Multidimensional Brownian Representation and Covariation

From One Brownian Coordinate to Many

In a multidimensional Brownian filtration, information arrives through several independent noise coordinates. A square-integrable martingale can therefore have several stochastic-integral contributions at once. The representation theorem says that these contributions account for all its randomness, provided the filtration is generated by the Brownian motion itself. Covariation with each Brownian coordinate then identifies the corresponding contribution separately.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard $d$-dimensional Brownian motion $\mathbf{B}_t=(B_t^1,\ldots,B_t^d)$. For this tutorial, assume $(\mathcal{F}_t)$ is the usual completion of the natural filtration generated by $\mathbf{B}$. In particular, $\mathcal{F}_T$ contains no additional randomness beyond the Brownian paths up to time $T$. The coordinates are independent Brownian motions, so $dB_t^i\,dB_t^j=\mathbf{1}_{\{i=j\}}dt$, while $dB_t^i\,dt=0$ and $dt\,dt=0$.

Definition: A vector integrand $H=(H^1,\ldots,H^d)$ is in $\mathcal{L}^2([0,T];\mathbb{R}^d)$ if each coordinate is predictable (and hence progressively measurable) and $\mathbb{E}\int_0^T\sum_{j=1}^d(H_s^j)^2\,ds<\infty$. Its Ito integral is $\int_0^t H_s\cdot d\mathbf{B}_s=\sum_{j=1}^d\int_0^t H_s^j\,dB_s^j$. All stochastic integrals below are Ito integrals under $\mathbb{P}$.

The Ito isometry gives $\mathbb{E}[(\int_0^T H_s\cdot d\mathbf{B}_s)^2]=\mathbb{E}\int_0^T\sum_j(H_s^j)^2ds$. The integral process is a square-integrable martingale: its increments have conditional expectation zero given the past, first for predictable step integrands by independent Brownian increments and then for $\mathcal{L}^2$ integrands by the isometry and approximation. This is a true martingale, not merely a local martingale.

The Multidimensional Representation Theorem

Theorem: Fix $T<\infty$. If $X\in L^2(\Omega,\mathcal{F}_T,\mathbb{P})$, then there is a predictable vector integrand $H\in\mathcal{L}^2([0,T];\mathbb{R}^d)$ such that, for every $0\le t\le T$, $$ \mathbb{E}[X\mid\mathcal{F}_t]=\mathbb{E}[X]+\int_0^t H_s\cdot d\mathbf{B}_s. $$ The integrand is unique up to equality outside a set of measure zero for the product measure $dt\,d\mathbb{P}$. For a square-integrable $m$-dimensional martingale, apply the result to each coordinate: the integrand is an $m\times d$ matrix of predictable processes.

Proof. Consider the Hilbert space of square-integrable predictable vector integrands, with squared norm $\mathbb{E}\int_0^T\sum_j(H_s^j)^2ds$. By the Ito isometry, the map taking $H$ to $\int_0^T H_s\cdot d\mathbf{B}_s$ is an isometry into $L^2(\mathcal{F}_T)$. Its range is therefore closed: a convergent sequence of integrals corresponds, by the isometry, to a Cauchy sequence of integrands, whose limit has an integral equal to the original $L^2$ limit. Each integral has mean zero, so this closed range is contained in the mean-zero subspace of $L^2(\mathcal{F}_T)$.

To show that the range is the whole mean-zero subspace, let $Y\in L^2(\mathcal{F}_T)$ have mean zero and be orthogonal to every such integral. For a deterministic, piecewise constant vector function $h:[0,T]\to\mathbb{R}^d$, define $Z_t=\exp(\int_0^t h_s\cdot d\mathbf{B}_s-\tfrac12\int_0^t|h_s|^2ds)$. Ito's formula gives $dZ_t=Z_t h_t\cdot d\mathbf{B}_t$. Moreover, $\mathbb{E}[Z_t^2]=\exp(\int_0^t|h_s|^2ds)$, by the normal moment-generating function. Thus $\mathbb{E}\int_0^T Z_s^2|h_s|^2ds<\infty$, and $Z_T-1$ is an Ito integral in the range. Orthogonality and $\mathbb{E}[Y]=0$ imply $\mathbb{E}[YZ_T]=0$.

