Covariance of Two Simple Stochastic Integrals

Why Covariance Needs More Than Two Isometries

The Ito isometry gives the second moment of one simple stochastic integral. For two different integrands, the corresponding question is how their integrals co-vary. The answer is the expected time integral of the product of the integrands. The key point is that coefficients chosen using the information available at the start of an interval are independent of the Brownian increment on that interval; increments from different intervals contribute no cross terms.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a Brownian motion $B_t$ with respect to this filtration. In particular, for $s<t$, $B_t-B_s$ is independent of $\mathcal{F}_s$ and has distribution $N(0,t-s)$. All integrals below are Ito integrals on $[0,T]$, where $T>0$. The integrands are simple predictable processes: on a deterministic partition, each coefficient is measurable with respect to the information at the left endpoint of its interval. We require each coefficient to be square-integrable, so these processes are progressively measurable and satisfy $\mathbb{E}\int_0^T H_s^2\,ds<\infty$.

Definition: Given a partition $0=t_0<t_1<\cdots<t_n=T$, a simple predictable process has the form $H_s=\sum_{i=0}^{n-1}H_i\mathbf{1}_{(t_i,t_{i+1}]}(s)$, where $H_i$ is $\mathcal{F}_{t_i}$-measurable and $\mathbb{E}[H_i^2]<\infty$. Its Ito integral is $I(H)=\int_0^T H_s\,dB_s=\sum_{i=0}^{n-1}H_i(B_{t_{i+1}}-B_{t_i})$. Values at the finitely many partition endpoints do not affect the time integral.

For square-integrable random variables $X,Y$, their covariance is $\operatorname{Cov}(X,Y)=\mathbb{E}[XY]-\mathbb{E}[X]\mathbb{E}[Y]$. Each simple integral here has mean zero, so its covariance with another such integral is just their expected product. We will verify this mean-zero fact as part of the argument.

The Covariance Identity

Theorem: Let $H$ and $K$ be simple predictable processes on $[0,T]$ with square-integrable coefficients. Then their Ito integrals are square-integrable, have mean zero, and satisfy $$ \mathbb{E}\left[\left(\int_0^T H_s\,dB_s\right)\left(\int_0^T K_s\,dB_s\right)\right] =\mathbb{E}\left[\int_0^T H_sK_s\,ds\right]. $$ Consequently, the left-hand side is $\operatorname{Cov}(I(H),I(K))$.

Proof. First place $H$ and $K$ on a common deterministic partition $0=t_0<\cdots<t_n=T$. This is done by taking the ordered union of their partition points and carrying each coefficient unchanged across any subintervals created by refinement. The coefficient remains measurable at the refined left endpoint because the filtration is increasing. Write $\Delta B_i=B_{t_{i+1}}-B_{t_i}$, $\Delta t_i=t_{i+1}-t_i$, and let $H_i,K_i$ be the coefficients on $(t_i,t_{i+1}]$.

For each $i$, independence of $\Delta B_i$ from $\mathcal{F}_{t_i}$ gives $\mathbb{E}[H_i^2(\Delta B_i)^2]=\Delta t_i\mathbb{E}[H_i^2]<\infty$. Thus every summand $H_i\Delta B_i$ is square-integrable, and the finite sum $I(H)$ is square-integrable. Moreover, $\mathbb{E}[H_i\Delta B_i]=0$, since the conditional mean of $\Delta B_i$ given $\mathcal{F}_{t_i}$ is zero. Therefore $\mathbb{E}[I(H)]=0$; the same reasoning applies to $I(K)$.

Expand the product of the two finite sums. For a diagonal term, $H_iK_i$ is $\mathcal{F}_{t_i}$-measurable and integrable, and $\mathbb{E}[(\Delta B_i)^2\mid\mathcal{F}_{t_i}]=\Delta t_i$. Hence $\mathbb{E}[H_iK_i(\Delta B_i)^2]=\Delta t_i\mathbb{E}[H_iK_i]$. This expectation is finite because $\mathbb{E}|H_iK_i|\le(\mathbb{E}H_i^2)^{1/2}(\mathbb{E}K_i^2)^{1/2}$.

