背景知識

定義

設總體$X$(不管服從什麼分佈, 只要均值和方差存在)的均值為$\mu$, 方差為$\sigma^2$, $X_1, X_2, \cdots, X_n$是來自$X$的一個樣本,

均值

樣本均值定義如下 $\overline{X} = \sum_{i=1}^n X_i$ 具有以下性質

1. $E(\overline{X}) = \mu$ 證明: $\begin{aligned} E(\overline{X}) &= \frac{1}{n} E(\sum_{i=1}^n X_i) \\ &= \frac{1}{n} \sum_{i=1}^n E(X_i) \\ &= \frac{1}{n} \cdot n \mu \\ &= \mu \qquad 證畢 \end{aligned}$
2. $D(\overline{X})=\sigma^2/n$ 證明: $\begin{aligned} D(\overline{X}) &= D(\frac{1}{n} \sum_{i=1}^n X_i) \\ &= \frac{1}{n^2} D(\sum_{i=1}^n X_i) \end{aligned}$ 根據方差的性質4有 $D(X+Y)=D(X)+D(Y)+2E\{(X-E(X))(Y-E(Y))\}$ 且根據定義, $X_1, X_2, \cdots, X_n$之間相互獨立, 因此 $\begin{aligned} D(\overline{X}) &= \frac{1}{n^2} D(\sum_{i=1}^n X_i) \\ &= \frac{1}{n^2} \cdot nD(X_i) \\ &= \frac{1}{n^2} \cdot n\sigma^2 \\ &= \sigma^2/n \qquad 證畢 \end{aligned}$

樣本方差

樣本方差為定義如下 $\begin{aligned} S^2 &= \frac{1}{n-1} \sum_{i=1}^n \left( X_i - \overline{X} \right)^2 \\ &= \frac{1}{n-1} \sum_{i=1}^n \left( X_i^2 - 2 X_i \overline{X} + \overline{X}^2 \right) \\ &= \frac{1}{n-1} \left( \sum_{i=1}^n X_i^2 - 2n \bar{X}^2 + n\bar{X}^2 \right) \\ &= \frac{1}{n-1} \left( \sum_{i=1}^n X_i - n \bar{X}^2 \right) \end{aligned}$

樣本方差具有如下性質 $E(S^2)=\sigma^2$ 證明: $\begin{aligned} E(S^2) &= E \left[ \frac{1}{n-1} \left( \sum_{i=1}^n X_i^2 - n \bar{X}^2 \right) \right] \\ &= \frac{1}{n-1} \left[ \sum_{i=1}^n E(X_i^2) - n E(\bar{X}^2) \right] \\ \end{aligned}$ 根據根據方差的性質1有 $\begin{aligned} &D(X)=E(X^2)-[E(X)]^2 \\ \Rightarrow &E(X^2) = DX + \left[ E(X) \right]^2 \end{aligned}$