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Probability spaces

Let us start with the most basic definition. Let Ω\Omega be a non-empty set. A function f:ΩRf: \Omega \rightarrow {\mathbb R} is said to be a probability distribution on Ω\Omega if it satisfies the following two conditions.

  • For all aΩa \in \Omega, we have f(a)0f(a) \geq 0
  • aΩf(a)=1\sum_{a \in \Omega} f(a) = 1

We additionally say that ff is uniform if f(a)=1/Ωf(a) = 1/|\Omega|.

Definition 3

A finite probability space is a pair (Ω,P)(\Omega, {\mathbb P}) where Ω\Omega is a non-empty finite set and P{\mathbb P} is a probability distribution on Ω\Omega.

We refer to Ω\Omega as a sample space (containing every possible outcome of an experiment). Each element xΩx \in \Omega is called atomic event.

Definition 4

An event of a probability space is a subset AΩA \subseteq \Omega. Atomic events are of the form {a}\{a\} for aΩa \in \Omega.

For AΩA \subseteq \Omega, we define

P[A]=aAP[a]{\mathbb P}[A] = \sum_{a \in A} {\mathbb P}[a]
Example 32

This definition implies that P[Ω]=aΩP[a]=1{\mathbb P}[\Omega] = \sum_{a \in \Omega} {\mathbb P}[a]=1. Moreover, P[]=0{\mathbb P}[\emptyset] = 0.

Example 33

From this definition, in a uniform probability space, calculating the probability of event AA amounts to dividing A|A| by Ω|\Omega| (i.e., the number of cases we are interested in divided by the number of all cases).

Definition 5

An event AA is said to be trivial if P[A]=0{\mathbb P}[A] = 0 or P[A]=1{\mathbb P}[A]= 1.

In the following exercise, you will investigate some abstract mathematical structure of finite probability spaces.

Exercise 83

Prove: The number of trivial events (in a finite probability space) is a power of two.

Theorem 36 (Union bound)

Let A1,,AnΩA_1,\ldots, A_n \subseteq \Omega be events. Then P[A1An]i=1nP[Ai]{\mathbb P}[A_1 \cup \cdots \cup A_n] \leq \sum_{i=1}^n {\mathbb P}[A_i]

Since we have already done a lot of tedious inductions in the lecture notes, this is again your turn to do it 😊

Exercise 84

Prove the union bound by induction on nn. Argue that the equality holds if and only if every pair AiA_i and AjA_j is almost disjoint (P[AiAj]=0{\mathbb P}[A_i \cap A_j] = 0).