Section03_大数定律和中心极限定理

大数定律

• 记住 $$\frac{1}{n}\sum_{i=1}^{n}Y_{i} \xrightarrow{P} E\bigg(\frac{1}{n}\sum_{i=1}^{n}Y_{i}\bigg)$$

例题

1. 设随机变量序列 $\{X_{n}\}$ 相互独立，且服从于参数为 $2$ 的指数分布，则当 $n\to \infty$ 时，$\frac{1}{n}\sum\limits_{i=1}^{n}X^{2}_{i}$ 依概率收敛于          $$\begin{array}{ll} \because & \frac{1}{n}\sum\limits_{i=1}^{n}X_{i}^{2}\xrightarrow{P} \frac{1}{n}E(\sum\limits_{i=1}^{n}X^{2}_{i}) = \frac{1}{n}\sum\limits_{i=1}^{n}EX_{i}^{2}\\ & X_{i}\sim E(2) \\ \therefore & EX_{i}^{2} = DX + (EX)^{2} = \frac{1}{2^{2}} + \frac{1}{2^{2}} = \frac{1}{2} \\ \therefore & \frac{1}{n}\sum\limits_{i=1}^{n}EX_{i}^{2} = \frac{1}{n}\times n\times \frac{1}{2} = \frac{1}{2} \\ \therefore & \frac{1}{n}\sum\limits_{i=1}^{n}X_{i}^{2}\xrightarrow{P} \frac{1}{2} \\ \end{array}$$
2. 设随机变量序列 $X_{1},X_{2},\cdots,X_{n},\cdots$ 独立同分布，且 $X_{i}$ 的概率密度为 $f(x) = \begin{cases} 1 - \vert x \vert, & \vert x \vert <1 \\ 0, & \text{其他} \end{cases}$，则 $n\to \infty$ 时，$\frac{1}{n}\sum\limits_{i=1}^{n}X_{i}^{2}$ 依概率收敛于         $$\begin{array}{ll} \because & f(x) = \begin{cases} 1-\vert x \vert, & \vert x \vert <1 \\ 0, & \text{其他} \end{cases} \\ \therefore & \displaystyle EX^{2} = \int_{-1}^{1}x^{2}(1-\vert x \vert)\cdot dx = 2 \int_{0}^{1}x^{2}(1-x)\cdot dx = \frac{1}{6} \\ \because & \frac{1}{n}\sum\limits_{i=1}^{n}X_{i}^{2}\xrightarrow{P} \frac{1}{n}E(\sum\limits_{i=1}^{n}X^{2}_{i}) = \frac{1}{n}\sum\limits_{i=1}^{n}EX_{i}^{2}\\ & X_{1},\cdots, X_{n} \text{独立同分布} \\ \therefore & \frac{1}{n}\sum\limits_{i=1}^{n}EX_{i}^{2} = \frac{1}{n}\times n\times \frac{1}{6} = \frac{1}{6} \\ \therefore & \frac{1}{n}\sum\limits_{i=1}^{n}X_{i}^{2}\xrightarrow{P}\frac{1}{6} \\ \end{array}$$

中心极限定理

• 记住 $$\sum_{i=1}^{n}X_{i} \overset{\text{approx}}{\sim} \mathcal{N}(E \sum_{i=1}^{n}X_{i}, D \sum_{i=1}^{n}X_{i})$$
  • 大量随机变量之和，近似正态分布

例题

1. 设随机变量 $X_{1},X_{2},\cdots, X_{32}$ 相互独立同分布，且 $X_{i}\sim E(2)$，记 $X = \sum\limits_{i=1}^{32}X_{i}$，$p_{1}= \mathbb{P}\{X<16\}, p_{2}=\mathbb{P}\{X>12\}$，用中心极限定理近似计算可得( ) A. $p_{1} = p_{2}$ B. $p_{1}<p_{2}$ C. $p_{1}>p_{2}$ D. $p_{1},p_{2}$ 大小不能确定 $$\begin{array}{ll} \because & X = \sum\limits_{i=1}^{32}X_{i}\overset{\text{approx}}{\sim} \mathcal{N}(E\sum\limits_{i=1}^{32}X_{i},D\sum\limits_{i=1}^{32}X_{2}) \\ & E\sum\limits_{i=1}^{32}X_{i} = \sum\limits_{i=1}^{32}EX_{i} = 32\times \frac{1}{2} = 16 \\ & D \sum\limits_{i=1}^{32}X_{i} = \sum\limits_{i=1}^{32}DX_{i} = 32\times \frac{1}{2^{2}} = 8 \\ \therefore & X \overset{\text{approx}}{\sim} \mathcal{N}(16,8) \\ \therefore & p_{1} = \mathbb{P}\{X<16\} < \mathbb{P}\{X>12\} = p_{2} \\ \therefore & \text{选} B \\ \end{array}$$