证明容量为n的一维正态分布样本的离差平方和的总体方差等分服从自由度为n-1的卡方分布，即nS_n^2/σ^2~χ^2(n-1)

设一组样本$(X_1,X_2,\cdots ,X_n)\sim N(\mu ,{\sigma }^2)$，记其期望和离差平方和为

$$\begin{array}{c}\overline{X}=\frac{1}{n}\sum _{i=1}^n{X}_i\\ SS(X)=\sum _{i=1}^n(X_i-\overline{X}{)}^2\end{array}$$

证明

$$\frac{SS(X)}{{\sigma }^2}\sim {\chi }^2(n-1)$$

证：对$(X_i-\overline{X}{)}^2$，凑出$(X_i-\mu {)}^2$并还原

$$SS(X)=\sum _{i=1}^n\left[(X_i-\mu {)}^2+2X_i\mu -{\mu }^2-2X_i\overline{X}+{\overline{X}}^2\right]$$

分解求和式并求和

$$\begin{array}{rl}SS(X)& =\sum _{i=1}^n(X_i-\mu {)}^2+\sum _{i=1}^n\left[{\overline{X}}^2-{\mu }^2+2X_i(\mu -\overline{X})\right]\\ & =\sum _{i=1}^n(X_i-\mu {)}^2+n\left({\overline{X}}^2-{\mu }^2\right)+2(\mu -\overline{X})\sum _{i=1}^n{X}_i\\ & =\sum _{i=1}^n(X_i-\mu {)}^2+n(\overline{X}+\mu )(\overline{X}-\mu )+2n(\mu -\overline{X})\overline{X}\\ & =\sum _{i=1}^n(X_i-\mu {)}^2+n(\overline{X}-\mu )(\overline{X}+\mu -2\overline{X})\\ & =\sum _{i=1}^n(X_i-\mu {)}^2-n(\overline{X}-\mu {)}^2\end{array}$$

两侧同除${\sigma }^2$

$$\begin{array}{rl}\frac{SS(X)}{{\sigma }^2}& =\frac{{\sum }_{i=1}^n(X_i-\mu {)}^2-n(\overline{X}-\mu {)}^2}{{\sigma }^2}\\ & =\sum _{i=1}^n(\frac{X_i-\mu }{\sigma }{)}^2-{\left[\frac{\sqrt{n}}{\sigma }(\overline{X}-\mu )\right]}^2\end{array}$$

$\frac{X_i-\mu }{\sigma }$是对$X_i$的归一化，因此$\frac{X_i-\mu }{\sigma }\sim N(0,1)$，故

$$\sum _{i=1}^n(\frac{X_i-\mu }{\sigma }{)}^2\sim {\chi }^2(n)$$

由于$\overline{X}\sim N\left(\mu ,\frac{{\sigma }^2}{n}\right)$，因此

$$\begin{array}{c}\frac{\sqrt{n}}{\sigma }(\overline{X}-\mu )\sim N(0,1)\\ {\left[\frac{\sqrt{n}}{\sigma }(\overline{X}-\mu )\right]}^2\sim {\chi }^2(1)\end{array}$$

故

$$\frac{SS(X)}{{\sigma }^2}\sim {\chi }^2(n-1)$$