299 lines
13 KiB
TeX
299 lines
13 KiB
TeX
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\documentclass[全部作业]{subfiles}
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\input{mysubpreamble}
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\begin{document}
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\setcounter{chapter}{5}
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\setcounter{section}{2}
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\section{统计量及其分布}
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\begin{enumerate}
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\questionandanswerSolution[1]{
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在一本书上我们随机地检查了10页，发现每页上的错误数为：
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$$
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4 \quad 5 \quad 6 \quad 0 \quad 3 \quad 1 \quad 4 \quad 2 \quad 1 \quad 4
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$$
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试计算其样本均值、样本方差和样本标准差。
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}{
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$$
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\bar{x}=\frac{4+5+6+0+3+1+4+2+1+4}{10} = 3
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$$
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$$
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s^{2}=\frac{1}{10-1}\sum_{i=1}^{10}(x_i-\bar{x})^{2}=\frac{34}{9} \approx 3.778
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$$
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$$
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s=\sqrt{\frac{34}{9}}\approx 1.944
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$$
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}
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\questionandanswerProof[2]{
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证明：对任意常数$c,d$，有
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$$
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\sum_{i=1}^{n}(x_i-c)(y_i-d)=\sum_{i=1}^{n}(x_i-\bar{x})(y_i-\bar{y})+n(\bar{x}-c)(\bar{y}-d)
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$$
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}{
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根据性质$\displaystyle \sum_{i=1}^{n}x_i=\sum_{i=1}^{n}\bar{x},\ \sum_{i=1}^{n}y_i=\sum_{i=1}^{n}\bar{y}$可得
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$$
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\begin{aligned}
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\text{右边}&=\sum_{i=1}^{n}(x_i-\bar{x})(y_i-\bar{y})+\sum_{i=1}^{n}(\bar{x}-c)(\bar{y}-d) \\
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&=\sum_{i=1}^{n}\left[ (x_i-\bar{x})(y_i-\bar{y})+(\bar{x}-c)(\bar{y}-d) \right] \\
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&=\sum_{i=1}^{n}(x_i y_i -\bar{x}y_i-\bar{y}x_i+\bar{x}\bar{y}+\bar{x}\bar{y}-\bar{x}d-\bar{y}c+cd) \\
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% &=\sum_{i=1}^{n}[(x_i-\bar{x})y_i - (x_i-\bar{x})\bar{y}+(\bar{x}-c)\bar{y}-(\bar{x}-c)d] \\
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% =\sum_{i=1}^{n}[(x_i-\bar{x}+\bar{x}-c)\bar{y}]
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&=\sum_{i=1}^{n}x_i y_i-\bar{x}\sum_{i=1}^{n}y_i -\bar{y}\sum_{i=1}^{n}x_i+n \bar{x}\bar{y} +n \bar{x}\bar{y}-n \bar{x}d- n\bar{y}c+ ncd \\
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&=\sum_{i=1}^{n}x_i y_i- n \bar{x}\bar{y} -n \bar{y}\bar{x}+n \bar{x}\bar{y}+n \bar{x}\bar{y}- \sum_{i=1}^{n}x_i d - \sum_{i=1}^{n}y_i c+\sum_{i=1}^{n}cd \\
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&=\sum_{i=1}^{n}(x_i y_i-x_id-y_ic+cd) \\
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&=\sum_{i=1}^{n}(x_i-c)(y_i-d) = \text{左边} \\
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\end{aligned}
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$$
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}
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\questionandanswerSolution[3]{
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设$x_1,x_2, \cdots ,x_n$和$y_1,y_2, \cdots ,y_n$是两组样本观测值，且有如下关系：
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$$
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y_i=3 x_i-4, i=1,2, \cdots ,n
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$$
