268 lines
15 KiB
TeX
268 lines
15 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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\section{最大似然估计与EM算法}
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\begin{enumerate}
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\questionandanswer[2]{
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设总体概率函数如下，$x_1,x_2, \cdots ,x_n$是样本，试求未知参数的最大似然估计。
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}{}
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\begin{enumerate}
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\questionandanswerSolution[]{
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$p(x;\theta)=c \theta^{c} x^{-(c+1)},x>\theta,\theta>0,c>0$已知；
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}{
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对数似然函数
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$$
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\begin{aligned}
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\ln L(\theta)&=\ln \prod_{i=1}^{n} p(x_i|\theta)=\ln \prod_{i=1}^{n} c \theta^{c} x_i^{-(c+1)} \\
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&= \sum_{i=1}^{n} \ln (c\theta^{c}x_i^{-(c+1)})=\sum_{i=1}^{n} (\ln c+c\ln \theta-(c+1)\ln x_i) \\
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&=n\ln c+nc\ln \theta-(c+1)\sum_{i=1}^{n} \ln x_i \\
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\end{aligned}
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$$
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只需要让$\theta$尽量大即可使似然函数取到最大值，又因为$\theta<x$，所以$\theta$的最大似然估计为$\hat{\theta}=x_{(1)}$。
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}
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\questionandanswerSolution[]{
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$p(x;\theta,\mu)=\displaystyle \frac{1}{\theta}e^{-\frac{x-\mu}{\theta}},x>\mu,\theta>0$；
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}{
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对数似然函数
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$$
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\begin{aligned}
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&\ln L(\theta,\mu)=\ln \prod_{i=1}^{n} p(x;\theta,\mu)=\ln \prod_{i=1}^{n} \frac{1}{\theta}e^{-\frac{x-\mu}{\theta}} \\
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&=\sum_{i=1}^{n} \ln \left( \frac{1}{\theta}e^{-\frac{x-\mu}{\theta}} \right) =\sum_{i=1}^{n} (-\ln \theta-\frac{x-\mu}{\theta}) \\
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&=-n\ln \theta - \frac{1}{\theta}\sum_{i=1}^{n} x_i+\frac{n\mu}{\theta} \\
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\end{aligned}
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$$
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对于$\mu$，由于$\ln L(\theta,\mu)$关于$\mu$是线性关系，所以只需要$\mu$尽量大即可使似然函数取到最大值，而$\mu<x$，所以$\hat{\mu}=x_{(1)}$。
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对于$\theta$，则需要求偏导，令
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$$
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\frac{\partial \ln L(\theta,\mu)}{\partial \theta}=-\frac{n}{\theta}+\frac{1}{\theta^{2}}\sum_{i=1}^{n} x_i-\frac{n\mu}{\theta^{2}}=0
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$$
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则可解得$\theta=\displaystyle \frac{1}{n}\sum_{i=1}^{n} x_i-\mu = \bar{x}-\mu$。此时$\ln L(\theta,\mu)$关于$\theta$最大。
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所以$\hat{\mu}=x_{(1)}$, $\hat{\theta}=\bar{x}-x_{(1)}$。
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}
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\questionandanswerSolution[]{
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$p(x;\theta)=(k\theta)^{-1}, \theta<x<(k+1)\theta, \theta>0,k>0$已知。
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}{
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对数似然函数
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$$
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\ln L(\theta)=\ln \prod_{i=1}^{n} (k\theta)^{-1}=\sum_{i=1}^{n} \ln (k\theta)^{-1}=\sum_{i=1}^{n} (-k\theta)=-nk\theta
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$$
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只要$\theta$尽量小即可使似然函数取得最大值。由于$\theta<x<(k+1)\theta$且$k>0$，所以$\frac{\theta}{k+1}<\frac{x}{k+1}<\theta$，所以$\theta$的最大似然估计为$\hat{\theta}=\dfrac{x_{(n)}}{k+1}$。
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}
