Quadratic Covariation for Continuous Semimartingales

Why Covariation Is Needed

For a single continuous process, quadratic variation records the accumulated effect of squared increments. For two processes, the corresponding quantity is quadratic covariation: it records whether their increments tend to move together or in opposite directions. This cross term is essential in the second-order change-of-variables formula. In particular, the product of two Brownian increments contributes at order $dt$, not at a negligible higher order.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying Brownian motion $B_t$. Assume the filtration satisfies the usual conditions. A continuous semimartingale $X$ has a decomposition $X=M+A$, where $M$ is a continuous local martingale and $A$ is a continuous adapted process of finite variation on every bounded time interval. We use the decomposition with $A_0=0$; in the usual canonical decomposition, $M_0=X_0$. A local martingale is a process that becomes a martingale after stopping along a sequence of stopping times increasing to infinity. It need not itself be a true martingale.

Definition: For continuous semimartingales $X$ and $Y$, their quadratic covariation $[X,Y]_t$ is the limit in probability, as the mesh tends to zero, of the sums $\sum_i (X_{t_i}-X_{t_{i-1}})(Y_{t_i}-Y_{t_{i-1}})$ over partitions of $[0,t]$. For a continuous local martingale $M$, its quadratic variation $[M]_t=[M,M]_t$ is continuous and increasing. For continuous local martingales $M,N$, $[M,N]$ is their local-martingale bracket.

The existence of the bracket for continuous local martingales, its characterization by convergence of these sums, and the property that $M^2-[M]$ is a local martingale are standard results from quadratic-variation theory. Their construction is not reproved here. The key point for general continuous semimartingales is that adding a continuous finite-variation process does not change the limit of the increment-product sums.

Finite Variation Does Not Contribute

Theorem: Let $X=M+A$ and $Y=N+C$ be continuous semimartingales, with continuous local martingales $M,N$ and continuous finite-variation processes $A,C$. Then $$ [X,Y]_t=[M,N]_t $$ for every $t\ge0$, up to indistinguishability. In particular, $[X]_t=[M]_t$.

Proof. Fix a bounded interval $[0,T]$ and partitions $\Pi=\{0=t_0<\cdots<t_n=T\}$ whose mesh tends to zero. Write $\Delta_i X=X_{t_i}-X_{t_{i-1}}$, and use the corresponding notation for the other processes. Expanding each product gives $$ \sum_i\Delta_iX\,\Delta_iY =\sum_i\Delta_iM\,\Delta_iN+\sum_i\Delta_iM\,\Delta_iC+\sum_i\Delta_iA\,\Delta_iN+\sum_i\Delta_iA\,\Delta_iC. $$ For a continuous path $M$, uniform continuity on $[0,T]$ implies $\max_i|\Delta_iM|\to0$. The total variation $\operatorname{Var}_{[0,T]}(C)$ is finite, and therefore $$ \left|\sum_i\Delta_iM\,\Delta_iC\right| \le \max_i|\Delta_iM|\,\operatorname{Var}_{[0,T]}(C)\longrightarrow0. $$ The same argument shows $\sum_i\Delta_iA\,\Delta_iN\to0$. For the final term, $$ \left|\sum_i\Delta_iA\,\Delta_iC\right| \le \max_i|\Delta_iA|\,\operatorname{Var}_{[0,T]}(C)\longrightarrow0, $$ because $A$ is continuous. These bounds hold pathwise. The first sum converges in probability to $[M,N]_T$ by the local-martingale bracket theorem. Hence the full sum converges in probability to $[M,N]_T$, which is the defining limit for $[X,Y]_T$. Applying the same argument at each time and using the continuous versions gives the asserted process equality. Taking $Y=X$ gives $[X]=[M]$. $\square$

This proof also explains why a drift has no effect on quadratic variation: its increments are small in total, and multiplying them by the largest increment of a continuous path makes the cross-sum vanish. The result is about the limit of sums, not an identity between individual increments.

Bracket Rules and Their Meaning

For continuous local martingales, polarization gives $$ [M,N]_t=\frac14\big([M+N]_t-[M-N]_t\big). $$ The bracket is bilinear, and $[M]$ is increasing. The bracket $[M,N]$ need not be increasing; it is a continuous finite-variation process that can have either sign. For semimartingales $X=M+A$ and $Y=N+C$, the theorem transfers these facts to their covariation through $[X,Y]=[M,N]$.

Takeaway: In differential notation, the continuous-semimartingale multiplication rules are $dX_t\,dY_t=d[X,Y]_t$, $dB_t\,dB_t=dt$, $dB_t\,dt=0$, and $dt\,dt=0$. They encode limits of increment sums; they are not ordinary algebraic identities for finite increments.

For example, the finite-variation contribution to a squared-increment sum vanishes because $\sum_i(\Delta_i A)^2\le \max_i|\Delta_iA|\,\operatorname{Var}(A)\to0$. The mixed time-Brownian sum vanishes because $\left|\sum_i\Delta t_i\Delta_iB\right|\le T\max_i|\Delta_iB|\to0$. Brownian quadratic variation is different: $\sum_i(\Delta_iB)^2$ converges in probability to $T$, not to zero.

