216 lines
11 KiB
TeX
216 lines
11 KiB
TeX
|
\documentclass[全部作业]{subfiles}
|
|||
|
|
\input{mysubpreamble}
|
|||
|
|
\begin{document}
|
|||
|
|
\renewcommand{\bar}{\xoverline}
|
|||
|
|
\renewcommand{\hat}{\xwidehat}
|
|||
|
|
\setcounter{chapter}{6}
|
|||
|
|
\section{点估计的概念与无偏性}
|
|||
|
|
\begin{enumerate}
|
|||
|
|
\questionandanswerProof[3]{
|
|||
|
|
设$\hat{\theta}$是参数$\theta$的无偏估计，且有$\operatorname{Var}(\hat{\theta})>0$，试证$(\hat{\theta})^{2}$不是$\theta^{2}$的无偏估计。
|
|||
|
|
}{
|
|||
|
|
由题意可知$E\hat{\theta}=\theta$，$\operatorname{Var}\hat{\theta}=E\hat{\theta}^{2}-(E\hat{\theta})^{2}>0$，所以$E\hat{\theta}^{2}>(E\hat{\theta})^{2}=\theta^{2}$，所以$\hat{\theta}^{2}$不是$\theta^{2}$的无偏估计。
|
|||
|
|
}
|
|||
|
|
\questionandanswerSolution[4]{
|
|||
|
|
设总体$X\sim N(\mu,\sigma^{2}), x_1,x_2, \cdots ,x_{n}$是来自该总体的一个样本。试确定常数$c$使$\displaystyle c\sum_{i=1}^{n-1} (x_{i+1}-x_{i})^{2}$为$\sigma^{2}$的无偏估计。
|
|||
|
|
}{
|
|||
|
|
$$
|
|||
|
|
\begin{aligned}
|
|||
|
|
Ec\sum_{i=1}^{n-1} (x_{i+1}-x_{i})^{2}&=Ec\sum_{i=1}^{n-1} (x_{i+1}^{2}-2x_{i}x_{i+1} + x_i^{2}) \\
|
|||
|
|
&=c\sum_{i=1}^{n-1} Ex_{i+1}^{2}-2c\sum_{i=1}^{n-1} Ex_{i}x_{i+1}+c\sum_{i=n}^{n-1} Ex_{i}^{2} \\
|
|||
|
|
\end{aligned}
|
|||
|
|
$$
|
|||
|
|
因为总体$X\sim N(\mu,\sigma)$，所以$\forall i=1,2, \cdots ,n$ ，$Ex_{i}=\mu, \operatorname{Var}x_i=Ex_{i}^{2}-(Ex_{i})^{2}=\sigma^{2}$，从而$Ex_{i}^{2}=\sigma^{2}+\mu^{2}$。由于$x_i$与$x_{i+1}$独立，所以$Ex_{i}x_{i+1}=Ex_{i}\cdot Ex_{i+1}=\mu^{2}$。所以
|
|||
|
|
$$
|
|||
|
|
\begin{aligned}
|
|||
|
|
\text{上式}&=c \sum_{i=1}^{n-1} (\sigma^{2}+\mu^{2})-2c\sum_{i=1}^{n-1} \mu^{2}+c\sum_{i=1}^{n-1} (\sigma^{2}+\mu^{2}) \\
|
|||
|
|
&=2c(n-1)(\sigma^{2}+\mu^{2})-2c(n-1)\mu^{2} \\
|
|||
|
|
&=2c(n-1)\sigma^{2} \\
|
|||
|
|
\end{aligned}
|
|||
|
|
$$
|
|||
|
|
所以当$\displaystyle c=\frac{1}{2(n-1)}$时，$\displaystyle c\sum_{i=1}^{n-1} (x_{i+1}-x_{i})^{2}$为$\sigma^{2}$的无偏估计。
|
|||
|
|
}
|
|||
|
|
\questionandanswerProof[5]{
|
|||
|
|
设$x_1,x_2, \cdots ,x_n$是来自下列总体的简单样本，
|
|||
|
|
$$
|
|||
|
|
p(x,\theta)=\begin{cases}
|
|||
|
|
1,\quad & \theta-\frac{1}{2}\leqslant x\leqslant \theta+\frac{1}{2} \\
|
|||
|
|
0,\quad & \text{其他} \\
|
|||
|
|
\end{cases}\quad -\infty<\theta<\infty
|
|||
|
|
$$
|
|||
|
|
证明样本均值$\bar{x}$及$\frac{1}{2}(x_{(1)}+x_{(n)})$都是$\theta$的无偏估计，问何者更有效？
|
|||
|
|
}{
|
|||
