Quadratic Variation as a Diagnostic for Representing Integrands

What the Brackets Reveal

A Brownian martingale representation expresses a martingale as a stochastic integral. Its integrand is not merely a formal coefficient: it determines the martingale’s accumulated quadratic variation and its covariation with each Brownian component. These quantities provide a practical diagnostic. The quadratic variation checks the integrand’s total squared size, while componentwise covariations retain its signs and directions.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a $d$-dimensional Brownian motion $\mathbf{B}_t=(B_t^1,\ldots,B_t^d)$, with respect to the given filtration. Fix $T>0$. Let $H=(H^1,\ldots,H^d)$ be progressively measurable and satisfy $\mathbb{E}\int_0^T |H_s|^2\,ds<\infty$, where $|H_s|^2=\sum_{k=1}^d(H_s^k)^2$. These conditions make $H$ an admissible square-integrable Ito integrand. Consider $M_t=M_0+\int_0^t H_s\cdot d\mathbf{B}_s$, where $M_0$ is square-integrable and $\mathcal{F}_0$-measurable.

Definition: The quadratic variation $[M]_t$ records the accumulated squared increments of a continuous martingale. The covariation $[M,B^j]_t$ records the corresponding cross increments with the $j$th Brownian component. For represented martingales, these processes have the identities proved below.

The Ito integral is a square-integrable martingale: for $0\le s\le t\le T$, its increment has conditional expectation zero given $\mathcal{F}_s$. Consequently $M$ is a square-integrable martingale, and its bracket processes can be used to inspect the representation. In differential shorthand, the relevant multiplication rules are $dB_t^j\,dB_t^k=\mathbf{1}_{\{j=k\}}dt$, $dB_t^j\,dt=0$, and $dt\,dt=0$. The bracket identities below are the precise integrated statements, not ordinary chain-rule calculations.

The Bracket Identities

Theorem: Under the assumptions above, for every $t\in[0,T]$, almost surely, $$ [M]_t=\int_0^t\sum_{k=1}^d(H_s^k)^2\,ds, \qquad [M,B^j]_t=\int_0^t H_s^j\,ds,\quad j=1,\ldots,d. $$

Proof. First consider quadratic variation. By the Ito isometry, the stochastic integral has conditional second moments $$ \mathbb{E}\left[\left(\int_s^t H_u\cdot d\mathbf{B}_u\right)^2\middle|\mathcal{F}_s\right] =\mathbb{E}\left[\int_s^t|H_u|^2\,du\middle|\mathcal{F}_s\right]. $$ The increment has conditional mean zero. Expanding $M_t^2=(M_s+(M_t-M_s))^2$ and taking conditional expectations therefore gives $$ \mathbb{E}[M_t^2\mid\mathcal{F}_s] =M_s^2+\mathbb{E}\left[\int_s^t|H_u|^2\,du\middle|\mathcal{F}_s\right]. $$ Thus $M_t^2-\int_0^t|H_u|^2du$ is a martingale. It is integrable because $\mathbb{E}[M_t^2]<\infty$ and $\mathbb{E}\int_0^t|H_u|^2du<\infty$. The quadratic variation characterization says that $M_t^2-[M]_t$ is a local martingale. The finite-variation compensator in this decomposition is unique: the difference of two such compensators would be a continuous finite-variation local martingale starting at zero, and such a process is identically zero. Hence $[M]_t=\int_0^t|H_u|^2du$.

For the covariation, use polarization of quadratic variation: $$ [M,B^j]_t=\frac14\left([M+B^j]_t-[M-B^j]_t\right). $$ The integrands representing $M+B^j$ and $M-B^j$ are respectively $H+e_j$ and $H-e_j$, where $e_j$ is the vector with $j$th coordinate $1$ and all other coordinates $0$. Applying the quadratic variation identity just proved, their bracket difference is the integral of $$ |H_s+e_j|^2-|H_s-e_j|^2 =(|H_s|^2+2H_s^j+1)-(|H_s|^2-2H_s^j+1)=4H_s^j. $$ Dividing by $4$ proves the covariation identity. All identities hold simultaneously for all $t\in[0,T]$ after choosing the continuous versions of the bracket processes. $\square$

The theorem gives two different kinds of information. The pathwise derivative of $[M]_t$, where it exists, is $|H_t|^2$: this gives the squared magnitude of the vector integrand, not its individual coordinates. In contrast, the derivative of $[M,B^j]_t$ is $H_t^j$. Since each covariation is an integral in time, its derivative recovers that coordinate for almost every time, in the product-measure sense $dt\,d\mathbb{P}$. Derivatives here are not asserted to exist at every time on every sample path.