Computing Brackets in Ito Processes

For the examples below, an Ito integral $\int_0^t H_s\,dB_s$ is taken with a progressively measurable integrand $H$ in $\mathcal{L}^2$ on each stated finite time interval; that is, $\mathbb{E}\int_0^T H_s^2\,ds<\infty$. The bracket rule for Brownian Ito integrals is $$ \left[\int_0^\cdot H_s\,dB_s,\int_0^\cdot K_s\,dB_s\right]_t =\int_0^t H_sK_s\,ds, $$ when the integrands satisfy these conditions. This standard bracket identity follows from the construction of the Ito integral and its isometry. For integrals against independent Brownian motions, the cross bracket is zero. These rules identify the martingale brackets; the theorem shows that adding drift terms does not alter them.

Worked Example: Brownian Motion with Drift

Let $X_t=3+2t+4B_t$. Find its quadratic variation on $[0,2]$.

The finite-variation part is $3+2t$, and the local-martingale part is $4B_t$. The constant initial value contributes no increments. Since $[B]_t=t$, scaling the local martingale gives $[4B]_t=4^2[B]_t=16t$. Thus $[X]_t=16t$, and the requested value is $16\cdot2=32$.

Check: For each partition, $\Delta_iX=2\Delta t_i+4\Delta_iB$. The squared-increment sum has a drift-square term $4\sum_i(\Delta t_i)^2$, which tends to zero; a mixed term $16\sum_i\Delta t_i\Delta_iB$, which tends to zero; and a Brownian term $16\sum_i(\Delta_iB)^2$, which tends to $16\cdot2=32$. This agrees with the bracket calculation.

Worked Example: Two Ito Processes Driven by the Same Brownian Motion

Let $X_t=1+t+2B_t$ and $Y_t=4-3t+5B_t$. Compute $[X,Y]_t$ and its value at $t=3$.

The local-martingale parts are $2B_t$ and $5B_t$. By bilinearity and $[B]_t=t$, $$ [X,Y]_t=[2B,5B]_t=2\cdot5[B]_t=10t. $$ At $t=3$, this is $10\cdot3=30$. The drift coefficients $1$ and $-3$ do not enter the bracket.

Check: The increments are $\Delta_iX=\Delta t_i+2\Delta_iB$ and $\Delta_iY=-3\Delta t_i+5\Delta_iB$. Their product is $-3(\Delta t_i)^2+(-3+10)\Delta t_i\Delta_iB+10(\Delta_iB)^2$. The first two sums vanish as the mesh tends to zero, while the last converges to $10t$, as claimed.

Worked Example: Independent Brownian Drivers

Let $X_t=2+2B_t$ and $Y_t=1+3C_t$, where $B$ and $C$ are independent Brownian motions with respect to the same filtration. Find $[X,Y]_t$.

The drift parts are constant, so only the martingale parts matter. Independent Brownian motions have zero cross bracket, hence $$ [X,Y]_t=[2B,3C]_t=2\cdot3[B,C]_t=0. $$ Thus the bracket is zero at every time, including $t=0$.

Check: On a deterministic partition, increments of $B$ and $C$ over each interval are independent and centered. Consequently, the expected value of each product $\Delta_iB\,\Delta_iC$ is zero. More precisely, independence of the two Brownian motions and the Brownian increment variances give $$ \mathbb{E}\left[\left(\sum_i\Delta_iB\,\Delta_iC\right)^2\right] =\sum_i(\Delta t_i)^2 \le t|\Pi|\longrightarrow0. $$ The cross-increment sum therefore converges to zero in $L^2$, confirming the bracket calculation.

More generally, suppose $$ X_t=X_0+\int_0^t\mu_s\,ds+\int_0^t\sigma_s\,dB_s,\qquad Y_t=Y_0+\int_0^t\nu_s\,ds+\int_0^t\tau_s\,dB_s, $$ where the drift integrals are finite on bounded time intervals and $\sigma,\tau$ are progressively measurable and in $\mathcal{L}^2$ there. Then the bracket rule gives $[X,Y]_t=\int_0^t\sigma_s\tau_s\,ds$. The product is integrable in time since $2|\sigma_s\tau_s|\le \sigma_s^2+\tau_s^2$. If the Brownian drivers are independent, the cross bracket is zero instead. The drift coefficients $\mu,\nu$ do not appear.

Common Pitfall: Multiplying Drifts into the Bracket

A tempting but incorrect calculation is to treat $dX_t=(2\,dt+4\,dB_t)$ as if ordinary algebra gave $(dX_t)^2=4\,dt+16\,dt$, counting the drift square as $4\,dt$. The correction is that the drift-square term is $4(dt)^2$, and $dt\,dt=0$; the mixed term is $16\,dt\,dB_t=0$. Only $(4\,dB_t)^2=16\,dt$ remains. Thus the quadratic variation rate is $16$, not $20$.

Quadratic covariation matters whenever a second-order formula involves two evolving quantities, including products and changes of variables for multidimensional processes. It also keeps distinct two ideas that are easy to conflate: the ordinary chain rule applies to differentiable paths, whereas the bracket records the non-vanishing second-order contribution of continuous local-martingale fluctuations.

Check Your Understanding

Use the bracket definition and the local-martingale parts to answer these questions.

If $X_t=\mu t+\sigma B_t$, what is $[X]_t$, and which part of $X$ contributes?
For $X_t=B_t$ and $Y_t=3B_t$, compute $[X,Y]_t$.
Why do the sums $\sum_i\Delta_iA\,\Delta_iM$ vanish when $A$ has continuous finite variation and $M$ is continuous?
If $X$ and $Y$ are driven by independent Brownian motions and have only deterministic drift in addition, what is $[X,Y]$?
State the integrability condition on $H$ that makes $\int_0^tH_s\,dB_s$ an Ito integral in $\mathcal{L}^2$ on $[0,T]$.