Information Available at Each Time
A stochastic process is indexed by time, but the information available to an observer generally grows as time passes. Adaptedness formalizes the requirement that the value assigned to time \(n\) can be determined from information available by time \(n\), rather than from future information. This property is a basic measurability condition used when studying martingales, but it does not by itself impose any condition on conditional means or expectations.
Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). A discrete-time filtration \((\mathcal{F}_n)_{n\ge0}\) is an increasing sequence of sub-\(\sigma\)-algebras of \(\mathcal{F}\): \(\mathcal{F}_n\subseteq\mathcal{F}_{n+1}\) for every \(n\). Each \(\mathcal{F}_n\) represents the events that can be determined using information available at time \(n\). In this tutorial, processes are real-valued, so their state space is \(S=\mathbb{R}\) with its Borel \(\sigma\)-algebra.
The requirement is made separately at every time. It does not say that \(X_n\) is independent of the future, nor that the process has any particular distribution. It says only that its time-\(n\) value is determined by the information represented by \(\mathcal{F}_n\). In particular, the same process may be adapted to one filtration and fail to be adapted to another.
A Measurability Test for Real-Valued Processes
Checking the inverse image of every Borel set can be cumbersome. For real-valued random variables, a countable family of sets is enough: it suffices to test half-lines with rational endpoints.
Proof. Fix \(n\). If \(X_n\) is \(\mathcal{F}_n\)-measurable, then the inverse image of every Borel set is in \(\mathcal{F}_n\), and \((-\infty,q]\) is Borel for each rational \(q\). Thus \(\{X_n\le q\}\in\mathcal{F}_n\). Conversely, suppose all these events belong to \(\mathcal{F}_n\). The sets \((-\infty,q]\), for rational \(q\), generate the Borel \(\sigma\)-algebra on \(\mathbb{R}\): open intervals can be formed using countable unions and complements of such half-lines, and every open subset of \(\mathbb{R}\) is a countable union of open intervals with rational endpoints. Therefore the \(\sigma\)-algebra generated by the inverse images of these half-lines contains the inverse image \(X_n^{-1}(B)\) of every Borel set \(B\). Each such inverse image belongs to \(\mathcal{F}_n\), so \(X_n\) is \(\mathcal{F}_n\)-measurable. This holds for every \(n\), proving the equivalence. \(\square\)
There is also a useful closure property: applying a measurable function to information already available does not create dependence on future information.
Proof. Fix \(n\). For each component \(r\), measurability of \(X_n^r\) means that the inverse image of every Borel set belongs to \(\mathcal{F}_n\). The vector \((X_n^1,\ldots,X_n^d)\) is therefore \(\mathcal{F}_n\)-measurable: inverse images of open rectangles are finite intersections of component inverse images, and open subsets of \(\mathbb{R}^d\) are countable unions of open rectangles with rational endpoints. For any Borel set \(B\subseteq\mathbb{R}\), the set \(g^{-1}(B)\) is Borel by the assumed measurability of \(g\). Hence \(\{Z_n\in B\}=\{(X_n^1,\ldots,X_n^d)\in g^{-1}(B)\}\in\mathcal{F}_n\). Thus \(Z_n\) is \(\mathcal{F}_n\)-measurable for every \(n\), as required. \(\square\)
For example, this proposition covers sums and products of finitely many adapted real-valued processes: addition and multiplication are Borel functions. The filtration itself also matters. If \((\mathcal{G}_n)\) is another filtration with \(\mathcal{F}_n\subseteq\mathcal{G}_n\) for all \(n\), then every process adapted to \((\mathcal{F}_n)\) is adapted to \((\mathcal{G}_n)\), because each of its time-\(n\) measurable events remains in the larger \(\sigma\)-algebra. The converse need not hold.
Worked Examples
Worked Example: A Random Walk with Revealed Increments
Let \(\Omega=\{-1,1\}^3\), with each of its eight outcomes assigned probability \(1/8\). Write an outcome as \((\xi_1,\xi_2,\xi_3)\), and let \(X_0=0\) and \(X_n=\sum_{k=1}^{n}\xi_k\) for \(1\le n\le3\). Define \(\mathcal{F}_0=\{\varnothing,\Omega\}\) and \(\mathcal{F}_n=\sigma(\xi_1,\ldots,\xi_n)\) for \(1\le n\le3\). Each \(\mathcal{F}_n\) records the first \(n\) increments.
At time \(0\), \(X_0\) is the constant \(0\), so it is \(\mathcal{F}_0\)-measurable. At time \(1\), the two possible values are \(X_1=1\) if \(\xi_1=1\), and \(X_1=-1\) if \(\xi_1=-1\); both events are in \(\mathcal{F}_1\). At time \(2\), the first two increments give these values:
Each row corresponds to an atom of \(\mathcal{F}_2\), so \(X_2\) is \(\mathcal{F}_2\)-measurable. At time \(3\), \(X_3=\xi_1+\xi_2+\xi_3\) is a function of the three increments and is therefore \(\mathcal{F}_3\)-measurable. For a direct path check, on \((1,-1,1)\) the successive values are \(X_0=0\), \(X_1=1\), \(X_2=0\), and \(X_3=1\), each determined by the increments revealed by that time. Thus \((X_n)\) is adapted.
Check: the time-\(2\) value on the path \((1,-1,1)\) is \(1+(-1)=0\), matching the table. The third increment is not needed to determine \(X_2\).
