Information as a Sigma-Field
Probability models describe not only which events can occur, but also what an observer can know at a given time. An event belongs to the information available at time \(t\) when it is measurable with respect to the sigma-field assigned to that time. As time advances, information may accumulate, but information already available is not lost. This is expressed by inclusion of sigma-fields.
Throughout, take a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Here \(\Omega\) is the set of outcomes, \(\mathcal{F}\) is a sigma-field of events, and \(\mathbb{P}\) is a probability measure on \(\mathcal{F}\). A sigma-field contains \(\Omega\), is closed under complements, and is closed under countable unions. In particular it contains the empty set and is closed under countable intersections. The ambient sigma-field \(\mathcal{F}\) specifies the events under consideration in the model.
The inclusion \(\mathcal{F}_s\subseteq\mathcal{F}_t\) means that every event distinguishable at time \(s\) remains distinguishable at time \(t\). A filtration need not reveal new information at every time: it may be constant on an interval. Nor does the definition require \(\mathcal{F}_0\) to be trivial, or require any continuity from the left or right. Such additional assumptions are imposed only when a result calls for them.
Proof. Fix \(s\le t\) and \(A\in\mathcal{F}_s\). By the defining inclusion of a filtration, \(\mathcal{F}_s\subseteq\mathcal{F}_t\). Membership of \(A\) in \(\mathcal{F}_s\) therefore implies \(A\in\mathcal{F}_t\). This applies also when \(A=\varnothing\) or \(A=\Omega\), since both belong to every sigma-field. \(\square\)
Building Information from Observations
An observation is a random variable, and the events it makes distinguishable form the sigma-field it generates. For a random variable \(X\) taking values in a measurable space, \(\sigma(X)\) denotes the smallest sigma-field on \(\Omega\) that makes \(X\) measurable. Equivalently, it is generated by the events \(\{X\in C\}\) as \(C\) ranges over measurable sets in the observation space. A history of observations accumulates these sigma-fields as more observations arrive.
Proof. Fix \(s\le t\). Every observation \(X_r\) used to generate \(\mathcal{H}_s\) has index \(r\in I_s\), and the assumption gives \(r\in I_t\). Thus each generator event \(\{X_r\in C\}\) used at time \(s\) is also among the generator events at time \(t\). The sigma-field \(\mathcal{H}_t\) contains all those events, so, by the defining minimality of the sigma-field generated by them, it contains \(\mathcal{H}_s\). Since each observation is assumed measurable with respect to the ambient \(\mathcal{F}\), the generated sigma-fields are sub-sigma-fields of \(\mathcal{F}\). \(\square\)
Worked Example: A Two-Stage Coin Experiment
Let \(\Omega=\{HH,HT,TH,TT\}\), with each outcome having probability \(1/4\), and let \(\mathcal{F}\) be the set of all subsets of \(\Omega\). At time 0 nothing has been observed; at time 1 the first toss is revealed; at time 2 both tosses are revealed. Specify the filtration and identify what can be known at each stage.
Set \(\mathcal{F}_0=\{\varnothing,\Omega\}\). At time 1 the outcomes are separated into the two groups with first toss \(H\) and first toss \(T\), so
At time 2 all four outcomes are distinguishable, hence \(\mathcal{F}_2=\mathcal{F}\). For a continuous time index, one possible filtration is \(\mathcal{F}_t=\mathcal{F}_0\) for \(0\le t<1\), \(\mathcal{F}_t=\mathcal{F}_1\) for \(1\le t<2\), and \(\mathcal{F}_t=\mathcal{F}_2\) for \(t\ge2\). These are nested sigma-fields, so this is a filtration.
The event \(A=\{\text{first toss is }H\}=\{HH,HT\}\) has probability \(2/4=1/2\) and belongs to \(\mathcal{F}_1\). The event \(C=\{HH\}\) has probability \(1/4\). It does not belong to \(\mathcal{F}_1\), because it separates \(HH\) from \(HT\), which are in the same atom \(\{HH,HT\}\) of that sigma-field. But \(C\in\mathcal{F}_2\). Check: the four singletons are all in \(\mathcal{F}_2\), and the set \(\{HH,HT\}\) is also in \(\mathcal{F}_2\), so \(\mathcal{F}_1\subseteq\mathcal{F}_2\) as required.