|
|
$E \bar{x}=\theta$，$\operatorname{Var}\bar{x}=\frac{1}{n}\times \frac{1}{12}=\frac{1}{12n}$。
|
|||
|
|
|
|||
|
|
而$E \frac{1}{2}(x_{(1)}+x_{(n)}) $为样本中最小值和最大值的平均，虽然计算不出，但理论上也应该是$\theta$。但是似乎不像样本均值一样覆盖了样本全部的信息，所以应该是$\operatorname{Var}\bar{x}\leqslant \operatorname{Var} \frac{1}{2}(x_{(1)}+x_{(n)})$，即$\bar{x}$更有效。
|
|||
|
|
}
|
|||
|
|
\questionandanswerSolution[9]{
|
|||
|
|
设有$k$台一起，已知用第$i$台仪器测量的标准差为$\sigma_i(i=1,2, \cdots ,k)$。用这些仪器独立地对某一物理量$\theta$各观察一次，分别得到$x_1,x_2, \cdots ,x_k$，设仪器都没有系统偏差。问$a_1,a_2, \cdots ,a_k$应取何值，方能使$\displaystyle \hat{\theta}=\sum_{i=1}^{k} a_i x_i$ 成为$\theta$的无偏估计，且方差达到最小？
|
|||
|
|
}{
|
|||
|
|
$$
|
|||
|
|
E \hat{\theta}=E\sum_{i=1}^{k} a_i x_i= \sum_{i=1}^{k} a_i E x_i=\theta \sum_{i=1}^{k} a_i=\theta \Longrightarrow \sum_{i=1}^{k} a_i=1
|
|||
|
|
$$
|
|||
|
|
$$
|
|||
|
|
\operatorname{Var} \hat{\theta}=\operatorname{Var} \sum_{i=1}^{k} a_i x_i=\sum_{i=1}^{k} a_i^{2} \operatorname{Var}x_i=\sum_{i=1}^{k} a_i^{2} \sigma_i^{2}
|
|||
|
|
$$
|
|||
|
|
所以原问题可以转化为
|
|||
|
|
$$
|
|||
|
|
\mathop{\arg\min}_{a_i}
|
|||
|
|
\quad \sum_{i=1}^{k} a_i^{2}\sigma_i^{2}
|
|||
|
|
\ \ ,\quad \text{s.t.}
|
|||
|
|
\ \ \sum_{i=1}^{k} a_i=1
|
|||
|
|
$$
|
|||
|
|
|
|||
|
|
对此可以使用拉格朗日乘数法。
|
|||
|
|
令
|
|||
|
|
$$
|
|||
|
|
f(a_1, \cdots ,a_n, \lambda)=\sum_{i=1}^{k} a_i^{2}\sigma_i^{2}+\lambda \left( \sum_{i=1}^{k} a_i - 1 \right)
|
|||
|
|
$$
|
|||
|
|
则
|
|||
|
|
$$
|
|||
|
|
\begin{cases}
|
|||
|
|
\forall i=1,2, \cdots ,k,\quad f'_{a_i}=2 a_i \sigma_i^{2}+\lambda=0 \\
|
|||
|
|
f'_{\lambda}=\sum_{i=1}^{k} a_i - 1=0 \\
|
|||
|
|
\end{cases}
|
|||
|
|
$$
|
|||
|
|
解得
|
|||
|
|
$$
|
|||
|
|
\begin{cases}
|
|||
|
|
\forall i=1,2, \cdots ,k, \quad a_i=\displaystyle \frac{\frac{1}{2\sigma_i^{2}}}{\sum_{i=1}^{k} \frac{1}{2\sigma_i^{2}}} \\
|
|||
|
|
\lambda=\displaystyle -\frac{1}{\sum_{i=1}^{k} \frac{1}{2\sigma_i^{2}}} \\
|
|||
|
|
\end{cases}
|
|||
|
|
$$
|
|||
|
|
|
|||
|
|
所以$\forall i=1,2, \cdots ,k, \quad a_i=\displaystyle \frac{\frac{1}{2\sigma_i^{2}}}{\sum_{i=1}^{k} \frac{1}{2\sigma_i^{2}}}$，方能使$\displaystyle \hat{\theta}=\sum_{i=1}^{k} a_i x_i$ 成为$\theta$的无偏估计，且方差达到最小。
|
|||
|
|
}
|
|||
|
|
\questionandanswerSolution[11]{
|
|||
|
|
设总体$X$服从正态分布$N(\mu,\sigma^{2})$，$x_1,x_2, \cdots ,x_n$为来自总体$X$的样本，为了得到标准差$\sigma$的估计量，考虑统计量：
|
