抽样分布定理证明

定理 ：$\frac{(n-1)S^{2}}{\sigma^{2}}$

证明

$$\begin{aligned} \frac{(n-1)S^2}{\sigma^2} =& \sum_{k=1}^n\frac{(x_i-\overline{x})^2}{\sigma^2}\\ =& \sum_{k=1}^n\frac{(x_i - \mu +\mu - \overline{x})^2}{\sigma^2}\\ =&\frac{1}{\sigma^2} \sum_{i=1}^{n}(( x_i-\mu)^2 - 2(x_i-\mu )(\bar{x} - \mu) + (\bar{x}-\mu)^2)\\ =& \frac{1}{\sigma^2} \sum_{i=1}^{n}( x_i-\mu)^2 - 2(\bar{x} - \mu)\sum_{i=1}^{n}(x_i-\mu ) + \sum_{i=1}^{n}(\bar{x}-\mu)^2\\ =& \frac{1}{\sigma^2} \sum_{i=1}^{n}( x_i-\mu)^2 - 2(\bar{x} - \mu)n(\frac{1}{n}\sum_{i=1}^{n}x_i -\mu)+ \sum_{i=1}^{n}(\bar{x}-\mu)^2\\ =& \frac{1}{\sigma^2} \sum_{i=1}^{n}( x_i-\mu)^2 - 2n(\bar{x} - \mu)^2+ \sum_{i=1}^{n}(\bar{x}-\mu)^2\\ =& \sum_{k=1}^n\frac{(x_i - \mu)^2}{\sigma^2} -\sum_{k=1}^n\frac{(\overline{x}-\mu)^2}{\sigma^2}\\ =& \sum_{k=1}^n\frac{(x_i - \mu)^2}{\sigma^2} - \frac{n(\overline{x}-\mu)^2 }{\sigma^2}\\ =& \sum_{k=1}^n\frac{(x_i - \mu)^2}{\sigma^2} - \frac{(\overline{x}-\mu)^2}{\sigma^2 / \sqrt{n}}\\ \end{aligned}$$ $$\sum_{k=1}^n\frac{(x_i - \mu)^2}{\sigma^2} \sim X_{n}^2$$ $$\frac{(\overline{x}-\mu)^2}{\sigma^2 / \sqrt{n}} \sim X_1^2$$

$(xi - \mu)^2 与 \overline{x}$相互独立，所以根据卡方分布可加性可得服从于$X{n-1}^2$