概率论 | 弱大数定律

2026-08-25 11:15:13 星期二
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弱大数定律

我们本节研究一族随机变量 \(\{X_n\}_{n=1}^{\infty}\)，\(S_n = X_1 + \cdots + X_n\)

\[\frac{S_n}{n} = \frac{1}{n}\sum_{i=1}^{n}EX_i + \underbrace{\frac{1}{n}\sum_{i=1}^{n}(X_i - EX_i)}_{n \to \infty \text{ 时，是否依概率收敛于 } 0？} \]

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Thm 1 (Markov 弱大数定律) 设 \(\{X_n\}_{n=1}^{\infty}\) 方差均有界，若 Markov 条件

\[\frac{1}{n^2}\operatorname{Var}\left(\sum_{i=1}^{n}X_i\right) \to 0 \]

成立，则有 WLLN：

\[\frac{1}{n}\sum_{i=1}^{n}(X_i - EX_i) \xrightarrow{P} 0. \]

证明.

\[\frac{1}{n}\sum_{i=1}^{n}(X_i - EX_i) = \frac{1}{n}(S_n - ES_n). \]

\(\forall \varepsilon > 0\)，

\[\begin{aligned} P\left( \left| \frac{1}{n}\sum_{i=1}^{n}(X_i - EX_i) \right| \geq \varepsilon \right) &= P\left( \left| \frac{1}{n}(S_n - ES_n) \right| \geq \varepsilon \right) \\ &\leq \frac{1}{\varepsilon^2} E\left[ \frac{1}{n}(S_n - ES_n) \right]^2 \\ &= \frac{1}{n^2 \varepsilon^2} \operatorname{Var}(S_n) \to 0. \end{aligned} \]

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Remark 1. 考虑一种特殊情况：\(\{X_n\}_{n=1}^{\infty}\) 两两不相关，即 \(\operatorname{Cov}(X_i, X_j) = 0\)，则

\[\operatorname{Var}(S_n) = \sum_{i=1}^{n} \operatorname{Var}(X_i). \]

Markov 条件写为

\[\frac{1}{n^2}\sum_{i=1}^{n}\operatorname{Var}(X_i) \to 0 \quad (n \to \infty). \]

2. \(\{X_n\}_{n=1}^{\infty}\) 两两不相关，且 \(\operatorname{Var}(X_n) \leq C\)，\(\forall n\)，则 Markov 条件成立，这时称为 Chebyshev WLLN.

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Prop 2
设 \(\{X_n\}_{n=1}^{\infty}\) 相互独立，且中心极限定理成立，则 WLLN \(\Leftrightarrow\) Markov 条件成立.

证明.
(\(\Leftarrow\)) 显然.

(\(\Rightarrow\)) 由于 CLT 成立，即

\[\frac{\sum_{i=1}^{n}(X_i - EX_i)}{\sqrt{\sum_{i=1}^{n}\operatorname{Var}(X_i)}} \xrightarrow{d} N(0,1), \]

\(\forall x \in \mathbb{R}^+\)，

\[\lim_{n \to \infty} P\left( \left| \frac{\sum_{i=1}^{n}(X_i - EX_i)}{\sqrt{\sum_{i=1}^{n}\operatorname{Var}(X_i)}} \right| < x \right) = \int_{-x}^{x} \frac{e^{-y^2/2}}{\sqrt{2\pi}} \, dy. \]

而由于 WLLN 成立，取 \(\varepsilon = 1\)，则有

\[\lim_{n \to \infty} P\left( \left| \frac{1}{n}\sum_{i=1}^{n}(X_i - EX_i) \right| \leq 1 \right) = 1. \]

改写一下：

\[\lim_{n \to \infty} P\left( \left| \frac{\sum_{i=1}^{n}(X_i - EX_i)}{\sqrt{\sum_{i=1}^{n}\operatorname{Var}(X_i)}} \right| \leq \frac{n}{\sqrt{\sum_{i=1}^{n}\operatorname{Var}(X_i)}} \right) = 1. \]

为了证明 Markov 条件成立，即

\[\frac{1}{n^2}\sum_{i=1}^{n}\operatorname{Var}(X_i) \to 0, \]

等价于

\[\frac{n^2}{\sum_{i=1}^{n}\operatorname{Var}(X_i)} \to \infty, \]

即

\[\lim_{n \to \infty} \frac{n}{\sqrt{\sum_{i=1}^{n}\operatorname{Var}(X_i)}} = +\infty. \]

反证法：若 \(\lim_{n \to \infty} a_n < +\infty\)，则存在 \(x_0 \in \mathbb{R}^+\) 和有界子列 \(\{a_{n_k}\}\)，使得 \(\forall k\)，\(a_{n_k} \leq x_0\).

定义

\[F_n(x) = P\left( \left| \frac{\sum_{i=1}^{n}(X_i - EX_i)}{\sqrt{\sum_{i=1}^{n}\operatorname{Var}(X_i)}} \right| < x \right), \]

\[F(x) = \frac{1}{\sqrt{2\pi}} \int_{-x}^{x} e^{-y^2/2} \, dy. \]

由于 \(\lim_{k \to \infty} F_{n_k}(a_{n_k}) = 1\)，且 \(F(x_0) < 1\)，取 \(\varepsilon = 1 - F(x_0) > 0\)，\(\exists k_0\)，\(\forall k > k_0\)，有

\[|F_{n_k}(a_{n_k}) - 1| < \frac{\varepsilon}{2}, \]

即

\[F_{n_k}(a_{n_k}) > 1 - \frac{\varepsilon}{2} = F(x_0) + \frac{\varepsilon}{2}. \]

而 \(x_0 \geq a_{n_k}\)，故

\[F_{n_k}(x_0) \geq F_{n_k}(a_{n_k}) > F(x_0) + \frac{\varepsilon}{2}. \]

令 \(k \to \infty\)，有

\[F(x_0) \geq F(x_0) + \frac{\varepsilon}{2} > F(x_0), \]

矛盾. 故 \(a_n \to +\infty\)，Markov 条件成立.