A Continuous Observation History and a Boundary Point
For a continuous-time example, let \(\Omega=[0,1]\), let \(\mathcal{F}\) be the Borel sigma-field, and let \(\mathbb{P}\) be the uniform probability measure. Write \(U(\omega)=\omega\), so \(U\) is uniform on \([0,1]\). At time \(t\), suppose the observer sees \(Y_t=U\wedge t=\min(U,t)\): the value is revealed up to \(t\), but if \(U\) has not yet been reached, the observation is just \(t\). Define the history sigma-field \(\mathcal{G}_t=\sigma(Y_s:0\le s\le t)\). These are sub-sigma-fields of \(\mathcal{F}\).
Proof. If \(s\le t\), every observation \(Y_r\) with \(r\le s\) is also among the observations \(Y_r\) with \(r\le t\). The generated sigma-fields therefore satisfy \(\mathcal{G}_s\subseteq\mathcal{G}_t\). For \(0<t<1\), \(Y_t=\min(U,t)\), and \(Y_t<t\) holds exactly when \(U<t\). Hence \(\{U<t\}=\{Y_t<t\}\in\mathcal{G}_t\), and its probability is \(t\) under the uniform law.
To prove nonmembership of \(\{U\le t\}\), observe that if \(u,v\ge t\), then for every \(r\le t\), \(Y_r(u)=r=Y_r(v)\). Thus all outcomes in \([t,1]\) give exactly the same history through time \(t\). More formally, the collection of Borel sets that either contain all of \([t,1]\) or contain none of it is a sigma-field. It contains every generator \(\{Y_r\in C\}\) of \(\mathcal{G}_t\), because \(Y_r\) is constant on \([t,1]\) for each \(r\le t\). Therefore every event in \(\mathcal{G}_t\) either contains all of \([t,1]\) or none of it. The set \(\{U\le t\}\) contains \(t\) but excludes every \(v>t\), so it has neither property and cannot belong to \(\mathcal{G}_t\). \(\square\)
Worked Example: Testing Events Against the Stopped Uniform History
Take \(t=1/4\). Determine which of the events \(\{U<1/4\}\) and \(\{U\le1/4\}\) can be determined from the observation history by time \(1/4\), and compute their probabilities.
The current observation is \(Y_{1/4}=\min(U,1/4)\). If \(U<1/4\), then \(Y_{1/4}=U<1/4\); if \(U\ge1/4\), then \(Y_{1/4}=1/4\). Consequently,
In contrast, when \(U=1/4\), the observation at every \(r\le1/4\) is \(Y_r=r\), just as it is for any \(U>1/4\). The history therefore cannot separate \(U=1/4\) from \(U>1/4\), whereas \(\{U\le1/4\}\) includes the former and excludes the latter. Thus \(\{U\le1/4\}\notin\mathcal{G}_{1/4}\). Its probability is nevertheless \(1/4\), since a single endpoint has probability zero under the uniform law. Check: the two events differ by \(\{U=1/4\}\), so their probabilities agree, but their exact set membership in \(\mathcal{G}_{1/4}\) does not. Equality up to a null set is not equality of events.
This example also shows why the word “history” matters. At time \(t\), the full history includes all \(Y_s\) for \(s\le t\), not merely the current value \(Y_t\). Here the earlier observations do not distinguish any two values \(U\ge t\); they reveal that the level has not been reached, but not whether \(U=t\) or \(U>t\). Completing a sigma-field by adjoining subsets of ambient null sets can change exact membership statements about a probability-zero boundary. The nonmembership above concerns the Borel history sigma-field \(\mathcal{G}_t\) as defined, without that completion.
