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抽样分布

一、三大抽样分布

1. $\chi^{2}$分布

定义

设$X_{1},X_{2},\cdots,X_{n}$为来自总体$N(0,1)$的简单随机样本，则称统计量$X^{2}=X_{1}^{2}+X_{2}^{2}+\cdots +X_{n}^{2}$服从于自由度为$n$的$\chi^{2}$分布，记作$X^{2}\sim \chi^{2}(n)$

性质

• $\chi^{2}$分布的可加性：设$X\sim \chi^{2}(m),Y\sim \chi^{2}(n)$，且$X$与$Y$相互独立，则$X+Y\sim \chi^{2}(m+n)$
• 期望与方差：若$X\sim \chi^{2}(n)$，则$E(X)=n),D(X)=2n$

$\chi^{2}$分布的上分位点

对于给定的正数$\alpha(0<\alpha<1)$，若满足$P\{X>\chi^{2}_{\alpha}(n)\}=\alpha$，则称点$\chi_{\alpha}^{2}(n)$为$\chi^{2}(n)$分布的上$\alpha$分位点

2. $F$分布

定义

设$X\sim \chi^{2}(m),Y\sim \chi^{2}(n)$，且$X$与$Y$互相独立，则称随机变量$F=\frac{X/m}{Y/n}$服从自由度为$(m,n)$的$F$分布，记为$F\sim F(m,n)$

性质

若$F\sim F(m,n)$，则$\frac{1}{F}\sim F(n,m)$

$F$分布的上分位点

对于给定的正数$\alpha(0<\alpha<1)$，若满足$P\{F>F_{\alpha}(m,n)\}$，则称点$F_{\alpha}(m,n)$为$F(m,n)$分布的上$\alpha$分位点

3. $t$分布

定义

设$X\sim N(0,1),Y\sim \chi^{2}(n)$，且$X$与$Y$相互独立，则称随机变量$t=\frac{X}{\sqrt{Y/n}}$服从自由度为$n$的$t$分布，记作$t\sim t(n)$

$t$分布的上分位点

对于给定的正数$\alpha(0<\alpha<1)$，若满足$P\{t>t_{\alpha}(n)\}=\alpha$，则称点$t_{\alpha}(n)$为$t(n)$分布的上$\alpha$分位点

二、六大统计量

1. 单正态总体

设$X_{1},X_{2},\cdots,X_{n}$为来自总体$N(\mu,\sigma^{2})$的样本，$\bar{X}$为样本均值，$S^{2}$为样本方差

$\bar{X}=\frac{1}{n}\sum\limits^{n}_{i=1}X_{i},S^{2}=\frac{1}{n-1}(\sum\limits^{n}_{i=1}X^{2}_{i}-n\bar{X}^{2})$ $\begin{aligned}E(\bar{X})&=E(\frac{1}{n}\sum\limits^{n}_{i=1}X_{i})\\&=\frac{1}{n}E(X_{1}+\cdots+X_{n})\\&=\frac{1}{n}(EX_{1}+\cdots+EX_{n})\\&=\mu\end{aligned}$ $\begin{aligned}D(\bar{X})&=D(\frac{1}{n}\sum\limits^{n}_{i=1}X_{i})\\&=\frac{1}{n^{2}}D(X_{1}+\cdots+D_{n})\\&=\frac{1}{n^{2}}(DX_{1}+\cdots+DX_{n})\\&=\frac{\sigma^{2}}{n}\end{aligned}$ $\begin{aligned}E(S^{2})&=E[\frac{1}{n-1}(\sum\limits^{n}_{i=1}X_{i}^{2}-n\bar{X}^{2})]\\&=\frac{1}{n-1}[\sum\limits^{n}_{i=1}E(X_{i}^{2}-nE(\bar{X}^{2}))]\\&=\frac{1}{n-1}\{\sum\limits^{n}_{i=1}[DX_{i}+(EX_{i})^{2}]-n[D\bar{X}+(E\bar{X})^{2}]\}\\&=\frac{1}{n-1}[n(\sigma^{2}+\mu^{2})-n(\frac{\sigma^{2}}{n}+\mu^{2})]\\&=\sigma^{2}\end{aligned}$