For an off-diagonal term, take $i<j$. The random variable $A=H_i\Delta B_i$ is $\mathcal{F}_{t_j}$-measurable and square-integrable. Also $C=K_j\Delta B_j$ is square-integrable and satisfies $\mathbb{E}[C\mid\mathcal{F}_{t_j}]=0$, by independence of the next Brownian increment from $\mathcal{F}_{t_j}$. To justify multiplying by possibly unbounded $A$, truncate it to $A_m=\max(-m,\min(A,m))$. Then $\mathbb{E}[A_mC]=0$ by conditional expectation. Since $\mathbb{E}[(A-A_m)^2]\to0$, Cauchy–Schwarz gives $\mathbb{E}[(A-A_m)C]\to0$. Thus $\mathbb{E}[AC]=0$. The terms with $j<i$ vanish by the same argument with the roles of the two intervals exchanged.

Only the diagonal terms remain, so $\mathbb{E}[I(H)I(K)]=\sum_i\Delta t_i\mathbb{E}[H_iK_i]$. On each interval $H_sK_s=H_iK_i$, and therefore the finite sum equals $\mathbb{E}\int_0^T H_sK_s\,ds$. Finally, both integral means are zero, so their expected product is their covariance. $\square$

A Second Derivation by Polarization

Proposition: For simple predictable $H,K$, the cross-covariance identity follows by applying the simple-process Ito isometry to $H+K$, $H$, and $K$.

Proof. The class of simple predictable processes is closed under addition, so $H+K$ is again simple predictable. The simple-process Ito isometry states that $\mathbb{E}[I(G)^2]=\mathbb{E}\int_0^T G_s^2\,ds$ for any such $G$. Linearity of the finite sums defining the integrals gives $I(H+K)=I(H)+I(K)$. Subtract the isometry identities for $H$ and $K$ from that for $H+K$. On the integral side, $\mathbb{E}[I(H+K)^2]-\mathbb{E}[I(H)^2]-\mathbb{E}[I(K)^2]=2\mathbb{E}[I(H)I(K)]$. On the time-integral side, expanding $(H_s+K_s)^2$ leaves $2\mathbb{E}\int_0^T H_sK_s\,ds$. Dividing by $2$ proves the identity. The integrals have mean zero by the conditional-increment argument in the theorem, so this is also the covariance formula. $\square$

The partition proof exposes why the formula holds: same-interval terms produce the product $H_iK_i$ times the interval length, and distinct intervals have zero expected cross term. Polarization is shorter, but it relies on the square-integral identity and linearity already established for simple processes.

Worked Calculations

Worked Example: Deterministic Coefficients on Three Intervals

Use the partition $0<\tfrac12<\tfrac32<2$. Let $H$ have coefficients $2,-1,3$, and let $K$ have coefficients $1,4,-2$, respectively. Compute the covariance of their integrals.

The interval lengths are $\Delta t_0=\tfrac12$, $\Delta t_1=1$, and $\Delta t_2=\tfrac12$. The integrals are $I(H)=2\Delta B_0-\Delta B_1+3\Delta B_2$ and $I(K)=\Delta B_0+4\Delta B_1-2\Delta B_2$. Each increment has mean zero, and distinct increments are independent, so cross-interval products have expectation zero. The three same-interval contributions are $2\cdot1\cdot\tfrac12=1$, $(-1)\cdot4\cdot1=-4$, and $3\cdot(-2)\cdot\tfrac12=-3$. Their sum is $-6$.

The theorem gives the same result from the integrands: $\mathbb{E}\int_0^2 H_sK_s\,ds=(2)(1)(\tfrac12)+(-1)(4)(1)+(3)(-2)(\tfrac12)=1-4-3=-6$. Check: expanding $I(H)I(K)$ yields the same three diagonal products; every other product contains increments from distinct intervals and has expectation zero. Thus the covariance is $-6$, despite both integral means being zero.

Worked Example: A Random Coefficient Chosen from the Past

On the partition $0<1<2$, set $H_0=1$, $H_1=B_1$, $K_0=2$, and $K_1=1$. The random coefficient $B_1$ is allowed on $(1,2]$ because it is known at time $1$. Find the covariance.

