随机算法 (Fall 2011)/Probability Space

Axioms of Probability

The axiom foundation of probability theory is laid by Kolmogorov, one of the greatest mathematician of the 20th century, who advanced various very different fields of mathematics.

Definition (Probability Space)

A probability space is a triple [math]\displaystyle{ (\Omega,\Sigma,\Pr) }[/math]. [math]\displaystyle{ \Omega }[/math] is a set, called the sample space. [math]\displaystyle{ \Sigma\subseteq 2^{\Omega} }[/math] is the set of all events, satisfying: (K1). [math]\displaystyle{ \Omega\in\Sigma }[/math] and [math]\displaystyle{ \empty\in\Sigma }[/math]. (The certain event and the impossible event.) (K2). If [math]\displaystyle{ A,B\in\Sigma }[/math], then [math]\displaystyle{ A\cap B, A\cup B, A-B\in\Sigma }[/math]. (Intersection, union, and diference of two events are events). A probability measure [math]\displaystyle{ \Pr:\Sigma\rightarrow\mathbb{R} }[/math] is a function that maps each event to a nonnegative real number, satisfying (K3). [math]\displaystyle{ \Pr(\Omega)=1 }[/math]. (K4). If [math]\displaystyle{ A\cap B=\emptyset }[/math] (such events are call disjoint events), then [math]\displaystyle{ \Pr(A\cup B)=\Pr(A)+\Pr(B) }[/math]. (K5*). For a decreasing sequence of events [math]\displaystyle{ A_1\supset A_2\supset \cdots\supset A_n\supset\cdots }[/math] of events with [math]\displaystyle{ \bigcap_n A_n=\emptyset }[/math], it holds that [math]\displaystyle{ \lim_{n\rightarrow \infty}\Pr(A_n)=0 }[/math].

The sample space [math]\displaystyle{ \Omega }[/math] is the set of all possible outcomes of the random process modeled by the probability space. An event is a subset of [math]\displaystyle{ \Omega }[/math]. The statements (K1)--(K5) are axioms of probability. A probability space is well defined as long as these axioms are satisfied.

Example

Consider the probability space defined by rolling a dice with six faces. The sample space is [math]\displaystyle{ \Omega=\{1,2,3,4,5,6\} }[/math], and [math]\displaystyle{ \Sigma }[/math] is the power set [math]\displaystyle{ 2^{\Omega} }[/math]. For any event [math]\displaystyle{ A\in\Sigma }[/math], its probability is given by [math]\displaystyle{ \Pr(A)=\frac{|A|}{6}. }[/math]

Remark

• In general, the set [math]\displaystyle{ \Omega }[/math] may be continuous, but we only consider discrete probability in this lecture, thus we assume that [math]\displaystyle{ \Omega }[/math] is either finite or countably infinite.
• In many cases (such as the above example), [math]\displaystyle{ \Sigma=2^{\Omega} }[/math], i.e. the events enumerates all subsets of [math]\displaystyle{ \Omega }[/math]. But in general, a probability space is well-defined by any [math]\displaystyle{ \Sigma }[/math] satisfying (K1) and (K2). Such [math]\displaystyle{ \Sigma }[/math] is called a [math]\displaystyle{ \sigma }[/math]-algebra defined on [math]\displaystyle{ \Omega }[/math].
• The last axiom (K5*) is redundant if [math]\displaystyle{ \Sigma }[/math] is finite, thus it is only essential when there are infinitely many events. The role of axiom (K5*) in probability theory is like Zorn's Lemma (or equivalently the Axiom of Choice) in axiomatic set theory.

Laws for probability can be deduced from the above axiom system. Denote that [math]\displaystyle{ \bar{A}=\Omega-A }[/math].

Proposition

[math]\displaystyle{ \Pr(\bar{A})=1-\Pr(A) }[/math].

Proof. Due to Axiom (K4), [math]\displaystyle{ \Pr(\bar{A})+\Pr(A)=\Pr(\Omega) }[/math] which is equal to 1 according to Axiom (K3), thus [math]\displaystyle{ \Pr(\bar{A})+\Pr(A)=1 }[/math]. The proposition follows. [math]\displaystyle{ \square }[/math]

Exercise: Deduce other useful laws for probability from the axioms. For example, [math]\displaystyle{ A\subseteq B\Longrightarrow\Pr(A)\le\Pr(B) }[/math].

Notation

An event [math]\displaystyle{ A\subseteq\Omega }[/math] can be represented as [math]\displaystyle{ A=\{a\in\Omega\mid \mathcal{E}(a)\} }[/math] with a predicate [math]\displaystyle{ \mathcal{E} }[/math].

The predicate notation of probability is

[math]\displaystyle{ \Pr[\mathcal{E}]=\Pr(\{a\in\Omega\mid \mathcal{E}(a)\}) }[/math].

Example

We still consider the probability space by rolling a six-face dice. The sample space is [math]\displaystyle{ \Omega=\{1,2,3,4,5,6\} }[/math]. Consider the event that the outcome is odd.

[math]\displaystyle{ \Pr[\text{ the outcome is odd }]=\Pr(\{1,3,5\}) }[/math].

During the lecture, we mostly use the predicate notation instead of subset notation.

The Union Bound

We are familiar with the principle of inclusion-exclusion for finite sets.