Now choose any finite partition of $[0,T]$ and let $U$ be the vector of all Brownian-coordinate increments on its intervals. Varying the constant values of $h$ on those intervals shows that $Y$ is orthogonal to every exponential of a linear function of $U$, with the Gaussian normalizing factor. Equivalently, the conditional expectation $g(U)=\mathbb{E}[Y\mid U]$ has zero Gaussian-weighted Laplace transform for every real parameter vector. Uniqueness of the Laplace transform of a finite-dimensional Gaussian-weighted integrable function implies $g(U)=0$ almost surely. Apply this to successive dyadic partitions: their increments generate the Brownian natural $\sigma$-field $\mathcal{F}_T$, and the $L^2$ martingale convergence theorem gives $\mathbb{E}[Y\mid\text{dyadic increments}]\to Y$ in $L^2$. Each conditional expectation is zero, hence $Y=0$.

Thus the orthogonal complement of the closed range inside the mean-zero subspace is trivial. The range is closed and dense there, so it equals that subspace. Apply this to $X-\mathbb{E}[X]$ to obtain the stated terminal representation. Taking conditional expectations at time $t$, the integral from $t$ to $T$ has conditional mean zero, giving the formula for every $t$. Coordinatewise application gives the vector-martingale statement. Uniqueness is proved below. $\square$

The existence proof uses only the Brownian-generated filtration assumption. If the filtration includes additional randomness, the orthogonality argument need not eliminate that randomness, and representation by $\mathbf{B}$ alone is not guaranteed. The theorem is stated under $\mathbb{P}$; it makes no change to a risk-neutral measure $\mathbb{Q}$. If a different measure is used, the Brownian motion and filtration hypotheses must be checked under that measure.

Worked Representations

Worked Example: A Linear Terminal Payoff

Let $T=2$ and $X=3B_2^1-2B_2^2+B_2^3$. Find the martingale $M_t=\mathbb{E}[X\mid\mathcal{F}_t]$ and its integrand.

Each future increment $B_2^j-B_t^j$ has conditional mean zero. Therefore $M_t=3B_t^1-2B_t^2+B_t^3$, and the vector integrand is $H_t=(3,-2,1)$. Indeed, $M_t=M_0+\int_0^t H_s\cdot d\mathbf{B}_s$, with $M_0=0$. Its integrand is square-integrable because $\mathbb{E}\int_0^2(3^2+(-2)^2+1^2)ds=2(9+4+1)=28$. This also equals $\mathbb{E}[X^2]$, by independence and the coordinate variances $\mathbb{E}[(B_2^j)^2]=2$.

Check: At $t=2$, the formula gives $M_2=X$. At $t=0$, it gives $M_0=0=\mathbb{E}[X]$. The Ito isometry gives $\mathbb{E}[(M_2-M_0)^2]=28$, agreeing with the direct variance calculation.

Worked Example: An Exponential Martingale in Three Dimensions

Set $T=1$ and $X=\exp(B_1^1+2B_1^2-B_1^3-3)$. Find its conditional-expectation martingale and integrand, and check both its expectation and integrability.

For $x=(x_1,x_2,x_3)$, define $f(t,x)=\exp(x_1+2x_2-x_3-3t)$. The conditional-expectation candidate is $M_t=f(t,\mathbf{B}_t)$. Its derivatives are $f_t=-3f$, $f_{x_1}=f$, $f_{x_2}=2f$, $f_{x_3}=-f$, and $f_{x_1x_1}=f$, $f_{x_2x_2}=4f$, $f_{x_3x_3}=f$. The multidimensional Ito correction is one half the sum of the three diagonal second derivatives; cross terms vanish because distinct Brownian coordinates have zero covariation. Hence $$ dM_t=\left(-3f+\tfrac12(f+4f+f)\right)dt +f\,dB_t^1+2f\,dB_t^2-f\,dB_t^3 =M_t\,dB_t^1+2M_t\,dB_t^2-M_t\,dB_t^3. $$ So $H_t=(M_t,2M_t,-M_t)$.

Since $B_1^1+2B_1^2-B_1^3$ is normal with mean zero and variance $1+4+1=6$, its exponential moment at parameter one is $e^{6/2}=e^3$. Thus $\mathbb{E}[X]=e^{-3}e^3=1=M_0$. Also $\mathbb{E}[M_t^2]=e^{6t}$, so $\mathbb{E}\int_0^1\sum_j(H_s^j)^2ds=6\int_0^1e^{6s}ds=e^6-1<\infty$. The integral is therefore a true square-integrable martingale.

Check: The Ito drift vanishes because $-3+\tfrac12(1+4+1)=0$. At $t=1$, the formula for $M_t$ gives exactly $X$, as required.

