Why Distinct Increments Can Be Orthogonal After Weighting
Brownian increments over disjoint time intervals are independent and have mean zero. But in a stochastic sum, an increment is multiplied by a random coefficient. That coefficient may depend on information from earlier intervals, so the weighted terms need not be independent. Orthogonality still holds when each coefficient is measurable at the left endpoint of its own interval: for two distinct terms, condition on the information available just before the later increment. The later increment then has conditional mean zero.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\) relative to this filtration. In particular, for \(s<t\), the increment \(B_t-B_s\) is independent of \(\mathcal{F}_s\) and has distribution \(N(0,t-s)\). Let \(0=t_0<t_1<\cdots<t_n=T\) be a deterministic partition, and write \(\Delta B_i=B_{t_{i+1}}-B_{t_i}\) for \(0\le i<n\).
The endpoint condition is the key: \(H_i\) is known at time \(t_i\), before the increment \(\Delta B_i\) occurs. The increment term \(Y_i=H_i\Delta B_i\) is square-integrable. Indeed, independence of \(\Delta B_i\) from \(\mathcal{F}_{t_i}\) gives
This factorization uses left-endpoint measurability. It need not hold if the coefficient depends on the increment it multiplies.
The Conditional Mean-Zero Principle
Proof. Put \(Z=B_t-B_s\). Brownian independent increments imply \(\mathbb{E}[Z\mid\mathcal{F}_s]=0\). If \(X\) is bounded and \(\mathcal{F}_s\)-measurable, the defining multiplication property of conditional expectation gives \(\mathbb{E}[XZ\mid\mathcal{F}_s]=X\mathbb{E}[Z\mid\mathcal{F}_s]=0\). For a possibly unbounded \(X\), define \(X_m=\max(-m,\min(X,m))\). Then each \(X_m\) is bounded and \(\mathcal{F}_s\)-measurable. Also, \(|(X_m-X)Z|\le |XZ|\) and \((X_m-X)Z\to0\) almost surely, so dominated convergence gives \(\mathbb{E}|(X_m-X)Z|\to0\). Conditional expectations contract in \(L^1\), hence \(\mathbb{E}[X_mZ\mid\mathcal{F}_s]\to\mathbb{E}[XZ\mid\mathcal{F}_s]\) in \(L^1\). The left side is zero for every \(m\), proving the claim. \(\square\)
The lemma clarifies which information matters. A coefficient can be random and can depend on the entire past; it is enough that it be measurable before the increment being conditioned on.
Worked Example: A Random Later Coefficient
Use the partition \(0<\tfrac12<1\), with \(H_0=2\) and \(H_1=B_{1/2}\). Compute the expectation of the product of the two weighted increments.
Here \(Y_0=2B_{1/2}\) and \(Y_1=B_{1/2}(B_1-B_{1/2})\). The coefficient \(H_1\) is random, but it is \(\mathcal{F}_{1/2}\)-measurable. Writing \(\Delta B_1=B_1-B_{1/2}\), the product is \(2B_{1/2}^2\Delta B_1\). Its expectation exists: \(B_{1/2}^2\) and \(\Delta B_1\) are independent, and both have finite moments of the required orders. Conditioning on \(\mathcal{F}_{1/2}\) gives
Thus \(\mathbb{E}[Y_0Y_1]=0\). As a direct check, \(B_{1/2}\sim N(0,\tfrac12)\), so \(\mathbb{E}[B_{1/2}^2]=\tfrac12\), while \(\mathbb{E}[\Delta B_1]=0\); independence yields \(\mathbb{E}[2B_{1/2}^2\Delta B_1]=2\cdot\tfrac12\cdot0=0\).
Orthogonality Theorem
Proof. Fix \(i<j\). Since \(Y_i\) is \(\mathcal{F}_{t_{i+1}}\)-measurable and \(t_{i+1}\le t_j\), it is \(\mathcal{F}_{t_j}\)-measurable. Also \(H_j\) is \(\mathcal{F}_{t_j}\)-measurable. Therefore \(X=Y_iH_j\) is \(\mathcal{F}_{t_j}\)-measurable. The product \(Y_iY_j\) is integrable by Cauchy–Schwarz, because \(Y_i,Y_j\in L^2(\mathbb{P})\). Apply the lemma with \(s=t_j\), \(t=t_{j+1}\), and this \(X\). Its product \(X(B_{t_{j+1}}-B_{t_j})\) is \(Y_iY_j\), so its conditional expectation given \(\mathcal{F}_{t_j}\) is zero. Taking expectations gives \(\mathbb{E}[Y_iY_j]=0\). \(\square\)
In particular, distinct weighted increments have zero covariance: their expectations are zero because \(\mathbb{E}[H_i\Delta B_i]=0\), by conditioning on \(\mathcal{F}_{t_i}\). Orthogonality is weaker than independence: the coefficients can share past information, and the weighted terms need not be independent. The proof needs only the conditional mean of the later increment and the required measurability.
