Why Future-Measurable Coefficients Are Not Predictable

Which Information Can a Coefficient Use?

For a simple Ito integral, the coefficient multiplying a Brownian increment must be chosen using information available before that increment begins. This timing condition is not a technical decoration: it is what lets the coefficient be separated from the independent, mean-zero increment. If the coefficient is allowed to depend on the increment itself, the product can have a nonzero expectation.

Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\) adapted to the filtration, with increments independent of the past. In particular, for \(s<t\), \(B_t-B_s\) is independent of \(\mathcal{F}_s\), has mean zero and variance \(t-s\). The Ito integrands considered here are progressively measurable processes \(H\) in \(\mathcal{L}^2\), meaning that \(\mathbb{E}\int_0^T H_u^2\,du<\infty\) for each finite \(T\). For simple integrands, the usual coefficient condition is explicit: the coefficient on an interval starting at \(t_i\) is \(\mathcal{F}_{t_i}\)-measurable.

Definition: A left-endpoint simple process on a partition \(0=t_0<t_1<\cdots<t_n=T\) has the form \(H_u=\sum_{i=0}^{n-1}\xi_i\mathbf{1}_{(t_i,t_{i+1}]}(u)\), where each \(\xi_i\) is square-integrable and \(\mathcal{F}_{t_i}\)-measurable. Its Ito integral is \(\int_0^T H_u\,dB_u=\sum_{i=0}^{n-1}\xi_i(B_{t_{i+1}}-B_{t_i})\). Such a process is progressively measurable and belongs to \(\mathcal{L}^2\).

By contrast, a coefficient measurable with respect to \(\mathcal{F}_{t_{i+1}}\) may use information revealed during the interval. Future measurability does not by itself prove that a coefficient is non-predictable: it might also have been measurable at the left endpoint. The issue is genuine dependence on information not available at \(t_i\). The resulting product with the increment can still be calculated as a random variable, but it is not thereby an Ito integral of an admissible integrand.

Why Left-Endpoint Coefficients Have Mean Zero

Theorem: For a left-endpoint simple process as defined above, every integral increment has mean zero. More precisely, for \(0\leq k<\ell\leq n\), \(\mathbb{E}\left[\sum_{i=k}^{\ell-1}\xi_i(B_{t_{i+1}}-B_{t_i})\mid\mathcal{F}_{t_k}\right]=0\).

Proof. Each product is integrable: by Cauchy–Schwarz, \(\mathbb{E}|\xi_i(B_{t_{i+1}}-B_{t_i})|\leq(\mathbb{E}\xi_i^2)^{1/2}(t_{i+1}-t_i)^{1/2}<\infty\). For the full interval starting at \(t_i\), independence of the Brownian increment from \(\mathcal{F}_{t_i}\) gives \(\mathbb{E}[\xi_i(B_{t_{i+1}}-B_{t_i})\mid\mathcal{F}_{t_i}]=\xi_i\mathbb{E}[B_{t_{i+1}}-B_{t_i}\mid\mathcal{F}_{t_i}]=0\). When conditioning on \(\mathcal{F}_{t_k}\), the term for \(i=k\) therefore has conditional expectation zero. For \(i>k\), the increment is independent of \(\mathcal{F}_{t_i}\), and \(\mathcal{F}_{t_k}\subseteq\mathcal{F}_{t_i}\); the tower property gives conditional expectation zero given \(\mathcal{F}_{t_k}\) as well. Summing the finitely many integrable terms proves the claim. Taking \(k=0\) and using that \(\mathcal{F}_0\)-conditional mean zero implies mean zero proves the unconditional statement. \(\square\)

The proof depends on two separate facts: the coefficient is known at the interval’s start, and the Brownian increment has conditional mean zero given the information then available. If the coefficient instead depends on the increment, the factorization in the proof is unavailable.

Worked Example: A Valid Coefficient Known at the Left Endpoint

On \([1,3]\), let \(\xi=\mathbf{1}_{\{B_1>0\}}\) and \(H_u=\xi\mathbf{1}_{(1,3]}(u)\). Compute the Ito integral’s mean and second moment.