Principle of Inclusion-Exclusion

Let [math]\displaystyle{ S_1, S_2, \ldots, S_n }[/math] be [math]\displaystyle{ n }[/math] finite sets. Then [math]\displaystyle{ \begin{align} \left|\bigcup_{1\le i\le n}S_i\right| &= \sum_{i=1}^n|S_i| -\sum_{i\lt j}|S_i\cap S_j| +\sum_{i\lt j\lt k}|S_i\cap S_j\cap S_k|\\ & \quad -\cdots +(-1)^{\ell-1}\sum_{i_1\lt i_2\lt \cdots\lt i_\ell}\left|\bigcap_{r=1}^\ell S_{i_r}\right| +\cdots +(-1)^{n-1} \left|\bigcap_{i=1}^n S_i\right|. \end{align} }[/math]

The principle can be generalized to probability events.

Principle of Inclusion-Exclusion for Probability

Let [math]\displaystyle{ \mathcal{E}_1, \mathcal{E}_2, \ldots, \mathcal{E}_n }[/math] be [math]\displaystyle{ n }[/math] events. Then [math]\displaystyle{ \begin{align} \Pr\left[\bigvee_{1\le i\le n}\mathcal{E}_i\right] &= \sum_{i=1}^n\Pr[\mathcal{E}_i] -\sum_{i\lt j}\Pr[\mathcal{E}_i\wedge \mathcal{E}_j] +\sum_{i\lt j\lt k}\Pr[\mathcal{E}_i\wedge \mathcal{E}_j\wedge \mathcal{E}_k]\\ & \quad -\cdots +(-1)^{\ell-1}\sum_{i_1\lt i_2\lt \cdots\lt i_\ell}\Pr\left[\bigwedge_{r=1}^\ell \mathcal{E}_{i_r}\right] +\cdots +(-1)^{n-1}\Pr\left[\bigwedge_{i=1}^n \mathcal{E}_{i}\right]. \end{align} }[/math]

We only prove the basic case for two events.

Lemma

For any two events [math]\displaystyle{ \mathcal{E}_1 }[/math] and [math]\displaystyle{ \mathcal{E}_2 }[/math], [math]\displaystyle{ \Pr[\mathcal{E}_1\vee\mathcal{E}_2]=\Pr[\mathcal{E}_1]+\Pr[\mathcal{E}_2]-\Pr[\mathcal{E}_1\wedge\mathcal{E}_2] }[/math].

Proof. The followings are due to Axiom (K4). [math]\displaystyle{ \begin{align} \Pr[\mathcal{E}_1] &=\Pr[\mathcal{E}_1\wedge\neg(\mathcal{E}_1\wedge\mathcal{E}_2)]+\Pr[\mathcal{E}_1\wedge\mathcal{E}_2];\\ \Pr[\mathcal{E}_2] &=\Pr[\mathcal{E}_2\wedge\neg(\mathcal{E}_1\wedge\mathcal{E}_2)]+\Pr[\mathcal{E}_1\wedge\mathcal{E}_2];\\ \Pr[\mathcal{E}_1\vee\mathcal{E}_2] &=\Pr[\mathcal{E}_1\wedge\neg(\mathcal{E}_1\wedge\mathcal{E}_2)]+\Pr[\mathcal{E}_2\wedge\neg(\mathcal{E}_1\wedge\mathcal{E}_2)]+\Pr[\mathcal{E}_1\wedge\mathcal{E}_2]. \end{align} }[/math] The lemma follows directly. [math]\displaystyle{ \square }[/math]

A direct consequence of the lemma is the following theorem, the union bound.

Theorem (Union Bound)

Let [math]\displaystyle{ \mathcal{E}_1, \mathcal{E}_2, \ldots, \mathcal{E}_n }[/math] be [math]\displaystyle{ n }[/math] events. Then [math]\displaystyle{ \begin{align} \Pr\left[\bigvee_{1\le i\le n}\mathcal{E}_i\right] &\le \sum_{i=1}^n\Pr[\mathcal{E}_i]. \end{align} }[/math]

The name of this inequality is Boole's inequality. It is usually referred by its nickname the "union bound". The bound holds for arbitrary events, even if they are dependent. Due to this generality, the union bound is extremely useful in probabilistic analysis.

Independence

Definition (Independent events)

Two events [math]\displaystyle{ \mathcal{E}_1 }[/math] and [math]\displaystyle{ \mathcal{E}_2 }[/math] are independent if and only if [math]\displaystyle{ \begin{align} \Pr\left[\mathcal{E}_1 \wedge \mathcal{E}_2\right] &= \Pr[\mathcal{E}_1]\cdot\Pr[\mathcal{E}_2]. \end{align} }[/math]

This definition can be generalized to any number of events:

Definition (Independent events)

Events [math]\displaystyle{ \mathcal{E}_1, \mathcal{E}_2, \ldots, \mathcal{E}_n }[/math] are mutually independent if and only if, for any subset [math]\displaystyle{ I\subseteq\{1,2,\ldots,n\} }[/math], [math]\displaystyle{ \begin{align} \Pr\left[\bigwedge_{i\in I}\mathcal{E}_i\right] &= \prod_{i\in I}\Pr[\mathcal{E}_i]. \end{align} }[/math]

Note that in probability theory, the "mutual independence" is not equivalent with "pair-wise independence", which we will learn in the future.