficient in the first equation depend on the second state?

Solutions. (1) $GG^\mathsf{T}=\operatorname{diag}(4,9)$. (2) The diffusion column is $(a,b)^\mathsf{T}$, so its covariance matrix is $\begin{pmatrix}a^2&ab\\ab&b^2\end{pmatrix}$; the cross-covariance rate is $ab$. (3) No. Zero cross-quadratic variation describes the direct martingale terms' instantaneous covariation. Drift coupling and state-dependent coefficients can still make states or finite-time increments dependent. (4) Yes. Diagonal structure specifies which Wiener increment enters each equation, not which state variables its coefficient may use.

Independent Noise For Multiple Compartments

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