Covariation With the Driving Brownian Motion

Why Covariation Remembers the Integrand

Let $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ be a filtered probability space satisfying the usual conditions, and let $B_t$ be a Brownian motion adapted to this filtration. Consider the Ito integral $M_t=\int_0^t H_s\,dB_s$. Its quadratic variation records the squared size of its increments and is $[M]_t=\int_0^t H_s^2\,ds$, after localization. Its covariation with the Brownian motion is different: it records the signed alignment of the increments of $M$ and $B$. The result is $[M,B]_t=\int_0^t H_s\,ds$.

For a finite time horizon $T$, the Ito integral is defined for progressively measurable $H$ in the class $\mathcal{L}^2$, meaning $\mathbb{E}\int_0^T H_s^2\,ds<\infty$. We will also allow progressively measurable $H$ that is locally square-integrable: $\int_0^T H_s^2\,ds<\infty$ almost surely for every finite $T$. In the latter case the integral is defined by localization. The resulting process $M$ is a continuous local martingale; without additional integrability, it need not be a true martingale.

Definition: For continuous local martingales $X$ and $Y$, their quadratic covariation $[X,Y]_t$ is the limit in probability of the cross-increment sums $\sum_i (X_{t_i}-X_{t_{i-1}})(Y_{t_i}-Y_{t_{i-1}})$, as the partition mesh tends to zero. Equivalently, polarization gives $[X,Y]=\tfrac12([X+Y]-[X]-[Y])$. In particular, $[B]_t=t$.

The polarization identity follows directly from the algebraic identity $2\Delta X\,\Delta Y=(\Delta X+\Delta Y)^2-(\Delta X)^2-(\Delta Y)^2$. It converts a cross-increment limit into quadratic variation limits, which are available for localized Ito integrals.

The Covariation Formula

Theorem: Suppose $H$ is progressively measurable and locally square-integrable, and define $M_t=\int_0^t H_s\,dB_s$. Then, for every $t\ge0$, almost surely, $$ [M,B]_t=\int_0^t H_s\,ds. $$

Proof. Fix $T<\infty$. Define the accumulated energy $A_t=\int_0^t H_s^2\,ds$, and let $\tau_n=\inf\{t\ge0:A_t\ge n\}$, with $\inf\varnothing=\infty$. These are stopping times because $A$ is adapted and continuous. Local square-integrability implies that, almost surely, $\tau_n>T$ for all sufficiently large $n$. Since $A$ is continuous and nondecreasing, $A_{T\wedge\tau_n}\le n$. The stopped integrand $H_s\mathbf{1}_{\{s\le\tau_n\}}$ is progressively measurable and satisfies $\mathbb{E}\int_0^T H_s^2\mathbf{1}_{\{s\le\tau_n\}}\,ds\le n$.

Write $M^n_t=M_{t\wedge\tau_n}$ and $B^n_t=B_{t\wedge\tau_n}$. The stopped process $M^n+B^n$ is the Ito integral with integrand $(H_s+1)\mathbf{1}_{\{s\le\tau_n\}}$. This integrand belongs to $\mathcal{L}^2$ on $[0,T]$, because $$ \mathbb{E}\int_0^T (H_s+1)^2\mathbf{1}_{\{s\le\tau_n\}}\,ds \le 2n+2T<\infty. $$ By the quadratic-variation formula for a localized Ito integral, applied to the three stopped integrals, $$ [M^n]_t=\int_0^{t\wedge\tau_n}H_s^2\,ds,\qquad [B^n]_t=t\wedge\tau_n,\qquad [M^n+B^n]_t=\int_0^{t\wedge\tau_n}(H_s+1)^2\,ds $$ for $0\le t\le T$. Polarization now gives $$ [M^n,B^n]_t =\tfrac12\int_0^{t\wedge\tau_n}\big((H_s+1)^2-H_s^2-1\big)\,ds =\int_0^{t\wedge\tau_n}H_s\,ds. $$ For almost every outcome, $\tau_n>T$ eventually. Thus, on that event, the stopped covariation and integral agree with the unstopped ones throughout $[0,T]$ for all sufficiently large $n$. This proves the asserted identity on $[0,T]$. Taking integer $T$ gives the identity for all finite times, almost surely. $\square$

This proof uses the earlier quadratic-variation result for localized Ito integrals, but not a formula for the covariation of two general Ito integrals. The special role of the driving Brownian motion is that adding it to $M$ simply changes the integrand from $H$ to $H+1$.

Worked Applications

Worked Example: A Constant Integrand

Let $H_s=3$ and $M_t=\int_0^t 3\,dB_s$. Find $[M,B]_t$, and check it directly using quadratic variation sums.

The formula gives $$ [M,B]_t=\int_0^t 3\,ds=3t. $$ Also $M_t=3B_t$, so each cross-increment sum is $\sum_i(3\Delta B_i)(\Delta B_i)=3\sum_i(\Delta B_i)^2$. Since Brownian quadratic variation sums converge in probability to $t$, this cross-increment sum converges to $3t$ as claimed.

Check: the quadratic variation of $M$ is $[M]_t=9t$, not $3t$. The cross-covariation is $3t$ because one factor in each cross-increment is $\Delta B_i$, rather than another copy of $\Delta M_i$.

