Adaptedness: What Is Knowable at Each Time?
For a stochastic process to describe information available as time passes, its value at time \(t\) must be measurable using only the information available by time \(t\). This requirement is called adaptedness. It is a separate check at every time: a process can be adapted even when its value depends on randomness, and a random variable can be well-defined on the probability space yet unavailable to the filtration at the time it is used.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\), where the sigma-fields increase with time. When Brownian motion is used below, \(B_t\) denotes a standard Brownian motion adapted to this filtration, with increments after time \(s\) independent of \(\mathcal{F}_s\). In the examples involving the natural filtration, we take \(\mathcal{F}_t^B=\sigma(B_s:0\le s\le t)\). No stochastic integrals are needed here.
The definition is pointwise in time. It does not, by itself, assert that the map \((t,\omega)\mapsto X_t(\omega)\) is jointly measurable. That is a distinct measurability question. For a process with values in \(\mathbb{R}^d\), adaptedness means that \(X_t\) is an \(\mathcal{F}_t\)-measurable random vector for each \(t\); equivalently, each coordinate is \(\mathcal{F}_t\)-measurable.
Basic Tests for Adaptedness
A useful way to check adaptedness is to fix a time \(t\), identify the random variable \(X_t\), and ask whether the events determined by that random variable belong to \(\mathcal{F}_t\). The filtration's increase matters: if a random variable is \(\mathcal{F}_s\)-measurable and \(s\le t\), then it is also \(\mathcal{F}_t\)-measurable.
Proof. Fix any \(t\ge0\). Since \(X\) is adapted, \(X_t\) is \(\mathcal{F}_t\)-measurable. For every Borel set \(G\subseteq\mathbb{R}\), Borel measurability of \(g\) implies that \(g^{-1}(G)\) is Borel. Therefore \(\{\omega:g(X_t(\omega))\in G\}=\{\omega:X_t(\omega)\in g^{-1}(G)\}\) belongs to \(\mathcal{F}_t\). Thus \(g(X_t)\) is \(\mathcal{F}_t\)-measurable for every \(t\), as required. \(\square\)
Worked Example: Deterministic Processes
Let \(a:[0,\infty)\to\mathbb{R}\) be any deterministic function and set \(X_t(\omega)=a(t)\) for all \(\omega\). Check whether \(X\) is adapted to an arbitrary filtration \((\mathcal{F}_t)\).
Fix \(t\) and a Borel set \(G\). The event \(\{\omega:X_t(\omega)\in G\}\) is \(\Omega\) if \(a(t)\in G\), and is \(\varnothing\) if \(a(t)\notin G\). Both \(\Omega\) and \(\varnothing\) belong to every sigma-field \(\mathcal{F}_t\). Hence \(X_t\) is \(\mathcal{F}_t\)-measurable for every \(t\), so \(X\) is adapted to every filtration.
Check: for \(a(t)=t^2\), at time \(t=3\) the random variable \(X_3\) is the constant \(9\). For any Borel set \(G\), its preimage is \(\Omega\) when \(9\in G\) and \(\varnothing\) otherwise. The same reasoning works at every other time; no time-measurability assumption on \(a\) was needed for adaptedness.
For a random step process, the time at which a coefficient becomes active must be no earlier than the time at which that coefficient is known. The half-open interval convention makes the value at each switching time unambiguous.
Proof. Suppose first that each \(\xi_i\) is \(\mathcal{F}_{t_i}\)-measurable. If \(t_i\le t<t_{i+1}\), then \(X_t=\xi_i\), and \(\mathcal{F}_{t_i}\subseteq\mathcal{F}_t\); thus \(X_t\) is \(\mathcal{F}_t\)-measurable. If \(t\ge t_n\), then \(X_t=\xi_n\) and \(\mathcal{F}_{t_n}\subseteq\mathcal{F}_t\), so it is again \(\mathcal{F}_t\)-measurable. This proves adaptedness at every time. Conversely, suppose \(X\) is adapted. At the switching time \(t_i\), the definition gives \(X_{t_i}=\xi_i\), including \(i=n\). Therefore \(\xi_i\) is \(\mathcal{F}_{t_i}\)-measurable for each \(i\). Both implications hold for all coefficients, including deterministic or almost surely constant ones. \(\square\)
Worked Example: A Brownian Step Process
Use the natural filtration \(\mathcal{F}_t^B\) and define \(X_t=1\) for \(0\le t<1\), \(X_t=B_1\) for \(1\le t<2\), and \(X_t=B_2\) for \(t\ge2\). Determine whether \(X\) is adapted.
The coefficient on \([0,1)\) is the constant \(1\), which is \(\mathcal{F}_0^B\)-measurable. The next coefficient \(B_1\) is \(\mathcal{F}_1^B\)-measurable by the definition \(\mathcal{F}_1^B=\sigma(B_s:0\le s\le1)\). The final coefficient \(B_2\) is \(\mathcal{F}_2^B\)-measurable for the same reason. The step-process theorem therefore proves that \(X\) is adapted.
Check at the boundaries: \(X_0=1\), \(X_1=B_1\), and \(X_2=B_2\). At a time such as \(t=\tfrac32\), \(X_t=B_1\), which is already known at time \(1\) and hence belongs to \(\mathcal{F}_{3/2}^B\). At \(t=\tfrac52\), \(X_t=B_2\), which is known by time \(2\) and hence belongs to \(\mathcal{F}_{5/2}^B\). The checks at the switching times agree with the interval convention in the definition.