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\end{enumerate}
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\questionandanswerSolution[4]{
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一地质学家为研究密歇根湖的湖滩地区的岩石成分，随机地自该地区取100个样品，每个样品有10块石子，记录了每个样品中属石灰石的石子数。假设这100次观察柜互独立，求这地区石子中石灰石的比例$p$的最大似然估计。该地质学家所得的数据如下：
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\begin{tabular}{c|ccccccccccc}
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样本中的石子数 & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\
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\hline
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样品个数 & 0 & 1 & 6 & 7 & 23 & 26 & 21 & 12 & 3 & 1 & 0 \\
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\end{tabular}
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}{
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当已知石灰石的比例为$p$时，并且如果每次抽样都是随机抽样，那么每个石子是石灰石的概率就是$p$，由于每个样品有10块石子，所以一次抽样服从二项分布$b(10,p)$，则概率函数为
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$$
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p(k;p)=\mathrm{C}_{10}^{k}p^{k}(1-p)^{10-k}
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$$
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设表格中的第一行为$x_i(i=0,1, \cdots ,10)$，第二行为$a_i(i=0,1, \cdots, 10)$，则对数似然函数为
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$$
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\begin{aligned}
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&\ln L(p)=\ln \prod_{i=1}^{n} \left( \mathrm{C}_{10}^{x_i} p^{x_i}(1-p)^{10-x_i} \right) ^{a_i} \\
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&=\sum_{i=1}^{n} a_i\left( \ln \mathrm{C}_{10}^{x_i}+x_i\ln p+(10-x_i)\ln (1-p) \right) \\
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&=\sum_{i=1}^{n} a_i \ln \mathrm{C}_{10}^{x_i} +\ln p \sum_{i=1}^{n} a_i x_i+\ln (1-p)\sum_{i=1}^{n} a_i(10-x_i) \\
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\end{aligned}
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$$
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令
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$$
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\frac{\mathrm{d}\ln L(p)}{\mathrm{d}p} = \frac{\sum_{i=1}^{n} a_i x_i}{p}-\frac{\sum_{i=1}^{n} a_i(10-x_i)}{1-p}=0
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$$
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解得
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$$
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p=\frac{\sum_{i=1}^{n} a_i x_i}{10 \sum_{i=1}^{n} a_i}= \frac{\sum_{i=1}^{n} a_i \frac{x_i}{10}}{\sum_{i=1}^{n} a_i}
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$$
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即以样品个数为权重，样品中石灰石比例的加权平均值。
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所以
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$$
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\hat{p}=\frac{\sum_{i=1}^{n} a_i x_i}{10 \sum_{i=1}^{n} a_i} = \frac{
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\begin{split}
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0\times 0+1\times 1+6\times 2+7\times 3+23\times 4+26\times 5 \\+21\times 6+12\times 7+3\times 8+1\times 9+0\times 10
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\end{split}
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}{10\times 100} = 0.499
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$$
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}
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\questionandanswerSolution[5]{
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在遗传学研究中经常要从截尾二项分布中抽样，其总体概率函数为
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$$
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p(X=k;p)=\frac{\displaystyle \binom{m}{k}p^{k}(1-p)^{m-k}}{1-(1-p)^{m}},\quad k=1,2, \cdots ,m
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$$
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若已知$m=2,x_1,x_2, \cdots ,x_n$是样本，试求$p$的最大似然估计。
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}{
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对数似然函数为
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$$
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\begin{aligned}
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\ln L(p)&= \ln \prod_{i=1}^{n} \frac{\displaystyle \binom{m}{x_i}p^{x_i}(1-p)^{m-x_i}}{1-(1-p)^{m}} \\