Worked Example: Conditioning Despite Dependence on the Past
Take the partition \(0<1<2\), with \(H_0=1\) and \(H_1=B_1^2\). Find the expectation of \((H_0\Delta B_0)(H_1\Delta B_1)\), where \(\Delta B_0=B_1-B_0\) and \(\Delta B_1=B_2-B_1\).
The coefficient \(H_1=B_1^2\) depends on the first increment, so the earlier weighted term and the coefficient of the later term are not independent. Nevertheless, \(H_1\) is \(\mathcal{F}_1\)-measurable. The terms are square-integrable: \(\mathbb{E}[(H_0\Delta B_0)^2]=1\), and independence of \(\Delta B_1\) from \(\mathcal{F}_1\) gives \(\mathbb{E}[(H_1\Delta B_1)^2]=\mathbb{E}[B_1^4]\mathbb{E}[\Delta B_1^2]=3\cdot1=3\). Their product is integrable by Cauchy–Schwarz. Conditioning now yields
Consequently the unconditional expectation is zero as well. Independently, \(\Delta B_1\) is independent of \((\Delta B_0,B_1)\) and has mean zero, so \(\mathbb{E}[\Delta B_0B_1^2]\mathbb{E}[\Delta B_1]=0\). Both calculations confirm the result without incorrectly requiring the weighted terms to be independent.
Why the Endpoint Condition Cannot Be Dropped
The conditional argument fails if the coefficient on the later interval is allowed to depend on that interval's increment. This is not an admissible simple Ito integrand: the process would not be adapted on that interval. The failure is substantive, not merely a technical gap.
Worked Example: An Anticipative Coefficient Breaks Orthogonality
On \(0<1<2\), set \(H_0=1\), but take \(H_1=\Delta B_0\Delta B_1\) for the interval \([1,2)\). This \(H_1\) is not \(\mathcal{F}_1\)-measurable, because it depends on \(\Delta B_1=B_2-B_1\). Compute the product of the two algebraic terms.
The first term is \(H_0\Delta B_0=\Delta B_0\), and the second is \(H_1\Delta B_1=\Delta B_0(\Delta B_1)^2\). Their product is \((\Delta B_0)^2(\Delta B_1)^2\). The increments are independent, with \(\mathbb{E}[(\Delta B_0)^2]=1\) and \(\mathbb{E}[(\Delta B_1)^2]=1\). Therefore
Both algebraic terms are square-integrable, but their product expectation is not zero. This is not a counterexample to the theorem: the coefficient for the second interval violates its left-endpoint measurability hypothesis, so the second term is not an Ito increment for an admissible simple integrand. The computation also shows the error in conditioning as though \(H_1\) were already known at time \(1\).
Common mistake. One might claim that the product expectation vanishes just because \(\mathbb{E}[\Delta B_j]=0\). That argument is incomplete: the other factors must be measurable with respect to the sigma-field on which the conditional mean-zero statement is based. In the theorem, \(Y_iH_j\) is \(\mathcal{F}_{t_j}\)-measurable, which is exactly why conditioning works. In the anticipative example, the factor includes \(\Delta B_j\) itself, and cannot be taken outside the conditional expectation.
The orthogonality result is an algebraic foundation for computing the second moment of a simple stochastic sum: when the square of \(\sum_iY_i\) is expanded, every cross term has expectation zero. The diagonal terms remain, and evaluating their sum is the next step in the development of the Ito isometry. The integrals here are Ito integrals defined by left-endpoint sums; no Stratonovich correction is being used.
Check Your Understanding
For each question, identify the sigma-field that contains the coefficient multiplying the increment being conditioned on.
- For \(i<j\), explain why \(Y_iH_j\) is \(\mathcal{F}_{t_j}\)-measurable.
- Which assumption allows \(\mathbb{E}[\Delta B_j\mid\mathcal{F}_{t_j}]=0\)?
- Can \(H_j\) depend on \(B_{t_j}\) and still satisfy the theorem's measurability condition? Explain.
- Why does square-integrability of \(Y_i\) and \(Y_j\) guarantee that their product is integrable?
- In the anticipative example, which hypothesis fails, and what is the computed cross-product expectation?