The Multidimensional Levy Characterization

From Pairwise Quadratic Covariation to Vector Brownian Motion

In one dimension, Lévy's characterization identifies Brownian motion among continuous local martingales by its quadratic variation. In several dimensions, the diagonal quadratic variations are not enough: one must also control the covariations between distinct components. Those cross terms determine whether the coordinates have independent increments or are correlated.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ satisfying the usual conditions. No Brownian motion driving the processes below is assumed in advance. A continuous local martingale is an adapted continuous process that becomes a martingale when stopped at a suitable increasing sequence of stopping times. Its quadratic covariation $[M^i,M^j]_t$ is the limit of sums of products of increments along refining partitions; in particular, $[M^i]_t=[M^i,M^i]_t$. Ito integrals, when used, are defined for progressively measurable integrands in $\mathcal{L}^2$, meaning $\mathbb{E}\int_0^T H_s^2\,ds<\infty$; locally square-integrable integrands are handled by localization.

Definition: A $d$-dimensional standard Brownian motion relative to $(\mathcal{F}_t)$ is a continuous adapted process $\mathbf{B}_t=(B_t^1,\ldots,B_t^d)$, with $\mathbf{B}_0=0$, such that for every $0\le s<t$, the increment $\mathbf{B}_t-\mathbf{B}_s$ is independent of $\mathcal{F}_s$ and has the centered Gaussian distribution with covariance matrix $(t-s)I_d$.

The independence from $\mathcal{F}_s$, not merely the Gaussian distribution of each increment, is essential: it makes the process Brownian relative to the given filtration.

The Multidimensional Characterization

Theorem: Let $\mathbf{M}_t=(M_t^1,\ldots,M_t^d)$ be a continuous local martingale with $\mathbf{M}_0=0$. Then $\mathbf{M}$ is a $d$-dimensional standard Brownian motion relative to $(\mathcal{F}_t)$ if and only if $$ [M^i,M^j]_t=\delta_{ij}t\qquad\text{for every }i,j\in\{1,\ldots,d\}\text{ and }t\ge0, $$ where $\delta_{ij}=1$ if $i=j$ and $0$ otherwise.

Proof. First suppose that $\mathbf{M}$ is a standard Brownian motion. Its components are continuous martingales. For a partition of $[0,T]$, write $\Delta t_k=t_k-t_{k-1}$ and $\Delta M_k^i=M_{t_k}^i-M_{t_{k-1}}^i$. The increments on distinct partition intervals are independent. For one interval, $\mathbb{E}[(\Delta M_k^i)^2]=\Delta t_k$ and $\operatorname{Var}((\Delta M_k^i)^2)=2(\Delta t_k)^2$, so the sum of squared increments converges in $L^2$ to $T$, since its variance is $2\sum_k(\Delta t_k)^2\le2T|\Pi|$. For $i\ne j$, the coordinate increments on each interval are independent and centered, so $\mathbb{E}[\Delta M_k^i\Delta M_k^j]=0$ and $\operatorname{Var}(\Delta M_k^i\Delta M_k^j)=(\Delta t_k)^2$. The sum of their products therefore converges in $L^2$ to zero, since its variance is $\sum_k(\Delta t_k)^2\le T|\Pi|$. These partition limits give $[M^i]_T=T$ and $[M^i,M^j]_T=0$ for $i\ne j$; applying the same argument on every interval $[0,t]$ gives the asserted identities.

Conversely, suppose the stated covariations hold. Fix $\theta=(\theta_1,\ldots,\theta_d)\in\mathbb{R}^d$, and put $q=\theta^\mathsf{T}\theta=\sum_j\theta_j^2$. Consider $f(t,x)=\exp(i\theta\cdot x+\tfrac12qt)$, where $\theta\cdot x=\sum_j\theta_jx_j$. Its derivatives are $f_t=\tfrac12qf$, $f_{x_j}=i\theta_jf$, and $f_{x_ix_j}=-\theta_i\theta_jf$. Ito's formula, including its second-order correction, gives $$ df(t,\mathbf{M}_t)=\tfrac12qf\,dt+\sum_j i\theta_jf\,dM_t^j +\tfrac12\sum_{i,j}(-\theta_i\theta_jf)\,d[M^i,M^j]_t =\sum_j i\theta_jf\,dM_t^j. $$ The time drift cancels because $d[M^i,M^j]_t=\delta_{ij}\,dt$. Thus $Z_t=f(t,\mathbf{M}_t)$ is a local martingale.

