Predictability and Admissible Integrands

Why Integrand Measurability Matters

For an Ito integral, it is not enough that the integrand be a random quantity at each time. Its information must be organized in time: at time \(t\), the integrand cannot use information that has not yet arrived. Predictability formalizes this requirement and includes the simple processes used to construct the integral. Progressive measurability is a closely related condition that also ensures joint measurability in time and outcome.

Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\), with the filtration satisfying the usual conditions. A process \(H=(H_t)_{t\ge0}\) is an Ito integrand on \([0,T]\) in the class \(\mathcal{L}^2\) when it is progressively measurable and \(\mathbb{E}\int_0^T H_t^2\,dt<\infty\). For such an integrand, the Ito integral \(\int_0^T H_t\,dB_t\) is defined by completion from simple predictable processes. Predictable processes are the canonical measurability class in that construction; the density and extension details are developed in the construction of the \(L^2\) integral.

Definition: A process \(H\) is adapted if \(H_t\) is \(\mathcal{F}_t\)-measurable for every \(t\). It is progressively measurable if, for every \(t\), the map \((s,\omega)\mapsto H_s(\omega)\) restricted to \([0,t]\times\Omega\) is measurable with respect to \(\mathcal{B}([0,t])\otimes\mathcal{F}_t\). The predictable sigma-field is generated by \(\{0\}\times A\), \(A\in\mathcal{F}_0\), and the sets \((s,t]\times A\), \(A\in\mathcal{F}_s\), for \(0\le s<t\). A process is predictable if it is measurable with respect to this sigma-field.

Adaptedness is a condition at each fixed time; by itself it does not assert joint measurability in time and outcome. Progressive measurability adds that joint measurability on every time interval and uses only \(\mathcal{F}_t\)-information up to the endpoint \(t\). Predictability is generated by sets whose random event is already known at the left endpoint \(s\), while the time interval is \((s,t]\). This is exactly the time orientation of a simple Ito integrand: its coefficient is known before the Brownian increment it multiplies.

How the Classes Relate

Proposition: Every predictable process is progressively measurable.

Proof. Fix \(u\ge0\). It is enough to check that each generator of the predictable sigma-field, restricted to \([0,u]\times\Omega\), belongs to \(\mathcal{B}([0,u])\otimes\mathcal{F}_u\). For \(\{0\}\times A\), where \(A\in\mathcal{F}_0\), the restriction is measurable because \(A\in\mathcal{F}_u\). For a generator \((s,t]\times A\), with \(A\in\mathcal{F}_s\), its restriction is \((s,t]\cap[0,u]\times A\). The time set is Borel, and \(A\in\mathcal{F}_s\subseteq\mathcal{F}_u\) whenever the restriction is nonempty. Thus every generator has a measurable restriction. The collection of sets whose restrictions are measurable is a sigma-field, so it contains the sigma-field generated by these sets. This proves progressive measurability. \(\square\)

In the reverse direction, adaptedness alone does not imply predictability or even progressive measurability. In particular, predictability is not just a synonym for “the value at time \(t\) is known at time \(t\).” It encodes measurability across time in a way that aligns with the integral's left-endpoint construction.

Worked Example: A Predictable Step Integrand

On \([0,2]\), define \(H_t=2\) for \(0<t\le1\), \(H_t=-1\) for \(1<t\le2\), and \(H_0=0\). Verify predictability and square-integrability.

The first nonzero piece is \(2\mathbf{1}_{(0,1]\times\Omega}\), whose time interval is a predictable generator with event \(\Omega\in\mathcal{F}_0\). The second is \(-\mathbf{1}_{(1,2]\times\Omega}\), also a predictable generator. The value at time zero is deterministic and measurable with respect to the initial sigma-field. Hence \(H\) is predictable and therefore progressive.

Its square-integrability calculation is

\mathbb{E}\int_0^2 H_t^2\,dt =\int_0^1 2^2\,dt+\int_1^2(-1)^2\,dt =4(1-0)+1(2-1)=5.

The value is finite, so \(H\in\mathcal{L}^2\) on this interval. As a check, the two intervals have lengths \(1\) and \(1\), and their squared coefficient contributions are \(4\) and \(1\); their sum is \(5\).

Continuous Adapted Processes Are Predictable

Theorem: If \(X_t\) is adapted and has continuous paths, then \(X\) is predictable.

Proof. Fix a finite horizon \(T\). For each positive integer \(n\), set \(\delta_n=2^{-n}\). For \(t\in(k\delta_n,(k+1)\delta_n]\), define \(X^{(n)}_t=X_{k\delta_n}\), and set \(X^{(n)}_0=X_0\). Each coefficient \(X_{k\delta_n}\) is \(\mathcal{F}_{k\delta_n}\)-measurable by adaptedness. Thus each step piece is measurable with respect to a predictable generator of the form \((k\delta_n,(k+1)\delta_n]\times A\); equivalently, the step process is predictable. For every \(t>0\), the left grid point \(k\delta_n\) in the interval containing \(t\) tends to \(t\) as \(n\to\infty\). Continuity of the path gives \(X^{(n)}_t\to X_t\). At \(t=0\), the values equal \(X_0\). A pointwise limit of measurable functions is measurable, so \(X\) is predictable on \([0,T]\). Since \(T\) was arbitrary, it is predictable on the full time domain. \(\square\)

The argument uses continuity to recover the value at time \(t\) from values at earlier grid points. The approximating coefficients are therefore known before the interval on which each approximation is used. This is why continuous adapted processes fit naturally into the predictable class.