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试求样本均值$\bar{x}$和$\bar{y}$间的关系以及样本方差$s_{x}^{2}$和$s_{y}^{2}$间的关系。
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}{
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$$
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\bar{y}=\frac{1}{n}\sum_{i=1}^{n}y_i=\frac{1}{n}\sum_{i=1}^{n}(3 x_i-4)=3\cdot \frac{1}{n}\sum_{i=1}^{n} x_i -4=3\bar{x}-4
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$$
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$$
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\begin{aligned}
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s_{y}^{2}&=\frac{1}{n-1}\sum_{i=1}^{n} (y_i-\bar{y})^{2}=\frac{1}{n-1}\sum_{i=1}^{n} (3 x_i-4-(3 \bar{x}-4))^{2}=\frac{1}{n-1}\sum_{i=1}^{n} [3(x_i-\bar{x})]^{2} \\
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&=9 \cdot \frac{1}{n-1}\sum_{i=1}^{n} (x_i-\bar{x})^{2}=9 s_{x}^{2} \\
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\end{aligned}
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$$
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}
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\questionandanswerProof[5]{
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从同一总体中抽取两个容量分别为$n,m$的样本，样本均值分别为$\bar{x}_1, \bar{x}_2$，样本方差分别为$s_1^{2}, s_2^{2}$，将两组样本合并，其均值、方差分别为$\bar{x}, s^{2}$，证明：
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$$
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\bar{x}=\frac{n \bar{x}_1+m \bar{x}_2}{n+m}
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$$
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$$
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s^{2}=\frac{(n-1)s_1^{2}+(m-1)s_2^{2}}{n+m-1}+\frac{nm(\bar{x}_1-\bar{x}_2)^{2}}{(n+m)(n+m+1)}
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$$
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}{
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$$
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\bar{x}=\frac{1}{n+m}\left( \sum_{i=1}^{n} x_{1_{i}} +\sum_{i=1}^{m} x_{2_{i}} \right) =\frac{n \bar{x}_1+m \bar{x}_2}{n+m}
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$$
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$$
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\begin{aligned}
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s^{2}&=\frac{1}{n+m-1} \left( \sum_{i=1}^{n} (x_{1i}-\bar{x})^{2}+\sum_{j=1}^{m} \left( x_{2j}-\bar{x} \right) ^{2} \right) \\
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&=\frac{1}{n+m-1}\left( \sum_{i=1}^{n} \left( x_{1i}-\frac{n \bar{x}_1+m \bar{x}_2}{n+m} \right) ^{2}+\sum_{i=1}^{n} \left( x_{2j}-\frac{n \bar{x}_1+m \bar{x}_2}{n+m} \right) ^{2} \right) \\
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&=\frac{(n-1)s_1^{2}+(m-1)s_2^{2}}{n+m-1}+\frac{nm(\bar{x}_1-\bar{x}_2)^{2}}{(n+m)(n+m+1)} \\
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\end{aligned}
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$$
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}
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\questionandanswerSolution[8]{
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设$x_1,x_2, \cdots ,x_n$是来自$U(-1,1)$的样本，试求$E(\bar{x})$和$\operatorname{Var}(\bar{x})$。
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}{
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设随机变量$X \sim U(-1,1)$，则
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$$
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E(\bar{x})=EX=\frac{-1+1}{2}=0
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$$
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$$
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\operatorname{Var}(\bar{x})=\frac{\operatorname{Var}(X)}{n}=\frac{\frac{(-1-1)^{2}}{12}}{n}=\frac{1}{3n}
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$$
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}
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\questionandanswerProof[9]{
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设总体二阶矩存在，$x_1,x_2, \cdots ,x_n$是样本，证明$x_i-\bar{x}$与$x_j-\bar{x}\ (i\neq j)$的相关系数为$-(n-1)^{-1}$。
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}{
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根据样本均值的性质，$E(x_i-\bar{x})=E (x_j- \bar{x})=0$。
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设随机变量$X$表示从总体中抽出的一个样本，则$EX^{2}$存在。
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$$