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&=\sum_{i=1}^{n} \left[ \ln \binom{m}{x_i}+x_i\ln p+(m-x_i)\ln (1-p)-\ln (1-(1-p)^{m}) \right] \\
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&=\sum_{i=1}^{n} \ln \binom{m}{x_i}+\ln p \sum_{i=1}^{n} x_i+\ln (1-p)\sum_{i=1}^{n} (m-x_i)-n\ln (1-(1-p)^{m}) \\
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\end{aligned}
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$$
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令
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$$
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\frac{\mathrm{d}\ln L(p)}{\mathrm{d}p}=\frac{\sum_{i=1}^{n} x_i}{p}-\frac{\sum_{i=1}^{n} (m-x_i)}{1-p}-n \frac{-m(1-p)^{m-1}}{1-(1-p)^{m}}=0
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% m=2
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% solve(latex2sympy(r"\frac{\sum_{i=1}^{n} x_i}{p}-\frac{\sum_{i=1}^{n} (m-x_i)}{1-p}-n \frac{-m(1-p)^{m-1}}{1-(1-p)^{m}}=0"), p)
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$$
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由于$m=2$，所以
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$$
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\frac{\sum_{i=1}^{n} x_i}{p}-\frac{\sum_{i=1}^{n} (2-x_i)}{1-p}+\frac{2n(1-p)}{1-(1-p)^{2}}=0
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% [ p = - \frac{\sqrt{(- 8 n x_{i} + 16 n + x_{i}^{2} \sum_{i=1}^{n} 1 - 8 x_{i} \sum_{i=1}^{n} 1 + 16 \sum_{i=1}^{n} 1) \sum_{i=1}^{n} 1}}{2 \cdot (2 n + 2 \sum_{i=1}^{n} 1)} + \frac{4 n + 2 \sum_{i=1}^{n} 2 + 2 \sum_{i=1}^{n} - x_{i} + 3 \sum_{i=1}^{n} x_{i}}{2 \cdot (2 n + \sum_{i=1}^{n} 2 + \sum_{i=1}^{n} - x_{i} + \sum_{i=1}^{n} x_{i})}, \ p = \frac{\sqrt{(- 8 n x_{i} + 16 n + x_{i}^{2} \sum_{i=1}^{n} 1 - 8 x_{i} \sum_{i=1}^{n} 1 + 16 \sum_{i=1}^{n} 1) \sum_{i=1}^{n} 1}}{2 \cdot (2 n + 2 \sum_{i=1}^{n} 1)} + \frac{4 n + 2 \sum_{i=1}^{n} 2 + 2 \sum_{i=1}^{n} - x_{i} + 3 \sum_{i=1}^{n} x_{i}}{2 \cdot (2 n + \sum_{i=1}^{n} 2 + \sum_{i=1}^{n} - x_{i} + \sum_{i=1}^{n} x_{i})}, \ n = \frac{- p (p - 1)^{2} (\sum_{i=1}^{n} x_{i} + \sum_{i=1}^{n} (2 - x_{i})) + p (\sum_{i=1}^{n} x_{i} + \sum_{i=1}^{n} (2 - x_{i})) + (p - 1)^{2} \sum_{i=1}^{n} x_{i} - \sum_{i=1}^{n} x_{i}}{2 p (p - 1)^{2}}]
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$$
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即
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$$
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\frac{n \bar{x}}{p}- \frac{2-n \bar{x}}{1-p}+\frac{2n(1-p)}{1-(1-p)^{2}}=0
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$$
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解得$p$的最大似然估计为
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$$
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\hat{p} = \frac{\bar{x} n + 4 n + 4}{4 (n + 1)} \pm \frac{\sqrt{\bar{x}^{2} n^{2} - 8 \bar{x} n^{2} - 8 \bar{x} n + 16 n + 16}}{4 (n + 1)}
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$$
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}
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\questionandanswerSolution[6]{
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已知在文学家萧伯纳的 "The Intelligent Woman's Guide to Socialism and Capitalism" 一书中 ，一个句子的单词数$X$近似地服从对数正态分布，即$Z=\ln X\sim N(\mu,\sigma^{2})$。今从该书中随机地取20个句子，这些句子中的单词数分别为
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$$
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52\quad24\quad15\quad67\quad15\quad22\quad63\quad26\quad16\quad32\quad7\quad33\quad28\quad14\quad7\quad29\quad10\quad6\quad59\quad30
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$$
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求该书中一个句子单词数均值$E(X)=e^{\mu+\frac{\sigma^{2}}{2}}$的最大似然估计。
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}{}
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{\kaishu
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根据题意，由于$Z=\ln X \sim N(\mu,\sigma^{2})$，可以将一个句子的单词数先取自然对数，此时即可使用正态分布的最大似然估计来估计$\mu$和$\sigma^{2}$。
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\begin{minted}[frame=single]{python}
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import numpy as np
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a = np.array([52,24,15,67,15,22,63,26,16,32,7,33,28,14,7,29,10,6,59,30])
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print(np.log(a))
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# [3.95124372 3.17805383 2.7080502 4.20469262 2.7080502 3.09104245
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# 4.14313473 3.25809654 2.77258872 3.4657359 1.94591015 3.49650756
