297 lines
17 KiB
TeX
297 lines
17 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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\renewcommand{\bar}{\xoverline}
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\renewcommand{\hat}{\xwidehat}
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\setcounter{chapter}{6}
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\setcounter{section}{4}
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\begin{enumerate}
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\questionandanswerSolution[11]{
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设$x_1,x_2, \cdots ,x_m \overset{\text{i.i.d.}}{\sim} N(a, \sigma^{2}), y_1,y_2, \cdots y_n\overset{\text{i.i.d.}}{\sim}N(a,2\sigma^{2})$，求$a$和$\sigma^{2}$的UMVUE。
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}{
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根据贝叶斯估计的方法，$\hat{a}$和$\widehat{\sigma^{2}}$应为两个信息源的加权平均，权重为方差的倒数，即
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$$
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\hat{a}= \frac{\frac{1}{m\sigma^{2}}}{\frac{1}{m\sigma^{2}}+\frac{1}{2n\sigma^{2}}} \bar{x} + \frac{\frac{1}{2n\sigma^{2}}}{\frac{1}{m\sigma^{2}}+\frac{1}{2n\sigma^{2}}} \bar{y} = \frac{2 \bar{x} n + \bar{y} m}{m + 2 n}
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$$
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$$
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\widehat{\sigma^{2}}=\frac{\frac{1}{m\sigma^{2}}}{\frac{1}{m\sigma^{2}}+\frac{1}{2n\sigma^{2}}} s_{x}^{2} + \frac{\frac{1}{2n\sigma^{2}}}{\frac{1}{m\sigma^{2}}+\frac{1}{2n\sigma^{2}}} s_{y}^{2} = \frac{m s_{y}^{2} + 2 n s_{x}^{2}}{m + 2 n}
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$$
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对$0$的任一无偏估计$\varphi(x_1,x_2, \cdots ,x_m,y_1,y_2, \cdots ,y_n)$，$\operatorname{Cov}(\hat{a},\varphi)=0, \operatorname{Cov}(\widehat{\sigma^{2}},\varphi)=0$，所以$\hat{a}$和$\widehat{\sigma^{2}}$是UMVUE。
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}
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\questionandanswerProof[12]{
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设$x_1,x_2, \cdots ,x_n\overset{\text{i.i.d.}}{\sim}N(\mu,1)$，求$\mu^{2}$的UMVUE。证明此UMVUE，达不到C-R不等式的下界，即它不是有效估计。
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}{
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直观上来看，$\mu^{2}$的UMVUE应该是$\bar{x}^{2}$。接下来计算C-R不等式的下界，由于$I(\mu)=1$，所以C-R不等式的下界为
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$$
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\frac{[g'(\mu)]^{2}}{n I(\mu)}=\frac{(2\mu)^{2}}{n} = \frac{4\mu^{2}}{n}
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$$
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由于$\bar{x}\sim N(\mu, \frac{1}{n})$，所以$(n(\bar{x}-\mu)) \sim \chi^{2}(1)$
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$$
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\operatorname{Var} \bar{x}^{2} = E \bar{x}^{4} - (E \bar{x}^{2})^{2} =\text{实在是不会算了} > \frac{4\mu^{2}}{n}
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$$
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所以此UMVUE达不到C-R不等式的下界，即它不是有效估计。
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}
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\questionandanswer[14]{
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设$x_1,x_2, \cdots x_n$为独立同分布变量，$0<\theta<1$，
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$$
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P(x_1=-1)=\frac{1-\theta}{2}, \quad P(x_1=0)=\frac{1}{2}, \quad P(x_1=1)=\frac{\theta}{2}
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$$
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}{}
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\begin{enumerate}
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\questionandanswerSolution[]{
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求$\theta$的MLE $\hat{\theta}_1$并问$\hat{\theta}_1$是否是无偏的；
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}{
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设在$x_1,x_2, \cdots ,x_n$中有$n_{-1}$个$-1$，$n_{0}$个$0$，$n_1$个$1$，则对数极大似然函数为
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$$
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\begin{aligned}
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\ln L(n_{-1},n_{0},n_1;\theta)&=\ln \left[ \left( \frac{1-\theta}{2} \right) ^{n_{-1}} \left( \frac{1}{2} \right) ^{n_0} \left( \frac{\theta}{2} \right) ^{n_1} \right] \\
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&=n_{-1}\ln \left( \frac{1-\theta}{2} \right) +n_0 \ln \frac{1}{2}+n_1 \ln \frac{\theta}{2} \\
