Information Accumulates Over Time
A stochastic process is observed over time, but the observer may not see every detail of the underlying outcome. At time \(n\), the available information is represented by a sigma-field: a collection of events whose occurrence can be determined from what has been observed by then. A filtration formalizes the requirement that information is retained. What is known at time \(n\) must still be known at time \(n+1\).
Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\), where \(\Omega\) is the outcome space, \(\mathcal{F}\) is the collection of events under consideration, and \(\mathbb{P}\) is the probability measure. A sigma-field is a collection containing \(\Omega\), closed under complements and countable unions. In particular, it contains the empty set and is closed under finite unions and intersections. A sigma-field \(\mathcal{F}_n\) represents the events that can be decided using information available at discrete time \(n\).
A sigma-field records more than the observed value itself. If an observation has several possible values, the sigma-field it generates contains every event that can be decided from that observation, including unions of observation outcomes. For finite sample spaces, this is especially concrete: outcomes that give the same observation are grouped together, and events in the generated sigma-field are unions of those groups.
Finite Observations and Their Atoms
For an observation \(Y\), the notation \(\sigma(Y)\) means the smallest sigma-field making \(Y\) observable. More explicitly, it is generated by the events \(\{Y\in C\}\), where \(C\) ranges over subsets of the observation's value space. In a finite sample space, the atoms of a sigma-field are its smallest nonempty events. They form a partition: every outcome belongs to exactly one atom, and every event in the sigma-field is a union of atoms.
Proof. Outcomes \(\omega,\omega'\) with \(Y(\omega)=Y(\omega')\) have the same membership in every event of the form \(\{Y\in C\}\), and therefore in every event of \(\sigma(Y)\): the collection of events that are unions of the sets \(\{Y=y\}\) is a sigma-field containing all the generators \(\{Y\in C\}\), while each fiber \(\{Y=y\}\) itself is one of those generators. Thus \(\sigma(Y)\) consists exactly of unions of the nonempty fibers. Each nonempty fiber cannot be split into smaller nonempty events in this sigma-field, since every event is a union of whole fibers. Hence these fibers are exactly its atoms. \(\square\)
For several observations, use their joint history. The observation history through time \(n\) is the vector \((Y_0,\ldots,Y_n)\), and its sigma-field is \(\sigma(Y_0,\ldots,Y_n)\). The definition of a filtration follows from the way the history is assembled.
Proof. Each \(\mathcal{F}_n\) is a sigma-field contained in \(\mathcal{F}\). Every generator of \(\mathcal{F}_n\) is an event determined by one of \(Y_0,\ldots,Y_n\); each is also a generator of \(\mathcal{F}_{n+1}\). Since \(\mathcal{F}_{n+1}\) is a sigma-field containing all generators of \(\mathcal{F}_n\), it contains the sigma-field they generate, so \(\mathcal{F}_n\subseteq\mathcal{F}_{n+1}\). For minimality, \(\mathcal{G}_n\) contains \(\sigma(Y_k)\) for each \(k\le n\): the assumed inclusion gives \(\sigma(Y_k)\subseteq\mathcal{G}_k\), and repeated nesting gives \(\mathcal{G}_k\subseteq\mathcal{G}_n\). As \(\mathcal{G}_n\) is a sigma-field containing all those observation sigma-fields, it contains \(\sigma(Y_0,\ldots,Y_n)=\mathcal{F}_n\). \(\square\)
Worked Coin-Toss Histories
Worked Example: Information After the First and Second Tosses
Let \(\Omega=\{HH,HT,TH,TT\}\) be the outcomes of two fair coin tosses, with \(\mathbb{P}(\{\omega\})=\tfrac14\) for each outcome and \(\mathcal{F}\) the power set of \(\Omega\). Let \(X_1\) and \(X_2\) denote the first and second toss. Suppose both tosses are observed as they occur. Set \(\mathcal{F}_0=\{\varnothing,\Omega\}\), \(\mathcal{F}_1=\sigma(X_1)\), and \(\mathcal{F}_2=\sigma(X_1,X_2)\).