Information Arriving in Brownian Time
Let \((B_t)_{t\ge0}\) be a standard Brownian motion on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\). In particular, \(B_0=0\), its paths are continuous, and its increments over disjoint time intervals are independent; \(B_t-B_s\) has distribution \(N(0,t-s)\) for \(s<t\). To describe an observer who sees the Brownian path only up to the current time, define \(\mathcal{B}_t=\sigma(B_s:0\le s\le t)\). The same generator-inclusion argument shows that \(\mathcal{B}_s\subseteq\mathcal{B}_t\) for \(s\le t\). This is an observation history; it need not equal the ambient \(\mathcal{F}_t\) of every model.
Worked Example: When a Future Brownian Increment Becomes Known
Let \(A=\{B_2-B_1>0\}\). Show that \(A\) belongs to \(\mathcal{B}_2\) but not to \(\mathcal{B}_1\), and compute its probability.
Since \(B_2\) and \(B_1\) are observed by time 2, their difference is measurable with respect to \(\mathcal{B}_2\), so \(A\in\mathcal{B}_2\). The increment \(B_2-B_1\) has distribution \(N(0,1)\), a continuous distribution symmetric about zero. Therefore \(\mathbb{P}(A)=1/2\).
By independent increments, \(B_2-B_1\) is independent of the Brownian observations up to time 1, and hence \(A\) is independent of \(\mathcal{B}_1\). If \(A\) were in \(\mathcal{B}_1\), it would be independent of itself, which would force \(\mathbb{P}(A)=\mathbb{P}(A\cap A)=\mathbb{P}(A)^2\). But \(1/2\ne(1/2)^2=1/4\), a contradiction. Hence \(A\notin\mathcal{B}_1\). Check: \(A\) is determined once both endpoint observations are available, while the independent, nondegenerate increment is not determined by the earlier history.
What Filtrations Do—and Do Not—Assert
A filtration is a bookkeeping structure for information, not a claim that every event in the ambient model is already observable. It also does not specify how quickly information is revealed: a filtration can stay unchanged over an interval and then increase. A typical finite-observation filtration has the stepwise form in the coin example; a path-observation filtration can add information continuously, as in the Brownian example.
Keep exact event membership distinct from probability. The uniform example has two events with equal probability, one measurable from the observed history and the other not. Their difference is a null event, but the raw sigma-field distinguishes exact sets, not merely their probabilities. If a model uses completed sigma-fields, state that convention: completion can add subsets of null sets and so alter such a membership conclusion.
A further distinction is between the observation history and the ambient filtration. The history \(\mathcal{B}_t\) records only the values \(B_s\) for \(s\le t\); a model may specify a larger \(\mathcal{F}_t\), provided it still satisfies the filtration property. Conversely, a process being defined on a probability space does not by itself say what information is available at each time. The filtration supplies that temporal structure, which later definitions of adapted processes and stopping times use.
| Feature | Meaning | Illustration |
|---|---|---|
| \(\mathcal{F}_s\subseteq\mathcal{F}_t\) | Earlier events remain available later | First toss remains known after the second toss |
| \(\sigma(\text{observations through }t)\) | Events distinguishable from the observed history | \(\mathcal{B}_t=\sigma(B_s:0\le s\le t)\) |
| Equal probabilities | Does not imply equal events or equal measurability | \(\{U<t\}\) and \(\{U\le t\}\) |
Check Your Understanding
For each question, distinguish the event itself from its probability and use the stated observation history.
- In the two-toss experiment, which atoms make up \(\mathcal{F}_1\), and why is \(\{HT\}\notin\mathcal{F}_1\)?
- For the uniform observation \(Y_s=U\wedge s\), show that \(\{U<t\}\in\mathcal{G}_t\) for \(0<t<1\), and calculate its probability.
- Explain why \(\{U\le t\}\notin\mathcal{G}_t\) even though it has the same probability as \(\{U<t\}\).
- For Brownian observations, why does \(\{B_2-B_1>0\}\) belong to \(\mathcal{B}_2\) but not \(\mathcal{B}_1\)?
- Does the definition of a filtration require \(\mathcal{F}_0\) to be trivial or the family to be constant-free on every time interval? Explain.