Three Ingredients of a Probability Model
A probability model does not assign a probability to every subset of a sample space automatically. It specifies which subsets count as events, then assigns probabilities to those events. Keeping these choices separate is essential: a set that is not in the event sigma-algebra has no probability assigned by that model.
A sigma-algebra \(\mathcal{F}\) contains \(\Omega\), is closed under complements relative to \(\Omega\), and is closed under countable unions. A measure \(\mathbb{P}\) on it takes values in \([0,\infty]\), assigns \(0\) to the empty set, and is countably additive: for pairwise disjoint events \(A_1,A_2,\ldots\) in \(\mathcal{F}\), the union is in \(\mathcal{F}\) and \(\mathbb{P}(\bigcup_{n=1}^{\infty}A_n)=\sum_{n=1}^{\infty}\mathbb{P}(A_n)\). In a probability space, the normalization \(\mathbb{P}(\Omega)=1\) ensures that all event probabilities are finite.
The event algebra is part of the model, not a technical afterthought. If \(\mathcal{F}\) is the power set of a finite \(\Omega\), every subset is an event. In other models, only certain unions of outcomes may be events. The same sample space can therefore support different probability spaces when the event sigma-algebra or measure changes.
Finite Partitions and Their Events
A partition of \(\Omega\) is a collection of pairwise disjoint, nonempty subsets whose union is \(\Omega\). For a finite partition, the sigma-algebra it generates consists exactly of unions of partition blocks. This gives a useful way to specify which outcomes the model distinguishes.
Proof. Call the collection of unions \(\mathcal{U}\). It contains \(\Omega=\bigcup_{j=1}^{k}B_j\) and \(\varnothing\). The complement of a union of partition blocks is the union of the remaining blocks, so it belongs to \(\mathcal{U}\). A countable union of members of \(\mathcal{U}\) is still a union of partition blocks: include a block precisely when it occurs in at least one of the sets being united. Hence \(\mathcal{U}\) is a sigma-algebra containing each \(B_j\), and therefore \(\sigma(\{B_1,\ldots,B_k\})\subseteq\mathcal{U}\). Conversely, every union of blocks belongs to any sigma-algebra containing all the blocks, since there are only finitely many blocks. Thus \(\mathcal{U}\subseteq\sigma(\{B_1,\ldots,B_k\})\), proving equality. \(\square\)
Worked Example: A Fair Die
Let \(\Omega=\{1,2,3,4,5,6\}\), let \(\mathcal{F}\) be its power set, and define \(\mathbb{P}(A)=|A|/6\) for each \(A\in\mathcal{F}\). Identify the event of rolling an even number and calculate its probability.
Here every subset of \(\Omega\) is an event, so \(A=\{2,4,6\}\) is an event. Its probability is
The complementary event is \(A^c=\{1,3,5\}\), with \(\mathbb{P}(A^c)=3/6=1/2\). The probabilities of the six singleton outcomes are each \(1/6\), and their sum is \(6(1/6)=1=\mathbb{P}(\Omega)\). This is a finite probability space; it is complete because its only probability-zero event is the empty set.
Building a Probability Measure from Finite Weights
On a finite sample space with the full power set as its event sigma-algebra, nonnegative outcome weights summing to one define a probability measure. The next result verifies this construction, including countable additivity rather than only normalization.
Proof. For every \(A\subseteq\Omega\), the finite sum defining \(\mathbb{P}(A)\) is nonnegative, and \(\mathbb{P}(\varnothing)=0\) because the empty sum is zero. The total mass is \(\mathbb{P}(\Omega)=\sum_{i=1}^{N}w(x_i)=1\). Now let \(A_1,A_2,\ldots\) be pairwise disjoint subsets of \(\Omega\). Each \(x\in\Omega\) belongs to at most one \(A_n\). Consequently, in the sum \(\sum_{n=1}^{\infty}\mathbb{P}(A_n)\), the weight \(w(x)\) occurs once if \(x\in\bigcup_n A_n\), and does not occur otherwise. Since \(\Omega\) is finite, this sum reduces to the finite sum of the weights of the points in the union. It therefore equals \(\mathbb{P}(\bigcup_n A_n)\). This proves countable additivity and all the required properties. \(\square\)
Weights can also be used when the event sigma-algebra does not contain every singleton, but the measure must then be defined on events, not on the unavailable singleton events. The following example makes that distinction explicit.