The coefficient is \(\mathcal{F}_1\)-measurable. The process is progressively measurable and square-integrable, since \(\mathbb{E}\int_0^3 H_u^2\,du=2\mathbb{E}[\xi^2]=2\mathbb{P}(B_1>0)=1<\infty\). Its Ito integral is \(\xi(B_3-B_1)\). The increment is independent of \(\mathcal{F}_1\), has mean zero, and has variance \(3-1=2\). Thus

\mathbb{E}[\xi(B_3-B_1)]=\mathbb{E}[\xi]\mathbb{E}[B_3-B_1]=\tfrac12\cdot0=0, \qquad \mathbb{E}[\xi^2(B_3-B_1)^2]=\mathbb{E}[\xi^2]\cdot2=\tfrac12\cdot2=1.

Check: the conditional expectation of the integral given \(\mathcal{F}_1\) is \(\xi\mathbb{E}[B_3-B_1\mid\mathcal{F}_1]=0\), consistent with the computed mean. The second moment is positive even though the mean is zero.

What Changes When the Coefficient Uses the Increment?

On an interval \((s,t]\), write \(\Delta B=B_t-B_s\). If a random coefficient \(\eta\) depends on \(\Delta B\), then \(\eta\Delta B\) need not have mean zero. The following product is a perfectly well-defined random variable whenever it is integrable. But if the process is \(H_u=\eta\) throughout \((s,t]\) and \(\eta\) is not \(\mathcal{F}_s\)-measurable, that process is not an admissible left-endpoint simple Ito integrand. Its algebraic product must not be assigned the usual Ito integral’s properties.

Proposition: Future measurability alone does not preserve the zero-mean property. For an interval of positive length \(h=t-s\), setting \(\eta=\mathbf{1}_{\{\Delta B>0\}}\) gives \(\mathbb{E}[\eta\Delta B]=\sqrt{h}/\sqrt{2\pi}>0\).

Proof. Since \(\Delta B\) has distribution \(N(0,h)\), its density at \(x\) is \((2\pi h)^{-1/2}e^{-x^2/(2h)}\). Therefore

\mathbb{E}[\mathbf{1}_{\{\Delta B>0\}}\Delta B] =\int_0^\infty x\frac{1}{\sqrt{2\pi h}}e^{-x^2/(2h)}\,dx =\frac{\sqrt{h}}{\sqrt{2\pi}}\int_0^\infty e^{-y}\,dy =\frac{\sqrt{h}}{\sqrt{2\pi}},

where \(y=x^2/(2h)\), so \(x\,dx=h\,dy\). The final integral equals \(1\), making the expectation strictly positive for \(h>0\). \(\square\)

Worked Example: Selecting Positive Increments

On \([0,1]\), take \(\eta=\mathbf{1}_{\{B_1>0\}}\). Calculate the mean and second moment of the future-dependent product \(\eta B_1\).

Here \(h=1\), so the proposition gives \(\mathbb{E}[\eta B_1]=1/\sqrt{2\pi}\). For the second moment, symmetry of the standard normal density gives \(\mathbb{E}[\eta B_1^2]=\tfrac12\mathbb{E}[B_1^2]=\tfrac12\). Hence the variance is \(\tfrac12-(1/\sqrt{2\pi})^2=\tfrac12-\tfrac1{2\pi}\).

The process \(H_u=\mathbf{1}_{\{B_1>0\}}\) on \((0,1]\) has finite expected time-square norm: \(\mathbb{E}\int_0^1H_u^2\,du=\mathbb{P}(B_1>0)=\tfrac12\). Nevertheless, for \(u<1\), \(H_u\) is not \(\mathcal{F}_u\)-measurable. Indeed, independence of \(B_1-B_u\) from \(\mathcal{F}_u\) gives \(\mathbb{E}[\mathbf{1}_{\{B_1>0\}}\mid\mathcal{F}_u]=\Phi(B_u/\sqrt{1-u})\), which is strictly between zero and one for finite \(B_u\). If the indicator were \(\mathcal{F}_u\)-measurable, its conditional expectation would equal that indicator and take only the values zero and one, a contradiction. Thus the product is not an Ito integral of this interval-step process.

Check: substituting \(h=1\) into the proposition gives \(\sqrt{1}/\sqrt{2\pi}=1/\sqrt{2\pi}\), exactly the computed mean. The nonzero value is caused by selecting only positive increments; it is not the mean of an admissible Ito integral.