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E(x_i-\bar{x})(x_j-\bar{x})=E(x_i x_j - \bar{x} x_i - \bar{x} x_j + \bar{x}^{2})= E x_i x_j - \bar{x}E x_i - \bar{x} E x_j + \bar{x}^{2}
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$$
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将$x_i$与$x_j$看作独立的两次抽样，则$x_i,x_j\overset{\text{i.i.d.}}{\sim}X $，所以$E x_i x_j=E x_i E x_j=(EX)^{2},$\\
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$E x_i=EX, E x_j=EX$。
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所以
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$$
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E(x_i-\bar{x})(x_j-\bar{x})=(EX)^{2}-2 \bar{x}EX + \bar{x}^{2}=\frac{1}{1-n}=-(n-1)^{-1}
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$$
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}
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\questionandanswerProof[10]{
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设$x_1,x_2, \cdots ,x_n$为一个样本，$\displaystyle s^{2}=\frac{1}{n-1}\sum_{i=1}^{n} (x_i-\bar{x})^{2}$是样本方差，试证：
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$$
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\frac{1}{n(n-1)}\sum_{i<j}(x_i-x_j)^{2} =s^{2}
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$$
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}{
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$$
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\begin{aligned}
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% s^{2}= \frac{1}{n-1}\sum_{i=1}^{n} (x_i-\bar{x})^{2}=
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&\frac{1}{n(n-1)}\sum_{i<j}(x_i-x_j)^{2}=\frac{1}{n(n-1)} \sum_{i<j}(x_i-\bar{x}+\bar{x}-x_j)^{2} \\
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&=\frac{1}{n(n-1)}\sum_{i<j} [(x_i-\bar{x})^{2}+2(x_i-\bar{x})(\bar{x}-x_j)+(\bar{x}-x_j)^{2}] \\
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&=\frac{1}{n(n-1)}\cdot \frac{1}{2}\sum_{i=1,2, \cdots ,n;j=1,2, \cdots ,n} [(x_i-\bar{x})^{2}+2(x_i-\bar{x})(\bar{x}-x_j)+(\bar{x}-x_j)^{2}] \\
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&=\frac{1}{2n(n-1)}\left[ n \sum_{i=1}^{n} (x_i-\bar{x})^{2}+0+n\sum_{j=1}^{n} (x_j-\bar{x})^{2} \right] \\
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&=\frac{1}{n-1}\sum_{i=1}^{n} (x_i-\bar{x})^{2} = s^{2} \\
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\end{aligned}
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$$
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}
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\questionandanswerProof[11]{
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设总体4阶中心距$\nu_4=E[x-E(x)]^{4}$存在，试证：对样本方差$\displaystyle s^{2}=\frac{1}{n-1} \sum_{i=1}^{n} (x_i-\bar{x})^{2}$，有
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$$
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\operatorname{Var}(s^{2})=\frac{n(\nu-\sigma^{4})}{(n-1)^{2}}-\frac{2(\nu_4-2\sigma^{4})}{(n-1)^{2}}+\frac{\nu-3\sigma^{4}}{n(n-1)^{2}}
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$$
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其中$\sigma^{2}$为总体$X$的方差。
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}{
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$$
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\begin{aligned}
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&\text{右边}=\frac{n^{2}\nu_4-n^{2}\sigma^{4}-2n\nu_4+4n\sigma^{4}+\nu_4-3\sigma^{4}}{n(n-1)^{2}} \\
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&=\frac{\nu_4(n^{2}-2n+1)-\sigma^{4}(n^{2}-4n+3)}{n(n-1)^{2}} \\
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&=\frac{\nu_4(n-1)^{2}-\sigma^{4}(n-1)(n-3)}{n(n-1)^{2}} \\
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&=\frac{\nu_4}{n}-\frac{\sigma^{4}(n-3)}{n(n-1)} \\
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\end{aligned}
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$$
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$$
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\begin{aligned}
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\text{左边}=E(s^{2})^{2}-(Es ^{2})^{2}=Es ^{4}-(Es ^{2})^{2}=E s^{4}-\sigma^{4}
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\end{aligned}
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$$
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实在证明不出来了。
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}