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\end{aligned}
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$$
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对$\theta$求偏导并令其为0
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$$
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\frac{\partial L}{\partial \theta}=-\frac{1}{2}\frac{2 n_{-1}}{1-\theta}+ \frac{1}{2}\frac{2n_1}{\theta} = \frac{n_1}{\theta}-\frac{n_{-1}}{1-\theta} = 0
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$$
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则最大似然估计为
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$$
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\hat{\theta}_1 = \frac{n_{1}}{n_{1} + n_{-1}}
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$$
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根据重期望公式，
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$$
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E \hat{\theta}_1 = E\left( E\left( \frac{n_1}{n_1+n_{-1}} \middle| n_1+n_{-1} \right) \right)
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$$
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其中
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$$
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E\left( \frac{n_1}{n_1+n_{-1}}\middle| n_1+n_{-1} \right) =E\left( \frac{n_1}{m}\middle| n_1+n_{-1}=m \right) = \frac{1}{m} \times m \frac{\frac{\theta}{2}}{\frac{1-\theta}{2}+\frac{\theta}{2}} = \theta
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$$
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所以$E \hat{\theta}_1 = E(\theta)= \theta$，即$\hat{\theta}_1$是无偏估计。
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}
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\questionandanswerSolution[]{
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求$\theta$的矩估计$\hat{\theta}_2$；
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}{
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设总体为$X$，则
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$$
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EX = -1 \times \frac{1-\theta}{2}+0\times \frac{1}{2}+1\times \frac{\theta}{2} = \theta - \frac{1}{2}
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$$
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所以矩估计$\hat{\theta}_2 = \bar{x}+\frac{1}{2}$。
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}
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\questionandanswerSolution[]{
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计算$\theta$的无偏估计的方差的C-R下界。
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}{
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$$
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p(x;\theta)=\begin{cases}
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\frac{1-\theta}{2},\quad & x=-1 \\
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\frac{1}{2},\quad & x=0 \\
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\frac{\theta}{2},\quad & x=1 \\
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0, \quad &\text{其他} \\
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\end{cases}, \quad \ln p(x;\theta)=\begin{cases}
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\ln (1-\theta)-\ln 2,\quad & x=-1 \\
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-\ln 2,\quad & x=0 \\
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\ln \theta-\ln 2,\quad & x=1 \\
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0,\quad & \text{其他} \\
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\end{cases},
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$$
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$$
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\frac{\partial \ln p(x;\theta)}{\partial \theta}=\begin{cases}
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-\frac{1}{1-\theta},\quad & x=-1 \\
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0,\quad & x=0 \\
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\frac{1}{\theta},\quad & x=1 \\
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0,\quad & \text{其他} \\
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\end{cases},\quad \left( \frac{\partial \ln p(x;\theta)}{\partial \theta} \right) ^{2} = \begin{cases}
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\frac{1}{(1-\theta)^{2}},\quad & x=-1 \\
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\frac{1}{\theta^{2}},\quad & x=1 \\
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0,\quad & \text{其他} \\
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\end{cases}
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$$