Worked Example: A Coarse Event Sigma-Algebra
Let \(\Omega=\{a,b,c,d,e\}\), partitioned into \(B_1=\{a,b\}\), \(B_2=\{c\}\), and \(B_3=\{d,e\}\). Take \(\mathcal{F}=\sigma(\{B_1,B_2,B_3\})\), the eight unions of these blocks. Assign outcome weights \(w(a)=1/10\), \(w(b)=2/10\), \(w(c)=3/10\), \(w(d)=1/5\), and \(w(e)=1/5\). Define \(\mathbb{P}(A)=\sum_{x\in A}w(x)\) for \(A\in\mathcal{F}\).
The block weights are \(\mathbb{P}(B_1)=1/10+2/10=3/10\), \(\mathbb{P}(B_2)=3/10\), and \(\mathbb{P}(B_3)=1/5+1/5=2/5\). Their sum is \(3/10+3/10+2/5=6/10+4/10=1\). For the event \(A=B_1\cup B_3=\{a,b,d,e\}\),
The complement is \(B_2=\{c\}\), whose probability is \(3/10\); adding gives \(7/10+3/10=1\). The set \(\{a\}\) is not an event in this model, so \(\mathbb{P}(\{a\})\) is not defined. The weights \(w(a)\) and \(w(b)\) are outcome weights used to calculate probabilities of events; they are not values of \(\mathbb{P}\) on singleton events in this sigma-algebra. Because all three blocks have positive probability, the only probability-zero event is \(\varnothing\), so this finite probability space is complete.
Probability on an Interval
For an interval-based model, take \(\Omega=[0,1]\), let \(\mathcal{F}=\mathcal{L}\cap[0,1]\) be the Lebesgue-measurable subsets of \([0,1]\), and define \(\mathbb{P}(A)=m(A)\). This is the restriction of Lebesgue measure \(m\) to \([0,1]\). It is a measure because disjoint countable unions of members of \(\mathcal{F}\) remain in \(\mathcal{F}\), and \(m\) is countably additive. Also \(m([0,1])=1\), so the restriction is normalized. This probability space is finite and complete, since the Lebesgue sigma-algebra includes every subset of a Lebesgue-null set.
Worked Example: An Event Made from Two Intervals
In the interval probability space above, let \(A=[0,1/4]\cup[1/2,1]\). These two intervals are disjoint, and both are Lebesgue measurable. Their lengths are \(1/4\) and \(1-1/2=1/2\), respectively, so countable additivity (here, finite additivity for two disjoint sets) gives
The complement in \([0,1]\) is \(A^c=(1/4,1/2)\), of length \(1/2-1/4=1/4\). Directly, \(\mathbb{P}(A)+\mathbb{P}(A^c)=3/4+1/4=1=m([0,1])\), which checks the calculation. Including or excluding either endpoint does not change these lengths, since a singleton has Lebesgue measure zero.
Interpreting the Components Correctly
When specifying a probability model, state each component explicitly:
- Sample space \(\Omega\): the set of possible outcomes in the model.
- Event sigma-algebra \(\mathcal{F}\): the collection of subsets to which the model assigns probabilities.
- Probability measure \(\mathbb{P}\): the countably additive measure on \(\mathcal{F}\) with total mass one.
A common mistake is to treat every subset of \(\Omega\) as an event, even when the chosen sigma-algebra is smaller. For the partition model, the calculation \(\mathbb{P}(\{a\})=1/10\) is not a valid probability calculation: \(\{a\}\notin\mathcal{F}\), so the probability measure has no value there. The correction is to use the outcome weight \(w(a)=1/10\) only as part of the sum defining \(\mathbb{P}(A)\) for an actual event \(A\in\mathcal{F}\). For example, \(\mathbb{P}(\{a,b\})=w(a)+w(b)=3/10\) is defined because \(\{a,b\}=B_1\) is an event.
Conversely, on the fair-die model, \(\mathcal{F}\) is the full power set, so every subset is an event and the singleton probabilities are defined. Thus, before asking for the probability of a set, check membership in the specified event sigma-algebra. A probability space is the full triple, not merely a list of outcomes or a formula for some probabilities.
Check Your Understanding
For each question, identify which components of the probability space are being used and verify the relevant event membership before calculating.
- For the fair-die model, write the event that the outcome is at most \(2\) and calculate its probability.
- For the partition model on \(\{a,b,c,d,e\}\), list the events contained in \(B_1\cup B_2\), and determine whether \(\{b,c\}\) is an event.
- In the interval model, calculate the probability of \([1/3,2/3]\) and its complement in \([0,1]\).
- What conditions must a finite family of outcome weights satisfy to define a probability measure on the full power set?
- Explain why a weight assigned to an outcome need not be the probability of a singleton event in every probability space.