For each finite $T$, $|Z_t|=\exp(\tfrac12qt)\le\exp(\tfrac12qT)$ for $0\le t\le T$. A bounded local martingale is a true martingale: stop along a localizing sequence, use the martingale conditional-expectation identity for each stopped process, and pass to the limit by dominated convergence, using this deterministic bound. Therefore, for $0\le s<t$, the conditional expectation of $Z_t$ given $\mathcal{F}_s$ equals $Z_s$. Dividing by the nonzero $Z_s$ yields $$ \mathbb{E}\!\left[\exp\!\big(i\theta\cdot(\mathbf{M}_t-\mathbf{M}_s)\big)\mid\mathcal{F}_s\right] =\exp\!\left(-\tfrac12q(t-s)\right). $$ The right side is the characteristic function of the centered Gaussian distribution with covariance $(t-s)I_d$, and it is deterministic. To see that this also gives independence from $\mathcal{F}_s$, multiply by $\mathbf{1}_C$ for any $C\in\mathcal{F}_s$ and take expectations: the joint characteristic function factors as $\mathbb{P}(C)\exp(-q(t-s)/2)$. Uniqueness of characteristic functions gives the required independence and Gaussian law. Together with continuity, adaptedness, and $\mathbf{M}_0=0$, this proves that $\mathbf{M}$ is standard Brownian motion. $\square$

The proof uses a scalar exponential for each vector $\theta$. The full matrix of covariations matters because it supplies the quadratic term $\theta^\mathsf{T}\theta$ for every direction $\theta$.

Constant Covariance Matrices

Theorem: Let $A$ be a fixed symmetric positive semidefinite $d\times d$ matrix, and let $\mathbf{M}$ be a continuous local martingale with $\mathbf{M}_0=0$. If $[M^i,M^j]_t=A_{ij}t$ for all $i,j,t$, then for every $s<t$, $\mathbf{M}_t-\mathbf{M}_s$ is independent of $\mathcal{F}_s$ and is centered Gaussian with covariance $(t-s)A$. Conversely, a continuous vector Brownian motion with this increment law has these covariations.

Proof. For a fixed $\theta\in\mathbb{R}^d$, set $q=\theta^\mathsf{T}A\theta$, which is nonnegative because $A$ is positive semidefinite. Apply Ito's formula to $f(t,x)=\exp(i\theta\cdot x+\tfrac12qt)$. Its derivatives are $f_t=\tfrac12qf$, $f_{x_j}=i\theta_jf$, and $f_{x_ix_j}=-\theta_i\theta_jf$. The correction is $-\tfrac12\sum_{i,j}\theta_i\theta_jA_{ij}f\,dt=-\tfrac12qf\,dt$, which cancels $f_t\,dt$. The resulting local martingale is bounded in modulus on each finite time interval, so the localization and dominated-convergence argument in the preceding proof makes it a true martingale. Consequently the conditional characteristic function of $\mathbf{M}_t-\mathbf{M}_s$ is $\exp(-\tfrac12(t-s)\theta^\mathsf{T}A\theta)$. This is the characteristic function of the stated centered Gaussian law, including when $A$ is singular; the same indicator argument proves independence from $\mathcal{F}_s$.

For the converse, the partition calculation in the first proof applies with the increment covariance $\Delta t\,A$: the expected product of coordinates $i,j$ over an interval is $A_{ij}\Delta t$, and the variance of each product is a constant depending on $A$ times $(\Delta t)^2$. Summing these variances gives a bound proportional to $\sum_k(\Delta t_k)^2$, which tends to zero with the mesh. Thus the covariation limit is $A_{ij}t$. $\square$

If $A$ is positive definite, the process $A^{-1/2}\mathbf{M}$ has covariations $\delta_{ij}t$, so the first theorem makes it a standard vector Brownian motion. If $A$ is singular, the Gaussian increment law is degenerate in the null directions of $A$; it is a Brownian motion with constant covariance $A$, but it is not a nondegenerate standard $d$-dimensional Brownian motion. Symmetry and positive semidefiniteness are necessary: a matrix of pairwise covariations must be symmetric, and its quadratic forms are nonnegative.

Worked Covariation Checks

Worked Example: A Correlated Pair

Let $B^1,B^2$ be independent standard Brownian motions on the filtered space, and define $X_t=B_t^1$ and $Y_t=\tfrac35B_t^1+\tfrac45B_t^2$. Find the covariance matrix of their increments and classify the pair.

