Information Available at Each Time
A stochastic process is a collection of random variables indexed by time. To ask whether its value at time \(n\) is available by time \(n\), we must specify what information is available then. A filtration \((\mathcal{F}_n)_{n\ge0}\) is an increasing sequence of sigma-algebras on a probability space \((\Omega,\mathcal{F},\mathbb{P})\): if \(m\le n\), then \(\mathcal{F}_m\subseteq\mathcal{F}_n\). The sigma-algebra \(\mathcal{F}_n\) represents information available by time \(n\).
Adaptedness is a measurability condition, not a condition about the distribution of a process or its dependence across times. It asks whether each value can be determined using the information available at that same time. The same process can be adapted to one filtration and fail to be adapted to another.
For a real-valued process, the state space is \(\mathbb{R}\) with its Borel sigma-algebra. The definition then requires that every Borel question about \(X_n\), such as whether \(X_n\le a\), can be answered from \(\mathcal{F}_n\). Measurability is required at every time, including time \(0\). If a process is only equal almost surely to an \(\mathcal{F}_n\)-measurable random variable, it is not necessarily \(\mathcal{F}_n\)-measurable as defined here; the distinction disappears if the filtration is complete and the appropriate measurable version is used.
Tests and Consequences
Proof. If \(X_n\) is \(\mathcal{F}_n\)-measurable, then the inverse image of the Borel set \((-\infty,q]\) belongs to \(\mathcal{F}_n\), so the stated condition holds. Conversely, suppose that \(\{X_n\le q\}\in\mathcal{F}_n\) for every rational \(q\). For any real \(a\), density of the rationals gives \(\{X_n<a\}=\bigcup_{q\in\mathbb{Q},\,q<a}\{X_n\le q\}\), which belongs to \(\mathcal{F}_n\) by closure under countable unions. Every open interval \((a,b)\) has inverse image \(\{X_n>a\}\cap\{X_n<b\}\); the first event is the complement of \(\{X_n\le a\}\) when \(a\) is rational, and in general can be written as a countable union of rational sublevel events’ complements. The second is the countable union just established. Thus inverse images of open intervals belong to \(\mathcal{F}_n\). Open intervals with rational endpoints form a countable base for the open sets of \(\mathbb{R}\), so inverse images of all open sets, and hence all Borel sets, belong to \(\mathcal{F}_n\). This is precisely \(\mathcal{F}_n\)-measurability. \(\square\)
Proof. Fix a time \(n\) and a measurable set \(B\) in \(T\). By composition, \(\{Y_n\in B\}=\{g_n(X_n)\in B\}=\{X_n\in g_n^{-1}(B)\}\). Since \(g_n\) is measurable, \(g_n^{-1}(B)\) is measurable in \(S\). Since \(X_n\) is \(\mathcal{F}_n\)-measurable, the last event belongs to \(\mathcal{F}_n\). This holds for every \(B\) and every \(n\), so \((Y_n)\) is adapted. \(\square\)
For real-valued processes, this result shows that applying a measurable function at each time preserves adaptedness. Examples include taking absolute values, indicators of Borel conditions, and sums of finitely many adapted real-valued variables at the same time. It does not say that a process can use a future value: a function of information available at time \(n\) remains available then, but applying a function cannot make an unobserved future value available.
Worked Examples
Worked Example: A Running Count of Observed Heads
Let \(Y_1,Y_2,\ldots\) be independent fair-coin indicators on a probability space, where \(Y_k=1\) for heads and \(Y_k=0\) for tails. Let \(\mathcal{F}_0=\{\varnothing,\Omega\}\) and \(\mathcal{F}_n=\sigma(Y_1,\ldots,Y_n)\) for \(n\ge1\). Define \(X_0=0\) and \(X_n=\sum_{k=1}^nY_k\). Show that \((X_n)\) is adapted, and list its possible values at time \(2\).
At time \(0\), \(X_0\) is constant, so the inverse image of any Borel set is either \(\varnothing\) or \(\Omega\), both in \(\mathcal{F}_0\). For \(n\ge1\), \(X_n\) is a function of \(Y_1,\ldots,Y_n\), all of which are \(\mathcal{F}_n\)-measurable. Equivalently, for each possible integer \(r\), the event \(\{X_n=r\}\) is determined by those \(n\) observations. Thus \(X_n\) is \(\mathcal{F}_n\)-measurable at every time.
At time \(2\), there are four equally likely ordered outcomes. The sums are \(0+0=0\), \(0+1=1\), \(1+0=1\), and \(1+1=2\), so \(\mathbb{P}(X_2=0)=1/4\), \(\mathbb{P}(X_2=1)=2/4=1/2\), and \(\mathbb{P}(X_2=2)=1/4\). Each event is determined by \((Y_1,Y_2)\), so it belongs to \(\mathcal{F}_2\). The probabilities sum to \(1/4+1/2+1/4=1\), as required for the three possible values.
Worked Example: A Future Observation Is Not Yet Available
Use the same independent fair-coin indicators and filtration as in the previous example. Define \(Z_n=Y_{n+1}\) for \(n\ge0\). Test whether \((Z_n)\) is adapted. Then compare it with the one-step lagged process \(W_0=0\) and \(W_n=Y_n\) for \(n\ge1\).