Write $\Delta B_1=B_2-B_1$. The two integrals are $I(H)=B_1+B_1\Delta B_1$ and $I(K)=2B_1+\Delta B_1$. Their means are zero: $B_1$ is centered, and $\mathbb{E}[B_1\Delta B_1]=\mathbb{E}[B_1]\mathbb{E}[\Delta B_1]=0$, since the second increment is independent of $\mathcal{F}_1$.

Expand their product: $I(H)I(K)=2B_1^2+B_1\Delta B_1+2B_1^2\Delta B_1+B_1(\Delta B_1)^2$. The expectations of these four terms are, in order, $2\mathbb{E}[B_1^2]=2$, $0$, $2\mathbb{E}[B_1^2]\mathbb{E}[\Delta B_1]=0$, and $\mathbb{E}[B_1]\mathbb{E}[(\Delta B_1)^2]=0$. Here $\mathbb{E}[B_1^2]=1$, $\mathbb{E}[\Delta B_1]=0$, and $\mathbb{E}[(\Delta B_1)^2]=1$. Hence the covariance is $2$.

Check: the theorem gives $\mathbb{E}\int_0^2H_sK_s\,ds=1\cdot2\cdot1+\mathbb{E}[B_1\cdot1]\cdot1=2+0=2$, in agreement with the direct product expansion.

Worked Example: Disjoint Time Supports

On $0<1<2$, take $H_0=1,H_1=0$, and $K_0=0,K_1=B_1$. Compute the covariance and explain why the second coefficient is admissible.

The integrals are $I(H)=B_1$ and $I(K)=B_1(B_2-B_1)=B_1\Delta B_1$. Since $B_1$ is $\mathcal{F}_1$-measurable, it is an admissible coefficient on the second interval. Both integral means are zero. Their product has expectation $\mathbb{E}[B_1^2\Delta B_1]=\mathbb{E}[B_1^2]\mathbb{E}[\Delta B_1]=1\cdot0=0$.

The time-integrand product is zero on each interval: on the first, $K=0$; on the second, $H=0$. Thus $\mathbb{E}\int_0^2H_sK_s\,ds=0$, confirming the covariance. The example shows that an interval where only one integrand is nonzero contributes no covariance, even if its coefficient is random.

Common Pitfall: Using Information from the End of an Interval

A coefficient in a simple predictable process must be known at the interval's left endpoint. For example, $B_2$ is not an admissible coefficient on $(1,2]$, because it includes the increment $B_2-B_1$ that occurs during that interval. Treating $B_2$ as if it were an allowed coefficient can create a false covariance contribution. Indeed, the ordinary product calculation gives $\mathbb{E}[B_2(B_2-B_1)]=\mathbb{E}[(B_1+\Delta B_1)\Delta B_1]=0+1=1$. That value does not contradict the theorem: the expression uses a future-measurable coefficient and is not the simple predictable Ito integral with coefficient chosen at time $1$. Replacing $B_2$ by the admissible left-endpoint coefficient $B_1$ gives $\mathbb{E}[B_1\Delta B_1]=0$.

The covariance identity is the bilinear companion to the Ito isometry. For simple predictable processes it is obtained from finite sums, Brownian independence, and conditional means and variances. In stochastic differential notation, its structure is consistent with $dB_t\,dB_t=dt$; it does not use the ordinary chain rule or assert that a random increment squared equals a time increment.

Check Your Understanding

Use the common-partition formula and the left-endpoint measurability condition in each question.

On $0<1<3$, let $H$ have coefficients $1,2$ and $K$ have coefficients $-1,3$. What is the covariance of the two integrals?
Why is $B_1$ an admissible coefficient on $(1,2]$, but $B_2$ is not?
For simple predictable $H,K$, explain why the integrals have mean zero.
Derive the covariance identity by expanding the isometry for $H+K$ and subtracting the identities for $H$ and $K$.
If $H$ and $K$ have disjoint time supports, what does the covariance identity imply, provided their product is zero at almost every time?