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# 3.33220451 2.63905733 1.94591015 3.36729583 2.30258509 1.79175947
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# 4.07753744 3.40119738]
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print(np.mean(np.log(a)))
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# 3.0890326915239807
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print(np.var(np.log(a)))
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# 0.5081312851436304
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\end{minted}
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所以$\hat{\mu}\approx 3.0890326915239807$, $\widehat{(\sigma^{2})}\approx 0.5081312851436304$。
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再根据最大似然估计的不变性，直接计算$\displaystyle e^{\hat{\mu}+\frac{\widehat{(\sigma^{2})}}{2}}$。
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\begin{minted}[frame=single]{python}
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np.exp(np.mean(np.log(a)) + np.var(np.log(a)) / 2)
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# 28.306694575039742
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\end{minted}
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则该书中一个句子单词数均值$E(X)=e^{\mu+\frac{\sigma^{2}}{2}}$的最大似然估计约为$28.306694575039742$。
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}
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\questionandanswer[7]{
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设总体$X\sim U(\theta,2\theta)$，其中$\theta>0$是未知参数，$x_1,x_2, \cdots ,x_n$为取自该总体的样本，$\bar{x}$为样本均值。
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}{}
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\begin{enumerate}
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\questionandanswerProof[]{
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证明$\hat{\theta}=\dfrac{2}{3} \bar{x}$是参数$\theta$的无偏估计和相合估计；
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}{
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$$
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E \hat{\theta}=E \frac{2}{3} \bar{x}= E \frac{2}{3} \frac{1}{n}\sum_{i=1}^{n} x_i=\frac{2}{3} \frac{1}{n} \sum_{i=1}^{n} EX=\frac{2}{3} \frac{1}{n} n \frac{\theta+2\theta}{2}=\theta
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$$
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$$
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\operatorname{Var} \hat{\theta}=\operatorname{Var} \frac{2}{3} \bar{x}=\frac{2}{3} \frac{n \operatorname{Var}X}{n^{2}}=\frac{2\operatorname{Var}X}{3n} \xrightarrow{n \to \infty} 0
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$$
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所以$\hat{\theta} = \dfrac{2}{3} \bar{x}$是参数$\theta$的无偏估计和相合估计。
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}
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\questionandanswerSolution[]{
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求$\theta$的最大似然估计，它是无偏估计吗？是相合估计吗？
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}{
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$$
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\ln L(\theta)= \ln \prod_{i=1}^{n} 1_{[\theta,2\theta]}(x_i) \cdot \frac{1}{\theta}=\frac{1}{\theta} \sum_{i=1}^{n} \ln 1_{[\theta,2\theta]}(x_i)
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$$
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要使似然函数最大，则需要$\theta$尽量小，同时要满足$\theta\leqslant x_i\leqslant 2\theta$，即$\frac{\theta}{2}\leqslant \frac{x_i}{2}\leqslant \theta$，所以$\theta$的最大似然估计为$\hat{\theta}=\dfrac{x_{(n)}}{2}$。
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下面验证无偏性。
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$$
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E \hat{\theta}=\frac{1}{2} \int_{\theta}^{2\theta} x \frac{n}{\theta} \left( \frac{x-\theta}{\theta} \right) ^{n-1} \mathrm{d}x = \frac{\theta (2 n + 1)}{2 (n + 1)} \xrightarrow{n \to \infty} \theta
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$$
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所以$\hat{\theta}$不是无偏估计，但是是渐近无偏估计。
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下面验证相合性。
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$$
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E \hat{\theta}^{2} = \frac{1}{4} \int_{\theta}^{2\theta} x^{2} \frac{n}{\theta} \left( \frac{x-\theta}{\theta} \right) ^{n-1} \mathrm{d}x=\frac{\theta^{2} (n^{2} + 2 n + \frac{1}{2})}{n^{2} + 3 n + 2}