The atoms at time \(0\) consist of the single set \(\Omega\). At time \(1\), the two fibers of \(X_1\) are \(A_H=\{HH,HT\}\) and \(A_T=\{TH,TT\}\). Thus \(\mathcal{F}_1=\{\varnothing,A_H,A_T,\Omega\}\). At time \(2\), the pair \((X_1,X_2)\) distinguishes all four outcomes, so \(\mathcal{F}_2\) is the power set of \(\Omega\).
Check the filtration: \(\mathcal{F}_0\subseteq\mathcal{F}_1\) because both \(\varnothing\) and \(\Omega\) belong to \(\mathcal{F}_1\). Also \(\mathcal{F}_1\subseteq\mathcal{F}_2\), since \(A_H=\{HH,HT\}\) and \(A_T=\{TH,TT\}\) are unions of singleton events in \(\mathcal{F}_2\). The event \(\{X_2=H\}=\{HH,TH\}\) is not in \(\mathcal{F}_1\): it contains \(HH\) but not \(HT\), so it splits the atom \(A_H\). Its probability is \(\tfrac14+\tfrac14=\tfrac12\). At time \(2\), the event is known because it is a union of two singleton atoms.
Worked Example: A Coarse Observation History
Let \(\Omega=\{H,T\}^3\) be three fair independent tosses, with each of the eight outcomes having probability \(\tfrac18\), and let \(\mathcal{F}\) be the power set. At time \(1\), observe whether the first toss is \(H\). At time \(2\), observe whether at least one of the first two tosses is \(H\). At time \(3\), observe whether the third toss is \(H\). Write these observations as \(Y_1,Y_2,Y_3\), and define \(\mathcal{F}_0=\{\varnothing,\Omega\}\) and \(\mathcal{F}_n=\sigma(Y_1,\ldots,Y_n)\) for \(n=1,2,3\).
The time-1 atoms are \(\{HHH,HHT,HTH,HTT\}\) and \(\{THH,THT,TTH,TTT\}\), each containing four outcomes and therefore having probability \(\tfrac48=\tfrac12\). At time \(2\), the possible histories give three atoms: \(Y_1=H,Y_2=1\) gives \(\{HHH,HHT,HTH,HTT\}\); \(Y_1=T,Y_2=0\) gives \(\{TTH,TTT\}\); and \(Y_1=T,Y_2=1\) gives \(\{THH,THT\}\). Their probabilities are respectively \(\tfrac48=\tfrac12\), \(\tfrac28=\tfrac14\), and \(\tfrac28=\tfrac14\).
At time \(3\), the additional observation splits each time-2 atom according to the third toss. The resulting six atoms are \(\{HHH,HTH\}\), \(\{HHT,HTT\}\), \(\{TTH\}\), \(\{TTT\}\), \(\{THH\}\), and \(\{THT\}\). Their probabilities are \(\tfrac28=\tfrac14\), \(\tfrac28=\tfrac14\), and four instances of \(\tfrac18\). Each time-2 atom is the union of its two time-3 atoms; for example, \(\{THH,THT\}=\{THH\}\cup\{THT\}\). Thus \(\mathcal{F}_2\subseteq\mathcal{F}_3\), and \(\mathcal{F}_1\subseteq\mathcal{F}_2\) because each time-1 atom is unchanged or split into time-2 atoms.
The event \(\{X_2=H\}\) is not known at time \(2\). The atom \(\{HHH,HHT,HTH,HTT\}\) contains outcomes with both \(X_2=H\) and \(X_2=T\), so the event is not a union of time-2 atoms. Its probability is \(\tfrac48=\tfrac12\). This illustrates that an observation history can retain less information than the full underlying toss history.
Worked Example: A Coarser First Observation
Use the same three-toss space, but suppose the first observation is only whether the first toss was \(H\), the second observation reports whether the first two tosses are equal, and the third observation again reports whether the third toss is \(H\). Call these observations \(Z_1,Z_2,Z_3\), and set \(\mathcal{G}_0=\{\varnothing,\Omega\}\), \(\mathcal{G}_n=\sigma(Z_1,\ldots,Z_n)\).
