总体方差区间估计推导

作者: pdnbplus | 发布时间: 2024/07/14 | 阅读量: 174

总体方差区间估计推导 -- 潘登同学的数理统计笔记

来源

在做总体方差的区间估计的时候遇到了，对于来自正态总体的样本容量为n的简单随机样本，统计量服从卡方分布
(n−1)s2σ2∼χ2(n−1)\frac{(n-1)s^2}{\sigma^2} \sim \chi^2(n-1)
老师上课讲了相关推导，感觉比网上用什么正交矩阵推导的简单自然，做下记录

推导

由恒等式开始

(n−1)Sn2=(n−2)Sn−12+n−1n(xn−xˉn−1)2(n-1)S^2_n = (n-2)S^2_{n-1} + \frac{n-1}{n}(x_n-\bar{x}_{n-1})^2

证明:
左边=(n−1)Sn2=∑i=1n(xi−xˉn)2=∑i=1n−1(xi−xˉn−1+xˉn−1−xˉn)2+(xn−xˉn)2=∑i=1n−1(xi−xˉn−1)2+2∑i=1n−1(xi−xˉn−1)(xˉn−1−xˉn)+∑i=1n−1(xˉn−1−xˉn)2+(xn−xˉn)2=∑i=1n−1(xi−xˉn−1)2+∑i=1n−1(xˉn−1−xˉn)2+(xn−xˉn)2=(n−2)sn−12+(n−1)(xˉn−1−xˉn)2+(xn−xˉn)2=(n−2)sn−12+n−1n2(xˉn−1−xn)2+(n−1n)2(xn−xˉn−1)2=(n−2)sn−12+n−1n(xˉn−1−xn)2=右边\begin{aligned} 左边 &= (n-1)S^2_n = \sum_{i=1}^n(x_i-\bar{x}_n)^2 \\ &= \sum_{i=1}^{n-1}(x_i-\bar{x}_{n-1}+\bar{x}_{n-1}-\bar{x}_n)^2 + (x_n - \bar{x}_n)^2 \\ &= \sum_{i=1}^{n-1}(x_i-\bar{x}_{n-1})^2 +2\sum_{i=1}^{n-1}(x_i-\bar{x}_{n-1})(\bar{x}_{n-1}-\bar{x}_n) + \sum_{i=1}^{n-1}(\bar{x}_{n-1}-\bar{x}_n)^2 + (x_n - \bar{x}_n)^2 \\ &= \sum_{i=1}^{n-1}(x_i-\bar{x}_{n-1})^2 + \sum_{i=1}^{n-1}(\bar{x}_{n-1}-\bar{x}_n)^2 + (x_n - \bar{x}_n)^2 \\ &= (n-2)s_{n-1}^2 + {(n-1)}(\bar{x}_{n-1}-\bar{x}_n)^2 + (x_n - \bar{x}_n)^2 \\ &= (n-2)s_{n-1}^2 + \frac{n-1}{n^2}(\bar{x}_{n-1}-{x}_n)^2 + (\frac{n-1}{n})^2(x_n-\bar{x}_{n-1})^2 \\ &= (n-2)s_{n-1}^2 + \frac{n-1}{n}(\bar{x}_{n-1}-{x}_n)^2 = 右边 \end{aligned}

数学归纳法

假设
(n−1)s2σ2∼χ2(n−1)\frac{(n-1)s^2}{\sigma^2} \sim \chi^2(n-1)

当n=2时
s22=(x1−xˉ2)2+(x2−xˉ2)22−1=(x2−x1)22s_2^2 = \frac{(x_1-\bar{x}_2)^2 + (x_2-\bar{x}_2)^2}{2-1} = \frac{(x_2-x_1)^2}{2}
根据正态分布可加性
(x2−x1)∼N(0,2σ2)(x_2-x_1) \sim N(0,2\sigma^2)
根据χ\chi定义，是标准正态的平方和
(2−1)s22σ2=(x2−x12σ)2∼χ2(1)\frac{(2-1)s_2^2}{\sigma^2} = (\frac{x_2-x_1}{\sqrt{2}\sigma})^2 \sim \chi^2(1)

当n=k时，假设
(k−1)s2σ2∼χ2(k−1)\frac{(k-1)s^2}{\sigma^2} \sim \chi^2(k-1)

当n=k+1时
ksk+12σ2=(k−1)sk2+kk+1(xk+1−xˉk)2σ2=(k−1)sk2σ2+(xk+1−xˉk)2k+1kσ2\begin{aligned} \frac{ks_{k+1}^2}{\sigma^2} &= \frac{(k-1)s_k^2 +\frac{k}{k+1}(x_{k+1}-\bar{x}_k)^2}{\sigma^2} \\ &= \frac{(k-1)s_k^2}{\sigma^2} + \frac{(x_{k+1}-\bar{x}_k)^2}{\frac{k+1}{k}\sigma^2} \end{aligned}
根据抽样分布
xˉk∼N(0,σ2k)\bar{x}_k \sim N(0,\frac{\sigma^2}{k})
根据正态分布可加性
xk+1−xˉk∼N(0,k+1kσ2)x_{k+1}-\bar{x}_k \sim N(0,\frac{k+1}{k}\sigma^2)
(xk+1−xˉk)2k+1kσ2=(xk+1−xˉkk+1kσ)2∼χ2(1)\frac{(x_{k+1}-\bar{x}_k)^2}{\frac{k+1}{k}\sigma^2} = (\frac{x_{k+1}-\bar{x}_k}{\sqrt{\frac{k+1}{k}}\sigma})^2 \sim \chi^2(1)
而前面部分根据归纳假设
(k−1)sk2σ2∼χ2(k−1)\frac{(k-1)s_k^2}{\sigma^2} \sim \chi^2(k-1)
根据卡方分布定义
ksk+12σ2∼χ2(k−1)+χ2(1)∼χ2(k)\frac{ks_{k+1}^2}{\sigma^2} \sim \chi^2(k-1) + \chi^2(1) \sim \chi^2(k)

回到方差估计上

正是因为
(n−1)s2σ2∼χ2(n−1)\frac{(n-1)s^2}{\sigma^2} \sim \chi^2(n-1)
我们就能用卡方分布去估计总体方差σ\sigma

在给定置信度1−α1-\alpha下
P(χ1−α22≤(n−1)s2σ2≤χα22)=1−αP(\chi^2_{1-\frac{\alpha}{2}}\leq\frac{(n-1)s^2}{\sigma^2}\leq\chi^2_{\frac{\alpha}{2}}) = 1-\alpha
将σ\sigma调整到不等式中间
P((n−1)s2χα22≤σ2≤(n−1)s2χ1−α22)=1−αP(\frac{(n-1)s^2}{\chi^2_{\frac{\alpha}{2}}}\leq\sigma^2\leq\frac{(n-1)s^2}{\chi^2_{1-\frac{\alpha}{2}}}) = 1-\alpha
不等式作用两端的就是估计区间了...