Worked Example: An Integrand That Stops at a Fixed Time

Fix $a>0$, take $H_s=\mathbf{1}_{\{s\le a\}}$, and compute the covariation for all $t\ge0$. Here the integrand is deterministic, progressively measurable, and has finite energy on every bounded interval.

Integrating the indicator gives $$ [M,B]_t=\int_0^t\mathbf{1}_{\{s\le a\}}\,ds =\begin{cases} t,&0\le t\le a,\\ a,&t>a. \end{cases} $$ The Ito integral itself is $M_t=B_{t\wedge a}$. Before time $a$, its increments agree with Brownian increments; after time $a$, its increments are zero. Thus the cross-increment sums accumulate Brownian quadratic variation only up to $a$, and their limit is $t\wedge a$, matching the displayed result.

Check: at $t=0$ the covariation is $0$; at $t=a$ it is $a$; and for $t>a$ it remains $a$, because the stopped process has no further increments.

Worked Example: The Brownian Motion as Integrand

Take $H_s=B_s$ on a finite interval $[0,T]$. This integrand is progressively measurable and belongs to $\mathcal{L}^2$, since $$ \mathbb{E}\int_0^T B_s^2\,ds=\int_0^T s\,ds=\frac{T^2}{2}<\infty. $$ Compute the integral and its covariation with $B$.

The covariation formula gives $$ \left[\int_0^\cdot B_s\,dB_s,B\right]_t=\int_0^t B_s\,ds. $$ To compute the Ito integral, apply Ito's formula to $f(x)=x^2$. Its derivatives are $f'(x)=2x$ and $f''(x)=2$, so the correction term is $\tfrac12 f''(B_t)\,dt=dt$. Therefore $$ d(B_t^2)=2B_t\,dB_t+dt,\qquad \int_0^t B_s\,dB_s=\frac12(B_t^2-t). $$

Check: the integral starts at zero because $B_0=0$, and its expectation is $\tfrac12(\mathbb{E}[B_t^2]-t)=\tfrac12(t-t)=0$. The covariation is generally random; for example, it is not obtained by squaring the integrand. The corresponding quadratic variation is instead $\left[\int_0^\cdot B_s\,dB_s\right]_t=\int_0^t B_s^2\,ds$.

What the Formula Says—and Does Not Say

The integrand appears to the first power in $[M,B]$, while it appears squared in $[M]$. This is the covariation analogue of the multiplication rules $dB_t\,dB_t=dt$, $dB_t\,dt=0$, and $dt\,dt=0$: formally, $dM_t=H_t\,dB_t$, so $dM_t\,dB_t=H_t\,dt$. The bracket formula makes that shorthand precise through limits of increment products.

Proposition: If $H_s\ge0$ for almost every $s$, almost surely, then $[M,B]$ is nondecreasing. Moreover, $[M,B]_t=0$ for every $t\ge0$, almost surely, if and only if $H=0$ almost everywhere with respect to time, almost surely.

Proof. By the theorem, $[M,B]_t=\int_0^t H_s\,ds$. A time integral of a nonnegative function is nondecreasing, proving the first claim. If $H=0$ almost everywhere in time, then every integral $\int_0^tH_s\,ds$ is zero. Conversely, suppose these integrals vanish for every $t$, on a single probability-one event. For each outcome in that event, the indefinite Lebesgue integral of the locally integrable function $s\mapsto H_s$ is identically zero. The fundamental theorem for Lebesgue integrals says its derivative equals $H_s$ for almost every $s$; since the derivative of the zero function is zero, $H_s=0$ for almost every $s$. This proves the equivalence. $\square$

The nonnegativity condition matters for the monotonicity claim. If $H$ changes sign, its accumulated integral may increase or decrease, even though the theorem remains valid. Also, the theorem concerns covariation, not ordinary covariance of random variables at a fixed time.

A Common Mistake

A common error is to report $[M,B]_t=\int_0^t H_s^2\,ds$. That is the quadratic variation $[M]_t$, not the covariation with $B$. For the constant integrand $H_s=3$, the incorrect calculation gives $9t$; the correct cross-increment calculation is $\sum_i(3\Delta B_i)(\Delta B_i)=3\sum_i(\Delta B_i)^2$, whose limit is $3t$. Squaring $H$ is appropriate only when both increments in the product come from $M$.

Finally, localization establishes the bracket identity but does not by itself make $M$ a true martingale. The result asserts a pathwise bracket identity up to indistinguishability; it does not assert that expectations of $M_t$ are preserved without further integrability. Distinguishing a continuous local martingale from a true martingale is therefore just as important here as distinguishing quadratic variation from covariation.

Check Your Understanding

Use the covariation formula and, where indicated, the definition by cross-increment sums.

If $H_s=-2$, what are $[M,B]_t$ and $[M]_t$? Explain why their signs differ in no way from the corresponding integrand and its square.
For $H_s=\mathbf{1}_{\{s\le a\}}$, what is $[M,B]_t$ when $t>a$, and why does it remain constant thereafter?
For $H_s=B_s$, state the Ito formula correction term used to compute $\int_0^t B_s\,dB_s$.
Does $[M,B]_t=0$ at one particular time imply $H=0$ almost everywhere before that time if $H$ can change sign? Explain.
Which bracket is given by $\int_0^t H_s^2\,ds$: $[M]$ or $[M,B]$?