Mutual Covariation of Two Ito Integrals

Why Mutual Covariation Matters

For one Ito integral, quadratic variation records the accumulated squared size of its integrand. For two Ito integrals driven by the same Brownian motion, mutual covariation records how their integrands overlap, including their signs. The key idea is polarization: the cross term in the square of a sum can be recovered from the quadratic variations of a sum and a difference.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$, and assume the filtration satisfies the usual conditions. An integrand is predictable, hence progressively measurable. For a fixed $T>0$, the square-integrable class $\mathcal{L}^2$ consists of such processes $H$ satisfying $\mathbb{E}\int_0^T H_s^2\,ds<\infty$. The Ito integral $\int_0^t H_s\,dB_s$ is defined for $H\in\mathcal{L}^2$; for locally square-integrable integrands it is defined by localization.

Definition: A predictable process $H$ is locally square-integrable if $\int_0^T H_s^2\,ds<\infty$ almost surely for every finite $T$. For continuous local martingales $M,N$, their mutual covariation is denoted by $[M,N]_t$. For the Ito integrals considered here, it is the limit in probability of sums of products of increments along partitions whose mesh tends to zero. Quadratic variation is the special case $[M,M]_t$.

The earlier quadratic-variation result for localized Ito integrals gives $[\int_0^\cdot H_s\,dB_s]_t=\int_0^t H_s^2\,ds$. We use that result and the elementary polarization identity to obtain the mutual formula. Here all integrals are Ito integrals: no ordinary chain rule is being applied. The shorthand $dB_t\,dB_t=dt$, $dB_t\,dt=0$, and $dt\,dt=0$ records the corresponding quadratic-variation bookkeeping; it is not ordinary multiplication of differentials.

Polarization for Square-Integrable Integrands

Theorem: Suppose $H,K\in\mathcal{L}^2$ on every finite time interval, and put $M_t=\int_0^t H_s\,dB_s$, $N_t=\int_0^t K_s\,dB_s$. Then, for every $t\ge0$, almost surely, $$ [M,N]_t=\int_0^t H_sK_s\,ds. $$ The equality can be chosen to hold simultaneously for all $t$, up to indistinguishability.

Proof. Fix a finite horizon $T$. Since $(H_s+K_s)^2\le2H_s^2+2K_s^2$ and $(H_s-K_s)^2\le2H_s^2+2K_s^2$, both $H+K$ and $H-K$ belong to $\mathcal{L}^2$ on $[0,T]$. Linearity of the Ito integral gives $M+N=\int_0^\cdot(H_s+K_s)\,dB_s$ and $M-N=\int_0^\cdot(H_s-K_s)\,dB_s$. By the quadratic-variation result, their brackets are the integrals of the squared integrands. The algebraic identity $4xy=(x+y)^2-(x-y)^2$, applied to increments and then to the defining limits for covariation, gives $[M,N]=\tfrac14([M+N]-[M-N])$. Therefore, for $t\le T$, $$ [M,N]_t=\frac14\int_0^t\big((H_s+K_s)^2-(H_s-K_s)^2\big)\,ds =\int_0^t H_sK_s\,ds. $$ The integrals on the right are finite almost surely: Cauchy–Schwarz gives $\int_0^T|H_sK_s|\,ds\le(\int_0^T H_s^2\,ds)^{1/2}(\int_0^T K_s^2\,ds)^{1/2}$. Taking integer horizons and using continuity of both sides gives the asserted simultaneous-in-$t$ equality. $\square$

The polarization identity is valid for any pair of continuous local martingales, not only Ito integrals. What makes the present result especially useful is that the quadratic variations of $M+N$ and $M-N$ can each be computed directly from their integrands. Notice that $HK$ may be negative: mutual covariation is not required to increase, unlike a quadratic variation.

Extending the Formula by Localization

Theorem: If $H$ and $K$ are predictable and locally square-integrable, and $M_t=\int_0^tH_s\,dB_s$, $N_t=\int_0^tK_s\,dB_s$ are their localized Ito integrals, then $$ [M,N]_t=\int_0^t H_sK_s\,ds $$ almost surely for all $t\ge0$, up to indistinguishability.

