Justifying Zero Means for Stochastic Integrals

When Can an Ito Integral Be Given Mean Zero?

In stochastic calculations, an Ito integral is often discarded after taking expectations: the random fluctuation is said to have mean zero. That step is valid under a specific integrability condition. Adaptedness alone is not enough to justify every expectation argument, and a local martingale cannot automatically be treated as an integrable martingale. Here we isolate the square-integrable case, prove the conditional-expectation statement that supports it, and work through examples where both the mean and the second moment can be checked.

Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\) relative to the filtration: in particular, for \(s<t\), \(B_t-B_s\) is independent of \(\mathcal{F}_s\) and has distribution \(N(0,t-s)\). Fix a finite horizon \(T>0\). All expectations below are under \(\mathbb{P}\).

Progressive measurability ensures that the integrand uses information available up to the time at which it is evaluated. The finite expectation of its integrated square ensures that the integral is square-integrable. These conditions are sufficient for the zero-mean conclusion; they are not a claim that every integrand outside \(\mathcal{L}^2\) must have a nonzero mean.

Zero Expectation and Conditional Expectation

Proof. First take a simple adapted integrand \(H_u=\sum_{i=0}^{n-1}\xi_i\mathbf{1}_{(t_i,t_{i+1}]}(u)\), where \(0=t_0<\cdots<t_n=T\) and each \(\xi_i\) is square-integrable and \(\mathcal{F}_{t_i}\)-measurable. Its integral up to \(t\) is a sum of terms \(\xi_i(B_{t_{i+1}\wedge t}-B_{t_i\wedge t})\), with terms after \(t\) interpreted as zero. Each nonzero Brownian increment is independent of \(\mathcal{F}_{t_i}\), has mean zero, and is multiplied by an \(\mathcal{F}_{t_i}\)-measurable coefficient. Therefore each term has expectation zero, so the finite sum has expectation zero.

For general \(H\in\mathcal{L}^2\), the construction of the Ito integral supplies simple adapted \(H^{(n)}\) such that \(\mathbb{E}\int_0^T|H^{(n)}_u-H_u|^2\,du\to0\). By the Ito isometry, their integrals up to \(t\) converge to \(M_t\) in \(L^2(\mathbb{P})\), hence in \(L^1(\mathbb{P})\), since \(\mathbb{E}|X|\le\sqrt{\mathbb{E}[X^2]}\). Thus their expectations converge to \(\mathbb{E}[M_t]\). Each approximating integral has expectation zero, so \(\mathbb{E}[M_t]=0\). Integrability also follows from the isometry: \(\mathbb{E}[|M_t|]\le\sqrt{\mathbb{E}\int_0^tH_u^2\,du}<\infty\). \(\square\)

Proof. For a simple adapted integrand, write \(M_t-M_s\) as a finite sum of coefficients times Brownian increments over the portions of its time intervals lying in \((s,t]\). If a portion starts at a time \(u\ge s\) that is a left endpoint of an integrand interval, its coefficient is measurable with respect to \(\mathcal{F}_u\). The Brownian increment after \(u\) has conditional mean zero given \(\mathcal{F}_u\); taking conditional expectation again given \(\mathcal{F}_s\subseteq\mathcal{F}_u\) gives zero. If the portion begins at \(s\) inside an interval whose left endpoint is earlier than \(s\), its coefficient is already \(\mathcal{F}_s\)-measurable, and the increment from \(s\) has conditional mean zero given \(\mathcal{F}_s\). Hence \(\mathbb{E}[M_t-M_s\mid\mathcal{F}_s]=0\) for simple integrands.

Now approximate \(H\) in the integrated square norm by simple adapted integrands. By the Ito isometry, the corresponding integrals at \(s\) and \(t\) converge in \(L^2\). Conditional expectation is a contraction in \(L^2\), so the conditional-expectation identity passes to the limit and gives \(\mathbb{E}[M_t\mid\mathcal{F}_s]=M_s\). The Ito isometry gives \(\mathbb{E}[M_t^2]=\mathbb{E}\int_0^tH_u^2\,du<\infty\); the integral is adapted by its construction. These are exactly the required properties. \(\square\)

The proposition gives more than the unconditional mean-zero statement. It says that from time \(s\), the conditional mean of the accumulated integral through time \(t\) is the amount already accumulated at \(s\). In particular, setting \(s=0\) gives the conditional statement \(\mathbb{E}[M_t\mid\mathcal{F}_0]=M_0=0\) almost surely.

Worked Examples

Common Mistake: Using Information from the Future

It is not valid to argue that any random coefficient times a Brownian increment has mean zero merely because the increment has mean zero. The coefficient must be measurable with respect to the information available before that increment. For a concrete counterexample to that mistaken calculation, let \(A=\{B_1>0\}\) and consider the expression \(\mathbf{1}_A B_1\). A careless calculation might factor its expectation as \(\mathbb{E}[\mathbf{1}_A]\mathbb{E}[B_1]=0\). That factorization is invalid: \(\mathbf{1}_A\) and \(B_1\) are not independent.

In fact, for \(Z\sim N(0,1)\), direct integration against the normal density gives \(\mathbb{E}[Z\mathbf{1}_{\{Z>0\}}]=\int_0^\infty z(2\pi)^{-1/2}e^{-z^2/2}\,dz=1/\sqrt{2\pi}\), so \(\mathbb{E}[\mathbf{1}_A B_1]=1/\sqrt{2\pi}\ne0\). But \(\mathbf{1}_A\), used as an integrand on \((0,1]\), is not adapted: it depends on \(B_1\), which is future information at times before \(1\). Thus this product is not an Ito integral with an admissible integrand, and the mean-zero theorem does not apply.

A second error is to use the mean-zero conclusion without checking integrability. The hypothesis \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\) provides both a finite second moment and the \(L^2\) approximation needed in the proof. If a calculation establishes only a local martingale, the martingale conditional-expectation argument above has not yet been justified. The safe procedure is to verify progressive measurability and the integrated-square bound, then use the isometry and conditional expectation.

Practical Checklist

Before removing an Ito integral from an expectation, check these points:

• The integral is an Ito integral under the stated probability measure \(\mathbb{P}\), with \(B_t\) Brownian relative to the filtration.
• The integrand is progressively measurable, so it does not use future information.
• The integrated square has finite expectation on the time interval in question.
• The isometry applies, giving a finite second moment; then the mean-zero and conditional-expectation conclusions are valid.

The central conclusion is precise: for \(H\in\mathcal{L}^2\), the Ito integral is not merely a local martingale but a square-integrable martingale, and its expectation at every time is zero. In estimates, this lets one discard the integral term after taking expectation, while the isometry records its contribution to the second moment.