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所以
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$$
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I(\theta)=E\left( \frac{\partial \ln p(x;\theta)}{\partial \theta} \right) ^{2}=\frac{1}{(1-\theta)^{2}}\times \frac{1-\theta}{2}+\frac{1}{\theta^{2}}\times \frac{\theta}{2} = \frac{1}{2 \theta (1-\theta )}
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$$
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所以$\theta$的无偏估计的方差的C-R下界为
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$$
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\frac{1}{n I(\theta)}=\frac{2\theta(1-\theta)}{n}
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$$
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}
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\end{enumerate}
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\end{enumerate}
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\section{贝叶斯估计}
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\begin{enumerate}
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\questionandanswerSolution[2]{
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设总体为均匀分布$U(\theta,\theta+1)$，$\theta$的先验分布是$U(10,16)$。现有三个观测值：$11.7, 12.1, 12.0$。求$\theta$的后验分布。
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}{
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$$
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p(X|\theta)=\begin{cases}
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1^{3},\quad & \theta\in [11.1,11.7] \\
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0,\quad & \theta \not \in [11.1,11.7] \\
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\end{cases}=1_{[11.1,11.7]}(\theta), \quad \pi(\theta)=\frac{1}{6}1_{[10,16]}(\theta)
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$$
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所以$h(X,\theta)=p(X|\theta)\pi(\theta)=\frac{1}{6} 1_{[11.1,11.7]}(\theta)$,\quad $m(X)=\int_{-\infty}^{+\infty} \frac{1}{6}1_{[11.1,11.7]}(\theta) \mathrm{d}\theta = \frac{1}{6}\times 0.7$。
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所以$\theta$的后验分布为
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$$
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\pi(\theta|X)=\frac{h(X,\theta)}{m(X)}=\frac{1}{0.7} 1_{[11.1,11.7]}(\theta) = \frac{10}{7} 1_{[11.1,11.7]}(\theta)
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$$
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}
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\questionandanswer[3]{
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设$x_1,x_2, \cdots ,x_n$是来自几何分布的样本，总体分布列为
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$$
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P(X=k|\theta)=\theta(1-\theta)^{k}, \quad k=0,1,2, \cdots ,
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$$
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$\theta$的先验分布是均匀分布$U(0,1)$。
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}{}
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\begin{enumerate}
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\questionandanswerSolution[]{
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求$\theta$的后验分布；
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}{
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$$
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p(\theta|x_1,x_2, \cdots ,x_n)=\frac{p(x_1,x_2, \cdots ,x_n|\theta)\pi(\theta)}{\int_{0}^{1} p(x_1,x_2, \cdots ,x_n|\theta)\pi(\theta) \mathrm{d}\theta} = \frac{\prod_{i=1}^{n} \left[ \theta(1-\theta)^{x_i} \right] 1_{[0,1]}(\theta)}{\int_{0}^{1} \prod_{i=1}^{n} \left[ \theta(1-\theta)^{x_i} \right] \mathrm{d}\theta}
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$$
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}
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\questionandanswerSolution[]{
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若$4$次观测值为$4,3,1,6$，求$\theta$的贝叶斯估计。
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}{
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$$
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E(\theta|4,3,1,6) = \int_{0}^{1} \theta p(\theta|4,3,1,6) \mathrm{d}\theta = \text{实在算不出来了}
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$$
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}
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\end{enumerate}
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\questionandanswerProof[5]{
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验证：正态总体方差（均值已知）的共轭先验分布是倒伽马分布（称$X$服从倒伽马分布，如果$\frac{1}{X}$服从倒伽马分布。
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}{
