PT_正态总体的抽样分布

正态总体的抽样分布

• $$\begin{gathered} 设X_1,X_2,\cdots ,X_n是取自总体正态X\sim N(\mu ,{\sigma }^2)的样本 \\ \overline{X}=\frac{1}{n}\sum _{i=1}^n{X}_i \\ {S}^2=\frac{1}{n-1}\sum _{i=1}^n(X_i-\overline{X}{)}^2为样本方差 \end{gathered}$$

性质

样本均值

• $\overline{X}\sim N(\mu ,\frac{{\sigma }^2}{n})$
  • 可以有独立正态分布可加性和随机变量线性函数的规律得到
• 标准化:${\overline{x}}^{\ast }=\frac{\overline{X}-\mu }{\sigma /\sqrt{n}}\sim N(0,1)$

一个正态总体的抽样分布

${\chi }^2分布$

cases1:

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

  • $$\begin{gathered} X\sim N(\mu ,{\sigma }^2) \\ {X}_i\sim N(\mu ,{\sigma }^2) \\ Z=X_i-\mu \sim N(0,{\sigma }^2),(a=1,b=-\mu ) \\ \frac{1}{\sigma }(X_i-\mu )=\frac{1}{\sigma }Z\sim N(0,1),(a=\frac{1}{\sigma },b=0) \\ 得证 \end{gathered}$$

cases2:

• $\overline{X}和S^2相互独立$

• $$\xi =\frac{(n-1)}{{\sigma }^2}{S}^2\sim {\chi }^2(n-1)$$

自由度说明

• 上式自由度:

  • $$\begin{gathered} S^2=\frac{1}{n-1}\sum _{i=1}^n(X_i-\overline{X}{)}^2 \\ {S}^2(n-1)=\sum _{i=1}^n(X_i-\overline{X}{)}^2 \\ {S}^2(n-1)\frac{1}{{\sigma }^2}=\frac{1}{{\sigma }^2}\sum _{i=1}^n(X_i-\overline{X}{)}^2 \\ 记\zeta =\sum _{i=1}^n(X_i-\overline{X}{)}^2 \\ \zeta 是n个随机变量(X) \\ A=\sum _{i=1}^n{X}_i \\ \overline{X}=\frac{1}{n}A \\ 并且存在约束:\sum _{i=1}^n(X_i-\overline{X})=0 \\ (\sum _{i=1}^n{X}_i-\sum _{i=1}^n(\overline{X})=\sum _{i=1}^n{X}_i-n(\overline{X})=A-A=0) \end{gathered}$$

t分布:

• $$\eta =\frac{{\overline{X}}^{\ast }}{\sqrt{\xi }/\sqrt{n-1}}=\frac{\overline{X}-\mu }{S/\sqrt{n}}\sim t(n-1)$$

  • $$\begin{gathered} t=\frac{X}{\sqrt{Y/n}};X\sim N(0,1);Y\sim t(n) \\ t=\frac{X}{\sqrt{Y/(n-1)}};X\sim N(0,1);Y\sim t(n-1) \\ t=\frac{{\overline{X}}^{\ast }}{\sqrt{\xi /(n-1)}}; \\ {\overline{x}}^{\ast }=\frac{\overline{X}-\mu }{\sigma /\sqrt{n}}\sim N(0,1) \\ \xi =\frac{(n-1)}{{\sigma }^2}{S}^2\sim t(n-1) \\ t=\frac{\frac{\overline{X}-\mu }{\sigma /\sqrt{n}}}{\sqrt{(\frac{(n-1)}{{\sigma }^2}{S}^2)/(n-1)}}=\frac{\overline{X}-\mu }{\frac{S}{\sigma }\sigma /\sqrt{n}} \\ t=\frac{\overline{X}-\mu }{S/\sqrt{n}}\sim t(n-1) \end{gathered}$$

两个正态总体的抽样分布

F分布

• $$\begin{gathered} {\mathcal{S}}_1=(X_1,X_2,\cdots ,X_{n_1}) \\ {\mathcal{S}}_2=(Y_1,Y_2,\cdots ,Y_{n_2}) \\ {\mathcal{S}}_i是总体X\sim N({\mu }_i,{\sigma }_i^2)的样本(i=1,2) \\ {\mathcal{S}}_1与{\mathcal{S}}_2相互独立 \end{gathered}$$

  • $$\begin{gathered} \overline{X}=\frac{1}{n_1}\sum _{i=1}^{n_1}{X}_i \\ \overline{Y}=\frac{1}{n_2}\sum _{i=1}^{n_2}{X}_i \\ {S}_1^2=\frac{1}{n_1-1}\sum _{i=1}^{n_1}(X_i-\overline{X}{)}^2 \\ {S}_2^2=\frac{1}{n_2-1}\sum _{i=1}^{n_2}(X_i-\overline{X}{)}^2 \end{gathered}$$

• $$F=\frac{S_1^2/S_2^2}{{\sigma }_1^2/{\sigma }_2^2}=\frac{S_1^2/{\sigma }_1^2}{S_2^2/{\sigma }_2^2}\sim F(n_1-1,n_2-1)$$

• 推导:

  • $$\begin{gathered} F=\frac{X/n_1}{Y/n_2}\sim F(n_1,n_2); \\ X\sim {\chi }^2(n_1);Y\sim {\chi }^2(n_2) \\ 根据F分布的定义:对于 \\ {\xi }_i=\frac{(n_i-1)}{{\sigma }_i^2}{S}_i^2\sim {\chi }^2(n_i-1) \\ 则: \\ \frac{{\xi }_1/(n_1-1)}{{\xi }_2/(n_2-1)}\sim F((n_1-1),(n_2-1)) \\ F=\frac{S_1^2/S_2^2}{{\sigma }_1^2/{\sigma }_2^2}=\frac{{\xi }_1/(n_1-1)}{{\xi }_2/(n_2-1)} \\ 从而证得:上述结论 \end{gathered}$$

${\sigma }_1={\sigma }_2$的t分布

$$\begin{gathered} T=\frac{\overline{X}-\overline{Y}-({\mu }_1-{\mu }_2)}{S_w\sqrt{\frac{n_1+n_2}{n_1{n}_2}}}\sim t(n_1+n_2-2) \\ {S}_w^2=\frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2} \end{gathered}$$