1. $\begin{aligned}\frac{\sqrt{n}(\bar{X}-\mu)}{\sigma}\sim N(0,1)\end{aligned}$ $\because\bar{X}\sim N(\mu,\frac{\sigma^{2}}{n})$ $\therefore \frac{\bar{X}-\mu}{\sqrt{\frac{\sigma^{2}}{n}}}\sim N(0,1)$ 得证
2. $\begin{aligned}\frac{(n-1)S^{2}}{\sigma^{2}}\sim \chi^{2}(n-1)\end{aligned}$，且$\bar{X}$与$S^{2}$相互独立
3. $\begin{aligned}\frac{\bar{X}-\mu}{S /\sqrt{n}}\sim t(n-1)\end{aligned}$ $\because\frac{\sqrt{n}(\bar{X}-\mu)}{\sigma}\sim N(0,1),\frac{(n-1)S^{2}}{\sigma^{2}}\sim \chi^{2}(n-1)$ $\begin{aligned}\therefore\frac{\frac{\sqrt{n}(\bar{X}-\mu)}{\sigma}}{\sqrt{\frac{(n-1)S^{2}}{\sigma^{2}}}/(n-1)}\sim t(n-1)\end{aligned}$ 得证

2. 双正态总体

设$X_{1},X_{2},\cdots,X_{n1}$与$Y_{1},Y_{2},\cdots,Y_{n2}$分别是来自正态总体$N(\mu_{1},\sigma_{1}^{2})$和$N(\mu_{2},\sigma_{2}^{2})$的样本，且这两个样本相互独立，设$\bar{X}=\frac{1}{n_{1}}\sum\limits^{n_{1}}_{i=1}X_{i},\bar{Y}=\frac{1}{n_{2}}\sum\limits^{n_{2}}_{i=1}Y_{i}$分别是这两个样本的样本均值；$S_{1}^{2}=\frac{1}{n_{1}-1}\sum\limits^{n_{1}}_{i=1}(X_{i}-\bar{X})^{2},S_{2}^{2}=\frac{1}{n_{2}-1}\sum\limits^{n_{2}}_{i=1}(X_{i}-\bar{X})^{2}$

4. $\begin{aligned}\frac{(\bar{X}-\bar{Y})-(\mu_{1}-\mu_{2})}{\sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}}\sim N(0,1)\end{aligned}$ $\because \bar{X}\sim N(\mu_{1},\frac{\sigma_{1}^{2}}{n_{1}}),\bar{Y}\sim N(\mu_{2},\frac{\sigma_{2}^{2}}{n_{2}})$ $\therefore \bar{X}-\bar{Y}\sim N(\mu_{1}-\mu_{2},\frac{\sigma_{1}^{2}}{n_{1}}+\frac{\sigma_{2}^{2}}{n_{2}})$ $\begin{aligned}\therefore\frac{(\bar{X}-\bar{Y})-(\mu_{1}-\mu_{2})}{\sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}}\sim N(0,1)\end{aligned}$ 得证
5. $\begin{aligned}\frac{S^{2}_{1}/\sigma_{2}^{1}}{S^{2}_{2}/\sigma^{2}_{2}}\sim F(n_{1}-1,n_{2}-1)\end{aligned}$ $\begin{aligned}\because\frac{(n_{1}-1)S_{1}^{2}}{\sigma_{1}^{2}}\sim \chi^{2}(n_{1}-1),\frac{(n_{2}-1)S_{2}^{2}}{\sigma_{2}^{2}}\sim \chi^{2}(n_{2}-1)\end{aligned}$ $\begin{aligned}\therefore\frac{\frac{(n_{1}-1)S_{1}^{2}}{\sigma_{1}^{2}}/(n_{1}-1)}{\frac{(n_{2}-1)S_{2}^{2}}{\sigma_{2}^{2}}/(n_{2}-1)}\sim F(n_{1}-1,n_{2}-1)\end{aligned}$

得证 6. 若$\sigma_{1}^{2}=\sigma_{2}^{2}=\sigma^{2}$，则$\begin{aligned}\frac{(\bar{X}-\bar{Y})-(mu_{1}-\mu_{2})}{S_{w}\sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}}\sim t(n_{1}+n_{2}-2)\end{aligned}$，其中$\begin{aligned}S_{w}=\sqrt{\frac{(n_{1}-1)S_{1}^{2}+(n_{2}-1)S_{2}^{2}}{n_{1}+n_{2}-2}}\end{aligned}$ $\begin{aligned}\because\frac{(n_{1}-1)S_{1}^{2}}{\sigma_{1}^{2}}+\frac{(n_{2}-1)S_{2}^{2}}{\sigma_{2}^{2}}=\frac{(n_{1}-1)S_{1}^{2}+(n_{2}-1)S_{2}^{2}}{\sigma^{2}}\sim \chi^{2}(n_{1}+n_{2}-2)\end{aligned}$ 故$\begin{aligned}\frac{(\bar{X}-\bar{Y})-(\mu_{1}-\mu_{2})}{\sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}}\Big/\sqrt{\frac{(n_{1}-1)S_{1}^{2}+(n_{2}-1)S_{2}^{2}}{\sigma^{2}}\Big/(n_{1}+n_{2}-2)}\sim t(n_{1}+n_{2}-2)\end{aligned}$ 得证