区间估计

Confidence Interval Estimation
设总体 $X$ 含有一个待估计参数 $θ$
给定常数 $α (0 < α < 1)$ , 找出统计量 $θ_{1} θ_{2}$ 使得：

\begin{array}{r} P {θ_{1} \leq θ \leq θ_{2}} = 1 - α 0 < α < 1 \end{array}

则称 $[θ_{1}, θ_{2}]$ 为参数 $θ$ 的置信度为 $1 - α$ 的置信区间

\begin{array}{r} P {| \hat{θ} - θ | \leq δ} = 1 - α \end{array}

对于对称分布的正态分布和 t 分布: $Z_{1 - α} = - Z_{α}$
所以： $Z_{1 - α / 2} = - Z_{α / 2}$

正态总体： $X \sim N (μ, σ^{2})$
$\overset{―}{X} \sim N (μ, \frac{σ^{2}}{n})$

\begin{array}{r} Z = U = \frac{\overset{―}{X} - μ}{\frac{σ}{\sqrt{n}}} \sim N (0, 1) \end{array}

\begin{aligned} P {| \frac{\overset{―}{X} - μ}{σ / \sqrt{n}} | \leq z_{\frac{α}{2}}} & = 1 - α \end{aligned}

分位点 $z_{α / 2}$

置信度为 $1 - α$ 的参数 $μ$ 的置信区间

\begin{array}{r} [\overset{―}{X} - \frac{δ}{\sqrt{n}} z_{\frac{α}{2}}, \overset{―}{X} + \frac{δ}{\sqrt{n}} z_{\frac{α}{2}}] \end{array}

置信区间长度 $δ = \frac{σ}{\sqrt{n}} z_{\frac{α}{2}}$

\begin{array}{r} T = \frac{\overset{―}{X} - μ}{\frac{S}{\sqrt{n}}} \sim t (n - 1) \end{array}

分位点 $t_{α / 2} (n - 1)$

\begin{array}{r} P {| \frac{\overset{―}{X} - μ}{S / \sqrt{n}} | \leq t_{α / 2} (n - 1)} = 1 - α \end{array}

置信区间

\begin{array}{r} [\overset{―}{X} - \frac{S}{\sqrt{n}} t_{α / 2} (n - 1), \overset{―}{X} + \frac{S}{\sqrt{n}} t_{α / 2} (n - 1)] \end{array}

\begin{aligned} χ^{2} & = \frac{\sum_{i = 1}^{n} (X_{i} - μ)^{2}}{σ^{2}} \sim χ^{2} (n) \end{aligned}

分位数：
$χ_{1 - α / 2}^{2}$ $χ_{α / 2}^{2}$

\begin{array}{r} P {χ_{1 - α / 2}^{2} (n) \leq χ^{2} \leq χ_{α / 2}^{2} (n)} = 1 - α \end{array}

\begin{array}{r} P {\frac{\sum_{i = 1}^{n} (X_{i} - μ_{0})^{2}}{χ_{α / 2}^{2} (n)} \leq σ^{2} \leq \frac{\sum_{i = 1}^{n} (X_{i} - μ_{0})^{2}}{χ_{1 - α / 2}^{2} (n)}} = 1 - α \end{array}

\begin{array}{r} \frac{(n - 1) S^{2}}{σ^{2}} \sim χ^{2} (n - 1) \end{array}

分位数
$χ_{1 - α / 2}^{2} (n - 1)$ $χ_{α / 2}^{2} (n - 1)$

\begin{array}{r} P {χ_{1 - α / 2}^{2} (n - 1) \leq \frac{(n - 1) S^{2}}{σ^{2}} \leq χ_{α / 2}^{2} (n - 1)} = 1 - α \end{array}

\begin{array}{r} P {\frac{(n - 1) S^{2}}{χ_{α / 2}^{2} (n - 1)} \leq σ^{2} \leq \frac{(n - 1) S^{2}}{χ_{1 - α / 2}^{2} (n - 1)}} \end{array}

在实际工作生活中，要根据样本分析两个不同正态总体之间的差异
$X \sim N (μ_{1}, σ_{1}^{2}), Y \sim N (μ_{2}, σ_{2}^{2})$

$\overset{―}{X} = \frac{1}{n_{1}} \sum_{i = 1}^{n_{1}} X_{i} \overset{―}{Y} = \frac{1}{n_{2}} \sum_{i = 1}^{n_{2}} Y_{i}$

\begin{aligned} S_{1}^{2} & = \frac{1}{n_{1} - 1} \sum_{i = 1}^{n_{1}} (X_{i} - \overset{―}{X})^{2} \\ S_{2}^{2} & = \frac{1}{n_{2} - 1} \sum_{i = 1}^{n_{2}} (Y_{i} - \overset{―}{Y})^{2} \end{aligned}

\begin{aligned} U & = \frac{(\overset{―}{X} - \overset{―}{Y}) - (μ_{1} - μ_{2})}{\sqrt{\frac{σ_{1}^{2}}{n_{1}} + \frac{σ_{2}^{2}}{n_{2}}}} \\ \sim N (0, 1) \end{aligned}

\begin{array}{r} P {| U | \leq u_{α / 2}} = 1 - α \end{array}

\begin{array}{r} [(\overset{―}{X} - \overset{―}{Y}) - u_{α / 2} \sqrt{\frac{σ_{1}^{2}}{n_{1}} + \frac{σ_{2}^{2}}{n_{2}}}, (\overset{―}{X} - \overset{―}{Y}) + u_{α / 2} \sqrt{\frac{σ_{1}^{2}}{n_{1}} + \frac{σ_{2}^{2}}{n_{2}}}] \end{array}

\begin{aligned} T & = \frac{\overset{―}{X} - \overset{―}{Y} - (μ_{1} - μ_{2})}{S_{w} \sqrt{\frac{1}{n_{1}} + \frac{1}{n_{2}}}} \\ \sim t (n_{1} + n_{2} - 2) \end{aligned}

\begin{array}{r} S_{w}^{2} = \frac{(n_{1} - 1) S_{1}^{2} + (n_{2} - 1) S_{2}^{2}}{n_{1} + n_{2} - 2} \end{array}

\begin{array}{r} P {| T | \leq t_{α / 2} (n_{1} + n_{2} - 2)} = 1 - α \end{array}

$X \sim N (μ_{1}, σ_{1}^{2}), Y \sim N (μ_{2}, σ_{2}^{2})$ 均值方差均未知

\begin{array}{r} F = \frac{S_{1}^{2} / S_{2}^{2}}{σ_{1}^{2} / σ_{2}^{2}} \sim F (n_{1} - 1, n_{2} - 1) \end{array}

分位数
$F_{1 - α / 2} (n_{1} - 1, n_{2} - 1)$ $F_{α / 2} (n_{1} - 1, n_{2} - 1)$

置信区间：

\begin{array}{r} [\frac{S_{1}^{2} / S_{2}^{2}}{F_{α / 2} (n_{1} - 1, n_{2} - 1)}, \frac{S_{1}^{2} / S_{2}^{2}}{F_{1 - α / 2} (n_{1} - 1, n_{2} - 1)}] \end{array}