Predictable Processes And The Predictable Sigma-Field

Predictability as Information Available Beforehand

Adaptedness asks whether a process value at time $t$ is measurable using the information $\mathcal{F}_t$ available at time $t$. Predictability records a more specific timing convention: on an interval that begins at $s$, an elementary predictable process uses a coefficient already measurable with respect to $\mathcal{F}_s$. This distinction matters when a process is used as an integrand, because a coefficient chosen using information from later in the interval would use the future.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$, where the filtration is increasing: $\mathcal{F}_s\subseteq\mathcal{F}_t$ for $s\le t$. Let this space carry a Brownian motion $B_t$ with respect to the filtration. The definition below does not require Brownian motion or any further conditions on the filtration; Brownian motion provides a familiar setting for the examples.

Definition: A real-valued process $X=(X_t)_{t\ge0}$ is adapted if $X_t$ is $\mathcal{F}_t$-measurable for every $t$. It is left-continuous if, for every $t>0$, $X_s\to X_t$ as $s\uparrow t$. The predictable sigma-field, denoted $\mathcal{P}$, is the sigma-field on $\Omega\times[0,\infty)$ generated by the sets $X^{-1}(C)=\{(\omega,t):X_t(\omega)\in C\}$, as $X$ ranges over adapted left-continuous real-valued processes and $C$ ranges over Borel subsets of $\mathbb{R}$. A process is predictable if it is measurable from $(\Omega\times[0,\infty),\mathcal{P})$ into the Borel subsets of its state space.

Left-continuity imposes a convention at jump times: the value at $t$ agrees with values immediately before $t$, not necessarily values immediately after it. The predictable sigma-field can also be described using rectangles that specify both a time interval and the information permitted to determine a process there.

A Rectangle Description

Theorem: The predictable sigma-field is $$ \mathcal{P} = \sigma\left( \mathcal{F}_0\times\{0\}, \; A\times(s,t] : 0\le s<t<\infty,\ A\in\mathcal{F}_s \right). $$ Here $\mathcal{F}_0\times\{0\}$ means the collection of sets $A\times\{0\}$ with $A\in\mathcal{F}_0$.

Proof. Let $\mathcal{G}$ denote the sigma-field on the right-hand side. First, each generator of $\mathcal{G}$ belongs to $\mathcal{P}$. For $A\in\mathcal{F}_0$, the process equal to $\mathbf{1}_A$ at time zero and zero at every positive time is adapted and left-continuous. Its inverse image of $\{1\}$ is $A\times\{0\}$. For $A\in\mathcal{F}_s$ and $s<t$, define $X_u=\mathbf{1}_A\mathbf{1}_{(s,t]}(u)$. At times $u\le s$ this process is zero; at times $s<u\le t$, its possibly nonzero value is measurable with respect to $\mathcal{F}_s\subseteq\mathcal{F}_u$; and after $t$ it is zero. It is left-continuous, including at $s$ and $t$, and $X^{-1}(\{1\})=A\times(s,t]$. Thus $\mathcal{G}\subseteq\mathcal{P}$.

For the reverse inclusion, take any adapted left-continuous real-valued process $X$. Set $h_n=2^{-n}$, and define $$ X^{(n)}_t = X_0\mathbf{1}_{\{0\}}(t) + \sum_{k=0}^{\infty}X_{kh_n}\mathbf{1}_{(kh_n,(k+1)h_n]}(t). $$ For each fixed $t$, at most one term in the sum is nonzero. Each coefficient $X_{kh_n}$ is $\mathcal{F}_{kh_n}$-measurable, so its product with the indicator of $(kh_n,(k+1)h_n]$ is $\mathcal{G}$-measurable as a function on $\Omega\times[0,\infty)$. The initial term is $\mathcal{G}$-measurable because $X_0$ is $\mathcal{F}_0$-measurable. Hence $X^{(n)}$ is $\mathcal{G}$-measurable. At any $t>0$, the coefficient used by $X^{(n)}_t$ is sampled at a grid point less than $t$ and at distance at most $h_n$ from $t$. These grid points approach $t$ from the left as $n\to\infty$. Left-continuity gives $X^{(n)}_t\to X_t$; at $t=0$, $X^{(n)}_0=X_0$. Therefore $X$ is $\mathcal{G}$-measurable. Its inverse images of Borel sets belong to $\mathcal{G}$, so $\mathcal{P}\subseteq\mathcal{G}$. The two inclusions prove the equality. $\square$

The interval $(s,t]$ is important: its right endpoint is included, while its left endpoint is excluded. The coefficient on that interval is allowed to depend on $A\in\mathcal{F}_s$, not on information that first becomes available after $s$.

Proposition: Every predictable process is adapted.

