测度论讲义 | 7.1 独立性 0-1律

2026-07-30 12:07:00 星期四
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独立性 0-1律

\((\Omega, \mathcal{F}, \mathbb{P})\)，\(\mathbb{P}\)为概率测度，即\(\mathbb{P}\)为测度，且\(\mathbb{P}(\Omega)=1\). \(\mathcal{F}\)中的元素称为（随机）事件。由于\(\Omega\in\mathcal{F}\)，记\(\Omega\)为必然事件。\(X: \Omega\to\mathbb{R}\)是可测的：即\(\forall B\in\mathcal{B}(\mathbb{R})\)，\(X^{-1}(B)\in\mathcal{F}\)，称\(X\)为随机变量。对于可测函数，我们定义过积分，套用过来期望定义为：若\(X\)关于\(\mathbb{P}\)积分存在（即\(\int_{\Omega} X^+ d\mathbb{P}<\infty\)或\(\int_{\Omega} X^- d\mathbb{P}<\infty\)），则称\(\int_{\Omega} X d\mathbb{P}\)为\(X\)的期望，记为\(\mathbb{E}X\).

注意 a.e. 成立与 a.s. 成立的区别：若\(\mu([f\ne g])=0\)，则\(f=g\) a.e.，\(\mu\)为任一测度；若\(\mathbb{P}([f=g])=1\)，则\(f=g\) a.s.，\(\mathbb{P}\)为概率测度.

独立性

Def 7.1.1 设\(A,B\in\mathcal{F}\)，若\(\mathbb{P}(A\cap B)=\mathbb{P}(A)\mathbb{P}(B)\)，称\(A\)与\(B\)独立。

设\(A_1,\cdots,A_n\in\mathcal{F}\)，仅有

\[\mathbb{P}\left(\bigcap_{k=1}^{n} A_k\right)=\prod_{k=1}^{n} \mathbb{P}(A_k) \]

是不足以推出\(A_1,\cdots,A_n\)独立的，须有\(\forall m\le n,\ \forall\ 1\le k_1<k_2<\cdots<k_m\le n\)，成立

\[\mathbb{P}\left(\bigcap_{j=1}^{m} A_{k_j}\right)=\prod_{j=1}^{m} \mathbb{P}(A_{k_j}) \]

才可以说\(A_1,\cdots,A_n\)相互独立。

以上都是有限个事件，若事件数不可数，该如何定义？

Def 7.1.2 设\(\mathcal{D}=\{A_t\mid t\in T\}\subset\mathcal{F}\)，如果\(\forall\)非空有限子集\(S\subset T\)，有

\[\mathbb{P}\left(\bigcap_{s\in S} A_s\right)=\prod_{s\in S} \mathbb{P}(A_s) \]

则称\(\mathcal{D}\)中事件相互独立。

设\(\{\mathcal{C}_t\}_{t\in T}\)为事件族，如果\(\forall t\in T\)，从\(\mathcal{C}_t\)中取一个事件\(A_t\)，组成\(\{A_t\}_{t\in T}\)是相互独立的，则称\(\{\mathcal{C}_t\}_{t\in T}\)为独立事件族。

设\(\{X_t\}_{t\in T}\)为一族随机变量，若\(\{\sigma(X_t)\}_{t\in T}\)为独立事件族，则\(\{X_t\}_{t\in T}\)相互独立。

Thm 7.1.3（独立类的扩张） 设\(\{\mathcal{C}_t\}_{t\in T}\)为独立事件族，若\(\mathcal{C}_t\)都为\(\pi\)类，则\(\{\sigma(\mathcal{C}_t)\}_{t\in T}\)也为独立事件族。

任取有限子集\(\{s_1,\cdots,s_n\} \subset T\)。证明：首先写出“\(\{\sigma(\mathcal{C}_t)\}_{t\in T}\)为独立事件族”的定义：\(\forall\)非空有限子集\(S\subset T\)，\(\{\sigma(\mathcal{C}_t)\}_{t\in S}\)为独立事件族，即\(\{\sigma(\mathcal{C}_{s_1}),\cdots,\sigma(\mathcal{C}_{s_n})\}\)为独立事件族。

令

\[D=\left\{A\in\mathcal{F}\mid \mathbb{P}\left(A\cap\left(\bigcap_{j=2}^{n} C_j\right)\right)=\mathbb{P}(A)\cdot\prod_{j=2}^{n} \mathbb{P}(C_j),\ C_j\in\mathcal{C}_{s_j},\ j=2,\cdots,n\right\}. \]

① \(\mathcal{C}_{s_1} \subset D\)，因为 \(\{\mathcal{C}_t\}_{t \in T}\) 为独立事件族.