Recovering and Checking Integrands

Worked Example: A Constant Integrand in Two Dimensions

Let $d=2$, $M_0=0$, and $M_t=B_t^1+2B_t^2$ on $[0,2]$. Recover the integrand from the brackets, then check the result independently.

The proposed vector is $H=(1,2)$, which is progressively measurable and satisfies $\mathbb{E}\int_0^2|H_s|^2ds=\int_0^2(1+4)ds=10<\infty$. The theorem gives $$ [M]_t=\int_0^t(1^2+2^2)\,ds=5t,\quad [M,B^1]_t=\int_0^t1\,ds=t,\quad [M,B^2]_t=\int_0^t2\,ds=2t. $$ Differentiating the covariations recovers $H^1=1$ and $H^2=2$; differentiating $[M]$ only recovers $1^2+2^2=5$.

Check: Independence of the Brownian components gives $\mathbb{E}[M_t]=0$ and $\mathbb{E}[M_t^2]=\mathbb{E}[(B_t^1)^2]+4\mathbb{E}[(B_t^2)^2]=t+4t=5t$, since the cross term has expectation zero. This agrees with $\mathbb{E}[M]_t=5t$. At $t=2$, the bracket is $5\cdot2=10$, matching the integral $\int_0^2(1+4)ds=10$.

Worked Example: Recovering a Random Integrand

In one dimension, take $M_0=0$ and $M_t=\int_0^t B_s\,dB_s$ on $[0,2]$. Compute its brackets, recover the integrand, and check the representation.

Here $H_s=B_s$, and $\mathbb{E}\int_0^2B_s^2ds=\int_0^2s\,ds=2<\infty$. The identities give $$ [M]_t=\int_0^tB_s^2\,ds,\qquad [M,B]_t=\int_0^tB_s\,ds=\frac12(B_t^2-t). $$ The last equality follows from Ito's formula applied to $f(t,x)=\tfrac12(x^2-t)$: $f_t=-\tfrac12$, $f_x=x$, $f_{xx}=1$, so the drift contribution is $f_t+\tfrac12 f_{xx}=0$, and $df(t,B_t)=B_t\,dB_t$. Thus differentiating $[M,B]$ recovers $B_t$ for almost every time.

Check: The same Ito calculation verifies $M_t=\tfrac12(B_t^2-t)$. For a standard normal variable $B_t$ with variance $t$, $\mathbb{E}[B_t^2]=t$, so $\mathbb{E}[M_t]=\tfrac12(t-t)=0$, as required for this square-integrable Ito integral. Also $\mathbb{E}[M]_2=\int_0^2\mathbb{E}[B_s^2]ds=\int_0^2s\,ds=2$. Using the representation, $\mathbb{E}[M_2^2]=\tfrac14\operatorname{Var}(B_2^2)=\tfrac14(2\cdot2^2)=2$, in agreement with the Ito isometry.

Worked Example: Same Quadratic Variation, Wrong Integrand

Let $M_t=B_t^1-B_t^2$ on $[0,1]$. Test the candidate $G=(1,1)$, which would represent $N_t=B_t^1+B_t^2$. Why does checking quadratic variation alone fail?

The true integrand is $H=(1,-1)$, so $[M]_t=\int_0^t(1^2+(-1)^2)ds=2t$. The candidate also has $[N]_t=\int_0^t(1^2+1^2)ds=2t$. Thus their quadratic variations agree at every time. But the covariations differ: $$ [M,B^1]_t=t,\quad [M,B^2]_t=-t,\qquad [N,B^1]_t=t,\quad [N,B^2]_t=t. $$ In particular, at $t=1$, the second component is $-1$ for $M$ and $1$ for $N$. Componentwise covariation rejects the candidate.