Proof. Fix $t\ge0$, and let $\mathcal{D}_t$ be the collection of sets $E\subseteq\Omega\times[0,\infty)$ whose section $E_t=\{\omega:(\omega,t)\in E\}$ belongs to $\mathcal{F}_t$. Sections commute with complements and countable unions, so $\mathcal{D}_t$ is a sigma-field. It contains each generator $A\times(s,u]$: its section at $t$ is either empty or $A$, and in the latter case $s<t$, so $A\in\mathcal{F}_s\subseteq\mathcal{F}_t$. It also contains the time-zero generators, whose sections at $t=0$ are in $\mathcal{F}_0$ and whose sections at $t>0$ are empty. The rectangle description therefore gives $\mathcal{P}\subseteq\mathcal{D}_t$. If $Y$ is predictable, then $\{Y_t\in C\}$ is the section at $t$ of the predictable set $\{Y\in C\}$; it belongs to $\mathcal{F}_t$ for every Borel set $C$. Thus $Y_t$ is $\mathcal{F}_t$-measurable at every $t$, which is adaptedness. $\square$

Elementary Predictable Processes

Definition: An elementary predictable process is a process of the form $$ H_t=\eta\,\mathbf{1}_{\{0\}}(t)+\sum_{i=0}^{n-1}\xi_i\,\mathbf{1}_{(t_i,t_{i+1}]}(t), \qquad 0=t_0<t_1<\cdots<t_n=T, $$ where $\eta$ is $\mathcal{F}_0$-measurable and each $\xi_i$ is $\mathcal{F}_{t_i}$-measurable. The process is taken to be zero for $t>T$.

Each summand is measurable with respect to $\mathcal{P}$, by the rectangle description (or by approximation of a measurable coefficient by simple random variables). Finite sums are therefore predictable. Conversely, an elementary process has a coefficient for each interval that is known at its left endpoint in the sense of $\mathcal{F}_{t_i}$-measurability. This is the defining information feature of the elementary class; it does not say that the coefficient is deterministic or already known at time zero.

Worked Example: A Deterministic Step Process

Let $T=2$, with partition $0<1<2$, and define $H_0=0$, $H_t=3$ for $0<t\le1$, $H_t=-2$ for $1<t\le2$, and $H_t=0$ for $t>2$. Verify predictability and check the endpoint values.

On $(0,1]$, the coefficient is $3$, which is measurable with respect to $\mathcal{F}_0$. On $(1,2]$, the coefficient is $-2$, which is measurable with respect to $\mathcal{F}_1$. The initial value $0$ is also $\mathcal{F}_0$-measurable. Thus this is elementary predictable. At $t=0$, $H_t=0$; at $t=1$, the first interval includes its right endpoint, so $H_1=3$; at $t=2$, $H_2=-2$; and at $t>2$, $H_t=0$.

Check: the interval indicators give $\mathbf{1}_{(0,1]}(1)=1$ and $\mathbf{1}_{(1,2]}(1)=0$, so the formula yields $H_1=3$. At $t=2$, they give $0$ and $1$, so $H_2=-2$, as specified.

Worked Example: Random Coefficients Known at the Start

On the Brownian filtered space, set $$ H_t = 2\mathbf{1}_{\{B_{1/2}\ge0\}}\mathbf{1}_{(1/2,1]}(t) - \mathbf{1}_{\{B_1<0\}}\mathbf{1}_{(1,2]}(t), $$ and set $H_0=0$ and $H_t=0$ for $t>2$. Verify predictability and compute $\mathbb{E}\int_0^2 H_t^2\,dt$, where this is an ordinary time integral of the displayed process.

The first coefficient is measurable with respect to $\mathcal{F}_{1/2}$, and the second is measurable with respect to $\mathcal{F}_1$. The intervals are $(1/2,1]$ and $(1,2]$, so each coefficient is available at its interval’s left endpoint. The rectangle description makes $H$ predictable.

For every outcome, the squared process equals $4\mathbf{1}_{\{B_{1/2}\ge0\}}$ on an interval of length $1/2$, and $\mathbf{1}_{\{B_1<0\}}$ on an interval of length $1$. Brownian symmetry gives $\mathbb{P}(B_{1/2}\ge0)=1/2$ and $\mathbb{P}(B_1<0)=1/2$. Consequently $$ \mathbb{E}\int_0^2 H_t^2\,dt = 4\left(\frac12\right)\left(\frac12\right) + 1\left(\frac12\right)(1) = 1+\frac12 = \frac32. $$

Check: if $B_{1/2}<0$ and $B_1\ge0$, both indicators vanish and the time integral is $0$. If $B_{1/2}\ge0$ and $B_1<0$, it is $4(1/2)+1(1)=3$. The expectation calculation weights each interval’s squared coefficient by its length and its event probability, giving $4(1/2)(1/2)+1(1/2)=3/2$.