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设总体$X\sim N(\mu,\sigma^{2})$，且$\sigma^{2}\sim \operatorname{IG}(\alpha,\gamma)$，则$\frac{1}{\sigma^{2}}\sim \operatorname{Ga}(\alpha,\lambda)$，所以
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$$
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h(X|\sigma^{2})=p(X|\sigma^{2})p(\sigma^{2})= \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{1}{2}\sum_{i=1}^{n} \left( \frac{x-\mu}{\sigma} \right) ^{2}} \cdot \frac{\lambda^{\alpha}}{\Gamma(\alpha)}\left( \frac{1}{\sigma^{2}} \right) ^{\alpha-1} e^{-\frac{1}{\sigma^{2}}}
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$$
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$$
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p(\sigma^{2}|X)=\frac{p(X|\sigma^{2})p(\sigma^{2})}{\int_{0}^{+\infty} p(X|\sigma^{2})p(\sigma^{2}) \mathrm{d}\sigma^{2}} = \frac{\frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{1}{2}\sum_{i=1}^{n} \left( \frac{x-\mu}{\sigma} \right) ^{2}} \cdot \frac{\lambda^{\alpha}}{\Gamma(\alpha)}\left( \frac{1}{\sigma^{2}} \right) ^{\alpha-1} e^{-\frac{1}{\sigma^{2}}}}{\int_{0}^{+\infty} \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{1}{2}\sum_{i=1}^{n} \left( \frac{x-\mu}{\sigma} \right) ^{2}} \cdot \frac{\lambda^{\alpha}}{\Gamma(\alpha)}\left( \frac{1}{\sigma^{2}} \right) ^{\alpha-1} e^{-\frac{1}{\sigma^{2}}} \mathrm{d}\sigma^{2}}
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$$
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计算可得$p(\sigma^{2}|X)$也是倒伽马分布的概率密度函数，因此$\sigma^{2}$的后验分布也是倒伽马分布，从而正态总体方差（均值已知）的共轭先验分布是倒伽马分布。
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}
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\questionandanswer[6]{
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设$x_1,x_2, \cdots ,x_n$是来自如下总体的一个样本
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$$
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p(x|\theta) = \frac{2x}{\theta^{2}}, \quad 0<x<\theta
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$$
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}{}
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\begin{enumerate}
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\questionandanswerSolution[]{
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若$\theta$的先验分布为均匀分布$U(0,1)$，求$\theta$的后验分布；
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}{
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$$
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\begin{aligned}
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&h(x_1,x_2, \cdots ,x_n,\theta)=P(x_1,x_2, \cdots ,x_n|\theta)\pi(\theta) \\
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=&\prod_{i=1}^{n} \frac{2 x_i}{\theta^{2}} 1_{[0,\theta]}(x_i) 1_{[0,1]}(\theta) =1_{0<x_{(1)}}1_{x_{(n)}<\theta} \frac{1}{\theta^{2n}} \prod_{i=1}^{n} 2
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x_i 1_{[0,1]}(\theta) \\
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\end{aligned}
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$$
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$$
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\begin{aligned}
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&m(x_1,x_2, \cdots x_n)=\int_{0}^{1} h(x_1,x_2, \cdots ,x_n,\theta) \mathrm{d}\theta = 1_{0<x_{(1)}} \prod_{i=1}^{n} 2 x_i \int_{0}^{1} 1_{x_{(n)}<\theta} \frac{1}{\theta^{2n}} \mathrm{d}\theta \\
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=&1_{0<x_{(1)}} \prod_{i=1}^{n} 2 x_i \int_{x_{(n)}}^{1} \frac{1}{\theta^{2n}} \mathrm{d}x = 1_{0<x_{(1)}} \left(-2n+1-(-2n+1)x_{(n)}^{-2n+1}\right) \prod_{i=1}^{n} 2 x_i \\
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\end{aligned}
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$$
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所以$\theta$的后验分布为
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$$
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\pi(\theta|x_1, \cdots ,x_n)=\frac{h(x_1, \cdots ,x_n,\theta)}{m(x_1, \cdots ,x_n)} = \frac{1_{x_{(n)}<\theta} 1_{[0,1]}(\theta)}{\theta^{2n} \left(-2n+1-(-2n+1)x_{(n)}^{-2n+1}\right)}
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$$
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}
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\questionandanswerSolution[]{
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若$\theta$的先验分布为$\pi(\theta)=3 \theta^{2}, 0<\theta<1$，求$\theta$的后验分布。
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}{
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$$
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\begin{aligned}
