Vector Ito Processes With Correlated Brownian Drivers

From Correlated Noise to Vector Covariance

A vector stochastic system can have several sources of noise, and those sources need not be independent. Even when each Brownian driver has variance rate one, their correlations affect the variance of each component and the covariance between components. The right way to keep track of these effects is to record the diffusion coefficients in a matrix and the driver correlations in a second matrix.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying an $m$-dimensional Brownian vector $\mathbf{B}_t=(B_t^1,\ldots,B_t^m)^\top$. Assume its components are standard Brownian motions with constant correlations $\rho_{ab}$: their joint increments are Gaussian, independent of the past, and satisfy $\mathbb{E}[(B_t^a-B_s^a)(B_t^b-B_s^b)]=\rho_{ab}(t-s)$ for $s<t$. The matrix $R=(\rho_{ab})_{a,b=1}^m$ is symmetric, positive semidefinite, and has diagonal entries equal to one. The case $R=I_m$ corresponds to independent drivers.

Definition: A $d$-dimensional Ito process driven by these $m$ Brownian motions has the form $$ \mathbf{X}_t=\mathbf{X}_0+\int_0^t\mathbf{b}_s\,ds+\int_0^t\Sigma_s\,d\mathbf{B}_s, \qquad d\mathbf{X}_t=\mathbf{b}_t\,dt+\Sigma_t\,d\mathbf{B}_t. $$ Here $\mathbf{b}_t\in\mathbb{R}^d$ is the drift vector and $\Sigma_t\in\mathbb{R}^{d\times m}$ is the diffusion matrix. Its entries are progressively measurable, and on every finite horizon $[0,T]$ assume $\mathbb{E}\int_0^T\|\Sigma_s\|_F^2\,ds<\infty$ and $\mathbb{E}\int_0^T\|\mathbf{b}_s\|\,ds<\infty$. The initial vector is $\mathcal{F}_0$-measurable. Each stochastic integral is an Ito integral with a progressively measurable, square-integrable integrand; the stated matrix condition ensures this for every entry.

In coordinates, $dX_t^i=b_t^i\,dt+\sum_{a=1}^m\Sigma_{ia}(t)\,dB_t^a$. The drift gives the finite-variation part of the motion. The diffusion matrix tells how each Brownian driver enters each component. The product rules for the drivers are $dB_t^a\,dB_t^b=\rho_{ab}\,dt$, $dB_t^a\,dt=0$, and $dt\,dt=0$. In particular, the diagonal rule is $dB_t^a\,dB_t^a=dt$. These are rules for quadratic covariation, not ordinary algebraic identities between infinitesimal random variables.

The Instantaneous Covariance Matrix

Theorem: For the Ito process above, the matrix of quadratic covariations satisfies $$ d[X^i,X^j]_t=A_{ij}(t)\,dt, \qquad A_t=\Sigma_t R\Sigma_t^\top. $$ Thus $A_t$ is the instantaneous covariance matrix contributed by the Brownian noise. The drift vector does not appear.

Proof. Write $M_t^i=\sum_a\int_0^t\Sigma_{ia}(s)\,dB_s^a$ for the martingale part of $X^i$. Its integrals are square-integrable martingales: the Ito isometry gives finite second moments, and the conditional expectation of each integral increment given $\mathcal{F}_s$ is zero. The finite-variation drift has zero quadratic covariation with continuous Ito integrals, so $[X^i,X^j]=[M^i,M^j]$. The quadratic-covariation rule for Ito integrals, together with $[B^a,B^b]_t=\rho_{ab}t$, gives $$ d[M^i,M^j]_t =\sum_{a=1}^m\sum_{b=1}^m\Sigma_{ia}(t)\Sigma_{jb}(t)\rho_{ab}\,dt =(\Sigma_tR\Sigma_t^\top)_{ij}\,dt. $$ The double sum is exactly the $(i,j)$-entry of the displayed matrix product. This proves the formula. $\square$

The diagonal entry $A_{ii}(t)$ is the instantaneous variance rate of component $X^i$; an off-diagonal entry $A_{ij}(t)$ is its instantaneous covariance rate with $X^j$. The accumulated bracket is $[X^i,X^j]_t=\int_0^t A_{ij}(s)\,ds$. A bracket is not, in general, the ordinary covariance of the finite-time random variables $X_t^i$ and $X_t^j$, especially when the coefficients depend on the current state.