When a Coefficient Uses the Future

Worked Example: Detecting an Unavailable Coefficient

Consider the tempting step process $\widetilde{H}_t=\mathbf{1}_{\{B_1>0\}}\mathbf{1}_{(0,1]}(t)$. Explain why it is not elementary predictable, show that it is not adapted, and give a predictable replacement.

The coefficient $\mathbf{1}_{\{B_1>0\}}$ is generally not $\mathcal{F}_0$-measurable, so it fails the elementary-process condition for the interval $(0,1]$. More strongly, at $t=1/2$ the value $\widetilde{H}_{1/2}=\mathbf{1}_{\{B_1>0\}}$ is not $\mathcal{F}_{1/2}$-measurable. Indeed, Brownian independent increments give $$ \mathbb{E}\!\left[\mathbf{1}_{\{B_1>0\}}\mid\mathcal{F}_{1/2}\right] = \Phi\!\left(\frac{B_{1/2}}{\sqrt{1/2}}\right), $$ where $\Phi$ is the standard normal CDF. Since $B_{1/2}$ is finite almost surely and $\Phi(x)$ lies strictly between $0$ and $1$ for finite $x$, this conditional expectation is strictly between $0$ and $1$ almost surely. If the indicator were $\mathcal{F}_{1/2}$-measurable, its conditional expectation would equal the indicator and hence take only the values $0$ and $1$, a contradiction.

A predictable replacement is $H_t=\mathbf{1}_{\{B_{1/2}>0\}}\mathbf{1}_{(1/2,1]}(t)$, with $H_0=0$. Its coefficient is $\mathcal{F}_{1/2}$-measurable, exactly the information allowed at the start of its interval. Its ordinary time integral over $[0,1]$ is $\tfrac12\mathbf{1}_{\{B_{1/2}>0\}}$, so its expectation is $\tfrac12\cdot\tfrac12=\tfrac14$, using Brownian symmetry.

Check: the replacement is zero through time $1/2$, equals $\mathbf{1}_{\{B_{1/2}>0\}}$ on $(1/2,1]$, and is zero after time $1$. The event used in its coefficient is known at $1/2$; the event $\{B_1>0\}$ in the original process was not known at time $1/2$.

What Predictability Does—and Does Not—Say

Predictability is a measurability condition that organizes information in time. It is stronger than simply allowing a process to depend on the observation at the end of an interval: an elementary predictable coefficient must be measurable at the interval’s beginning. The rectangle description also shows that predictable processes are adapted, while the converse is not automatic. The failed future-dependent example demonstrates why adaptedness must be checked at every time, not only at selected interval endpoints.

Predictability alone is not an integrability condition and does not assert that a stochastic integral exists. For example, when discussing an Itô integral $\int_0^T H_s\,dB_s$, the usual square-integrable integrand class $\mathcal{L}^2$ consists of progressively measurable processes satisfying $\mathbb{E}\int_0^T H_s^2\,ds<\infty$; predictability is a distinct measurability concept. No stochastic integral is needed for the definitions and examples here.

Common mistake. Writing a step process with coefficient $\mathbf{1}_{\{B_{t_{i+1}}>0\}}$ on $(t_i,t_{i+1}]$ and calling it predictable uses the wrong time in the measurability test. The required test is whether the coefficient belongs to $\mathcal{F}_{t_i}$, not whether it belongs to $\mathcal{F}_{t_{i+1}}$. The correction is to use a coefficient measurable with respect to $\mathcal{F}_{t_i}$, such as $\mathbf{1}_{\{B_{t_i}>0\}}$, on that interval. Endpoint conventions also matter: $(t_i,t_{i+1}]$ assigns the value at $t_{i+1}$ to the interval that ends there, but does not make information first observed at $t_{i+1}$ available at $t_i$.

Check Your Understanding

For each question, state which filtration sigma-field determines whether the proposed coefficient is allowed.

For $A\in\mathcal{F}_2$, is $\mathbf{1}_A\mathbf{1}_{(2,3]}$ elementary predictable?
What is the value at $t=1$ of $H_t=4\mathbf{1}_{(0,1]}(t)-\mathbf{1}_{(1,2]}(t)$?
Why does the rectangle $A\times(1,4]$ belong to the predictable sigma-field when $A\in\mathcal{F}_1$?
Can an adapted process fail to be elementary predictable because its proposed interval coefficient is measurable only at the interval’s right endpoint?
For the random-coefficient process in the second worked example, which sigma-field must contain each of its two coefficient events?