Section01_总体、样本和统计量

总体和样本

1. 总体 所研究对象的某一数量指标的全体 $X$
2. 简单随机样本 看到"$X_{1},X_{2},\cdots,X_{n}$ 是来自总体 $X$ 容量为 $n$ 的简单随机样本"
   1. $X_{1},X_{2},\cdots, X_{n}$ 相互独立
   2. $X_{1}, X_{2},\cdots, X_{n}$ 与总体 $X$ 同分布，即 $FX_{i}(x) = F_{X}(x), EX_{i} = EX, DX_{i} = DX$
   3. 注
      • $X_{1}, X_{2},\cdots, X_{n}$ 为随机变量 (观察前)
      • $x_{1}, x_{2},\cdots, x_{n}$ 为具体数值 (观察后)

例题

1. 设 $(X_{1},X_{2},X_{3},X_{4})$ 为来自总体 $X\sim \left( \begin{matrix} 0 & 1 & 2\\ 1/6 & 1/2 & 1/3 \end{matrix} \right)$ 的一个简单随机样本，求 $\mathbb{P}\{\min\limits_{1\le i \le 4}X_{i} \le 1\}$ $$\begin{array}{ll} & \mathbb{P}\{\min\limits_{1\le i \le 4}X_{i} \le 1\} = 1 - \mathbb{P}\{\min\limits_{1\le i \le 4} X_{i}> 1\} \\ & = 1- \mathbb{P}(X_{1}>1, X_{2}>1, X_{3}>1, X_{4}>1) \\ & = 1- \mathbb{P}\{X_{1}>1\}\mathbb{P}\{X_{2}>1\}\mathbb{P}\{X_{3}>1\}\mathbb{P}\{X_{4}>1\} \\ & = 1 - \frac{1}{3}\times \frac{1}{3}\times \frac{1}{3}\times \frac{1}{3} = \frac{80}{81} \end{array}$$

统计量

1. 定义 样本的不含任何未知参数的函数: $g(X_{1},X_{2},\cdots,X_{n})$ 称为统计量
2. 常用统计量
   1. 样本均值 $\displaystyle \bar{X} = \frac{1}{n}\sum_{i=1}^{n}X_{i}$
   2. 样本方差 $\displaystyle S^{2} = \frac{1}{n-1}\sum_{i=1}^{n}(X_{i}-\bar{X})^{2} = \frac{1}{n-1}\bigg(\sum_{i=1}^{n}X_{i}^{2} - n\bar{X}^{2}\bigg)$

      注 总体$X,EX = \mu, DX = \sigma^{2}$ (未必正态)

      • $E \bar{X} = \mu$
      • $D \bar{X} = \frac{\sigma^{2}}{n}$
      • $\begin{array}{ll} E S^{2} = \frac{1}{n-1}[\sum\limits_{i=1}^{n}EX_{i}^{2} - n\cdot E \bar{X}^{2}]\\= \frac{1}{n-1}[n(\mu^{2} + \sigma^{2}) - n(\mu^{2} + \frac{\sigma^{2}}{n})]=\sigma^{2} \end{array}$

   3. 样本标准差 $S = \sqrt{S^{2}}$
   4. $k$阶原点矩 $A_{k} = \frac{1}{n}\sum\limits_{1}^{n}Y_{i}^{k}, k = 1,2,\cdots$

      注 总体的 $k$ 阶原点矩 $EX^{k}$；由大数定律得 $\frac{1}{n}\sum\limits_{i=1}^{n}Y^{k}_{i}\xrightarrow{P}E(\frac{1}{n}\sum\limits_{i=1}^{n}Y^{k}_{i})$ 故

      • $k=1\quad \bar{X} = \frac{1}{n}\sum\limits_{i=1}^{n}X_{i} \xrightarrow{P} EX =\mu$ 故 $\bar{X}$ 大概率在 $EX$ 附近取值 $\bar{X}\approx EX$ (矩估计)
      • $k=2\quad A_{2}=\frac{1}{n}\sum\limits_{i=1}^{n}X^{2}_{i} \xrightarrow{P} EX^{2}\Rightarrow A_{2}\approx EX^{2}$

   5. 样本的$k$阶中心矩 $B_{k} = \frac{1}{n}\sum\limits_{i=1}^{n}(X_{i}-\bar{X})^{k}, k = 1,2,\cdots$
      • $B_{1} = 0$
      • $B_{2} = \frac{1}{n}\sum\limits_{i=1}^{n}(X_{i}-\bar{X})^{2}$

        注

        $\begin{array}{ll}B_{2} = \frac{1}{n}\sum\limits_{i=1}^{n}(X_{i}-\bar{X})^{2} = \frac{1}{n}\sum\limits_{i=1}^{n}X_{i}^{2} - (\frac{1}{n}\sum\limits_{i=1}^{n}X_{i})^{2} \\ \xrightarrow{P} \mu^{2} + \sigma^{2} - \mu^{2} = \sigma^{2} \end{array}$

        • 因此 $B_{2}\approx \sigma^{2}$ (矩估计)

   6. 顺序统计量
      1. $\max\{X_{1},X_{2},\cdots,X_{n}\}$
      2. $\min\{X_{1},X_{2},\cdots,X_{n}\}$
      3. 注 中央台去掉一个最高分去掉一个最低分 $\sum\limits_{i=1}^{n}X_{i} - \min\{X_{1},X_{2},\cdots,X_{n}\} - \max\{X_{1},X_{2},\cdots,X_{n}\}$