Worked Example: Using the Increment Itself as the Coefficient

On \([0,2]\), let \(\eta=B_2\). Find the mean and variance of the future-dependent product \(\eta(B_2-B_0)=B_2^2\).

Since \(B_0=0\), \(B_2\sim N(0,2)\), and the product is \(B_2^2\). Its mean is the variance of \(B_2\), namely \(2\). A centered normal variable with variance \(2\) has fourth moment \(3(2)^2=12\). Therefore \(\operatorname{Var}(B_2^2)=12-2^2=8\).

Here \(H_u=B_2\) on \((0,2]\) has \(\mathbb{E}\int_0^2H_u^2\,du=2\mathbb{E}[B_2^2]=2\cdot2=4<\infty\), but it is not adapted at times \(u<2\). A finite time-square norm alone does not make a process an admissible Ito integrand.

Check: if the interval length is replaced by \(h\), the product is \((\Delta B)^2\), whose mean is \(h\); substituting \(h=2\) gives the mean \(2\) above. Its variance is \(2h^2=2\cdot 2^2=8\), also matching the direct calculation.

The Boundary Between Measurability and Predictability

For an elementary process, the relevant rule is concrete: on \((t_i,t_{i+1}]\), choose a coefficient \(\xi_i\) using \(\mathcal{F}_{t_i}\), not merely \(\mathcal{F}_{t_{i+1}}\). Left-endpoint simple processes are predictable. A coefficient that genuinely uses the interval’s future increment generally makes the corresponding interval process non-adapted, so it cannot be predictable. But the converse phrasing needs care: a variable measurable at the right endpoint could also be measurable at the left endpoint. In that case it remains a valid coefficient.

The coefficient need not be deterministic. The first worked example used information about \(B_1\), and that randomness was allowed because it was already available at time \(1\). What is ruled out is choosing the coefficient after seeing the increment it multiplies. With a future-dependent choice, the coefficient can favor positive increments, or equal the increment itself, and thereby create a nonzero mean.

Common Mistake: Conditioning on Information You Do Not Have

A tempting but incorrect calculation is \(\mathbb{E}[\mathbf{1}_{\{B_1>0\}}B_1]=\mathbb{E}[\mathbf{1}_{\{B_1>0\}}]\mathbb{E}[B_1]=\tfrac12\cdot0=0\). This multiplication of expectations assumes independence. The indicator and \(B_1\) are functions of the same random variable, so they are not independent. Integrating over the positive half of the normal density instead gives \(\mathbb{E}[\mathbf{1}_{\{B_1>0\}}B_1]=1/\sqrt{2\pi}\), which is not zero.

Another error is to argue that any coefficient measurable at the right endpoint must be forbidden. The correct test is whether it was already measurable at the left endpoint. For example, a constant coefficient is measurable at both endpoints and is valid; a coefficient such as \(\mathbf{1}_{\{B_t-B_s>0\}}\) need not be known at time \(s\). The zero-mean proof applies to the first kind, not automatically to the second.

Coefficient on \((s,t]\)	Information available at \(s\)?	Conclusion
\(\xi\in\mathcal{F}_s\), square-integrable	Yes	Admissible simple Ito coefficient; product has mean zero
\(\mathbf{1}_{\{B_t-B_s>0\}}\)	Not in general	Future-dependent product has positive mean
\(B_t-B_s\)	Not in general	Product is \((B_t-B_s)^2\), with mean \(t-s\)
A coefficient measurable at both \(s\) and \(t\)	Yes	Allowed by the left-endpoint condition

Check Your Understanding

For each question, distinguish an admissible Ito integral from a future-dependent product, and state which information is available at the interval’s start.

If \(\xi\in\mathcal{F}_s\) is square-integrable, explain why \(\mathbb{E}[\xi(B_t-B_s)]=0\).
For \(\eta=\mathbf{1}_{\{B_t-B_s>0\}}\), compute the mean of \(\eta(B_t-B_s)\) in terms of \(h=t-s>0\).
Does finite \(\mathbb{E}\int_s^t H_u^2\,du\) alone ensure that \(H\) is an admissible Ito integrand? Explain.
Give an example of a coefficient measurable at \(t\) that is also measurable at \(s\), and explain why it is allowed.
Identify the independence error in claiming \(\mathbb{E}[\mathbf{1}_{\{B_1>0\}}B_1]=0\), and give the correct value.