Multidimensional Stratonovich Integrals and Covariance

Several Brownian Drivers, Several Covariations

With one Brownian driver, converting a Stratonovich integral to an Ito integral requires the covariation of its integrand with that driver. With several drivers, the same principle applies component by component. The correction for an integrand paired with one driver depends on its covariations with all the Brownian components; those covariations are determined in part by the drivers’ covariance matrix. Keeping the indices visible prevents a common mistake: confusing the covariance between two drivers with the correction for a particular integrand-driver pair.

Throughout, work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a $d$-dimensional Brownian motion $\mathbf{B}_t=(B_t^1,\ldots,B_t^d)^\mathsf{T}$, adapted to the filtration. Assume its increments are independent of the past and have covariance matrix $(t-s)C$ over $[s,t]$, where $C=(c_{k\ell})$ is a constant symmetric positive semidefinite matrix. Thus $[B^k,B^\ell]_t=c_{k\ell}t$. When $C$ is the identity matrix, the components are independent standard Brownian motions. A non-identity matrix allows correlated components; it need not be invertible.

Definition: Let $H=(H_1,\ldots,H_d)$ be a row of progressively measurable integrands with $\mathbb{E}\int_0^T\sum_{j=1}^d H_j(s)^2\,ds<\infty$. Its Ito integral against $\mathbf B$ is the scalar $\int_0^t H_s\,d\mathbf B_s:=\sum_{j=1}^d\int_0^t H_j(s)\,dB_s^j$. If each $H_j$ is also a continuous semimartingale and the covariations in the following expression are integrable, define its Stratonovich integral by $\int_0^t H_s\circ d\mathbf B_s:=\int_0^t H_s\,d\mathbf B_s+\tfrac12\sum_{j=1}^d[H_j,B^j]_t$. Each Ito integral here uses a progressively measurable integrand with finite expected integral of its square on $[0,T]$.

For an $m$-by-$d$ matrix process $H$, apply this definition to each row. The Ito integral is an $m$-vector, and the Stratonovich correction for row $a$ uses the covariation of entry $H_{aj}$ with driver $B^j$. In particular, it does not sum all covariations $[H_{aj},B^k]$ indiscriminately: the integral pairs column $j$ with driver $j$. Those other covariations can still enter when $H_{aj}$ itself depends on several Brownian components.

Covariances of Linear Brownian Combinations

Proposition: For constant vectors $u,v\in\mathbb{R}^d$, let $U_t=u^\mathsf{T}\mathbf B_t$ and $V_t=v^\mathsf{T}\mathbf B_t$. Then $[U,V]_t=(u^\mathsf{T}Cv)t$.

Proof. Covariation is bilinear, so $[U,V]_t=\sum_{k=1}^d\sum_{\ell=1}^d u_kv_\ell[B^k,B^\ell]_t$. Substituting $[B^k,B^\ell]_t=c_{k\ell}t$ gives $t\sum_{k,\ell}u_kc_{k\ell}v_\ell=(u^\mathsf{T}Cv)t$, including when $u$, $v$, or $C$ is zero. $\square$

This proposition gives the basic calculation for a conversion correction. If the continuous semimartingale integrand $H_j$ has a decomposition $dH_j(s)=a_j(s)\,ds+\sum_{k=1}^dG_{jk}(s)\,dB_s^k$, then its finite-variation part has zero covariation with Brownian motion, and bilinearity gives $d[H_j,B^j]_s=\sum_{k=1}^dG_{jk}(s)c_{kj}\,ds$. This identity requires the displayed stochastic integrals to be defined and the covariations to be integrable. It shows how correlations enter even though column $j$ is paired with $B^j$.

