Why State Probability Axioms?
A probability model does more than list possible outcomes: it assigns probabilities to events the model permits. The axioms specify the minimum consistency requirements for those assignments. On a finite sample space, they can be read concretely: probabilities cannot be negative, the whole sample space has probability one, and probabilities add across disjoint allowed events.
The distinction between possible outcomes and allowed events matters. A model may permit every subset of its sample space, or it may permit only a smaller collection of subsets. In the second case, a singleton such as \(\{a\}\) may not itself be an allowed event, so it cannot be assigned a probability by that model. The axioms must therefore be stated on the model's actual event collection.
The Three Axioms on a Finite Model
Nonnegativity: \(P(A)\geq 0\) for every \(A\in\mathcal F\).
Unit total probability: \(P(\Omega)=1\).
Additivity for disjoint events: For every finite collection \(A_1,\ldots,A_k\in\mathcal F\) that are pairwise disjoint, \[ P\left(\bigcup_{i=1}^k A_i\right)=\sum_{i=1}^k P(A_i). \]
An event algebra is an appropriate domain for the assignment: it includes the whole sample space, and its closure ensures that complements and finite unions of allowed events are also allowed. Since it contains \(\Omega\) and is closed under complements, it also contains the empty event. The axioms only assign probabilities to members of \(\mathcal F\), not automatically to every subset of \(\Omega\).
The additivity axiom is about disjointness: no outcome belongs to more than one of the events being combined. Its equality says that the probability of their union is the sum of their probabilities. It does not assert this equality for overlapping events. Here the axiom is stated for any finite number of pairwise disjoint allowed events, which is the useful finite-model formulation.
Worked Example: A Probability Assignment on a Die
Let \(\Omega=\{1,2,3,4,5,6\}\), let \(\mathcal F\) be the collection of all subsets of \(\Omega\), and assign probability \(1/6\) to each outcome. For an event \(A\), define \(P(A)\) as the sum of the assigned probabilities of its outcomes. Check the axioms and find the probability of an even result.
Nonnegativity. Each outcome contributes \(1/6\), so every event has probability equal to a sum of zero or more copies of \(1/6\). Such a sum is nonnegative.
Unit total probability. The six outcomes make up \(\Omega\), and \[ P(\Omega)=\frac16+\frac16+\frac16+\frac16+\frac16+\frac16=1. \]
Additivity. If events are disjoint, they contain no outcome in common. Summing \(1/6\) over their union therefore sums each outcome's contribution exactly once, which is the same as adding the separate event probabilities.
The even-result event is \(A=\{2,4,6\}\). Thus \(P(A)=1/6+1/6+1/6=3/6=1/2\). This example uses all subsets as allowed events, so singleton events are available.
Atoms: The Basic Pieces of a Finite Event Algebra
When the event algebra is smaller than the collection of all subsets, its atoms provide the right objects for checking probabilities. An atom is an allowed event that cannot be split into smaller nonempty allowed events. It need not be a single outcome of the sample space.
In a finite event algebra, the atoms partition the sample space, and every allowed event is a union of atoms. This fact is what makes atom probabilities useful: every probability assignment on the algebra is determined by its values on these basic pieces.
Proof. Because \(\Omega\) is finite, every nonempty event \(A\in\mathcal F\) contains an atom: among the nonempty members of \(\mathcal F\) contained in \(A\), choose one with the fewest outcomes. It cannot contain a smaller nonempty member of \(\mathcal F\), so it is an atom.
Suppose two atoms \(C\) and \(D\) intersect. Their intersection belongs to \(\mathcal F\): complements and finite unions are available, and \(C\cap D=(C^c\cup D^c)^c\). Since \(C\cap D\) is nonempty and is contained in \(C\), the definition of atom implies \(C\cap D=C\). The same reasoning applied to \(D\) gives \(C\cap D=D\). Hence \(C=D\), so distinct atoms are disjoint.
Every outcome \(x\in\Omega\) belongs to at least one atom, because \(\Omega\) is an event in \(\mathcal F\) and contains an atom, and the finite collection of atoms can be considered across all nonempty events. More directly, among events in \(\mathcal F\) containing \(x\), choose one with the fewest outcomes; it is an atom. Thus the atoms cover \(\Omega\), and their pairwise disjointness makes them a partition.
