Events Allowed by a Probability Model

Why a Probability Model Needs an Event Collection

A sample space lists the possible outcomes of an experiment, and an event is a subset of those outcomes. But specifying the sample space alone does not always settle which subsets the model will treat as events. A probability model also specifies a collection of subsets on which probabilities are to be defined. That collection is the subject of this tutorial.

For a finite sample space, it is usually natural to allow every subset as an event. On an infinite sample space, a model may instead use a smaller collection. The key requirement is that the collection be closed under the basic set operations needed to form complements and countable unions. These closure requirements describe the event structure; they do not yet assign numerical probabilities to events.

The term “countable” in the third condition matters even when \(\Omega\) itself is countable. The condition requires closure under a union of any sequence of events in the collection, not merely under finite unions. Finite unions follow by repeating events in a sequence and using the closure condition. The empty event also belongs to every sigma-algebra: since \(\Omega\in\mathcal{F}\), closure under complements gives \(\Omega^c=\varnothing\in\mathcal{F}\). Intersections are available as well, because De Morgan’s laws express intersections using complements and unions.

Finite Sample Spaces: Every Subset Can Be an Event

For a set \(\Omega\), its power set, written \(\mathcal{P}(\Omega)\), is the collection of all subsets of \(\Omega\). It is always a sigma-algebra: it contains \(\Omega\), complements of subsets of \(\Omega\) are still subsets of \(\Omega\), and unions of subsets are still subsets of \(\Omega\). In a finite model, taking \(\mathcal{F}=\mathcal{P}(\Omega)\) allows every possible event.

A model can also use a smaller event collection even when its sample space is finite. Doing so makes the model less detailed: outcomes grouped together by the collection cannot be separated by its events. This is useful for understanding the general role of \(\mathcal{F}\), though the full power set is the standard choice for elementary finite models.

Countable Spaces and Restricted Event Collections

A countably infinite sample space can also be equipped with its full power set. For instance, if \(\Omega=\{0,1,2,\ldots\}\), then every subset—including the even integers, the primes, or a single outcome—is an event under \(\mathcal{F}=\mathcal{P}(\Omega)\). The countability of \(\Omega\) does not force a model to restrict its event collection.

However, a model may choose a coarser collection. For \(\Omega=\{0,1,2,\ldots\}\), let \(E\) be the even integers and \(O\) the odd integers. Then \(\mathcal{F}=\{\varnothing,E,O,\Omega\}\) is a sigma-algebra. Complements interchange \(E\) and \(O\), while unions of sets chosen from these four possibilities remain one of the four possibilities. Individual outcomes such as \(\{0\}\) are not events in this collection.

What Every Finite Event Collection Looks Like

On a finite sample space, any sigma-algebra has a simple structure: its events are exactly the unions of blocks in a partition. A partition is a collection of nonempty, pairwise disjoint sets whose union is the whole sample space. The blocks are the smallest groups of outcomes that the event collection can distinguish.

Proof. Define two outcomes \(x,y\in\Omega\) to be equivalent when they belong to exactly the same events in \(\mathcal{F}\):

This is an equivalence relation: each outcome has the same membership pattern as itself; the condition is symmetric in \(x,y\); and if \(x\) and \(y\) have the same membership pattern, and \(y\) and \(z\) do too, then \(x\) and \(z\) do. Its equivalence classes therefore form a partition of \(\Omega\).

Fix \(x\in\Omega\). For each \(y\) not equivalent to \(x\), some event \(A_y\in\mathcal{F}\) contains exactly one of \(x\) and \(y\). Choose \(A_y\) if it contains \(x\), and choose \(A_y^c\) otherwise; call the chosen set \(D_y\). Then \(D_y\in\mathcal{F}\), it contains \(x\), and it excludes \(y\). Because \(\Omega\) is finite, there are only finitely many such \(y\), so the intersection \(C_x=\bigcap_{y\not\sim x}D_y\) belongs to \(\mathcal{F}\). (If there are no such \(y\), this intersection is \(\Omega\).) The outcomes in \(C_x\) are exactly those equivalent to \(x\): it contains \(x\)'s entire equivalence class, and it excludes every outcome outside that class. Thus every block belongs to \(\mathcal{F}\).

Any union of blocks belongs to \(\mathcal{F}\), because there are only finitely many blocks and a sigma-algebra is closed under finite unions. Conversely, every \(A\in\mathcal{F}\) is a union of whole blocks: if \(x\in A\) and \(y\sim x\), the definition of \(\sim\) implies \(y\in A\). So \(\mathcal{F}\) consists exactly of the unions of the partition blocks. \(\square\)

This theorem explains the partition example for the die: each block collects outcomes the event collection cannot tell apart. It also explains why the full power set corresponds to the partition into singletons. The finite result uses the finiteness of \(\Omega\) when it forms an intersection with only finitely many terms; it does not claim the same description for every event collection on an infinite space.

Why the Countable-Union Condition Cannot Be Dropped

A common mistake is to check only complements and finite unions. Those checks do not establish that a collection is a sigma-algebra on an infinite space. For example, the collection of finite and cofinite subsets of \(\Omega=\{0,1,2,\ldots\}\) contains \(\Omega\), is closed under complements, and is closed under finite unions. But it is not closed under countable unions: every singleton \(\{2k\}\) is finite, while their union is the infinite set of even integers, whose complement is also infinite. That union is neither finite nor cofinite.

The correction is to check the full defining condition: the union of every sequence of allowed events must itself be allowed. In contrast, the collection \(\{\varnothing,E,O,\Omega\}\) passes that check because any union of its members is still one of its four members.