At time \(n\), the event \(\{Z_n=1\}=\{Y_{n+1}=1\}\) concerns the next coin, not one of \(Y_1,\ldots,Y_n\). To show it is not in \(\mathcal{F}_n\), suppose it were. It is independent of \(\mathcal{F}_n\), because \(Y_{n+1}\) is independent of the first \(n\) coin outcomes. An event \(E\) that belongs to \(\mathcal{F}_n\) and is independent of \(\mathcal{F}_n\) must be independent of itself, which would give \(\mathbb{P}(E)=\mathbb{P}(E\cap E)=\mathbb{P}(E)^2\). Here \(E=\{Y_{n+1}=1\}\) has probability \(1/2\), but \(1/2\ne(1/2)^2=1/4\). Therefore \(E\notin\mathcal{F}_n\), and \(Z_n\) is not \(\mathcal{F}_n\)-measurable. This fails for each \(n\), including \(n=0\).
In contrast, \(W_0\) is constant, and for \(n\ge1\), \(W_n=Y_n\) is among the observations generating \(\mathcal{F}_n\). Therefore \(W_n\) is \(\mathcal{F}_n\)-measurable at every time and \((W_n)\) is adapted. The difference is the time index: the current filtration contains observations through \(n\), not through \(n+1\).
Worked Example: Recording Whether a First Success Has Occurred
Again let \(Y_1,Y_2,\ldots\) be independent fair-coin indicators, with the filtration \(\mathcal{F}_n=\sigma(Y_1,\ldots,Y_n)\) for \(n\ge1\) and \(\mathcal{F}_0=\{\varnothing,\Omega\}\). Let \(T=\inf\{k\ge1:Y_k=1\}\), with \(\inf\varnothing=\infty\), and define \(U_n=\mathbf{1}_{\{T\le n\}}\) for \(n\ge0\). Show that \((U_n)\) is adapted and compute \(\mathbb{P}(U_3=1)\).
For \(n=0\), \(U_0=0\), because \(T\ge1\), so \(U_0\) is \(\mathcal{F}_0\)-measurable. For \(n\ge1\), \(\{U_n=1\}=\{T\le n\}=\bigcup_{k=1}^{n}\{Y_k=1\}\). This is a finite union of events in \(\mathcal{F}_n\); its complement \(\{U_n=0\}\) is also in \(\mathcal{F}_n\). Thus \(U_n\) is measurable at time \(n\), and the process is adapted.
The event \(U_3=1\) means at least one of the first three flips is heads. Its complement is three tails. By independence, the probability of three tails is \((1/2)(1/2)(1/2)=1/8\). Hence \(\mathbb{P}(U_3=1)=1-1/8=7/8\), and \(\mathbb{P}(U_3=0)=1/8\). These values add to \(1\). Notice that deciding whether a success has occurred by time \(3\) uses only the first three observations; deciding whether a success will ever occur would ask a different, future-dependent question.
What Adaptedness Does—and Does Not—Say
Adaptedness is relative to a specified filtration. If \((\mathcal{G}_n)\) and \((\mathcal{F}_n)\) are filtrations with \(\mathcal{G}_n\subseteq\mathcal{F}_n\) for every \(n\), any process adapted to \((\mathcal{G}_n)\) is also adapted to \((\mathcal{F}_n)\): each event measurable with respect to \(\mathcal{G}_n\) is measurable with respect to the larger sigma-algebra \(\mathcal{F}_n\). The converse need not hold. A larger filtration can contain information that a smaller one does not.
An adapted process need not have independent values, stationary distributions, or any particular transition law. Conversely, a process can have simple marginal distributions and still fail to be adapted if its time-\(n\) value depends on an observation from after time \(n\). Adaptedness addresses the timing of information, not the probabilistic relationship between values.
Common pitfall: checking measurability at the wrong time. For the future-observation process \(Z_n=Y_{n+1}\), it is true that \(Z_n\) is \(\mathcal{F}_{n+1}\)-measurable, because \(\mathcal{F}_{n+1}\) includes \(Y_{n+1}\). But this does not establish adaptedness: the required condition is \(Z_n\in\mathcal{F}_n\)-measurable at time \(n\). In fact, the event \(\{Z_n=1\}\) has probability \(1/2\), is independent of \(\mathcal{F}_n\), and cannot belong to \(\mathcal{F}_n\), since membership would imply \(1/2=(1/2)^2\). Always match the process index \(n\) with the filtration \(\mathcal{F}_n\), not a later sigma-algebra.
Check Your Understanding
For each question, identify the information available at the time being tested.
- If \(X_n\) is adapted and \(g:\mathbb{R}\to\mathbb{R}\) is Borel measurable, why is \(g(X_n)\) measurable with respect to \(\mathcal{F}_n\)?
- With independent fair-coin indicators \(Y_k\) and \(\mathcal{F}_n=\sigma(Y_1,\ldots,Y_n)\), is \(Y_{n-1}\) measurable with respect to \(\mathcal{F}_n\) for \(n\ge2\)? Explain.
- For a real-valued random variable \(X_n\), what events need to be checked in the rational sublevel-set test?
- In the first-success example, write \(\{U_4=0\}\) in terms of \(Y_1,\ldots,Y_4\), and compute its probability.
- Can a process be adapted to a larger filtration but not to a smaller one? Explain how the information sets can lead to this difference.