Covariation of Stochastic Integrals

Why Brackets of Integrals Matter

A continuous stochastic integral changes its quadratic variation according to the size of its integrand. For two integrals, their mutual covariation depends on the product of the two integrands and the covariation of the processes being integrated. This rule lets us calculate brackets without returning to fine partition sums each time.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a Brownian motion $B_t$. The main result below concerns continuous semimartingales, not only Brownian motion. A continuous semimartingale $X$ has a decomposition $X=X_0+M+A$, where $M$ is a continuous local martingale starting at zero and $A$ is a continuous finite-variation process starting at zero. For $Y=Y_0+N+C$, its covariation with $X$ is $[X,Y]=[M,N]$; finite-variation parts contribute zero.

Definition: An integrand for a continuous semimartingale is predictable and semimartingale-integrable. In particular, it must be integrable against the finite-variation part pathwise and locally square-integrable against the continuous local martingale part. A sufficient local condition is that it is locally bounded. For Brownian integrals in $\mathcal{L}^2$, the integrand is progressively measurable and satisfies $\mathbb{E}\int_0^T H_s^2\,ds<\infty$. All integrals in this tutorial are Itô semimartingale integrals.

Here, “locally square-integrable against $M$” means that the integral of $H_s^2$ with respect to the increasing process $[M]_s$ is finite on bounded time intervals, after localization if necessary. Brackets and identities between processes below are understood up to indistinguishability: outside a single null set, they hold at every time.

The Covariation Identity

Theorem: Let $X,Y$ be continuous semimartingales and let $H,K$ be predictable integrands such that $H$ is integrable against $X$ and $K$ is integrable against $Y$. Then $HK$ is integrable against the covariation measure $[X,Y]$, locally, and $$ [H\mathbin{\cdot}X,K\mathbin{\cdot}Y]_t =\int_0^t H_sK_s\,d[X,Y]_s. $$ Here $H\mathbin{\cdot}X=\int_0^\cdot H_s\,dX_s$. In particular, $[H\mathbin{\cdot}X]_t=\int_0^t H_s^2\,d[X]_s$.

Proof. First consider continuous local martingales $M,N$. By localization, it is enough to work on a bounded time interval after stopping so that the relevant integrals and brackets are square-integrable. For elementary predictable integrands, the integral is a finite sum of martingale increments multiplied by coefficients known at the beginning of each interval. Expanding the product of two such sums, terms from distinct intervals have zero conditional expectation: the later martingale increment has conditional mean zero given the information at its starting time. On each interval, the conditional product of increments is governed by the bracket increment, because $MN-[M,N]$ is a local martingale. Thus the compensating finite-variation process for the product of the two integrals is $\int HK\,d[M,N]$.

For general integrands, use the construction of continuous semimartingale integrals: after localization, elementary predictable integrands approximate admissible integrands in the local square-integrability norm against the brackets. The Itô isometry gives convergence of the corresponding martingale integrals in $L^2$ after stopping. Applying the same approximation to pairs of integrands, the bilinear product identity for elementary integrands passes to the limit; the cross terms are controlled by Cauchy–Schwarz with respect to the increasing bracket measures. Consequently, the product of the two limiting integrals minus $\int HK\,d[M,N]$ is a local martingale. The continuous finite-variation compensator in this product characterization is unique, so it is the covariation of the integrals.

For semimartingales $X=M+A$ and $Y=N+C$, the integral $H\mathbin{\cdot}X$ is the sum of the local martingale integral $H\mathbin{\cdot}M$ and the finite-variation integral $\int H\,dA$; the analogous decomposition holds for $K\mathbin{\cdot}Y$. A continuous finite-variation process has zero bracket with every continuous semimartingale: along partitions its squared increments are bounded by its largest increment times its total variation, and its largest increment tends to zero by continuity. The same estimate, with Cauchy–Schwarz, makes its cross-covariation with a continuous semimartingale vanish. Hence only the local martingale parts contribute, and $[X,Y]=[M,N]$. Applying the local martingale identity proves the formula for $X,Y$. Taking $Y=X$ and $K=H$ gives the quadratic-variation identity. $\square$

The theorem uses the Itô integral. It does not replace the ordinary chain rule: the bracket records the second-order contribution of the martingale parts. In Brownian notation, $dB_t\,dB_t=dt$, whereas $dB_t\,dt=0$ and $dt\,dt=0$. These are shorthand for bracket and finite-variation rules, not ordinary products of differentials.

Deterministic Integrands Against Brownian Motion

Worked Example: A Time-Weighted Brownian Integral

Let $U_t=\int_0^t s\,dB_s$, where the integrand $s\mapsto s$ is progressively measurable and belongs to $\mathcal{L}^2$ on every bounded interval. Find $[U,B]_t$ and $[U]_t$.

Use $H_s=s$, $K_s=1$, and $[B]_s=s$. The theorem gives $$ [U,B]_t=\int_0^t s\,ds=\frac{t^2}{2}. $$ Taking both integrands equal to $s$ gives $$ [U]_t=\int_0^t s^2\,ds=\frac{t^3}{3}. $$

Check: for $t=2$, the bracket with $B$ is $\int_0^2s\,ds=[s^2/2]_0^2=2$; the quadratic variation is $\int_0^2s^2\,ds=[s^3/3]_0^2=8/3$. The latter is nonnegative and increases with time, as a quadratic variation must.

