Information That Accumulates
In a stochastic process, the index often represents time. At each time, we may know some facts about the outcome but not others. A filtration is a mathematical way to record how that information accumulates: every event knowable at an earlier index must remain knowable at a later one. The definition concerns collections of events, not the values of a process or the probabilities of those events.
Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Here \(\Omega\) is the set of possible outcomes, \(\mathcal{F}\) is a sigma-field of events, and \(\mathbb{P}\) assigns probabilities to those events. A sigma-field \(\mathcal{G}\) is a collection of subsets of \(\Omega\) that contains \(\Omega\), is closed under complements, and is closed under countable unions. It therefore also contains the empty set and is closed under countable intersections. When we use \(\mathcal{G}\) to represent information, an event \(A\) belongs to \(\mathcal{G}\) precisely when that information can distinguish whether \(A\) occurred.
The phrase “increasing information” refers exactly to these set inclusions. It does not require that new information arrive at every step: it is permitted that \(\mathcal{F}_{n+1}=\mathcal{F}_n\). Also, \(\mathcal{F}_0\) need not be the trivial sigma-field \(\{\varnothing,\Omega\}\) unless a particular model imposes that condition. In continuous time, the same definition uses a family \((\mathcal{F}_t)_{t\ge0}\) and requires \(\mathcal{F}_s\subseteq\mathcal{F}_t\) whenever \(s\le t\).
Finite Sigma-Fields as Partitions
On a finite sample space, a sigma-field can be described by a partition. A partition is a collection of nonempty, pairwise disjoint subsets, called cells, whose union is \(\Omega\). The sigma-field associated with a partition consists of all unions of its cells, including the empty union \(\varnothing\). Within a cell, the information represented by the partition does not distinguish individual outcomes; it can only determine which cell contains the outcome.
Proof. Suppose first that \(\mathcal{G}_1\subseteq\mathcal{G}_2\), and take a cell \(C\) of \(\Pi_1\). Since \(C\in\mathcal{G}_1\), the inclusion gives \(C\in\mathcal{G}_2\), so \(C\) is a union of cells of \(\Pi_2\). Conversely, suppose every cell of \(\Pi_1\) is a union of cells of \(\Pi_2\). Any event in \(\mathcal{G}_1\) is a union of cells of \(\Pi_1\); replacing each of those cells by its union of \(\Pi_2\)-cells shows that the event is a union of \(\Pi_2\)-cells. Thus it belongs to \(\mathcal{G}_2\), proving \(\mathcal{G}_1\subseteq\mathcal{G}_2\). \(\square\)
This criterion turns a sigma-field inclusion question into a partition check. The direction matters: a finer partition distinguishes more outcomes and therefore corresponds to a larger sigma-field.
Worked Example: Information from Two Coin Tosses
Let \(\Omega=\{HH,HT,TH,TT\}\) be the outcomes of two tosses of a fair coin, with \(\mathcal{F}\) the set of all subsets of \(\Omega\) and each outcome assigned probability \(\tfrac14\). Consider the following partitions at indices \(0,1,2\):
Let \(\mathcal{F}_n\) be the sigma-field of unions of cells of \(\Pi_n\). At index \(0\), the only events are \(\varnothing\) and \(\Omega\). At index \(1\), the events are \(\varnothing,\Omega,\{HH,HT\},\{TH,TT\}\): the information distinguishes the first toss but not the second. At index \(2\), every subset of \(\Omega\) is an event in \(\mathcal{F}_2\). The sole cell \(\Omega\) of \(\Pi_0\) is the union of all cells of \(\Pi_1\); each cell of \(\Pi_1\) is the union of two cells of \(\Pi_2\). The refinement criterion therefore gives \(\mathcal{F}_0\subseteq\mathcal{F}_1\subseteq\mathcal{F}_2=\mathcal{F}\), so this is a filtration.
For instance, the event \(\{HH,HT\}\), meaning that the first toss is heads, belongs to \(\mathcal{F}_1\) and remains in \(\mathcal{F}_2\). The event \(\{HH\}\), meaning that both tosses are heads, does not belong to \(\mathcal{F}_1\), because it is not a union of the two cells in \(\Pi_1\). It does belong to \(\mathcal{F}_2\). This illustrates how event membership records what can be determined at each index.
Checking the Inclusions
For any proposed family to be a filtration, every pair of indices in the required order must satisfy the inclusion condition. It is not enough that the partitions look similar, or that each has a valid sigma-field. A family of sigma-fields can fail to be a filtration if later information does not retain an earlier distinction.
Worked Example: Two Valid Sigma-Fields That Are Not Nested
Let \(\Omega=\{1,2,3,4\}\), with \(\mathcal{F}\) all subsets of \(\Omega\) and probability \(\tfrac14\) on each outcome. Consider the partitions \(\Pi_1=\{\{1,2\},\{3,4\}\}\) and \(\Pi_2=\{\{1,3\},\{2,4\}\}\), and let \(\mathcal{F}_1,\mathcal{F}_2\) be their respective sigma-fields. Each is a valid sigma-field, but neither contains the other.