Brownian Coordinates and Future Information
By construction, \(B_t\) is \(\mathcal{F}_t^B\)-measurable: it is one of the coordinates used to generate \(\mathcal{F}_t^B\). Applying the proposition above also shows that \(B_t^2\), or \(g(B_t)\) for any Borel function \(g\), is measurable at time \(t\). In contrast, a quantity involving Brownian motion after time \(t\) need not be available at time \(t\).
Worked Example: A Future Increment Is Not Adapted
Fix \(h>0\) and define \(Y_t=B_{t+h}-B_t\). Test whether \(Y\) is adapted to the natural filtration \((\mathcal{F}_t^B)\).
For any fixed \(t\), the Brownian independent-increments property says that \(Y_t\) is independent of \(\mathcal{F}_t^B\). Its distribution is \(N(0,h)\), so it is not almost surely constant. If \(Y_t\) were \(\mathcal{F}_t^B\)-measurable, then the event \(A=\{Y_t\le0\}\) would belong to \(\mathcal{F}_t^B\). Independence would then make \(A\) independent of itself, giving \(\mathbb{P}(A)=\mathbb{P}(A)^2\). But the \(N(0,h)\) distribution gives \(\mathbb{P}(A)=\tfrac12\), and \(\tfrac12\ne(\tfrac12)^2\). Thus \(Y_t\) is not \(\mathcal{F}_t^B\)-measurable. This holds for each \(t\), so \(Y\) is not adapted.
Check: at \(t=0\), \(Y_0=B_h-B_0=B_h\), and \(B_h\) has distribution \(N(0,h)\), not a constant distribution. The same independence argument applies. By contrast, the coordinate process \(B_t\) itself is adapted because its value at time \(t\) belongs to the information generated by the Brownian path up to \(t\).
When Information Arrives Later
A simple filtration can make the timing issue explicit. Let \(U\) be a random variable taking the values \(0\) and \(1\), each with probability \(\tfrac12\), and define \(\mathcal{F}_t=\{\varnothing,\Omega\}\) for \(t<1\) and \(\mathcal{F}_t=\sigma(U)\) for \(t\ge1\). Before time \(1\), the filtration contains no information about \(U\); from time \(1\) onward, it does.
Worked Example: Revealing a Random Variable at a Fixed Time
Compare the constant-in-time process \(V_t=U\) with the process \(Z_t=\mathbf{1}_{\{t\ge1\}}U\) under the filtration just defined.
For \(V\), at time \(t=0\) and Borel set \(G=\{1\}\), the preimage \(\{V_0\in G\}=\{U=1\}\) has probability \(\tfrac12\), so it is neither \(\varnothing\) nor \(\Omega\). It is not in the trivial sigma-field \(\mathcal{F}_0\). Therefore \(V\) is not adapted. For \(Z\), if \(t<1\), then \(Z_t=0\), a deterministic variable measurable in \(\mathcal{F}_t\). If \(t\ge1\), then \(Z_t=U\), which is \(\mathcal{F}_t=\sigma(U)\)-measurable. Thus \(Z\) is adapted.
Check: at \(t=\tfrac12\), \(Z_t=0\) and \(V_t=U\), so only \(Z_t\) is measurable in the trivial sigma-field. At \(t=1\), the indicator equals \(1\), so \(Z_1=U\), which is measurable in \(\mathcal{F}_1\). The inclusion of the reveal time is important: the process is defined to switch at \(t=1\), not immediately after it.
Common Pitfall: A Coefficient Can Be Known Too Late
Consider the altered Brownian step process \(\widetilde{X}_t=1\) for \(0\le t<1\), \(\widetilde{X}_t=B_2\) for \(1\le t<2\), and \(\widetilde{X}_t=B_2\) for \(t\ge2\). A tempting but wrong argument is: “\(B_2\) is a random variable, so using it as a step coefficient makes the process adapted.” That checks only that \(B_2\) exists, not that it is measurable at the time the coefficient first appears.
At \(t=1\), \(\widetilde{X}_1=B_2\). If \(B_2\) were \(\mathcal{F}_1^B\)-measurable, then \(B_2-B_1\) would also be \(\mathcal{F}_1^B\)-measurable, since \(B_1\) is. But \(B_2-B_1\) is independent of \(\mathcal{F}_1^B\) and has distribution \(N(0,1)\). The event \(\{B_2-B_1\le0\}\) would then be both in \(\mathcal{F}_1^B\) and independent of \(\mathcal{F}_1^B\), hence independent of itself. Its probability is \(\tfrac12\), contradicting \(\tfrac12\ne(\tfrac12)^2\). Thus \(\widetilde{X}_1\) is not \(\mathcal{F}_1^B\)-measurable and \(\widetilde{X}\) is not adapted.
The reliable test is always time-specific: at each \(t\), identify the actual value \(X_t\) and check its measurability in \(\mathcal{F}_t\). Deterministic values pass for any filtration; random values pass only when the filtration has enough information by the time they are used.
Check Your Understanding
For each question, specify the time at which measurability must be checked.
- Why is every deterministic process adapted to every filtration, even if its deterministic time function is not continuous?
- For a step process equal to \(\xi_i\) on \([t_i,t_{i+1})\), why does adaptedness require \(\xi_i\) to be measurable at \(t_i\)?
- Is \(g(B_t)\) adapted to the natural Brownian filtration when \(g\) is Borel measurable? Justify your answer.
- For \(h>0\), why does independence of \(B_{t+h}-B_t\) from \(\mathcal{F}_t^B\) rule out its \(\mathcal{F}_t^B\)-measurability?
- In the reveal-time example, what changes at \(t=1\) that makes \(Z_t=\mathbf{1}_{\{t\ge1\}}U\) adapted?