At time \(1\), the atoms are \(\{HHH,HHT,HTH,HTT\}\) and \(\{THH,THT,TTH,TTT\}\), each of probability \(\tfrac48=\tfrac12\). At time \(2\), each first-toss atom splits according to equality of the first two tosses. The four atoms are \(\{HHH,HHT\}\), \(\{HTH,HTT\}\), \(\{TTH,TTT\}\), and \(\{THH,THT\}\); each has probability \(\tfrac28=\tfrac14\). At time \(3\), each of these four atoms splits into two singleton outcomes according to the third toss. Each singleton has probability \(\tfrac18\), and every time-2 atom is the union of its two corresponding singletons. Therefore \(\mathcal{G}_0\subseteq\mathcal{G}_1\subseteq\mathcal{G}_2\subseteq\mathcal{G}_3\).
The probability of \(\{X_2=H\}\) is \(\tfrac48=\tfrac12\), but this event is not in \(\mathcal{G}_1\): each time-1 atom contains outcomes with both values of \(X_2\). It is in \(\mathcal{G}_2\), because the first toss and whether the first two tosses are equal determine the second toss. In particular, the atom \(\{HHH,HHT\}\) has \(X_2=H\), while \(\{HTH,HTT\}\) has \(X_2=T\); the two atoms starting with \(T\) have the corresponding values \(X_2=T\) and \(X_2=H\). Their union \(\{HHH,HHT,THH,THT\}\) is exactly \(\{X_2=H\}\). This checks membership by explicitly expressing the event as a union of atoms.
Why Nesting Matters
A filtration distinguishes the information available now from information that will become available later. This is essential whenever a stochastic process is analyzed conditionally on its history: conditioning on \(\mathcal{F}_n\) means conditioning on the events that can be resolved from observations through time \(n\). The filtration must therefore be specified, not inferred from the process alone. Two observers watching the same underlying outcomes can have different filtrations if they receive different reports.
A common mistake is to record only the sigma-field generated by the current observation and assume the resulting sequence is automatically a filtration. For instance, let \(R_n\) be the parity of the number of heads in the first \(n\) tosses, for \(n=1,2\), on the two-toss space \(\{HH,HT,TH,TT\}\). Then \(\sigma(R_1)\) has atoms \(\{HH,HT\}\) and \(\{TH,TT\}\). But \(\sigma(R_2)\) has atoms \(\{HH,TT\}\) and \(\{HT,TH\}\), since the total number of heads is even in the first pair and odd in the second. The event \(\{R_1=1\}=\{HH,HT\}\) is not a union of the atoms of \(\sigma(R_2)\), so \(\sigma(R_1)\not\subseteq\sigma(R_2)\). Using only the current parity has lost the earlier report.
The correction is to retain the history: define \(\mathcal{F}_1=\sigma(R_1)\) and \(\mathcal{F}_2=\sigma(R_1,R_2)\). The atoms of \(\mathcal{F}_2\) are \(\{HH\},\{HT\},\{TH\},\{TT\}\), because the two parity reports together distinguish all four outcomes. Each atom of \(\mathcal{F}_1\) is a union of atoms of \(\mathcal{F}_2\), so \(\mathcal{F}_1\subseteq\mathcal{F}_2\), as required.
Check Your Understanding
For each question, identify the relevant observation atoms before deciding which events are known.
- On \(\Omega=\{HH,HT,TH,TT\}\), what are the atoms of \(\sigma(X_1)\)? Is the event \(\{X_2=T\}\) in this sigma-field?
- Suppose \(Y_1\) records whether the first toss is \(H\), and \(Y_2\) records whether at least one of the first two tosses is \(H\). List the atoms of \(\sigma(Y_1,Y_2)\) on the three-toss space.
- In the coarse-observation example, why is \(\{X_2=H\}\) not in the time-1 sigma-field but in the time-2 sigma-field?
- Explain why \((\sigma(R_n))_{n\ge1}\), with \(R_n\) the running parity in the two-toss example, need not be a filtration.
- State the smallest filtration containing \(\sigma(Y_k)\) at time \(k\), and explain why it is increasing.