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$$
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$$
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\operatorname{Var}\hat{\theta} = E\hat{\theta}^{2} - (E \hat{\theta})^{2}=
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\frac{n \theta^{2}}{4 (n^{3} + 4 n^{2} + 5 n + 2)} \xrightarrow{n \to \infty} 0
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$$
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所以$\hat{\theta}$是相合估计。
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}
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\end{enumerate}
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\questionandanswer[8]{
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设$x_1,x_2, \cdots ,x_n$是来自密度函数为$p(x;\theta)=e^{-(x-\theta)},x>\theta$的总体的样本。
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}{}
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\begin{enumerate}
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\questionandanswerSolution[]{
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求$\theta$的最大似然估计$\hat{\theta}_1$，它是否是相合估计？是否是无偏估计？
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}{
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$$
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\ln L(\theta)= \ln \prod_{i=1}^{n} e^{-(x-\theta)}=\sum_{i=1}^{n} (-(x_i-\theta))= -\sum_{i=1}^{n} x_i+ n \theta
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$$
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要让似然函数最大，$\theta$要尽量大，同时$\theta<x$，所以$\theta$的最大似然估计为$\hat{\theta}=x_{(1)}$。
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|
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$\hat{\theta}=x_{(1)}$的概率函数为
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|
$$
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p(x)=n \left[1-\int_{\theta}^{x} e^{-(t-\theta)} \mathrm{d}t\right]^{n-1} e^{-(x-\theta)} = n (e^{\theta - x})^{n}
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|
$$
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|
则可以验证无偏性
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|
$$
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|
E \hat{\theta}_1= \int_{\theta}^{+\infty} x n(e^{\theta-x})^{n} \mathrm{d}x = \frac{1}{n} + \theta \xrightarrow{n \to \infty} \theta
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|
$$
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|
所以$\hat{\theta}_1$不是无偏估计，但是是渐近无偏估计。
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|
|
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|
下面验证相合性。
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|||
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|
$$
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|
E \hat{\theta}_1^{2}=\int_{\theta}^{+\infty} x^{2}n(e^{\theta-x})^{n} \mathrm{d}x=\frac{2}{n^{2}}+\frac{2}{n} \theta+\theta^{2}
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|
|
$$
|
|||
|
|
$$
|
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|
|
\operatorname{Var} \hat{\theta}_1=E \hat{\theta}_1^{2}-(E \hat{\theta})^{2}=\frac{2}{n^{2}}+\frac{2}{n}\theta+\theta^{2}- \left( \frac{1}{n}+\theta \right) ^{2} = \frac{1}{n^{2}} \xrightarrow{n \to \infty} 0
|
|||
|
|
$$
|
|||
|
|
所以$\hat{\theta}_1$是相合估计。
|
|||
|
|
}
|
|||
|
|
\questionandanswerSolution[]{
|
|||
|
|
求$\theta$的矩估计$\hat{\theta}_2$，它是否是相合估计？是否是无偏估计？
|
|||
|
|
}{
|
|||
|
|
$$
|
|||
|
|
EX=\int_{\theta}^{+\infty} x e^{-(x-\theta)} \mathrm{d}x = \theta + 1
|
|||
|
|
$$
|
|||
|
|
所以$\hat{\theta}_2=1- \bar{x}$。
|
|||
|
|
$$
|
|||
|
|
E \hat{\theta}_2=E(1-\bar{x})=1-EX=\theta
|
|||
|
|
$$
|
|||
|
|
所以$\hat{\theta}_2$是无偏估计。
|
|||
|
|
$$
|
|||
|
|
\operatorname{Var} \hat{\theta}_2=\operatorname{Var}(1-\bar{x})=\frac{\operatorname{Var}X}{n} \xrightarrow{n \to \infty} 0
|
|||
|
|
$$
|
|||
|
|
所以$\hat{\theta}_2$是相合估计。
|
|||
|
|
}
|
|||
|
|
\end{enumerate}
|
|||
|
|
\questionandanswerProof[10]{
|
|||
|
|
证明：对正态分布$N(\mu,\sigma^{2})$，若只有一个观测值，则$\mu,\sigma^{2}$的最大似然估计不存在。
|
|||
|
|
}{
|
|||
|
|
设此观测值为$x$，则似然函数为
|
|||
|
|
$$
|
|||
|
|
L(\mu, \theta)=\frac{1}{\sqrt{2\pi}} e^{-\left( \frac{x-\mu}{\sigma} \right) ^{2}}
|
|||
|
|
$$
|
|||
|
|
要使似然函数最大，则$\left( \frac{x-\mu}{\sigma} \right) ^{2}$应尽量小，则$\frac{(x-\mu)^{2}}{\sigma^{2}} \to 0$，所以$\mu =x, \sigma^{2}=\infty$，由于$\infty \not \in \mathbb{R}$，所以$\mu,\sigma^{2}$的最大似然估计不存在。
|
|||
|
|
}
|
|||
|
|
\end{enumerate}
|
|||
|
|
\end{document}
|