Worked Example: Brownian Motion as an L2 Integrand

Take \(H_t=B_t\) on \([0,1]\). Brownian motion is adapted and has continuous paths, so the theorem makes \(H\) predictable. To check the separate \(L^2\) requirement, use \(B_t\sim N(0,t)\), which gives \(\mathbb{E}[B_t^2]=t\). Tonelli's theorem applies to the nonnegative integrand \(B_t^2\), and therefore

\mathbb{E}\int_0^1 B_t^2\,dt =\int_0^1\mathbb{E}[B_t^2]\,dt =\int_0^1t\,dt =\left[\frac{t^2}{2}\right]_0^1 =\frac12.

Thus \(B_t\) is an admissible \(L^2\) integrand on this horizon. As an independent check, on a general finite interval \([0,T]\) the same calculation gives \(\int_0^Tt\,dt=T^2/2\), which is finite for every finite \(T\); substituting \(T=1\) recovers \(1/2\).

Measurability and Integrability Are Different Tests

Predictability settles the time-information requirement, but it does not guarantee square-integrability. The condition \(\mathbb{E}\int_0^T H_t^2\,dt<\infty\) must still be checked. Conversely, a formal time integral of \(H_t^2\) cannot be used as an \(L^2\) test until its required measurability is established.

Worked Example: Predictable but Not Square-Integrable

On \([0,1]\), set \(H_t=(1-t)^{-1/2}\) for \(0\le t<1\), and set \(H_1=0\). This is a deterministic Borel function of time, hence predictable: Borel time sets crossed with \(\Omega\) are measurable in the predictable sigma-field. But

\mathbb{E}\int_0^1 H_t^2\,dt =\int_0^1\frac{1}{1-t}\,dt =\lim_{b\uparrow1}\left[-\log(1-t)\right]_{0}^{b} =\lim_{b\uparrow1}-\log(1-b) =\infty.

The value assigned at \(t=1\) does not change the Lebesgue integral. This process is predictable but is not in \(\mathcal{L}^2\) on \([0,1]\). The check is the divergence of \(-\log(1-b)\) as \(b\) increases to \(1\); predictability alone cannot make that integral finite.

At the other extreme, adaptedness alone need not provide enough measurability even to form the \(dt\times\mathbb{P}\) integral in the definition of \(\mathcal{L}^2\). For example, choose a non-Borel set \(D\subseteq[0,1]\) and define the deterministic process \(H_t=\mathbf{1}_D(t)\). Each \(H_t\) is a constant random variable, so \(H\) is adapted to any filtration. But if \(H\) were progressively measurable, then \(D\times\Omega=\{(t,\omega):H_t(\omega)=1\}\) would be measurable in \(\mathcal{B}([0,1])\otimes\mathcal{F}_1\). Taking a section at any fixed \(\omega\) would make \(D\) Borel, a contradiction. Thus this adapted process is not progressive.

A Common Pitfall

Mistake: “The process is adapted, so its squared integral is an admissibility calculation.” For the process \(H_t=\mathbf{1}_D(t)\) just described, one might write \(\mathbb{E}\int_0^1H_t^2\,dt=\int_0^1\mathbf{1}_D(t)\,dt\) and then try to assign a length to \(D\). That calculation is invalid: \(D\) is non-Borel and not necessarily Lebesgue measurable, and \(H\) is not progressive. The correction is to check measurability first, then check the \(L^2\) condition. For a predictable or progressive process, the product-measurability needed for the time integral is available; finiteness of the expectation remains a separate question.

The practical sequence is therefore: verify the appropriate time-measurability condition, then verify the square-integrability condition on the time horizon in question. Continuous adapted processes pass the first test by the theorem above, but they must still pass the second test. In applications, Brownian motion itself passes both tests on every finite horizon, while the singular deterministic example passes predictability but fails square-integrability.

Property	What it checks	What it does not guarantee
Adapted	Each \(H_t\) is \(\mathcal{F}_t\)-measurable	Joint measurability or predictability
Progressively measurable	Joint measurability up to each time using \(\mathcal{F}_t\)	Finite expected squared time integral
Predictable	Measurability from information available at interval left endpoints	Finite expected squared time integral
\(\mathcal{L}^2\)	Progressive measurability and \(\mathbb{E}\int_0^T H_t^2\,dt<\infty\)	Any property not included in its definition

Check Your Understanding

For each question, distinguish the measurability condition from the square-integrability condition.

Why does the coefficient on \((s,t]\) in a simple predictable process need to be \(\mathcal{F}_s\)-measurable?
Use left grid-point approximations to explain why an adapted continuous process is predictable.
For \(H_t=B_t\) on \([0,2]\), compute \(\mathbb{E}\int_0^2H_t^2\,dt\) and state whether it is finite.
Can a predictable process fail to belong to \(\mathcal{L}^2\)? Identify the failed condition in the singular deterministic example.
Why does the deterministic process \(H_t=\mathbf{1}_D(t)\), for non-Borel \(D\), show that adaptedness alone is insufficient?