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&h(x_1,x_2, \cdots ,x_n,\theta)=P(x_1,x_2, \cdots ,x_n|\theta)\pi(\theta) \\
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=&\prod_{i=1}^{n} \frac{2 x_i}{\theta^{2}} 1_{[0,\theta]}(x_i) 3\theta^{2}1_{[0,1]}(\theta) =1_{0<x_{(1)}}1_{x_{(n)}<\theta} \frac{3\theta^{2}}{\theta^{2n}} \left(\prod_{i=1}^{n} 2 x_i\right) 1_{[0,1]}(\theta) \\
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\end{aligned}
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$$
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$$
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\begin{aligned}
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&m(x_1,x_2, \cdots x_n)=\int_{0}^{1} h(x_1,x_2, \cdots ,x_n,\theta) \mathrm{d}\theta = 1_{0<x_{(1)}} \prod_{i=1}^{n} 2 x_i \int_{0}^{1} 1_{x_{(n)}<\theta} \frac{3\theta^{2}}{\theta^{2n}} \mathrm{d}\theta \\
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=&1_{0<x_{(1)}} \prod_{i=1}^{n} 2 x_i \int_{x_{(n)}}^{1} \frac{3\theta^{2}}{\theta^{2n}} \mathrm{d}x = 1_{0<x_{(1)}} \left(9-6n-(9-6n)x_{(n)}^{3-2n}\right) \prod_{i=1}^{n} 2 x_i \\
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\end{aligned}
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$$
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所以$\theta$的后验分布为
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$$
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\pi(\theta|x_1, \cdots ,x_n)=\frac{h(x_1, \cdots ,x_n,\theta)}{m(x_1, \cdots ,x_n)} = \frac{1_{x_{(n)}<\theta} 1_{[0,1]}(\theta)}{\theta^{2n} \left(9-6n-(9-6n)x_{(n)}^{3-2n}\right)}
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$$
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}
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\end{enumerate}
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\questionandanswer[8]{
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设$x_1,x_2, \cdots ,x_n$是来自均匀分布$U(0,\theta)$的样本，$\theta$的先验分布是帕雷托分布，其密度函数为$\displaystyle \pi(\theta)=\frac{\beta\theta_0^{\beta}}{\theta^{\beta+1}}, \theta>\theta_0$，其中$\beta,\theta_0$是两个已知的常数。
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}{}
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\begin{enumerate}
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\questionandanswerProof[]{
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验证：帕雷托分布是$\theta$的共轭先验分布；
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}{
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令$X=\{ x_1,x_2, \cdots ,x_n \}$，则 $P(X|\theta)=\prod_{i=1}^{n} \frac{1}{\theta} 1_{[0,\theta]}(x_i)=\frac{1}{\theta^{n}}1_{x_{(1)}\geqslant 0} 1_{x_{(n)}\leqslant \theta}$
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$$
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h(X,\theta)=P(X|\theta)\pi(\theta)=\frac{\beta \theta_0^{\beta}}{\theta^{\beta+1+n}} 1_{x_{(1)}\geqslant 0} 1_{x_{(n)}\leqslant \theta}
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$$
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$$
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m(X)=\int_{x_{(n)}}^{+\infty} h(X,\theta) \mathrm{d}\theta= \beta \theta_0^{\beta} 1_{x_{(1)}\geqslant 0} \int_{x_{(n)}}^{+\infty} \theta^{-\beta-1-n} \mathrm{d}\theta = \frac{\beta \theta_0^{\beta} 1_{x_{(1)}\geqslant 0}}{\beta+n} x_{(n)}^{-\beta-n}
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$$
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% 所以
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$$
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P(\theta|X)=\frac{h(X,\theta)}{m(X)}=\frac{\frac{1_{x_{(n)}\leqslant \theta}}{\theta^{\beta+n+1}}}{\frac{x_{(n)}^{-\beta-n}}{\beta+n}} = \frac{(\beta+n)x_{(n)}^{\beta+n}}{\theta^{\beta+n-1}} 1_{x_{(n)}\leqslant \theta}
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$$
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所以$\theta$的后验分布为参数为$\beta+n$和$x_{(n)}$的帕雷托分布，从而帕雷托分布是$\theta$的共轭先验分布。
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}
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\questionandanswerSolution[]{
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求$\theta$的贝叶斯估计。
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}{
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$\theta$的贝叶斯估计为
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$$
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\begin{aligned}
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\hat{\theta} = \int_{x_{(n)}}^{+\infty} \theta p(\theta|X) \mathrm{d}\theta = \int_{x_{(n)}}^{+\infty} \frac{\theta (\beta+n) x_{(n)}^{\beta+n}}{\theta^{\beta+n+1}} \mathrm{d}\theta = \frac{\beta+n}{\beta+n-1} x_{(n)}
