Covariance Matrices of Multidimensional Brownian Motion

From Coordinate Covariations to a Matrix

For a one-dimensional Brownian motion, the quadratic variation over $[0,t]$ is $t$. In several dimensions, each coordinate has its own quadratic variation, but different coordinates may also accumulate nonzero covariation. Collecting these quantities into a matrix makes it possible to compute the covariations of any linear combinations by ordinary matrix multiplication.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$. A $d$-dimensional Brownian motion with covariance matrix $\Gamma$ is an adapted, continuous process $\mathbf{B}_t=(B_t^1,\ldots,B_t^d)^{\mathsf T}$, with $\mathbf{B}_0=0$, whose increments after time $s$ are independent of $\mathcal{F}_s$ and whose increment $\mathbf{B}_t-\mathbf{B}_s$ is a centered Gaussian vector with covariance $(t-s)\Gamma$, for $0\le s<t$. Here $\Gamma$ is a fixed symmetric positive semidefinite $d\times d$ matrix. Its entries satisfy $\mathbb{E}[(B_t^i-B_s^i)(B_t^j-B_s^j)]=(t-s)\Gamma_{ij}$.

Definition: The quadratic-covariation matrix of $\mathbf{B}$ at time $t$ is the matrix $[\mathbf{B}]_t$ whose $(i,j)$ entry is $[B^i,B^j]_t$, the limit of the sums $\sum_k \Delta B_k^i\Delta B_k^j$ as the mesh of partitions of $[0,t]$ tends to zero. The diagonal entries are coordinate quadratic variations; the off-diagonal entries are coordinate covariations.

When $\Gamma=I_d$, the coordinates are independent standard Brownian motions. If $\Gamma_{ii}=1$ for every $i$, the coordinates each have standard Brownian variance, but they need not be independent. For positive diagonal entries, their instantaneous correlation is $\Gamma_{ij}/\sqrt{\Gamma_{ii}\Gamma_{jj}}$. A singular $\Gamma$ is allowed: it means that some nonzero linear combination of the coordinates has zero variance.

The Covariation Matrix Theorem

Theorem: For a Brownian motion with covariance matrix $\Gamma$, for every $T\ge0$ and every pair $i,j$, $$ [B^i,B^j]_t=t\Gamma_{ij}\quad(0\le t\le T), \qquad [\mathbf{B}]_t=t\Gamma. $$ The convergence of the defining partition sums holds in $L^2(\mathbb{P})$ along every deterministic sequence of partitions whose mesh tends to zero.

Proof. Fix $T>0$ and a deterministic partition $0=t_0<\cdots<t_n=T$. Write $\Delta t_k=t_k-t_{k-1}$, $\Delta B_k^i=B_{t_k}^i-B_{t_{k-1}}^i$, and $S_\Pi^{ij}=\sum_k\Delta B_k^i\Delta B_k^j$. The Gaussian increment covariance gives $\mathbb{E}[\Delta B_k^i\Delta B_k^j]=\Gamma_{ij}\Delta t_k$, so $\mathbb{E}[S_\Pi^{ij}]=T\Gamma_{ij}$.

For jointly centered Gaussian variables $X,Y$, with variances $a,b$ and covariance $c$, the fourth-moment identity gives $\mathbb{E}[X^2Y^2]=ab+2c^2$, and hence $\operatorname{Var}(XY)=ab+c^2$. To verify the identity when $a>0$, write $X=\sqrt a\,U$ and $Y=(c/\sqrt a)U+\sqrt{b-c^2/a}\,V$, where $U,V$ are independent standard normal variables. Expanding $X^2Y^2$, using $\mathbb{E}[U^4]=3$, $\mathbb{E}[U^2V^2]=1$, and the vanishing of the odd terms gives $ab+2c^2$. If $a=0$, then $X=0$ almost surely and $c=0$, so the same identity holds.

Apply this identity to each increment pair. Since increments on distinct partition intervals are independent, the products $\Delta B_k^i\Delta B_k^j$ are independent across $k$. Their variances therefore add, giving $$ \operatorname{Var}(S_\Pi^{ij}) =\left(\Gamma_{ii}\Gamma_{jj}+\Gamma_{ij}^2\right)\sum_k(\Delta t_k)^2 \le \left(\Gamma_{ii}\Gamma_{jj}+\Gamma_{ij}^2\right)T|\Pi|. $$ Here $|\Pi|=\max_k\Delta t_k$, and the inequality follows from $\sum_k(\Delta t_k)^2\le|\Pi|\sum_k\Delta t_k=T|\Pi|$. Thus $\mathbb{E}[(S_\Pi^{ij}-T\Gamma_{ij})^2]=\operatorname{Var}(S_\Pi^{ij})\to0$ as $|\Pi|\to0$. Applying the same argument on $[0,t]$ gives the limit $t\Gamma_{ij}$ for each $t$. There are only finitely many entries, so the matrix of sums converges entry by entry in $L^2$ to $t\Gamma$. The case $T=0$ has empty sums and the stated limit is zero. $\square$

In differential notation, the matrix result is recorded as $dB_t^i\,dB_t^j=\Gamma_{ij}\,dt$. This is shorthand for the partition-sum limit, not an identity between individual increments. In particular, the diagonal rule is $dB_t^i\,dB_t^i=\Gamma_{ii}\,dt$; it equals $dt$ only when that coordinate has unit variance per unit time.