For an Itô integral against $B$, the same formula also follows from $[B]_t=t$: its quadratic variation is the accumulated squared integrand, not the square of the integral itself.

Two Brownian Motions and Correlation

Worked Example: Covariation with a Correlated Brownian Motion

Let $B_t$ and $W_t$ be independent standard Brownian motions and set $C_t=\tfrac35B_t+\tfrac45W_t$. Since $[B,W]_t=0$, bilinearity of covariation gives $[B,C]_t=\tfrac35t$. Compute the bracket of $U_t=2B_t$ and $V_t=3C_t$.

The integrands are the constants $2$ and $3$, so $$ [U,V]_t=\int_0^t 6\,d[B,C]_s =6\cdot\frac35t=\frac{18}{5}t. $$ Their individual quadratic variations are $[U]_t=4t$ and $[V]_t=9t$.

Check: $U_t+V_t=2B_t+3(\tfrac35B_t+\tfrac45W_t)=\tfrac{19}{5}B_t+\tfrac{12}{5}W_t$. Independence gives $$ [U+V]_t=\left(\frac{19}{5}\right)^2t+\left(\frac{12}{5}\right)^2t =\frac{361+144}{25}t=\frac{101}{5}t. $$ The bracket identity also gives $[U]_t+[V]_t+2[U,V]_t=4t+9t+\tfrac{36}{5}t=\tfrac{101}{5}t$, in agreement.

This example shows why the covariation of the underlying processes matters. If two Brownian motions are independent, their bracket is zero; if they are correlated, the cross-bracket contributes. The identity keeps that dependence explicit instead of treating the noises as interchangeable.

Finite-Variation Drift Does Not Add Quadratic Variation

Worked Example: A Brownian Motion with a Smooth Drift

Let $X_t=B_t+t^2$, and define $Z_t=\int_0^t(1+s)\,dX_s$. The integrand is deterministic and bounded on every bounded time interval. The decomposition of $X$ has local martingale part $B$ and finite-variation part $t^2$, so $[X]_t=[B]_t=t$. Therefore $$ [Z]_t=\int_0^t(1+s)^2\,ds =\left[s+s^2+\frac{s^3}{3}\right]_0^t =t+t^2+\frac{t^3}{3}. $$

Check: the integral itself can be written as $Z_t=\int_0^t(1+s)\,dB_s+\int_0^t(1+s)\,d(s^2)$. The second term is $\int_0^t2s(1+s)\,ds=t^2+\tfrac23t^3$, a continuous finite-variation process. It contributes no quadratic variation, leaving the squared Brownian integrand $\int_0^t(1+s)^2ds$, as calculated.

More generally, if $X_t=\int_0^t\mu_s\,ds+\int_0^t\sigma_s\,dB_s$, under conditions that make these integrals well-defined, then $[X]_t=\int_0^t\sigma_s^2\,ds$. The drift $\mu$ has finite variation on bounded intervals when it is pathwise integrable there; it is the diffusion coefficient $\sigma$ that contributes to the bracket.

Common Mistake: Omitting the Product of Integrands

A frequent error is to write $[H\mathbin{\cdot}X,K\mathbin{\cdot}Y]=\int H_s\,d[X,Y]_s$, leaving out $K_s$. For example, take $X=Y=B$, $H_s=2$, and $K_s=3$. The incorrect calculation gives $2t$. The correct calculation is $$ [2B,3B]_t=\int_0^t 2\cdot3\,d[B]_s=6t. $$

As a check, $[2B+3B]_t=[5B]_t=25t$, while $[2B]_t+[3B]_t+2[2B,3B]_t=4t+9t+12t=25t$. The incorrect value $2t$ would give $4t+9t+4t=17t$, contradicting the quadratic variation of $5B$.

Another useful distinction is between a bracket and a product of endpoint values. In general, $[H\mathbin{\cdot}X,K\mathbin{\cdot}Y]_t$ is an accumulated integral over time; it is not $(H\mathbin{\cdot}X)_t(K\mathbin{\cdot}Y)_t$. The theorem computes the former, using the integrands and the bracket of the underlying processes.

Check Your Understanding

In each question, assume all stated integrals are well-defined; equalities of brackets are understood almost surely for all times.

If $U_t=\int_0^t(1+s)\,dB_s$, calculate $[U,B]_t$ and $[U]_t$.
If $[X,Y]_t=\rho t$, where $\rho$ is constant, and $U=2\mathbin{\cdot}X$, $V=K\mathbin{\cdot}Y$ for a deterministic integrand $K_s=s$, calculate $[U,V]_t$.
Explain why adding a continuous finite-variation process to a continuous semimartingale does not change its quadratic variation.
For $U=H\mathbin{\cdot}B$ and $V=K\mathbin{\cdot}B$, state the condition on the integrands and write the formula for $[U,V]_t$.
Find the error in claiming $[2B,3B]_t=2t$, and check the corrected answer using $[2B+3B]_t$.