样本均值和样本方差的无偏性证明、正态分布样本方差的方差

样本均值和样本方差的无偏性

　　对于独立同分布的样本$x_1...x_n$来说，他们的均值为与方差分别为：

$ \begin{aligned}&\bar{x} = \frac{1}{n}\sum\limits_{i=1}^{n}x_i \\& s^2 = \frac{\sum\limits_{i=1}^{n} (x_i - \bar{x})^2}{n-1}\end{aligned} $

　　要证明样本方差的无偏性，首先要计算样本均值的方差。

样本均值的方差

$ \begin{aligned} D(\bar{x}) = D\left(\frac{\sum_{i=1}^{n}x_i}{n}\right) = \frac{1}{n^2}\sum\limits_{i=1}^{n}D(x_i) = \frac{1}{n^2}\sum\limits_{i=1}^{n}\sigma^2 = \frac{\sigma^2}{n} \end{aligned} $

样本均值和样本方差的无偏性证明

\begin{array}{lcl} \displaystyle E(\bar{x}) = E\left(\frac{1}{n}\sum\limits_{i=1}^{n}x_i\right) = \frac{1}{n}\sum\limits_{i=1}^{n}E(x_i) = \frac{1}{n}\sum\limits_{i=1}^{n}\mu = \mu \\\\ \begin{aligned} E(s^2) &= E\left(\frac{\sum_{i=1}^{n} (x_i - \bar{x})^2}{n-1}\right) \\ &= \frac{1}{n-1}\sum\limits_{i=1}^{n}E(x_i^2 + \bar{x}^2 - 2x_i\bar{x}) \\ &= \frac{1}{n-1}\sum\limits_{i=1}^{n}\left[D(x) + E(x)^2 + D(\bar{x}) + E(\bar{x})^2 - 2E\left(\frac{x_i^2+x_i\sum_{j=1,j\ne i}^n x_j}{n} \right)\right]\\ &= \frac{1}{n-1}\sum\limits_{i=1}^{n}\left[\sigma^2 + \mu^2 + \frac{\sigma^2}{n} + \mu^2 - 2\frac{\sigma^2 + \mu^2 + (n-1)\mu^2}{n}\right]\\ &=  \frac{1}{n-1}\sum\limits_{i=1}^{n}\left(\frac{n - 1}{n}\sigma^2\right)\\ &= \sigma^2 \end{aligned} \end{array}

正态分布样本方差的方差

　　对于服从正态分布的总体$X \sim N(\mu, \sigma^2)$，它的样本方差经过变换服从$\chi^2$分布，即：

　　$\displaystyle\frac{(n-1)s^2}{\sigma^2} \sim \chi^2(n - 1)$.

　　$\chi^2$分布的方差为它的自由度的两倍，所以：

$ \begin{aligned} &D\left(\frac{(n-1)s^2}{\sigma^2}\right) = 2(n - 1) \\ &D(s^2) = \frac{2\sigma^4}{n - 1} \end{aligned} $