Worked Example: Products of Revealed Variables
Let \(\Omega=\{\tfrac12,2\}^3\), with each of its eight outcomes assigned probability \(1/8\), and let \(Y_k\) be the \(k\)th coordinate. Set \(\mathcal{F}_0=\{\varnothing,\Omega\}\), \(\mathcal{F}_n=\sigma(Y_1,\ldots,Y_n)\) for \(1\le n\le3\), and define \(Z_0=1\) and \(Z_n=\prod_{k=1}^nY_k\). Each \(Y_k\) is revealed at time \(k\).
At time \(1\), \(Z_1=Y_1\), so its possible values are \(\tfrac12\) and \(2\), both determined by \(\mathcal{F}_1\). At time \(2\), the products for all possible pairs are
The event \(\{Z_2=1\}\) is the union of the two atoms \(\{Y_1=\tfrac12,Y_2=2\}\) and \(\{Y_1=2,Y_2=\tfrac12\}\), so it belongs to \(\mathcal{F}_2\); the other displayed value events are atoms as well. At time \(3\), the product \(Y_1Y_2Y_3\) is a function of the three variables generating \(\mathcal{F}_3\). Thus \(Z\) is adapted; alternatively, it follows from the measurable-function proposition.
Check: on \((Y_1,Y_2,Y_3)=(2,\tfrac12,2)\), the successive products are \(Z_0=1\), \(Z_1=2\), \(Z_2=2\cdot\tfrac12=1\), and \(Z_3=1\cdot2=2\). Each product uses only variables revealed by its time.
Worked Example: Running Data Summaries and a Future Observation
Let \(\Omega=\{0,1\}^3\), with each of its eight outcomes assigned probability \(1/8\). Let \(U_k\) be the \(k\)th coordinate, and take \(\mathcal{F}_0=\{\varnothing,\Omega\}\), \(\mathcal{F}_n=\sigma(U_1,\ldots,U_n)\) for \(1\le n\le3\). Define the running count \(C_0=0\), \(C_n=\sum_{k=1}^nU_k\), and the running average \(A_0=0\), \(A_n=C_n/n\) for \(1\le n\le3\).
Both \(C_n\) and \(A_n\) are functions of the first \(n\) observations, so they are \(\mathcal{F}_n\)-measurable. For the outcome \((1,0,1)\), the observed summaries are \(C_1=1\), \(A_1=1\); \(C_2=1\), \(A_2=1/2\); and \(C_3=2\), \(A_3=2/3\). At time \(2\), these values use \(U_1=1\) and \(U_2=0\), not the unrevealed value \(U_3=1\). The initial values \(C_0=A_0=0\) are constants and hence measurable with respect to \(\mathcal{F}_0\). Both processes are adapted.
By contrast, the process \(V_n=U_3\) for every \(n\) is not adapted to this filtration. In particular, the event \(\{V_1=1\}=\{U_3=1\}\) is not in \(\mathcal{F}_1\): the outcomes \((0,0,0)\) and \((0,0,1)\) agree on \(U_1\), so they belong to the same atom of \(\mathcal{F}_1\), but only the second belongs to \(\{U_3=1\}\). An event in \(\mathcal{F}_1\) must be a union of its atoms, so this event cannot belong to \(\mathcal{F}_1\).
Check: the two outcomes used in the nonmeasurability test have the same first coordinate, \(0\), but their third coordinates are \(0\) and \(1\). Therefore time-\(1\) information cannot determine \(V_1\).
Common Mistake: Using the Full Data Too Early
A tempting but incorrect argument is: “Every \(X_n\) is a function of the entire outcome \(\omega\), and therefore it is measurable at time \(n\).” Being a function on \(\Omega\) only guarantees measurability with respect to the full \(\sigma\)-algebra \(\mathcal{F}\); adaptedness requires measurability with respect to the particular, usually smaller, \(\sigma\)-algebra \(\mathcal{F}_n\). In the revealed-data example, \(\{U_3=1\}\in\mathcal{F}_3\), but \(\{U_3=1\}\notin\mathcal{F}_1\). The correction is to test the information available at the time in question, not the information available at the end.
The choice of filtration can change the answer. If one instead uses the constant filtration \(\mathcal{G}_n=\mathcal{F}_3\) at every time \(n\), then the process \(V_n=U_3\) is adapted to \((\mathcal{G}_n)\), because \(U_3\) is measurable with respect to \(\mathcal{F}_3\). This filtration makes all three observations available from time \(0\); it describes a different information structure. Adaptedness is always a relation between a process and a specified filtration.
Finally, adaptedness is only a measurability property. It does not assert that \(X_n\) has a finite expectation, that its increments have zero conditional mean, or that it satisfies any martingale condition. Those are additional questions. For now, the key test is precise: at each time \(n\), can every event \(\{X_n\in B\}\) be determined from \(\mathcal{F}_n\)?
Check Your Understanding
For each question, identify the filtration and the time-\(n\) information before deciding whether the process is adapted.
- If \(X_n\) is adapted and \(g:\mathbb{R}\to\mathbb{R}\) is Borel measurable, why is \(g(X_n)\) measurable with respect to \(\mathcal{F}_n\)?
- For observations \(U_1,U_2,\ldots\) and \(\mathcal{F}_n=\sigma(U_1,\ldots,U_n)\), is \(U_{n+1}\) necessarily \(\mathcal{F}_n\)-measurable? Explain what additional information would be needed to decide.
- For the random walk in the first example, compute \(X_1,X_2,X_3\) on the path \((-1,1,1)\).
- Why is the event \(\{U_3=1\}\) not in \(\sigma(U_1)\) when the sample space is \(\{0,1\}^3\)?
- Can a process be adapted to one filtration but not another? Give the revealed-data example that demonstrates this.