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\end{aligned}
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$$
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}
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\end{enumerate}
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\questionandanswerProof[12]{
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从正态总体$N(\theta,2^{2})$中随机抽取容量为$100$的样本，又设$\theta$的先验分布为正态分布，证明：不管先验分布的标准差为多少，后验分布的标准差一定小于$\frac{1}{5}$。
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}{
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设样本为$X$，$\theta$的先验分布为$N(\mu,\sigma^{2})$，则$\theta$的后验概率密度函数为
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$$
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\begin{aligned}
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&\pi(\theta|X) = c f(X|\theta) f(\theta) \\
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&=c \left( \prod_{i=1}^{n} \frac{1}{2\sqrt{2\pi}} e^{-\frac{1}{2} \left( \frac{x_i-\theta}{2} \right) ^{2}} \right) \cdot \frac{1}{\sqrt{2\pi} \sigma} e^{-\frac{1}{2} \left( \frac{\theta-\mu}{\sigma} \right) ^{2}} \\
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&=c e^{-\frac{1}{2} \left( \frac{\theta-\mu}{\sigma} \right) ^{2} - \frac{1}{2} \sum_{i=1}^{n} \left( \frac{x_i-\theta}{2} \right) ^{2}} \\
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&\geqslant ce^{-\frac{1}{2} \cdot 25 (\theta-\mu-\bar{x})^{2}} \\
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\end{aligned}
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$$
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所以后验分布的标准差一定小于$\frac{1}{5}$。
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}
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\questionandanswerProof[13]{
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设随机变量$X$服从负二项分布，其概率分布为
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$$
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f(x|p)=\binom{x-1}{k-1} p^{k} (1-p)^{x-k}, \quad x=k,k+1, \cdots
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$$
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证明其成功概率$p$的共轭先验分布族为贝塔分布族。
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}{
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设$X=\{ x_1,x_2, \cdots ,x_n \}$。设$p$的先验分布为贝塔分布$Be(a,b)$，则$\pi(p)=\frac{1}{B(a,b)} p^{a-1}(1-p)^{b-1}$，所以
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$$
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\begin{aligned}
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P(p|X)&= c \cdot h(X,p)=c \cdot P(X|p)\pi(p) =c \left(\prod_{i=1}^{n} \mathrm{C}_{x_i-1}^{k-1} p^{k} (1-p)^{x_i-k}\right) \frac{1}{B(a,b)}p^{a-1} (1-p)^{b-1} \\
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&=c p^{nk} (1-p)^{-nk} (1-p)^{\sum_{i=1}^{n} x_i} p^{a-1} (1-p)^{b-1} \\
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&=c p^{nk+a-1} (1-p)^{\sum_{i=1}^{n} x_i-nk+b-1} \\
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\end{aligned}
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|
$$
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其中$c$为与$p$无关的数。
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所以$p$的后验分布为$\displaystyle Be(nk+a, \sum_{i=1}^{n} x_i -nk +b)$，从而$p$的共轭先验分布族为贝塔分布族。
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}
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|
\questionandanswerSolution[14]{
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|
从一批产品中抽检$100$个，发现$3$个不合格，假定该产品不合格率$\theta$的先验分布为贝塔分布$Be(2,200)$，求$\theta$的后验分布。
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|
}{
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|
设总体为$X$，则$X\sim b(100, \theta)$，所以
|
|||
|
|
$$
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|||
|
|
\begin{aligned}
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|||
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|
P(\theta|X) &=c\cdot P(X|\theta) \pi(\theta) = c\cdot \mathrm{C}_{100}^{3} \theta^{3} (1-\theta)^{97} \frac{1}{B(2,200)} \theta^{1} (1-\theta)^{199} \\
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|
&=c \cdot \theta^{4} (1-\theta)^{296} \\
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|||
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|
\end{aligned}
|
|||
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|
$$
|
|||
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|
其中$c$为与$\theta$无关的数。
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|||
|
|
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|||
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|
所以$\theta$的后验分布为$Be(5,297)$。
|
|||
|
|
}
|
|||
|
|
\end{enumerate}
|
|||
|
|
\end{document}
|