Linear Combinations and Matrix Multiplication

Proposition: Let $A$ be a fixed $m\times d$ real matrix, and define $\mathbf{Y}_t=A\mathbf{B}_t$. Then $$ [\mathbf{Y}]_t=tA\Gamma A^{\mathsf T}. $$ In particular, for vectors $a,b\in\mathbb{R}^d$, $$ [a^{\mathsf T}\mathbf{B},b^{\mathsf T}\mathbf{B}]_t=t\,a^{\mathsf T}\Gamma b. $$

Proof. On any partition, the increment of coordinate $r$ of $\mathbf{Y}$ is $\Delta Y_k^r=\sum_i A_{ri}\Delta B_k^i$. For coordinates $r,s$, expand the partition sum exactly: $$ \sum_k\Delta Y_k^r\Delta Y_k^s =\sum_i\sum_j A_{ri}A_{sj}\sum_k\Delta B_k^i\Delta B_k^j. $$ Each inner sum converges in $L^2$ to $t\Gamma_{ij}$ by the theorem. The sums over $i,j$ are finite, so the full expression converges to $t\sum_{i,j}A_{ri}\Gamma_{ij}A_{sj}$, which is the $(r,s)$ entry of $tA\Gamma A^{\mathsf T}$. Taking $m=1$ for each of $a^{\mathsf T}\mathbf{B}$ and $b^{\mathsf T}\mathbf{B}$, or using the same expansion with two different coefficient vectors, gives the scalar formula. $\square$

The transformed matrix is again symmetric positive semidefinite. Indeed, for every $x\in\mathbb{R}^m$, $$ x^{\mathsf T}A\Gamma A^{\mathsf T}x=(A^{\mathsf T}x)^{\mathsf T}\Gamma(A^{\mathsf T}x)\ge0, $$ because $\Gamma$ is positive semidefinite. Thus the variance rate of every scalar combination $a^{\mathsf T}\mathbf{B}$ is $a^{\mathsf T}\Gamma a$, which cannot be negative.

Worked Example: Sum and Difference of Correlated Coordinates

Suppose $\Gamma=\begin{pmatrix}1&3/5\\3/5&1\end{pmatrix}$, and set $Y_t=B_t^1+B_t^2$ and $Z_t=B_t^1-B_t^2$. Find their quadratic variations and covariation.

The coefficient matrix is $A=\begin{pmatrix}1&1\\1&-1\end{pmatrix}$. First multiply: $$ A\Gamma =\begin{pmatrix}1&1\\1&-1\end{pmatrix} \begin{pmatrix}1&3/5\\3/5&1\end{pmatrix} =\begin{pmatrix}8/5&8/5\\2/5&-2/5\end{pmatrix}. $$ Then $$ A\Gamma A^{\mathsf T} =\begin{pmatrix}8/5&8/5\\2/5&-2/5\end{pmatrix} \begin{pmatrix}1&1\\1&-1\end{pmatrix} =\begin{pmatrix}16/5&0\\0&4/5\end{pmatrix}. $$ Consequently, $[Y]_t=16t/5$, $[Z]_t=4t/5$, and $[Y,Z]_t=0$.

Check: Direct expansion gives $[Y]_t=(1+1+2(3/5))t=16t/5$ and $[Z]_t=(1+1-2(3/5))t=4t/5$. The cross terms in $[Y,Z]$ cancel: $(1-3/5+3/5-1)t=0$. The original matrix is valid because its leading principal minors are $1$ and $1-9/25=16/25>0$.

Worked Example: Transforming Three Correlated Coordinates

Let the covariance matrix and transformation be $$ \Gamma=\begin{pmatrix}1&1/2&0\\1/2&1&1/2\\0&1/2&1\end{pmatrix}, \qquad A=\begin{pmatrix}1&1&0\\0&1&-1\end{pmatrix}. $$ For $Y_t=B_t^1+B_t^2$ and $Z_t=B_t^2-B_t^3$, compute their covariation matrix.

The leading principal minors of $\Gamma$ are $1$, $1-1/4=3/4$, and $\det(\Gamma)=1/2$, all positive, so this symmetric matrix is positive definite. Multiplication gives $$ A\Gamma =\begin{pmatrix}3/2&3/2&1/2\\1/2&1/2&1/2\end{pmatrix}, \qquad A\Gamma A^{\mathsf T} =\begin{pmatrix}3&1\\1&1\end{pmatrix}. $$ Hence $[Y]_t=3t$, $[Z]_t=t$, and $[Y,Z]_t=t$.