Covariation Identifies Each Coordinate

For continuous square-integrable martingales $M$ and $N$, $[M,N]_t$ denotes their quadratic covariation; it can be defined by polarization from quadratic variation. For an Ito integral $M_t=M_0+\sum_k\int_0^tH_s^k\,dB_s^k$, its covariation with a Brownian coordinate is $[M,B^j]_t=\int_0^tH_s^jds$. This follows first for predictable step integrands by summing products of increments and using $[B^k,B^j]_t=\mathbf{1}_{\{k=j\}}t$, and then for square-integrable integrands by the Ito-isometry approximation that defines the integral.

Proposition: If a square-integrable martingale has two representations with integrands $H$ and $K$, then $H^j=K^j$ for $dt\,d\mathbb{P}$-almost every $(t,\omega)$, for each $j=1,\ldots,d$. Thus the representing vector is unique up to product-measure almost-everywhere equality, not necessarily at every time and outcome.

Proof. Subtract the representations. The difference is the zero process, so its covariation with each $B^j$ is zero. The covariation formula gives, simultaneously for all $t$, $$ 0=\int_0^t(H_s^j-K_s^j)ds. $$ For almost every outcome, this integral is an absolutely continuous function of $t$ that is identically zero. Its derivative is therefore zero for Lebesgue-almost every $t$, giving $H_t^j=K_t^j$ for almost every $t$ on those outcomes. This is exactly equality $dt\,d\mathbb{P}$-almost everywhere. The argument holds for every coordinate, so the whole vector is unique in the stated sense. $\square$

Worked Example: Recovering a Matrix Integrand

Let $M^1_t=B_t^1+2B_t^2$ and $M^2_t=-B_t^1+3B_t^3$. Recover their integrand matrix from covariations and compute their mutual covariation.

The componentwise covariations are $[M^1,B^1]_t=t$, $[M^1,B^2]_t=2t$, $[M^1,B^3]_t=0$, and $[M^2,B^1]_t=-t$, $[M^2,B^2]_t=0$, $[M^2,B^3]_t=3t$. Since each bracket equals the time integral of the corresponding integrand entry, the matrix is $$ \begin{pmatrix} 1&2&0\\ -1&0&3 \end{pmatrix}. $$ Using the same-coordinate products in the covariation formula gives $[M^1,M^1]_t=(1^2+2^2)t=5t$, $[M^1,M^2]_t=(1)(-1)t+(2)(0)t+(0)(3)t=-t$, and $[M^2,M^2]_t=((-1)^2+0^2+3^2)t=10t$.

Check: The matrix product of the first row with the second row is $-1$, matching the coefficient of $t$ in $[M^1,M^2]_t$. The diagonal row products are $1+4=5$ and $1+9=10$, matching the two quadratic variations.

A Common Covariation Pitfall

For a scalar martingale represented by $H$, its quadratic variation is $[M]_t=\int_0^t\sum_j(H_s^j)^2ds$. This records the squared length of the integrand, not its direction. For example, $B^1$ and $B^2$ both have quadratic variation $t$, but they are different processes. It is therefore incorrect to identify a vector integrand from $[M]$ alone. If $M=B^1$, the candidate integrand $(0,1)$ has the same quadratic variation as $(1,0)$, but it produces $B^2$, not $B^1$. The componentwise check detects the error: $[B^1,B^1]_t=t$, whereas $[B^2,B^1]_t=0$.

In applications, componentwise covariations are useful both for constructing a candidate integrand and for checking uniqueness. They also explain why uniqueness is naturally an almost-everywhere statement: changing an integrand on a set of zero $dt\,d\mathbb{P}$ measure does not change its Ito integral. The representation theorem applies to square-integrable martingales on the Brownian-generated filtration; representation of more general local martingales is a separate question.

Check Your Understanding

Use the componentwise representation and covariation formulas in your answers.

For $X=2B_T^1-B_T^2$, write the conditional-expectation martingale and its vector integrand.
If $M_t=\int_0^t(H_s^1,H_s^2)\cdot d(B_s^1,B_s^2)$, express $[M,B^1]_t$ and $[M,B^2]_t$.
Why does equality of two scalar martingales imply equality of their representing integrands only $dt\,d\mathbb{P}$-almost everywhere?
Give two different two-dimensional integrands with the same instantaneous squared length, and explain why their quadratic-variation rates alone do not identify the integrand.
Which filtration assumption is needed for the stated representation theorem, and what can go wrong if the filtration contains additional randomness?