样本均值和样本方差

日期: 2023-10-01 11:28 查看: 29 次更新导出Word

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设总体 $X$ 的均值 $E(X)=\mu$ ，方差 $D(X)=\sigma^2 ，\left(X_1, X_2, \cdots, X_n\right)$ 为取自该总体的一个 样本，则
(1) $\quad E(\bar{x})=\mu, D(\bar{x})=\frac{\sigma^2}{n}$
(2) $E\left(S^2\right)=\sigma^2, E\left(S_n^2\right)=\frac{n-1}{n} \sigma^2, n \geq 2$
(3) $\bar{X} \stackrel{P}{\longrightarrow} \mu, S^2=\frac{1}{n-1} \sum_{i=1}^n\left(X_i-\bar{X}\right)^2 \stackrel{P}{\rightarrow} \sigma^2, S_n^2=\frac{1}{n} \sum_{i=1}^n\left(X_i-\bar{X}\right)^2 \stackrel{P}{\longrightarrow} \sigma^2$.
证 (1)

$$
\begin{aligned}
& E(\bar{X})=E\left(\frac{1}{n} \sum_{i=1}^n X_i\right)=\frac{1}{n} \sum_{i=1}^n E\left(X_i\right)=\mu, \\
& D(\bar{X})=D\left(\frac{1}{n} \sum_{i=1}^n X_i\right)=\frac{1}{n^2} \sum_{i=1}^n D\left(X_i\right)=\frac{1}{n} \sigma^2,
\end{aligned}
$$

证 (2)

$$
\begin{aligned}
(n-1) E\left(S^2\right) & =E\left(\sum_{i=1}^n X_i^2-n \bar{X}^2\right)=\left(\sum_{i=1}^n E\left(X_i^2\right)-n E\left(\bar{X}^2\right)\right) \\
& =n\left(D\left(X_1\right)+E^2\left(X_1\right)-D(\bar{X})-E^2(\bar{X})\right) \\
& =n\left(\sigma^2-\frac{\sigma^2}{n}\right)=(n-1) \sigma^2 \\
& \Rightarrow E\left(S^2\right)=\sigma^2, \quad E\left(S_n^2\right)=\frac{n-1}{n} \sigma^2
\end{aligned}
$$

证 (3)
独立同分布的大数定律，即得 $\bar{X} \stackrel{P}{\longrightarrow} \mu$

$$
\frac{1}{n} \sum_{i=1}^n X_i^2 \stackrel{p}{\rightarrow} \sigma^2+\mu^2,
$$

所以 $S_n^2=\frac{1}{n} \sum_{i=1}^n\left(X_i-\bar{X}\right)^2=\frac{1}{n} \sum_{i=1}^n X_i^2-\bar{X}^2 \stackrel{P}{\longrightarrow} \sigma^2+\mu^2-\mu^2=\sigma^2$

$$
S^2=\frac{n}{n-1} S_n^2 \stackrel{P}{\rightarrow} \sigma^2
$$

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