Check: Expanding the first variance gives $[Y]_t=(\Gamma_{11}+\Gamma_{22}+2\Gamma_{12})t=(1+1+1)t=3t$. The second gives $[Z]_t=(\Gamma_{22}+\Gamma_{33}-2\Gamma_{23})t=(1+1-1)t=t$. The cross term is $(\Gamma_{12}-\Gamma_{13}+\Gamma_{22}-\Gamma_{23})t=(1/2-0+1-1/2)t=t$, agreeing with the matrix entry.

Worked Example: A Singular Matrix and a Zero Combination

Let $U,V$ be independent standard Brownian motions on the same filtered probability space and define $\mathbf{B}_t=(U_t,V_t,U_t+V_t)^{\mathsf T}$. Find its covariance matrix and verify the covariation of $B^3-B^1-B^2$.

The coordinate covariations are $[U]_t=t$, $[V]_t=t$, and $[U,V]_t=0$. Since $[U,U+V]_t=t$, $[V,U+V]_t=t$, and $[U+V]_t=2t$, the matrix is $$ [\mathbf{B}]_t=t\Gamma,\qquad \Gamma=\begin{pmatrix}1&0&1\\0&1&1\\1&1&2\end{pmatrix}. $$ Take $a=(-1,-1,1)^{\mathsf T}$, so $a^{\mathsf T}\mathbf{B}_t=B_t^3-B_t^1-B_t^2$. Direct multiplication yields $\Gamma a=(0,0,0)^{\mathsf T}$, and therefore $a^{\mathsf T}\Gamma a=0$. The proposition gives $[a^{\mathsf T}\mathbf{B}]_t=0$.

Check: By the definition of the coordinates, $B_t^3-B_t^1-B_t^2=(U_t+V_t)-U_t-V_t=0$ at every time, so its quadratic variation is indeed zero. The matrix is positive semidefinite because $x^{\mathsf T}\Gamma x=(x_1+x_3)^2+(x_2+x_3)^2\ge0$ for every $x\in\mathbb{R}^3$, and it is singular because $\Gamma a=0$.

What the Matrix Records—and a Common Error

The covariance matrix at a fixed time and the quadratic-covariation matrix are related but distinct objects. The increment definition gives $\operatorname{Cov}(\mathbf{B}_t)=t\Gamma$; the theorem gives $[\mathbf{B}]_t=t\Gamma$ as the limit of products of increments along fine partitions. They have the same deterministic matrix in this Brownian setting, but one is a covariance of a random vector at a fixed time and the other is a process defined through path increments.

The matrix formulation also identifies which combinations have zero covariation. For $a\in\mathbb{R}^d$, the variance rate of $a^{\mathsf T}\mathbf{B}$ is $a^{\mathsf T}\Gamma a$. It is zero exactly when this scalar combination has zero quadratic variation. For a positive semidefinite matrix, $a^{\mathsf T}\Gamma a=0$ implies $\Gamma a=0$: diagonalizing $\Gamma$ shows that every component of $a$ in a positive-eigenvalue direction must vanish. Thus zero-variance combinations correspond to null directions of $\Gamma$.

Common mistake: For $Y=B^1+B^2$, it is incorrect to write $[Y]_t=([B^1]_t+[B^2]_t)$, because this drops the cross terms. The correct expansion is $$ [Y]_t=[B^1]_t+[B^2]_t+2[B^1,B^2]_t =t(\Gamma_{11}+\Gamma_{22}+2\Gamma_{12}). $$ For the first example this gives $t(1+1+6/5)=16t/5$, not $2t$. The correction is essential whenever coordinates are correlated, and the matrix product $A\Gamma A^{\mathsf T}$ includes all such cross terms automatically.

Takeaway: For correlated Brownian coordinates, the complete second-order information is the matrix $[\mathbf{B}]_t=t\Gamma$. Under a linear transformation $A$, it becomes $tA\Gamma A^{\mathsf T}$; diagonal entries give quadratic variations, while off-diagonal entries give covariations.

Check Your Understanding

Use the matrix formulas and state any conditions needed for your conclusions.

If $\Gamma=\begin{pmatrix}4&1\\1&9\end{pmatrix}$, find the variance rates of $B^1+B^2$ and $2B^1-B^2$.
For a two-dimensional Brownian motion with $\Gamma_{11}=\Gamma_{22}=1$ and $\Gamma_{12}=-1/3$, find $[B^1+B^2,B^1-B^2]_t$.
Explain why $A\Gamma A^{\mathsf T}$ is positive semidefinite whenever $\Gamma$ is positive semidefinite.
Give a reason that a covariance matrix with a nonzero null vector can correspond to coordinates that are linearly dependent.
Identify the missing term in the incorrect calculation $[B^1+B^2]_t=[B^1]_t+[B^2]_t$, and state the correct formula.

Covariance Matrices of Multidimensional Brownian Motion

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