Both processes are continuous martingales starting at zero. Using bilinearity of covariation and $[B^1,B^1]_t=[B^2,B^2]_t=t$, $[B^1,B^2]_t=0$, we obtain $[X]_t=t$ and $$ [Y]_t=\left(\tfrac35\right)^2t+\left(\tfrac45\right)^2t =\left(\tfrac9{25}+\tfrac{16}{25}\right)t=t. $$ The cross term is $$ [X,Y]_t=\tfrac35[B^1,B^1]_t+\tfrac45[B^1,B^2]_t=\tfrac35t. $$ Hence the covariance matrix per unit time is $\begin{pmatrix}1&3/5\\3/5&1\end{pmatrix}$. The constant-covariance theorem shows that $(X,Y)$ has Gaussian increments with this covariance. It is not standard two-dimensional Brownian motion, because its cross-covariation is not zero.

Check: the matrix is positive definite: its determinant is $1-(3/5)^2=1-9/25=16/25>0$. The computed diagonal entries are both $1$, while the off-diagonal entry $3/5$ matches the direct cross-covariation.

Worked Example: A Singular Covariance Matrix

Let $B$ be standard Brownian motion and set $\mathbf{M}_t=(B_t,2B_t)$. Compute every entry of its covariance matrix and explain why it is not standard two-dimensional Brownian motion.

The two coordinates have covariations $[M^1]_t=[B]_t=t$, $[M^1,M^2]_t=[B,2B]_t=2t$, and $[M^2]_t=[2B]_t=4t$. Thus $A=\begin{pmatrix}1&2\\2&4\end{pmatrix}$. It is positive semidefinite and singular: its determinant is $1\cdot4-2\cdot2=0$. The constant-covariance theorem gives Gaussian increments with covariance $(t-s)A$; the relation $M_t^2=2M_t^1$ shows directly why the law is degenerate.

Check: for any vector $(u,v)$, its quadratic form is $u^2+4uv+4v^2=(u+2v)^2\ge0$. Also $[M^2]_t=4t$, exactly the second diagonal entry times $t$.

Worked Example: Testing the Identity Covariation Condition

Suppose $\mathbf{N}_t=(N_t^1,N_t^2)$ is a continuous local martingale starting at zero and its covariations are $[N^1]_t=2t$, $[N^2]_t=3t$, and $[N^1,N^2]_t=0$. Which Brownian characterization applies?

The matrix is $A=\begin{pmatrix}2&0\\0&3\end{pmatrix}$, which is symmetric positive definite. The constant-covariance theorem applies, so increments are centered Gaussian, independent of the past, with covariance $(t-s)\begin{pmatrix}2&0\\0&3\end{pmatrix}$. The original $\mathbf{N}$ is not standard two-dimensional Brownian motion because its diagonal covariations are $2t$ and $3t$, rather than $t$.

Normalize the components by setting $U_t=N_t^1/\sqrt2$ and $V_t=N_t^2/\sqrt3$. Then $[U]_t=(1/2)2t=t$, $[V]_t=(1/3)3t=t$, and $[U,V]_t=(1/\sqrt6)0=0$. The multidimensional characterization shows that $(U,V)$ is standard two-dimensional Brownian motion.

Check: multiplying the normalized covariations by $\sqrt2$ and $\sqrt3$ recovers $[N^1]_t=2t$ and $[N^2]_t=3t$; the zero cross term remains zero.

Common Pitfall: Checking Only the Diagonal

A common error is to check $[M^i]_t=t$ for every coordinate and conclude that the vector is standard Brownian motion. That verifies the variance growth of each coordinate separately, but says nothing about dependence between the coordinates. For instance, with $X=B^1$ and $Y=\tfrac35B^1+\tfrac45B^2$, both diagonal covariations equal $t$. The incorrect calculation would stop there and call $(X,Y)$ standard Brownian motion. The correction is to compute the cross term: $[X,Y]_t=\tfrac35t$, not zero. The required identity matrix condition fails, so this pair is correlated Brownian motion with a nonidentity covariance matrix.

The distinction matters whenever a vector process is used as a noise source: cross-covariations encode instantaneous correlation. The full matrix condition, together with continuity and the local martingale assumption, yields conditional Gaussian increments; diagonal checks alone do not.

Check Your Understanding

For each question, distinguish the identity-covariance characterization from the general constant-covariance result.

Why does the proof use the exponential associated with every vector $\theta$, rather than checking the coordinates one at a time?
If $[M^1]_t=[M^2]_t=t$ but $[M^1,M^2]_t=\rho t$ for a fixed $\rho$ with $0<|\rho|<1$, what is the covariance matrix of an increment over $[s,t]$?
What is the role of the deterministic bound on the modulus of the exponential local martingale?
For $A=\begin{pmatrix}1&1\\1&1\end{pmatrix}$, identify a nonzero vector in the null space and explain what degeneracy it implies for increments.
Which hypotheses in the characterization ensure that the increment law is Brownian relative to the filtration, rather than merely Gaussian at each fixed time?