Componentwise Conversion for Matrix Integrals

Theorem: Suppose $H$ is an $m$-by-$d$ matrix of progressively measurable entries satisfying $\mathbb{E}\int_0^T\sum_{a=1}^m\sum_{j=1}^dH_{aj}(s)^2\,ds<\infty$. Suppose also that its entries are continuous semimartingales and that $\sum_j[H_{aj},B^j]_t$ is defined and integrable for every row $a$. Then, row by row, $$ \int_0^t H_s\circ d\mathbf B_s = \int_0^t H_s\,d\mathbf B_s +\frac12 \begin{pmatrix} \sum_{j=1}^d[H_{1j},B^j]_t\\ \vdots\\ \sum_{j=1}^d[H_{mj},B^j]_t \end{pmatrix}. $$

Proof. Fix a row $a$. By definition, its Stratonovich integral is its Ito integral plus one half of the sum of the covariations between each integrand entry $H_{aj}$ and its paired driver $B^j$. Summing these $d$ scalar identities gives the $a$th component of the displayed vector identity. This argument applies to every $a=1,\ldots,m$, proving the matrix identity. The Ito integrals exist by the stated square-integrability assumption; the Stratonovich corrections exist by the stated covariation assumption. $\square$

This componentwise result uses the scalar Stratonovich-to-Ito conversion and makes explicit how to apply it to multiple drivers. If an entry $H_{aj}$ is driven by several Brownian components, the preceding linear-combination calculation, or its semimartingale version, evaluates its covariation with the paired driver. The covariance matrix belongs in that covariation calculation, not as an extra matrix multiplication appended to the conversion formula.

Worked Example: A Cross-Integral with Correlated Drivers

Let $d=2$, $C=\begin{pmatrix}1&1/2\\1/2&1\end{pmatrix}$, and define $J_t=\int_0^t B_s^2\circ dB_s^1+\int_0^t B_s^1\circ dB_s^2$. Calculate its Ito form and its expectation at $t=2$. These integrands are progressively measurable and square-integrable on every finite interval because Brownian second moments are finite; they are continuous semimartingales.

The two relevant covariations are $[B^2,B^1]_t=\tfrac12t$ and $[B^1,B^2]_t=\tfrac12t$. Therefore the total correction is $\tfrac12(\tfrac12t+\tfrac12t)=\tfrac12t$, and $$ J_t=\int_0^t B_s^2\,dB_s^1+\int_0^t B_s^1\,dB_s^2+\frac12t. $$

To check this independently, apply Ito’s formula to $f(x,y)=xy$. Its first derivatives are $f_x=y$ and $f_y=x$; its second derivatives are $f_{xx}=0$, $f_{yy}=0$, and $f_{xy}=1$. The second-order term is therefore $d[B^1,B^2]_t=\tfrac12dt$, so $d(B_t^1B_t^2)=B_t^2\,dB_t^1+B_t^1\,dB_t^2+\tfrac12dt$. Integrating gives $J_t=B_t^1B_t^2$, in agreement with the conversion. At $t=2$, $\mathbb E[J_2]=\mathbb E[B_2^1B_2^2]=c_{12}\cdot2=\tfrac12\cdot2=1$.

Check: Each Ito integral has expectation zero: its integrand is square-integrable, so the Ito integral is square-integrable by the Ito isometry and is a martingale. Its conditional expectation at an earlier time equals its value then, by the martingale property of the Ito integral; in particular its expectation at time zero is zero. Thus the converted expression gives $\mathbb E[J_2]=\tfrac12\cdot2=1$, matching the product calculation.

Worked Example: Converting a Matrix-Valued Integrand

Use the same covariance matrix $C=\begin{pmatrix}1&1/2\\1/2&1\end{pmatrix}$, and set $$ H_t= \begin{pmatrix} B_t^1+2B_t^2&3B_t^1-B_t^2\\ 2B_t^1-B_t^2&B_t^1+B_t^2 \end{pmatrix}. $$ Convert $Y_t=\int_0^tH_s\circ d\mathbf B_s$ to Ito form. The entries are progressively measurable, continuous semimartingales, and square-integrable on finite intervals by the finite second moments of Brownian motion.

For row one, the first entry has covariation with $B^1$ equal to $[B^1+2B^2,B^1]_t=(1+2\cdot\tfrac12)t=2t$. The second entry has covariation with $B^2$ equal to $[3B^1-B^2,B^2]_t=(3\cdot\tfrac12-1)t=\tfrac12t$. Thus row one’s correction is $\tfrac12(2t+\tfrac12t)=\tfrac54t$.

