事件空间 Fsub2^(Omega)\mathcal{F} \subset 2^{\Omega} 是我们希望分配概率的样本空间的子集集合。一个集合 A inFA \in \mathcal{F} 称为事件;
P\mathbb{P} 是一种概率测度,它为每个事件 A inFA \in \mathcal{F} 指派一个概率 P(A)in[0,1]\mathbb{P}(A) \in[0,1] 。
示例 1.1.1. 考虑抛掷一个公平硬币两次的实验。我们模型 Omega={HH,HT,TH,TT}=\Omega=\{H H, H T, T H, T T\}={omega_(i):i=1,dots,4}\left\{\omega_{i}: i=1, \ldots, 4\right\} 和
{:[F=2^(Omega)={O/","Omega","{HH}","{HT}","{TH}","{TT}","{HH","HT}","{HH","TH}","{HH","TT}","{HT","TH}","{HT","TT}","],[{TH","TT}","{HH","HT","TH}","{HH","HT","TT}","{HH","TH","TT}","{HT","TH","TT}}]:}\begin{aligned}
& \mathcal{F}=2^{\Omega}=\{\varnothing, \Omega,\{H H\},\{H T\},\{T H\},\{T T\},\{H H, H T\},\{H H, T H\},\{H H, T T\},\{H T, T H\},\{H T, T T\}, \\
&\{T H, T T\},\{H H, H T, T H\},\{H H, H T, T T\},\{H H, T H, T T\},\{H T, T H, T T\}\}
\end{aligned}
由于硬币是公平的,每个 4 个基本事件 {HH},{HT},{TH},{TT}\{H H\},\{H T\},\{T H\},\{T T\} 具有相同的概率,即 1//41 / 4 ,对于所有 omega in Omega\omega \in \Omega 来说是 P({omega})=1//4\mathbb{P}(\{\omega\})=1 / 4 。其中一个不同结果 HTH T 和 THT H 发生的概率可以通过求和直观地获得,即 P({HT,TH})=\mathbb{P}(\{H T, T H\})=P({HT})+P({TH})=1//4+1//4=1//2\mathbb{P}(\{H T\})+\mathbb{P}(\{T H\})=1 / 4+1 / 4=1 / 2 。更一般地,对于 C sub{1,dots,n}C \subset\{1, \ldots, n\} ,我们可以设定
P({omega_(i):i in C})=P(uuu_(i in C){omega_(i)})=sum_(i in C)P({omega_(i)})=|C|//4.\mathbb{P}\left(\left\{\omega_{i}: i \in C\right\}\right)=\mathbb{P}\left(\bigcup_{i \in C}\left\{\omega_{i}\right\}\right)=\sum_{i \in C} \mathbb{P}\left(\left\{\omega_{i}\right\}\right)=|C| / 4 .
注意,然而根据这个定义,事件的并集的概率(意为“至少一个事件发生”)并不总是通过相加得到的:
P({HT}uu{HT,TH})=P({HT,TH})=1//2!=3//4=P({HT})+P({HT,TH}).\mathbb{P}(\{H T\} \cup\{H T, T H\})=\mathbb{P}(\{H T, T H\})=1 / 2 \neq 3 / 4=\mathbb{P}(\{H T\})+\mathbb{P}(\{H T, T H\}) .
原因在于:这些事件不是不相交的!在上述模型中,我们可以例如将“第一次掷出正面”识别为事件 {HH,HT}\{H H, H T\} ,其概率为 1//21 / 2 ,或者将“至少出现一次正面”识别为事件 {HT,TH,HH}\{H T, T H, H H\} ,其概率为 3//43 / 4 。对此进行重新表述可以是“反面未出现两次”,其概率可以表示为
引理 1.1.3。任何 sigma\sigma -代数 F\mathcal{F} 都具有以下性质: {i} O/ inF\varnothing \in \mathcal{F}
(ii) A,B inFLongrightarrow A uu B inFA, B \in \mathcal{F} \Longrightarrow A \cup B \in \mathcal{F}
(iii) A,B inFLongrightarrow A nn B inFA, B \in \mathcal{F} \Longrightarrow A \cap B \in \mathcal{F} (四) (A_(n))_(n inN)subFLongrightarrownnn_(n inN)A_(n)inF\left(A_{n}\right)_{n \in \mathbb{N}} \subset \mathcal{F} \Longrightarrow \bigcap_{n \in \mathbb{N}} A_{n} \in \mathcal{F}
(v) A,B inFLongrightarrow A\\B inFA, B \in \mathcal{F} \Longrightarrow A \backslash B \in \mathcal{F}
练习 1.1.6. 设 Omega\Omega 是一个可数集, E={{omega}:omega in Omega}\mathcal{E}=\{\{\omega\}: \omega \in \Omega\} 。证明 sigma(E)=2^(E)\sigma(\mathcal{E})=2^{\mathcal{E}} 。
For our statistical purposes the by far most important sigma\sigma-algebra is the Borel sigma\sigma-algebra B(R)\mathcal{B}(\mathbb{R}) over the real numbers R\mathbb{R}, which is defined as 对于我们的统计目的,迄今为止最重要的 sigma\sigma -代数是实数 R\mathbb{R} 上的 Borel sigma\sigma -代数 B(R)\mathcal{B}(\mathbb{R}) ,其定义为
B(R):=sigma({O subR:O" open "})\mathcal{B}(\mathbb{R}):=\sigma(\{O \subset \mathbb{R}: O \text { open }\})
and we will always implicitly equip R\mathbb{R} with this sigma\sigma-algebra when considering it as a measurable space. The Borel sigma\sigma-algebra has the following simpler characterisation. 我们在将 R\mathbb{R} 视为可测空间时,会始终隐式地为其装备这个 sigma\sigma -代数。Borel sigma\sigma -代数有以下更简单的表述。
Lemma 1.1.7. All of the following families of sets are generators of B(R)\mathcal{B}(\mathbb{R}) : 引理 1.1.7. 以下所有集合族都是 B(R)\mathcal{B}(\mathbb{R}) 的生成元:
(i) E_(1)={(a,b):-oo < a <= b < oo}\mathcal{E}_{1}=\{(a, b):-\infty<a \leq b<\infty\}E_(1)={(a,b):-oo < a <= b < oo}\mathcal{E}_{1}=\{(a, b):-\infty<a \leq b<\infty\}
(ii) E_(2)={[a,b]:-oo < a <= b < oo}\mathcal{E}_{2}=\{[a, b]:-\infty<a \leq b<\infty\}
(iii) E_(3)={(a,b]:-oo < a <= b < oo}\mathcal{E}_{3}=\{(a, b]:-\infty<a \leq b<\infty\}
(iv) E_(4)={(-oo,a):a inR}\mathcal{E}_{4}=\{(-\infty, a): a \in \mathbb{R}\}
(v) E_(5)={(-oo,a]:a inR}\mathcal{E}_{5}=\{(-\infty, a]: a \in \mathbb{R}\}
证明。我们只证明 sigma(E_(1))=B(R)\sigma\left(\mathcal{E}_{1}\right)=\mathcal{B}(\mathbb{R}) ,其余的陈述留作练习。由于任何开区间 (a,b)(a, b) 在 R\mathbb{R} 中都是开放的,我们有 E_(1)sub{O subR:O\mathcal{E}_{1} \subset\{O \subset \mathbb{R}: O 开放 }\} ,因此
sigma(E_(1))sub sigma({O subR:O" open "})=B(R)\sigma\left(\mathcal{E}_{1}\right) \subset \sigma(\{O \subset \mathbb{R}: O \text { open }\})=\mathcal{B}(\mathbb{R})