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\questionandanswerProof[12]{
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设总体$X$的3阶矩存在，若$x_1,x_2, \cdots ,x_n$是取自该总体的简单随机样本，$\bar{x}$为样本均值，$s^{2}$为样本方差，试证：$\operatorname{Cov}(\bar{x}, s^{2})=\dfrac{\nu_3}{n}$，其中$\nu_3=E[x-E(x)]^{3}$。
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}{
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$$
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E \bar{x}=EX, Es ^{2}=\operatorname{Var}X
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$$
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$$
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E(\bar{x} s^{2})=E\left( \frac{1}{n}\sum_{i=1}^{n} x_i+\frac{1}{n-1}\sum_{i=1}^{n} (x_i-\bar{x})^{2} \right)
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$$
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$$
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\operatorname{Var}\bar{x}=\frac{\operatorname{Var}X}{n}, \operatorname{Var}s ^{2}=\operatorname{Var}\left( \frac{1}{n-1}\sum_{i=1}^{n} (x_i-\bar{x})^{2} \right)
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$$
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也证明不出来了。
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}
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\questionandanswerSolution[15]{
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从指数总体$\operatorname{Exp}(\frac{1}{\theta})$抽取了40个样品，试求$\bar{x}$的渐近分布。
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}{
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设随机变量$X$表示从总体中抽出的一个样本，则
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$$
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EX=\frac{1}{\frac{1}{\theta}}=\theta,\ \operatorname{Var}X=\frac{1}{\left( \frac{1}{\theta} \right) ^{2}}=\theta^{2}
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$$
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所以$\bar{x}$的渐近分布为$N(\theta, \theta^{2})$。
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}
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\questionandanswerSolution[17]{
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设$x_1,x_2, \cdots x_{20}$是从二点分布$b(1,p)$抽取的样本，试求样本均值$\bar{x}$的渐近分布。
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}{
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设随机变量$X$表示从总体中抽出的一个样本，则
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$$
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EX=p,\ \operatorname{Var}X=p(1-p)
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$$
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所以$\bar{x}$的渐近分布为$N(p, p(1-p))$。
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}
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\questionandanswerSolution[23]{
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设总体$X$服从几何分布，即$P(X=k)=pq^{k-1}, k=1,2, \cdots $，其中$0<p<1,q=1-p,\\ x_1,x_2, \cdots ,x_n$为该总体的样本，求$x_{(n)}, x_{(1)}$的概率分布。
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}{
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设总体$X$的概率密度函数为$p(x)$，分布函数为$F(x)$，则
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$$
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F(x)=\sum_{k=1}^{\left\lfloor x \right\rfloor} pq^{k-1}=\frac{p-pq^{\left\lfloor x \right\rfloor}}{1-q}
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$$
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$$
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p_{x_{(n)}}(x)=\frac{n!}{(n-1)!}[F(x)]^{n-1}p(x)=n pq^{\left\lfloor x \right\rfloor-1}\left[ \frac{p-pq^{\left\lfloor x \right\rfloor}}{1-q} \right] ^{n-1}
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$$
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$$
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p_{x_{(1)}}=\frac{n!}{(n-1)!}[1-F(x)]^{n-1}p(x)=npq^{\left\lfloor x \right\rfloor-1}\left[ 1-\frac{p-pq^{\left\lfloor x \right\rfloor}}{1-q} \right] ^{n-1}
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$$
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}
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\questionandanswer[28]{
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设总体$X$的分布函数$F(x)$是连续的，$x_{(1)},x_{(2)}, \cdots ,x_{(n)}$为取自此总体的次序统计量，设$\eta_i=F(x_{(i)})$，试证：
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}{}
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\begin{enumerate}