Now take any \(A\in\mathcal F\). Each \(x\in A\) belongs to an atom \(C\). The intersection \(C\cap A\) is in \(\mathcal F\) and is nonempty. Since \(C\) is an atom, \(C\cap A=C\), so \(C\subseteq A\). Therefore \(A\) is exactly the union of the atoms it contains. Applying finite additivity to this disjoint union expresses \(P(A)\) as the sum of their probabilities. In particular, the atoms partition \(\Omega\), so their probabilities sum to \(P(\Omega)=1\); nonnegativity gives nonnegative atom probabilities. This also proves that the atom probabilities determine \(P(A)\) for every allowed event \(A\). \(\square\)
Worked Example: An Event Algebra Without Singleton Events
Let \(\Omega=\{a,b,c,d\}\), but suppose the allowed events are only \[ \mathcal F=\{\varnothing,\{a,b\},\{c,d\},\Omega\}. \] This is an event algebra: it contains \(\Omega\), and complements and finite unions of its members remain in the collection. Its atoms are \(C_1=\{a,b\}\) and \(C_2=\{c,d\}\). The singletons \(\{a\},\{b\},\{c\},\{d\}\) are not allowed events.
Assign \(P(C_1)=0.3\) and \(P(C_2)=0.7\). These atom probabilities are nonnegative and sum to \(0.3+0.7=1\). The allowed event probabilities are therefore \(P(\varnothing)=0\), \(P(C_1)=0.3\), \(P(C_2)=0.7\), and \(P(\Omega)=1\). The two atoms are disjoint, and their union is \(\Omega\), so their probabilities add to the unit total.
This assignment gives no probability to \(\{a\}\), for example, because \(\{a\}\notin\mathcal F\). It would be incorrect to treat this model as if it supplied separate probabilities for \(a\) and \(b\). The model distinguishes the block \(\{a,b\}\) from the block \(\{c,d\}\), but it does not distinguish outcomes within either block.
Two Useful Ways to Check an Assignment
The theorem gives a precise checking procedure for finite models. If every subset is allowed, the atoms are the singleton events, so check their assigned probabilities for nonnegativity and verify that they sum to one. The probability of any event is then the sum over its singleton outcomes. If only some subsets are allowed, identify the atoms of that event algebra instead; check the weights on those atoms and use them to determine probabilities of unions of atoms.
This is not a shortcut that permits arbitrary values for all events. Assignments must agree with the atom sums. For instance, in the two-block example, assigning \(P(C_1)=0.3\), \(P(C_2)=0.7\), and \(P(\Omega)=0.9\) would be inconsistent: the atoms are disjoint and have union \(\Omega\), so additivity requires \(P(\Omega)=0.3+0.7=1\).
Worked Example: Finding an Axiom Violation
Consider \(\Omega=\{u,v,w\}\) with all subsets allowed. A proposed assignment gives \(P(\{u\})=0.2\), \(P(\{v\})=-0.1\), and \(P(\{w\})=0.9\). It also gives \(P(\Omega)=1\). Determine whether it satisfies the axioms.
The singleton events are pairwise disjoint and their union is \(\Omega\). Their proposed probabilities sum to \(0.2+(-0.1)+0.9=1.0\), consistent with the proposed total probability. But nonnegativity fails because \(P(\{v\})=-0.1<0\). Thus the assignment is not a probability model, even though the total and this additivity check agree.
If instead the proposed values were \(0.2,0.1,0.9\), their sum would be \(1.2\), so additivity for the three disjoint singleton events would conflict with \(P(\Omega)=1\). Both checks are necessary; passing one does not guarantee passing the others.
Why the Axioms Matter
The axioms separate modeling choices from consistency requirements. A modeler chooses the sample space, the allowed events, and the probability assignments based on the experiment being represented. The axioms then test whether those assignments behave coherently: no event has negative probability, the model assigns unit probability to the whole space, and disjoint alternatives combine by addition.
A common mistake is to assume that every finite probability model assigns probabilities to individual outcomes. The reasoning might be: “The sample space is finite, so I can check that the singleton probabilities are nonnegative and sum to one.” The correction is that finiteness of \(\Omega\) does not imply that every subset belongs to \(\mathcal F\). Singleton checks are appropriate when singleton events are allowed; for a general finite event algebra, check the atoms. In the extreme case \(\mathcal F=\{\varnothing,\Omega\}\), there is only one atom, \(\Omega\) itself, even if \(\Omega\) contains many outcomes.
Check Your Understanding
Use the axioms and the event algebra actually specified in each question.
- State the three axioms for a probability assignment on a finite event algebra.
- For \(\Omega=\{1,2,3,4\}\) and \(\mathcal F=\{\varnothing,\{1,2\},\{3,4\},\Omega\}\), identify the atoms. Are singleton probabilities part of this model?
- Suppose the two atoms in Question 2 are assigned probabilities \(0.4\) and \(0.6\). What probability must their union have, and which axioms justify the value?
- Why is checking nonnegative singleton probabilities and their sum not a valid general procedure for every finite probability model?
- A model assigns probability \(0.25\) and \(0.75\) to two disjoint allowed events whose union is \(\Omega\), but assigns \(P(\Omega)=0.95\). Which axiom is violated?