Proposition: The matrix $A_t=\Sigma_tR\Sigma_t^\top$ is symmetric and positive semidefinite. It may be singular, including when some components have no noise or when the diffusion directions are linearly dependent.

Proof. Since $R^\top=R$, transposing the product gives $A_t^\top=\Sigma_tR^\top\Sigma_t^\top=A_t$. For any $c\in\mathbb{R}^d$, $$ c^\top A_t c=(\Sigma_t^\top c)^\top R(\Sigma_t^\top c)\ge0, $$ because $R$ is positive semidefinite. This proves positive semidefiniteness. If $\Sigma_t^\top c=0$, the displayed quadratic form is zero; this can happen for nonzero $c$, so strict positive definiteness is not guaranteed. The zero diffusion matrix is also allowed and gives $A_t=0$. $\square$

Worked Matrix Calculations

Worked Example: A Two-Component Process with Two Correlated Drivers

Suppose $\rho_{12}=\rho_{21}=\tfrac12$, and let $$ d\mathbf{X}_t= \begin{pmatrix}1\\-2\end{pmatrix}dt+ \begin{pmatrix}1&2\\-1&1\end{pmatrix} \begin{pmatrix}dB_t^1\\dB_t^2\end{pmatrix}. $$ Find the instantaneous covariance matrix.

Here $R=\begin{pmatrix}1&\tfrac12\\\tfrac12&1\end{pmatrix}$ and $\Sigma=\begin{pmatrix}1&2\\-1&1\end{pmatrix}$. First compute $$ \Sigma R= \begin{pmatrix}1&2\\-1&1\end{pmatrix} \begin{pmatrix}1&\tfrac12\\\tfrac12&1\end{pmatrix} = \begin{pmatrix}2&\tfrac52\\-\tfrac12&\tfrac12\end{pmatrix}. $$ Since $\Sigma^\top=\begin{pmatrix}1&-1\\2&1\end{pmatrix}$, multiplication gives $$ A=\Sigma R\Sigma^\top =\begin{pmatrix}2&\tfrac52\\-\tfrac12&\tfrac12\end{pmatrix} \begin{pmatrix}1&-1\\2&1\end{pmatrix} =\begin{pmatrix}7&\tfrac12\\\tfrac12&1\end{pmatrix}. $$ The noise variance rates are $7$ and $1$; the covariance rate is $\tfrac12$. As a check, the determinant is $7\cdot1-(\tfrac12)^2=\tfrac{27}{4}>0$, consistent with positive definiteness in this example. The drift $(1,-2)^\top$ does not affect these entries.

Worked Example: Negative Driver Correlation

Let $\rho_{12}=-\tfrac13$, and consider $dX_t=dB_t^1+2dB_t^2$ and $dY_t=3dB_t^1-dB_t^2$, with zero drift. Compute all entries of the instantaneous covariance matrix.

The driver matrix $R=\begin{pmatrix}1&-\tfrac13\\-\tfrac13&1\end{pmatrix}$ is valid: it is symmetric, has unit diagonal, and has determinant $1-\tfrac19=\tfrac89>0$. The diffusion matrix has rows $(1,2)$ and $(3,-1)$. Therefore $$ A_{11}=1^2+2^2+2(1)(2)(-\tfrac13)=5-\tfrac43=\tfrac{11}{3}, $$ and $$ A_{22}=3^2+(-1)^2+2(3)(-1)(-\tfrac13)=10+2=12. $$ For the cross-entry, the same-driver products contribute $1\cdot3+2\cdot(-1)=1$. The cross-driver products contribute $(-\tfrac13)\big(1\cdot(-1)+2\cdot3\big)=(-\tfrac13)(5)=-\tfrac53$. Hence $A_{12}=1-\tfrac53=-\tfrac23$, and symmetry gives $A_{21}=-\tfrac23$. Thus $$ A=\begin{pmatrix}\tfrac{11}{3}&-\tfrac23\\-\tfrac23&12\end{pmatrix}. $$ As a check, its determinant is $\tfrac{11}{3}\cdot12-(-\tfrac23)^2=44-\tfrac49=\tfrac{392}{9}>0$. The negative covariance rate is compatible with both individual variance rates being positive.

Worked Example: Recovering Independent Drivers

For the first example, represent the correlated Brownian drivers using independent standard Brownian motions $U_t,V_t$, and check the covariance matrix by recomputing it.