② \(D\) 为 \(\lambda\) 类.

由单调类定理，\(\sigma(\mathcal{C}_{s_1}) = \lambda(\mathcal{C}_{s_1}) \subset D\)，即 \(\{\sigma(\mathcal{C}_{s_1}), \mathcal{C}_{s_2}, \dots, \mathcal{C}_{s_n}\}\) 为独立事件族.

类似地，定义

\[D' = \left\{A \in \mathcal{F} \mid \mathbb{P}\left(A \cap C_{s_1} \cap \bigcap_{j=3}^n C_j\right) = \mathbb{P}(A)\mathbb{P}(C_{s_1})\prod_{j=3}^n \mathbb{P}(C_j)\right\}, \]

其中 \(C_{s_1} \in \sigma(\mathcal{C}_{s_1})\)，\(C_j \in \mathcal{C}_{s_j}\)，\(j = 3, \cdots, n\).

\(\mathcal{C}_{s_2} \subset D'\)，\(D'\) 为 \(\lambda\) 类. 用单调类定理即得.

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习题

习题7.1.2

题目：设\((X_{i},1\leqslant i\leqslant n)\)为独立随机变量序列，若每个\(X_{i}\)非负或可积，则有\(\mathbf{E}[\prod_{i=1}^{n}X_{i}]=\prod_{i=1}^{n}\mathbf{E}[X_{i}]\)（提示：从简单随机变量过渡到非负随机变量）.

这道题在durrett的书里面有一个小定理，贴在这里：
Theorem 2.1.13. If \(X_1,\ldots,X_n\) are independent and have (a) \(X_i \ge 0\) for all \(i\), or (b) \(E|X_i| < \infty\) for all \(i\), then

\[E\left( \prod_{i=1}^n X_i \right) = \prod_{i=1}^n E X_i, \]

i.e., the expectation on the left exists and has the value given on the right.

Proof. \(X = X_1\) and \(Y = X_2 \cdots X_n\) are independent by Theorem 2.1.10, so taking \(f(x) = |x|\) and \(g(y) = |y|\) we have

\[E|X_1 \cdots X_n| = E|X_1| \, E|X_2 \cdots X_n|, \]

and it follows by induction that if \(1 \le m \le n\),

\[E|X_m \cdots X_n| = \prod_{i=m}^n E|X_k|. \]

If the \(X_i \ge 0\), then \(|X_i| = X_i\) and the desired result follows from the special case \(m = 1\). To prove the result in general, note that the special case \(m = 2\) implies \(E|Y| = E|X_2 \cdots X_n| < \infty\), so using Theorem 2.1.12 with \(f(x) = x\) and \(g(y) = y\) shows

\[E(X_1 \cdots X_n) = E X_1 \cdot E(X_2 \cdots X_n), \]

and the desired result follows by induction. \(\square\)

这里面提到的定理2.1.12是：

Theorem \(2. 1. 12. \textit{ Suppose X and Y are independent and have distribu- }\) tions \(\mu\) and \(\nu .\) \(If\) \(h:\) \(\mathbf{R} ^{2}\rightarrow \mathbf{R}\) \(is\) \(a\) measurable function with \(h\geq 0\) \(or\) \(E| h( \dot{X} , Y) | < \infty \textit{ then}\)

\[Eh(X,Y)=\int\int h(x,y)\:\mu(dx)\:\nu(dy) \]

\(In\textit{ particular, if h}( x, y) = f( x) g( y) \textit{ where f, g}: \mathbf{R} \to \mathbf{R} \textit{ are measurable}\)
functions with \(f, g\geq 0\) \(or\) \(E| f( X) |\) and \(E| g( Y) | < \infty\) then

\[Ef(X)g(Y)=Ef(X)\cdot Eg(Y) \]