For row two, $[2B^1-B^2,B^1]_t=(2-\tfrac12)t=\tfrac32t$, while $[B^1+B^2,B^2]_t=(\tfrac12+1)t=\tfrac32t$. Its correction is $\tfrac12(\tfrac32t+\tfrac32t)=\tfrac32t$. Consequently, $$ Y_t=\int_0^tH_s\,d\mathbf B_s+ \begin{pmatrix}\tfrac54t\\[2pt]\tfrac32t\end{pmatrix}. $$

Check: The Ito integral has expectation zero componentwise. Indeed, each row is progressively measurable and square-integrable on $[0,T]$; the Ito isometry gives finite second moments, and the Ito integral’s martingale property gives zero expectation at time zero. Hence $\mathbb E[Y_t]=(\tfrac54t,\tfrac32t)^\mathsf{T}$, exactly the computed correction. In particular, at $t=1$ the mean is $(\tfrac54,\tfrac32)^\mathsf{T}$.

Worked Example: Covariance of a Constant Matrix Integral

Let $C=\begin{pmatrix}1&1/2\\1/2&1\end{pmatrix}$, $A=\begin{pmatrix}1&2\\-1&1\end{pmatrix}$, and $Z_t=\int_0^t A\,d\mathbf B_s$. Since the entries of $A$ are constant, their covariations with the drivers are zero, so the Ito and Stratonovich integrals agree. Find the covariance matrix of $Z_2$.

For a row $a=(a_1,a_2)$, the variance per unit time is $a_1^2+a_2^2+2(\tfrac12)a_1a_2$. The first row $(1,2)$ gives $1+4+2\cdot\tfrac12\cdot1\cdot2=7$. The second row $(-1,1)$ gives $1+1+2\cdot\tfrac12\cdot(-1)\cdot1=1$. The covariance per unit time between the rows $(1,2)$ and $(-1,1)$ is $1(-1)+2(1)+\tfrac12(1\cdot1+2\cdot(-1))=1-\tfrac12=\tfrac12$. Multiplying every entry by the time $2$ gives $$ \operatorname{Cov}(Z_2)=2ACA^\mathsf{T} =\begin{pmatrix}14&1\\1&2\end{pmatrix}. $$

Check: The off-diagonal entry can also be computed directly as $\operatorname{Cov}(B_2^1+2B_2^2,-B_2^1+B_2^2)$. Expanding gives $-2+1-2+4=1$: the four terms use, respectively, $\operatorname{Var}(B_2^1)=2$, $\operatorname{Cov}(B_2^1,B_2^2)=1$, $\operatorname{Cov}(B_2^2,B_2^1)=1$, and $\operatorname{Var}(B_2^2)=2$. This agrees with the matrix calculation.

Common Pitfall: Losing the Cross Terms

In the last example, a wrong calculation might replace the cross term $\tfrac12(1\cdot1+2\cdot(-1))$ by $0$, as if the two products canceled. They do not: $1\cdot1+2\cdot(-1)=1-2=-1$, so the cross term is $-\tfrac12$. The covariance per unit time is therefore $-1+2-\tfrac12=\tfrac12$, and over time $2$ it is $1$, not $2$. Write out both cross products before simplifying.

Two distinctions help organize calculations. First, the Ito-versus-Stratonovich correction for a matrix integral pairs each entry $H_{aj}$ with its own driver $B^j$. Second, the covariance of the resulting vector uses all pairings between the rows and therefore involves the full matrix $C$. A constant integrand has no Stratonovich correction, but its output covariance can still depend on every entry of $C$. For a state-dependent integrand, the correction can also depend on off-diagonal entries of $C$, through covariations between the integrand and its paired driver.

Check Your Understanding

In each question, assume the Brownian covariance matrix is constant and that the stated integrands meet the integrability and semimartingale conditions above.

For $C=\begin{pmatrix}1&\rho\\\rho&1\end{pmatrix}$, calculate $[B^1+B^2,B^1]_t$.
For $J_t=\int_0^t B_s^2\circ dB_s^1-\int_0^t B_s^1\circ dB_s^2$, find the total correction in terms of $\rho$ and $t$.
State the correction for row $a$ of a matrix Stratonovich integral with $d$ Brownian drivers.
For a constant row $a$, express the variance per unit time of $a^\mathsf{T}\mathbf B_t$ using $C$.
Explain why a zero Stratonovich correction for a constant matrix integrand does not imply that the output components are uncorrelated.

Multidimensional Stratonovich Integrals and Covariance

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