Proof. Define the accumulated energy $A_t=\int_0^t(H_s^2+K_s^2)\,ds$, and for integers $n\ge1$ let $\tau_n=\inf\{t\ge0:A_t\ge n\}$, with $\inf\varnothing=\infty$. The process $A$ is continuous and increasing, and is finite on every finite interval almost surely by local square-integrability. Thus $\tau_n$ is a stopping time, $\tau_n$ increases with $n$, and $\tau_n\to\infty$ almost surely. The stopped integrands $H_s\mathbf{1}_{\{s\le\tau_n\}}$ and $K_s\mathbf{1}_{\{s\le\tau_n\}}$ are predictable. On every finite horizon their expected energies are at most $n$, since their accumulated energy is $A_{t\wedge\tau_n}\le n$. They therefore belong to $\mathcal{L}^2$ there.

Apply the square-integrable theorem to these stopped integrands. Their Ito integrals are $M_{t\wedge\tau_n}$ and $N_{t\wedge\tau_n}$, so $$ [M^{\tau_n},N^{\tau_n}]_t=\int_0^{t\wedge\tau_n}H_sK_s\,ds. $$ Stopping a continuous local martingale stops its covariation at the same time, so the left side is $[M,N]_{t\wedge\tau_n}$. For any fixed finite $T$, almost surely $\tau_n\ge T$ for all sufficiently large $n$, because $A_T<\infty$. Thus on $[0,T]$ the stopped equality eventually becomes $[M,N]_t=\int_0^tH_sK_s\,ds$. Taking integer $T$ proves the identity for all finite times, up to indistinguishability. The integral is well-defined pathwise on finite horizons because Cauchy–Schwarz gives $\int_0^T|H_sK_s|\,ds<\infty$ almost surely. $\square$

This localization proof does not assert that $M$ or $N$ is a true martingale on the whole time interval. The stopped integrals have bounded expected energy and are square-integrable martingales by the Ito-integral construction; their conditional expectations follow first for adapted step integrands from the zero conditional mean of Brownian increments, and then by the $L^2$ construction. The unstopped integrals are continuous local martingales, which need not be true martingales without further integrability. The bracket identity concerns their pathwise covariation and does not require them to be true martingales.

Worked Applications

Worked Example: Piecewise-Constant Integrands

On $[0,2]$, let $H_s=1$ for $0\le s\le1$ and $H_s=2$ for $1<s\le2$. Let $K_s=3$ for $0\le s\le1$ and $K_s=-1$ for $1<s\le2$. Compute $[M,N]_t$, where $M=\int H\,dB$ and $N=\int K\,dB$.

On the first interval $H_sK_s=1\cdot3=3$, and on the second it is $2\cdot(-1)=-2$. Integrating these values gives $$ [M,N]_t= \begin{cases} 3t,&0\le t\le1,\\ 3-2(t-1)=5-2t,&1<t\le2. \end{cases} $$ In particular, $[M,N]_2=5-4=1$.

Check: $H+K$ equals $4$ on the first interval and $1$ on the second, so $[M+N]_2=4^2\cdot1+1^2\cdot1=17$. Also $H-K$ equals $-2$ and $3$ on those intervals, so $[M-N]_2=(-2)^2\cdot1+3^2\cdot1=13$. Polarization gives $([M+N]_2-[M-N]_2)/4=(17-13)/4=1$, as required.

Worked Example: A Random Amplitude

Let $A=\{B_1>0\}$, and on $[0,3]$ set $H_s=\mathbf{1}_A\mathbf{1}_{(1,3]}(s)$ and $K_s=2\mathbf{1}_{(1,3]}(s)$. The event $A$ belongs to $\mathcal{F}_1$, so its indicator used after time $1$ is predictable. Both integrands are bounded and square-integrable. Compute the bracket at each time.