Check: Independence and unit variances give $\mathbb{E}[M_1^2]=1+1=2$ and $\mathbb{E}[N_1^2]=1+1=2$, agreeing with their common bracket at $1$. The variance check cannot distinguish them; their covariation with $B^2$ does distinguish them because $\mathbb{E}[M_1B_1^2]=-1$ while $\mathbb{E}[N_1B_1^2]=1$.

When the Diagnostic Identifies the Representation

Proposition: Suppose $M$ has a representation with integrand $H$ satisfying the theorem’s assumptions, and $G$ is another progressively measurable integrand with $\mathbb{E}\int_0^T|G_s|^2ds<\infty$. If $[M,B^j]_t=\int_0^tG_s^jds$ for every $j$ and every $t\in[0,T]$, almost surely, then $G=H$ $dt\,d\mathbb{P}$-almost everywhere. In particular, the two integrands represent the same Ito integral.

Proof. The theorem gives $[M,B^j]_t=\int_0^tH_s^jds$. The assumed equalities imply $\int_0^t(H_s^j-G_s^j)ds=0$ for every $t$, almost surely. An absolutely continuous function that is identically zero has derivative zero almost everywhere, so $H_s^j=G_s^j$ for Lebesgue-almost every $s$, almost surely. Taking the finite union of the exceptional sets over $j=1,\ldots,d$ gives $H=G$ $dt\,d\mathbb{P}$-almost everywhere. The Ito isometry then yields $$ \mathbb{E}\left[\left(\int_0^T(H_s-G_s)\cdot d\mathbf{B}_s\right)^2\right] =\mathbb{E}\int_0^T|H_s-G_s|^2ds=0. $$ Thus the stochastic integrals agree almost surely at $T$, and the same argument on each time interval gives indistinguishable continuous integral processes. $\square$

The proposition makes the role of covariations precise: when a representation is known to exist, matching every component’s covariation identifies its integrand up to the unavoidable $dt\,d\mathbb{P}$-null sets. Quadratic variation supplies a useful additional consistency check, but it does not replace the componentwise tests. In a representation problem, the bracket formulas are necessary diagnostics; by themselves, they do not establish that an arbitrary process is a martingale or that it has a Brownian representation.

Common Pitfall

Using only the squared magnitude loses direction. In two dimensions, the wrong calculation “$[B^1-B^2]_t=2t$, while the proposed integrand $(1,1)$ also has squared magnitude $2$, so it is a valid representation” ignores the signs of the covariations. Both processes have bracket $2t$, but $[B^1-B^2,B^2]_t=-t$, whereas $[B^1+B^2,B^2]_t=t$. The correction is to compute $[M,B^j]$ for every component $j$, not only $[M]$.

Quantity	Information about $H$	Diagnostic use
$[M]_t$	$\int_0^t\|H_s\|^2ds$	Checks accumulated squared magnitude
$[M,B^j]_t$	$\int_0^tH_s^jds$	Recovers the $j$th coordinate, almost everywhere in time
All component covariations	The full vector integrand, up to $dt\,d\mathbb{P}$-null sets	Identifies and checks a represented martingale’s integrand

Check Your Understanding

Use the bracket identities and the stated integrability assumptions to answer each question.

If $M_t=\int_0^t(2,-3)\cdot d\mathbf{B}_s$, compute $[M]_t$ and both componentwise covariations.
For $M_t=\int_0^t H_s\cdot d\mathbf{B}_s$, what information about $H_t$ can be recovered from $[M]_t$ alone?
Explain why equality of all $[M,B^j]$ processes identifies two square-integrable representing integrands only up to $dt\,d\mathbb{P}$-almost-everywhere equality.
In one dimension, if $[M,B]_t=\int_0^t\sin(s)\,ds$, what integrand does the covariation diagnose for almost every time?
State the progressively measurable integrability condition that makes the Ito integral in this tutorial square-integrable on $[0,T]$.

Quantity	Information about \(H\)	Diagnostic use
\([M]_t\)	\(\int_0^t\|H_s\|^2ds\)	Checks accumulated squared magnitude
\([M,B^j]_t\)	\(\int_0^tH_s^jds\)	Recovers the \(j\)th coordinate, almost everywhere in time
All component covariations	The full vector integrand, up to \(dt\,d\mathbb{P}\)-null sets	Identifies and checks a represented martingale’s integrand