Take $B_t^1=U_t$ and $B_t^2=\tfrac12U_t+\tfrac{\sqrt3}{2}V_t$. The second driver has variance rate $(\tfrac12)^2+(\tfrac{\sqrt3}{2})^2=\tfrac14+\tfrac34=1$, and its covariance rate with $B^1$ is $\tfrac12$, as required. Substituting into the noise terms of the first example gives $$ dX_t^1=2\,dU_t+\sqrt3\,dV_t,\qquad dX_t^2=-\tfrac12\,dU_t+\tfrac{\sqrt3}{2}\,dV_t. $$ The independent-driver coefficient rows are now $(2,\sqrt3)$ and $(-\tfrac12,\tfrac{\sqrt3}{2})$. Their variances are $2^2+(\sqrt3)^2=4+3=7$ and $(-\tfrac12)^2+(\tfrac{\sqrt3}{2})^2=\tfrac14+\tfrac34=1$. Their covariance is $2(-\tfrac12)+\sqrt3\,\tfrac{\sqrt3}{2}=-1+\tfrac32=\tfrac12$. This reproduces every entry of $\begin{pmatrix}7&\tfrac12\\\tfrac12&1\end{pmatrix}$.

Linear Combinations and Interpretation

For a fixed vector $c\in\mathbb{R}^d$, the scalar process $c^\top\mathbf{X}_t$ has drift $c^\top\mathbf{b}_t$ and diffusion row $c^\top\Sigma_t$. Its instantaneous variance rate is $c^\top A_t c$. This follows directly by applying the matrix formula to the one-row diffusion matrix $c^\top\Sigma_t$. It is a useful check: every proposed covariance matrix must assign a nonnegative variance rate to every linear combination.

Worked Example: Variance Rate of a Portfolio of Components

In the first example, let $Z_t=2X_t^1-X_t^2$. Find its drift and instantaneous variance rate, and verify the variance by combining the Brownian terms directly.

The drift is $2(1)-(-2)=4$. With $c=(2,-1)^\top$ and $A=\begin{pmatrix}7&\tfrac12\\\tfrac12&1\end{pmatrix}$, the variance rate is $$ c^\top A c =2\left(2\cdot7-\tfrac12\right)-\left(2\cdot\tfrac12-1\right) =2\left(\tfrac{27}{2}\right)-0=27. $$ Directly, the Brownian part is $2(dB^1+2dB^2)-(-dB^1+dB^2)=3dB^1+3dB^2$. Its variance rate is $3^2+3^2+2(3)(3)(\tfrac12)=9+9+9=27$. The two calculations agree.

Common Mistake: Dropping the Cross Terms

A frequent error is to treat correlated drivers as if they were independent. In the first example, keeping only the squared coefficients would give the incorrect variance rate $1^2+2^2=5$ for $X^1$. The correct rate includes the cross term: $$ 1^2+2^2+2(1)(2)(\tfrac12)=1+4+2=7. $$ Likewise, the covariance between two components includes contributions from every pair of drivers, weighted by their correlations. Using only matching-driver products is valid when $R=I_m$, not in general.

Correlated drivers can also be represented through independent Brownian motions when a matrix factorization $R=LL^\top$ is available: if $\mathbf{U}$ has independent standard Brownian components, then $\mathbf{B}=L\mathbf{U}$ has driver covariance matrix $R$. The third example shows this explicitly in dimension two. A singular $R$ is permitted; it simply means the drivers contain fewer independent noise directions than their number suggests. The effective covariance of the state is governed jointly by $R$ and $\Sigma$, through $\Sigma R\Sigma^\top$.

Check Your Understanding

For each question, use the stated correlation matrix and include the cross-driver terms where required.

If $dX_t=2dB_t^1-dB_t^2$ and $\rho_{12}=\tfrac14$, what is the instantaneous variance rate of $X$?
For two components with diffusion rows $(1,1)$ and $(1,-1)$, and driver correlation $\rho_{12}=\tfrac12$, compute their full instantaneous covariance matrix.
Explain why a drift vector does not contribute to the quadratic-covariation matrix.
For a given matrix $A=\Sigma R\Sigma^\top$, what condition on $c^\top A c$ must hold for every vector $c$, and what does it represent?
In the independent-driver representation of the third example, verify the covariance rate between $X^1$ and $X^2$ by summing the products of their coefficients on $U$ and $V$.