Zero Mean of Simple Brownian-Increment Sums

Why Left-Endpoint Information Matters

A simple Ito integral is a finite sum of coefficients multiplied by Brownian increments. Its expectation is zero when each coefficient is known at the beginning of the interval whose increment it multiplies. Brownian motion has mean-zero future increments even after conditioning on the information available at that left endpoint. The conditioning step, rather than a cancellation between unrelated terms, is the key reason the whole sum has mean zero.

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 \(s<t\), the increment \(B_t-B_s\) is independent of \(\mathcal{F}_s\) and has distribution \(N(0,t-s)\). Fix a finite partition \(0=t_0<t_1<\cdots<t_n=T\). A simple predictable integrand on this partition has the form \(H_s=\sum_{k=0}^{n-1}\xi_k\mathbf{1}_{(t_k,t_{k+1}]}(s)\), where each \(\xi_k\) is \(\mathcal{F}_{t_k}\)-measurable. We require \(\mathbb{E}[\xi_k^2]<\infty\) for every \(k\). Thus \(H\) is progressively measurable and belongs to \(\mathcal{L}^2\), the class of progressively measurable processes satisfying \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\).

The square-integrability requirement is explicit here: since \(H_s=\xi_k\) on an interval of length \(t_{k+1}-t_k\), \(\mathbb{E}\int_0^T H_s^2\,ds=\sum_{k=0}^{n-1}(t_{k+1}-t_k)\mathbb{E}[\xi_k^2]<\infty\). We will prove the zero-mean statement directly for these finite sums. This is about the Ito integral defined by left-endpoint coefficients; it is not a claim about a Stratonovich integral, which uses a different convention. The result concerns the expectation at time \(T\), not a proof of the martingale property of the entire integral process.

The Conditional-Expectation Argument

Proof. Fix an interval \([t_k,t_{k+1}]\) and write \(\Delta B_k=B_{t_{k+1}}-B_{t_k}\) and \(\Delta t_k=t_{k+1}-t_k\). The Brownian increment has second moment \(\mathbb{E}[(\Delta B_k)^2]=\Delta t_k\). By Cauchy–Schwarz, \(\mathbb{E}[|\xi_k\Delta B_k|]\le\sqrt{\mathbb{E}[\xi_k^2]\mathbb{E}[(\Delta B_k)^2]}=\sqrt{\mathbb{E}[\xi_k^2]\Delta t_k}<\infty\). Thus each summand, and hence the finite sum, is integrable. Since \(\xi_k\) is \(\mathcal{F}_{t_k}\)-measurable and \(\Delta B_k\) is independent of \(\mathcal{F}_{t_k}\) with mean zero, \(\mathbb{E}[\Delta B_k\mid\mathcal{F}_{t_k}]=0\) almost surely. The defining conditional-expectation identity then gives \(\mathbb{E}[\xi_k\Delta B_k]=\mathbb{E}[\xi_k\mathbb{E}[\Delta B_k\mid\mathcal{F}_{t_k}]]=0\). (For an unbounded \(\xi_k\), this identity follows first for bounded truncations and then by integrability of the product.) Linearity of expectation for the finite sum yields \(\mathbb{E}[I_T(H)]=\sum_{k=0}^{n-1}\mathbb{E}[\xi_k\Delta B_k]=0\). \(\square\)

The proof uses two separate facts. The square-integrability assumption ensures the Ito sum is integrable; left-endpoint measurability and the conditional mean of the Brownian increment ensure each term has expectation zero. The size or sign of a coefficient can depend on the past in an arbitrary square-integrable way. It cannot, however, use information from the increment that it multiplies.

Proof. Set \(\Delta B=B_t-B_s\). Independence of \(\xi\), which is \(\mathcal{F}_s\)-measurable, and \(\Delta B\) gives \(\mathbb{E}[|\xi\Delta B|]=\mathbb{E}[|\xi|]\mathbb{E}[|\Delta B|]<\infty\); both factors are finite because \(\xi\) is integrable and a normal random variable has finite absolute first moment. Independence also gives \(\mathbb{E}[\xi\Delta B]=\mathbb{E}[\xi]\mathbb{E}[\Delta B]=\mathbb{E}[\xi]\cdot0=0\). \(\square\)

This proposition separates the expectation calculation from the stronger integrability condition used to construct the Ito integral. In the examples below, we check square-integrability as well, so each sum is an admissible simple Ito integral.

Worked Partition-Based Calculations

When the Zero-Mean Argument Fails

A common mistake is to apply the conditioning proof when a coefficient depends on the same Brownian increment that it multiplies. For example, on \((0,1]\), consider the expression \(B_1(B_1-B_0)=B_1^2\). Its expectation is \(\mathbb{E}[B_1^2]=1\), not zero. The coefficient \(B_1\) is not \(\mathcal{F}_0\)-measurable: it contains the future increment. Therefore this expression is not a simple predictable Ito integral term on that interval. The incorrect calculation would write \(\mathbb{E}[B_1(B_1-B_0)]=\mathbb{E}[B_1]\mathbb{E}[B_1-B_0]=0\); that factorization falsely treats dependent random variables as independent. The correction is to recognize the dependence and compute \(\mathbb{E}[B_1^2]=1\).

There is an important scope distinction. Zero expectation for each terminal simple sum does not, by itself, establish the martingale property of the process of integrals indexed by time; that property requires a conditional-expectation statement at arbitrary times. Here the conclusion is exactly the terminal expectation for finite sums with left-endpoint coefficients. Each term vanishes because its coefficient belongs to the information before the increment, not because every random coefficient has mean zero.