Indeed, \(\{1,2\}\in\mathcal{F}_1\). The cells of \(\Pi_2\) are \(\{1,3\}\) and \(\{2,4\}\); their possible unions are \(\varnothing,\{1,3\},\{2,4\},\Omega\). Since \(\{1,2\}\) is not among these unions, \(\{1,2\}\notin\mathcal{F}_2\), and \(\mathcal{F}_1\not\subseteq\mathcal{F}_2\). In the reverse direction, \(\{1,3\}\in\mathcal{F}_2\), but the unions of cells of \(\Pi_1\) are \(\varnothing,\{1,2\},\{3,4\},\Omega\). Thus \(\{1,3\}\notin\mathcal{F}_1\), and \(\mathcal{F}_2\not\subseteq\mathcal{F}_1\). If these sigma-fields were assigned to consecutive indices, the family would not be a filtration.
Worked Example: Information Revealed in Stages
Let \(\Omega=\{a,b,c,d\}\), with \(\mathcal{F}\) all subsets and probability \(\tfrac14\) on each outcome. Define partitions at indices \(0,1,2,3\) by
At index \(1\), the information distinguishes the pair \(\{a,b\}\) from the pair \(\{c,d\}\). At index \(2\), it distinguishes \(a\) from \(b\), while still grouping \(c\) with \(d\). At index \(3\), it distinguishes all four outcomes. To check the filtration condition, the only cell of \(\Pi_0\) is the union of all cells of \(\Pi_1\). Each cell of \(\Pi_1\) is a union of cells of \(\Pi_2\): \(\{a,b\}=\{a\}\cup\{b\}\) and \(\{c,d\}\) remains a cell. Each cell of \(\Pi_2\) is a cell of \(\Pi_3\), except that \(\{c,d\}=\{c\}\cup\{d\}\). Every step refines the preceding partition, so the associated sigma-fields satisfy \(\mathcal{F}_0\subseteq\mathcal{F}_1\subseteq\mathcal{F}_2\subseteq\mathcal{F}_3\).
For a concrete membership check, the event \(\{a,b\}\) belongs to \(\mathcal{F}_1\), since it is one cell of \(\Pi_1\). It remains in \(\mathcal{F}_2\) as \(\{a\}\cup\{b\}\), and remains in \(\mathcal{F}_3\) as the same union of two cells. By contrast, \(\{c\}\notin\mathcal{F}_2\): the cell \(\{c,d\}\) cannot be split using the information at index \(2\). It does belong to \(\mathcal{F}_3\).
What Inclusion Guarantees
Proof. The definition of a filtration requires \(\mathcal{F}_m\subseteq\mathcal{F}_n\) whenever \(m\le n\). Since \(A\in\mathcal{F}_m\), the definition of set inclusion gives \(A\in\mathcal{F}_n\). Thus every event recorded at index \(m\) is still recorded at index \(n\). \(\square\)
In particular, checking consecutive inclusions is enough for a discrete-time family: if \(\mathcal{F}_n\subseteq\mathcal{F}_{n+1}\) for every \(n\), then for \(m<n\), repeated inclusion gives \(\mathcal{F}_m\subseteq\mathcal{F}_{m+1}\subseteq\cdots\subseteq\mathcal{F}_n\). Conversely, the filtration definition includes each consecutive inclusion. This is a direct consequence of transitivity of set inclusion.
A filtration describes the organization of information; it does not say how likely an event is, nor does it require each step to add a new event. For example, repeated sigma-fields are allowed, and a filtration may begin with information already available at index \(0\). The definition alone also does not say how the information relates to a particular process. That relationship is specified separately when needed.
Common mistake. A frequent error is to reverse the refinement direction. Suppose the earlier partition is \(\{\{a,b\},\{c,d\}\}\) and the later partition is \(\{\{a\},\{b\},\{c\},\{d\}\}\). One might incorrectly say that the later sigma-field is smaller because each of its cells is smaller. But \(\{a,b\}\) is an event in the earlier sigma-field and it is the union \(\{a\}\cup\{b\}\) of later cells, so it is also in the later sigma-field. The later partition distinguishes more outcomes, and its sigma-field is larger: \(\mathcal{F}_{\text{earlier}}\subseteq\mathcal{F}_{\text{later}}\). Smaller partition cells mean more information, not less.
Check Your Understanding
For each question, decide which events belong to the stated sigma-field and justify any claimed inclusion using the definition or the partition criterion.
- On \(\Omega=\{1,2,3,4\}\), let \(\Pi_1=\{\{1,2\},\{3,4\}\}\) and \(\Pi_2=\{\{1\},\{2\},\{3,4\}\}\). Do the associated sigma-fields satisfy \(\mathcal{F}_1\subseteq\mathcal{F}_2\)?
- For the partitions in Question 1, is \(\{1\}\) in \(\mathcal{F}_1\), in \(\mathcal{F}_2\), in both, or in neither?
- Can a filtration satisfy \(\mathcal{F}_{n+1}=\mathcal{F}_n\) for every \(n\)? Explain using the definition.
- If \(A\in\mathcal{F}_2\) for a filtration, what can be concluded about membership of \(A\) in \(\mathcal{F}_5\)? State the inclusion that justifies your answer.
- Give a pair of finite partitions for which neither associated sigma-field contains the other, and identify one event witnessing each failure of inclusion.