Section03_正态总体下常见的抽样分布

单正态总体

设总体$X\sim \mathcal{N}(\mu,\sigma^{2})$，$X_{1},X_{2},\cdots,X_{n}$ 为样本

$$\begin{array}{ll} \displaystyle \bar{X} = \frac{1}{n}\sum_{i=1}^{n}X_{i}\quad & \displaystyle S^{2}=\frac{1}{n-1}\sum_{i=1}^{n}(X_{i}-\bar{X})^{2}\\ \displaystyle B_{2} = \frac{1}{n}\sum_{i=1}^{n}(X_{i}-\bar{X})^{2} \end{array}$$

1. $\displaystyle \bar{X}\sim \mathcal{N}(\mu, \frac{\sigma^{2}}{n});\quad \frac{\bar{x}- \mu}{\sigma/\sqrt{n}} \sim \mathcal{N}(0,1);\quad \bigg(\frac{\bar{X}-\mu}{\sigma/\sqrt{n}}\bigg)^{2}\sim \chi^{2}(1)$
2. $\displaystyle \sum\limits_{i=1}^{n}\bigg(\frac{X_{i}-\mu}{\sigma}\bigg)^{2} = \frac{\sum\limits_{i=1}^{n}(X_{i}-\mu)^{2}}{\sigma^{2}} \sim \chi^{2}(n)$
3. $\displaystyle \sum\limits_{i=1}^{n}\bigg(\frac{X_{i}- \bar{X}}{\sigma}\bigg)^{2} = \frac{\sum\limits_{i=1}^{n}(X_{i}-\bar{X})^{2}}{\sigma^{2}} = \frac{(n-1)S^{2}}{\sigma^{2}}= \frac{nB_{2}}{\sigma^{2}} \sim \chi^{2}(n-1)$
   • 且 $\bar{X}$ 与 $S^{2}$ 独立
4. $\displaystyle \frac{\bar{X} - \mu}{S/\sqrt{n}} = \frac{\frac{\bar{X} -\mu}{\sigma/\sqrt{n}}}{\sqrt{\frac{(n-1)S^{2}}{\sigma^{2}(n-1)}}}\sim t(n-1);\quad \bigg(\frac{\bar{X} - \mu}{S/\sqrt{n}}\bigg)^{2}\sim F(1,n-1)$

双正态分布

总体 $X\sim \mathcal{N}(\mu_{1},\sigma_{1}^{2})$，总体 $Y\sim \mathcal{N}(\mu_{2},\sigma_{2}^{2})$；$X_{1},X_{2},\cdots X_{n_{1}}$ 和 $Y_{1},Y_{2},\cdots, Y_{n_{2}}$ 分别是总体 $X$ 和 $Y$ 的样本

$$\begin{array}{ll} \displaystyle \bar{X} = \frac{1}{n_{1}}\sum_{i=1}^{n_{1}}X_{i} & \displaystyle \bar{Y} = \frac{1}{n_{2}}\sum_{j=1}^{n_{2}}Y_{j} \\ \displaystyle S_{1}^{2} = \frac{1}{n_{1}-1}\sum_{i=1}^{n_{1}}(X_{i}-\bar{X})^{2} & \displaystyle S_{2}^{2} = \frac{1}{n_{2}-1}\sum_{j=1}^{n_{2}}(Y_{j}-\bar{Y})^{2} \end{array}$$

且 $X,Y$ 独立，则

1. $\bar{X}\sim \mathcal{N}(\mu_{1}, \frac{\sigma_{1}^{2}}{n_{1}});\quad \bar{Y}\sim \mathcal{N}(\mu_{2}, \frac{\sigma_{2}^{2}}{n_{2}})$，若 $\sigma_{1}^{2} = \sigma_{2}^{2} = \sigma^{2}$，则

   • $\displaystyle \bar{X}-\bar{Y}\sim \mathcal{N}(\mu_{1}-\mu_{2}, \sigma^{2}(\frac{1}{n_{1}}+\frac{1}{n_{2}}))$
   • $\displaystyle \frac{(n_{1}-1)S_{1}^{2}}{\sigma^{2}} + \frac{(n_{2}-1)S_{2}^{2}}{\sigma^{2}} \sim\chi^{2}(n_{1}+n_{2}- 2)$
   • $\displaystyle \frac{\frac{\bar{X}- \bar{Y} - (\mu_{1} - \mu_{2})}{\sigma^{2}(\frac{1}{n_{1}} + \frac{1}{n_{2}})}}{\sqrt{\frac{(n_{1}-1)S_{1}^{2} + (n_{2}-1)S_{2}^{2}}{\sigma^{2}\times (n_{1}+n_{2} -2)}}} = \frac{\bar{X} - \bar{Y} - (\mu_{1} - \mu_{2})}{S_{\omega}\sqrt{\frac{1}{n_{1}}+ \frac{1}{n_{2}}}} \sim t(n_{1}+n_{2} -2)$
     • $\displaystyle S_{\omega}^{2} = \frac{(n_{1}-1)S_{1}^{2} + (n_{2}-1)S_{2}^{2}}{n_{1} + n_{2} - 2}$

2. $\displaystyle \frac{\frac{(n_{1}-1)S_{1}^{2}}{(n_{1}-1)\sigma_{1}^{2}}}{\frac{(n_{2}-1)S_{2}^{2}}{(n_{2}-2)\sigma_{2}^{2}}} = \frac{S_{1}^{2}/\sigma_{1}^{2}}{S_{2}^{2}/\sigma_{2}^{2}} \sim F(n_{1}-1, n_{2} - 1)$

例题

1. 设总体 $X\sim \mathcal{N}(0,1)$，$X_{1},X_{2},X_{3},X_{4},X_{5}$ 为来自总体 $X$ 的简单随机样本，$\bar{X},S$ 分别为样本均值和样本标准差，则下列结论中正确的是( ) $$\begin{array}{ll} \text{A. }\bar{X}\sim \mathcal{N}(0,1) & \text{B. }\sum\limits_{i=1}^{5}(X_{i}-\bar{X})^{2} \sim \chi^{2}(5) \\ \text{C. }\sqrt{\frac{3}{2}}\frac{X_{1}+X_{2}}{\sqrt{X_{3}^{2}+X_{4}^{2}+X_{5}^{2}}}\sim t(3) & \text{D. }\frac{X_{1}^{2} + 2X_{2}^{2}}{X_{3}^{2}+X_{4}^{2}+X_{5}^{2}}\sim F(3,3) \end{array}$$ $$\begin{array}{ll} &\bar{X} \sim \mathcal{N}(0, \frac{1}{5}) \\\\ &\sum\limits_{i=1}^{5}(X_{i} - \bar{X})^{2} = \frac{\sum\limits_{i=1}^{5}(X_{i} - \bar{X})^{2}}{1^{2}} \sim \chi^{2}(4) \\\\ &\sqrt{\frac{3}{2}}\frac{X_{1}+X_{2}}{\sqrt{X_{1}^{2}+X_{2}^{2}+X_{3}^{2}}} = \frac{(X_{1}+X_{2})/\sqrt{2}}{\sqrt{(X_{1}^{2}+X_{2}^{2}+X_{3}^{2})/3}} \sim t(3) \\\\ \because & X_{2}\text{同} X_{2} \text{不独立} \\ \therefore & X_{1}^{2}+2X_{2}^{2} \nsim \chi^{2}(3) \\ \therefore & \frac{X_{1}^{2} + 2X_{2}^{2}}{X_{3}^{2}+X_{4}^{2}+X_{5}^{2}}\nsim F(3,3)\\\\ \therefore & \text{选}C \\ \end{array}$$