The product is zero before or at time $1$, and equals $2\mathbf{1}_A$ for $1<s\le3$. Hence $$ [M,N]_t=2\mathbf{1}_A(t-1)\quad(1<t\le3),\qquad [M,N]_t=0\quad(0\le t\le1). $$ At $t=3$, this is $4\mathbf{1}_A$. Since $B_1$ has the continuous distribution $N(0,1)$, $\mathbb{P}(A)=\mathbb{P}(B_1>0)=\tfrac12$; the bracket is $4$ on $A$ and $0$ on its complement, up to the null event $B_1=0$.

Check: On $(1,3]$, $H+K=(\mathbf{1}_A+2)$ and $H-K=(\mathbf{1}_A-2)$. If $A$ occurs, their squared values are $9$ and $1$, and their difference divided by $4$ is $(9-1)/4=2$ per unit time. If $A$ does not occur, their squared values are $4$ and $4$, giving zero. Multiplying by the interval length $2$ gives $4\mathbf{1}_A$, agreeing with the formula.

Worked Example: Disjoint Time Supports

On $[0,2]$, let $H_s=\mathbf{1}_{(0,1]}(s)$ and $K_s=\mathbf{1}_{(1,2]}(s)$. Compute the covariation at every time and at the endpoint.

The two integrands are never simultaneously nonzero except possibly at a single endpoint, which has zero time measure. Thus $H_sK_s=0$ for Lebesgue-almost every $s$, and $[M,N]_t=0$ for every $t\in[0,2]$. At the endpoint, $M_2=B_1$ and $N_2=B_2-B_1$; their increments occupy separate time intervals.

Check: $H+K=1$ almost everywhere on $(0,2]$, so $[M+N]_2=\int_0^2 1\,ds=2$. Also $H-K$ is $1$ on $(0,1]$ and $-1$ on $(1,2]$, giving $[M-N]_2=1^2\cdot1+(-1)^2\cdot1=2$. Their difference divided by $4$ is $(2-2)/4=0$.

Interpretation and Common Pitfalls

The identity says that the instantaneous mutual covariation of the two Ito integrals is $H_tK_t\,dt$, in the same bookkeeping sense that $dB_t\,dB_t=dt$. If $H_t$ and $K_t$ have the same sign, their product contributes positively; opposite signs contribute negatively; and disjoint time supports contribute zero. This is a statement about integrals driven by the same Brownian motion. With different Brownian drivers, their mutual covariation depends on the covariation of those drivers and cannot be inferred from this formula alone.

A common mistake is to square each integrand separately and then add the results, writing $[M,N]_t=\int_0^t(H_s^2+K_s^2)\,ds$. That expression is $[M,M]_t+[N,N]_t$, not their mutual covariation. For example, in the first worked example its endpoint value would be $(1^2+2^2)+(3^2+(-1)^2)=15$, whereas the correct mutual covariation is $1$. The correction is to use polarization, or directly integrate $H_sK_s$.

Finally, the formula is for Ito integrals. For a continuous semimartingale integrand $Y$ for which the integrals exist, the Stratonovich integral is related by $\int_0^tY_s\circ dB_s=\int_0^tY_s\,dB_s+\tfrac12[Y,B]_t$. That conversion does not change the definition of the Ito mutual covariation proved here; it supplies an additional correction when converting an integral convention.

Check Your Understanding

Use the product formula and polarization, and state any integrability condition needed in each case.

If $H=2$ and $K=-3$ on $[0,T]$, what is $[M,N]_t$ for $0\le t\le T$?
For the random-amplitude example, what is $\mathbb{E}[[M,N]_3]$, and how does it follow from $\mathbb{P}(B_1>0)=\tfrac12$?
Explain why local square-integrability suffices for $\int_0^tH_sK_s\,ds$ to be finite almost surely on finite horizons.
State the stopping times used in the localization proof and explain why they tend to infinity almost surely.
Which quantity is obtained from $\int_0^t(H_s^2+K_s^2)\,ds$: mutual covariation, or the sum of the two quadratic variations?