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\questionandanswerProof[-]{
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\item $\eta_1\leqslant \eta_2\leqslant \cdots\leqslant \eta_n$，且$\eta_i$是来自均匀分布$U(0,1)$总体的次序统计量。
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}{
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因为$x_{(1)}\leqslant x_{(2)}\leqslant \cdots\leqslant x_{(n)}$且$F(x)$单调，$\eta_i=F(x_{(i)})$，所以$\eta_1\leqslant \eta_2\leqslant \cdots\leqslant \eta_n$。
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}
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\questionandanswerProof[-]{
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\item $\displaystyle E(\eta_i)=\frac{i}{n+1}, \ \operatorname{Var}(\eta_i)=\frac{i(n+1-i)}{(n+1)^{2}(n+2)},1\leqslant i\leqslant n$
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}{
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设总体的概率密度函数为$p(x)$，则$\eta_i$的分布函数为
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$$
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p_{(i)}(x)=\frac{n!}{(i-1)!(n-i)!}[F(x)]^{i-1}[1-F(x)]^{n-i}p(x)
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$$
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$$
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\begin{aligned}
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E(\eta_i)&=\sum_{i=1}^{n} F(x_{(i)})p_{(i)}(x_{(i)}) \\
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&=\sum_{i=1}^{n} F(x_{(i)}) \frac{n!}{(i-1)!(n-i)!}[F(x_{(i)})]^{i-1}[1-F(x_{(i)})]^{n-i}p(x) \\
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&=\sum_{i=1}^{n} \frac{n!}{(i-1)!(n-i)!}[F(x_{(i)})]^{i}[1-F(x_{(i)})]^{n-i}p(x) \\
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\end{aligned}
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$$
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$$
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\operatorname{Var}(\eta_i)=
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$$
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实在是不会了。
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}
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\questionandanswerProof[-]{
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\item $\eta_i$和$\eta_j$的协方差矩阵为
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$
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\begin{bmatrix}
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\frac{a_1(1-a_1)}{n+2} & \frac{a_1(1-a_2)}{n+2} \\
|
|||
|
|
\frac{a_1(1-a_2)}{n+2} & \frac{a_2(1-a_2)}{n+2} \\
|
|||
|
|
\end{bmatrix}
|
|||
|
|
$
|
|||
|
|
，其中$\displaystyle a_1=\frac{i}{n+1}, a_2=\frac{j}{n+1}$。
|
|||
|
|
}{
|
|||
|
|
$$
|
|||
|
|
E(\eta_i)=\frac{i}{n+1},\ E(\eta_j)=\frac{j}{n+1},\ E(\eta_i \eta_j)=
|
|||
|
|
$$
|
|||
|
|
实在是不会了。
|
|||
|
|
}
|
|||
|
|
\end{enumerate}
|
|||
|
|
\questionandanswerProof[32]{
|
|||
|
|
设总体$X$的密度函数为
|
|||
|
|
$
|
|||
|
|
p(x)=\begin{cases}
|
|||
|
|
3x^{2},\quad & 0<x<1, \\
|
|||
|
|
0,\quad & \text{其他}, \\
|
|||
|
|
\end{cases}
|
|||
|
|
$
|
|||
|
|
，$x_{(1)}\leqslant x_{(2)}\leqslant \cdots\leqslant x_{(5)}$为容量为5的取自此总体的次序统计量，试证$\dfrac{x_{(2)}}{x_{(4)}}$与$x_{(4)}$相互独立。
|
|||
|
|
}{
|
|||
|
|
根据相互独立的定理，需要证明$\displaystyle \forall x,y,\ p_{\frac{x_{(2)}}{x_{(4)}}}(x)\cdot p_{x_{(4)}}(y)=p_{\frac{x_{(2)}}{x_{(4)}},x_{(4)}}(x,y)$，之后就不会了。
|
|||
|
|
}
|
|||
|
|
\questionandanswer[35]{
|
|||
|
|
对下列数据构造箱线图：\\
|
|||
|
|
472 \quad
|
|||
|
|
425 \quad
|
|||
|
|
447 \quad
|
|||
|
|
377 \quad
|
|||
|
|
341 \quad
|
|||
|
|
369 \quad
|
|||
|
|
412 \quad
|
|||
|
|
419 \quad
|
|||
|
|
400 \quad
|
|||
|
|
382 \quad
|
|||
|
|
366 \quad
|
|||
|
|
425 \quad
|
|||
|
|
399 \quad
|
|||
|
|
398 \quad
|
|||
|
|
423 \quad
|
|||
|
|
384 \quad
|
|||
|
|
418 \quad
|
|||
|
|
392 \quad
|
|||
|
|
372 \quad
|
|||
|
|
418 \quad
|
|||
|
|
374 \quad
|
|||
|
|
385 \quad
|
|||
|
|
439 \quad
|
|||
|
|
428 \quad
|
|||
|
|
429 \quad
|
|||
|
|
428 \quad
|
|||
|
|
430 \quad
|
|||
|
|
413 \quad
|
|||
|
|
405 \quad
|
|||
|
|
381 \quad
|
|||
|
|
403 \quad
|
|||
|
|
479 \quad
|
|||
|
|
381 \quad
|
|||
|
|
443 \quad
|
|||
|
|
441 \quad
|
|||
|
|
433 \quad
|
|||
|
|
419 \quad
|
|||
|
|
379 \quad
|
|||
|
|
386 \quad
|
|||
|
|
387 \quad
|
|||
|
|
}{
|
|||
|
|
\begin{center}
|
|||
|
|
\includegraphics[width=0.5\linewidth]{imgs/5.3.35.png}
|
|||
|
|
\end{center}
|
|||
|
|
}
|
|||
|
|
\end{enumerate}
|
|||
|
|
\end{document}
|