|||
|
|
$$
|
|||
|
|
y_1=\frac{1}{n}\sum_{i=1}^{n} \left\vert x_i-\bar{x} \right\vert ,\quad \bar{x}=\frac{1}{n}\sum_{i=1}^{n} x_i, \quad n\geqslant 2
|
|||
|
|
$$
|
|||
|
|
$$
|
|||
|
|
y_2=\frac{1}{n(n-1)} \sum_{i=1}^{n} \sum_{j=1}^{n} \left\vert x_i-x_j \right\vert ,\quad n\geqslant 2
|
|||
|
|
$$
|
|||
|
|
求常数$C_1$与$C_2$，使得$C_1y_1$与$C_2y_2$都是$\sigma$的无偏估计。
|
|||
|
|
}{
|
|||
|
|
由于$\forall i,j=1,2, \cdots ,n(i\neq j), \quad x_i\sim N(\mu,\sigma^{2}),x_j\sim N(\mu,\sigma^{2}), \bar{x}\sim N(\mu,\frac{\sigma^{2}}{n})$且它们应该相互独立。
|
|||
|
|
所以
|
|||
|
|
$$
|
|||
|
|
x_i-\bar{x}\sim N(0, \sigma^{2}+\frac{\sigma^{2}}{n}), \quad x_i-x_j\sim N(0, 2\sigma^{2})
|
|||
|
|
$$
|
|||
|
|
因为$Y\sim N(0,\sigma^{2})$时$E\left\vert Y \right\vert =\sigma \sqrt{\frac{2}{\pi}}$，所以
|
|||
|
|
$$
|
|||
|
|
E\left\vert x_i-\bar{x} \right\vert =\sqrt{\sigma^{2}+\frac{\sigma^{2}}{n}}\sqrt{\frac{2}{\pi}}=\sigma\sqrt{1+\frac{1}{n}}\sqrt{\frac{2}{\pi}}, \quad E\left\vert x_i-x_j \right\vert = \sqrt{2}\sigma\sqrt{\frac{2}{\pi}}=\frac{2\sigma}{\sqrt{\pi}}
|
|||
|
|
$$
|
|||
|
|
所以
|
|||
|
|
$$
|
|||
|
|
EC_1y_1=C_1\frac{1}{n}\times n \cdot \sigma\sqrt{1+\frac{1}{n}}\sqrt{\frac{2}{\pi}}=\sigma \Longrightarrow C_1=\frac{1}{\sqrt{1+\frac{1}{n}}\sqrt{\frac{2}{\pi}}} = \sqrt{\frac{n\pi}{2n+2}}
|
|||
|
|
$$
|
|||
|
|
$$
|
|||
|
|
EC_2y_2=C_2 \frac{1}{n(n-1)} (n^{2}-n) \cdot \frac{2\sigma}{\sqrt{\pi}}=\sigma \Longrightarrow C_2=\frac{\sqrt{\pi}}{2}
|
|||
|
|
$$
|
|||
|
|
|
|||
|
|
所以
|
|||
|
|
$$
|
|||
|
|
C_1=\sqrt{\frac{n\pi}{2n+2}}, \quad C_2=\frac{\sqrt{\pi}}{2}
|
|||
|
|
$$
|
|||
|
|
}
|
|||
|
|
\end{enumerate}
|
|||
|
|
\section{矩估计及相合性}
|
|||
|
|
\begin{enumerate}
|
|||
|
|
\questionandanswer[3]{
|
|||
|
|
设总体分布列如下，$x_1,x_2, \cdots ,x_n$是样本，试求未知参数的矩估计：
|
|||
|
|
}{}
|
|||
|
|
\begin{enumerate}
|
|||
|
|
\questionandanswerSolution[]{
|
|||
|
|
$P(X=k)=\frac{1}{N}, k=0,1,2, \cdots ,N-1$，$N$（正整数）是未知参数；
|
|||
|
|
}{
|
|||
|
|
$$
|
|||
|
|
EX = \sum_{k=0}^{N-1} k \cdot \frac{1}{N}=\frac{1}{N}\cdot \frac{N(N-1)}{2}=\frac{N-1}{2}
|
|||
|
|
$$
|
|||
|
|
所以$N=2EX+1$，所以$N$的矩估计为
|
|||
|
|
$$
|
|||
|
|
\hat{N}=2 \bar{x}+1
|
|||
|
|
$$
|
|||
|
|
}
|
|||
|
|
\questionandanswerSolution[]{
|
|||
|
|
$P(X=k)=(k-1)\theta^{2}(1-\theta)^{k-2},\quad k=2,3, \cdots ,\quad 0<\theta<1$。
|
|||
|
|
}{
|
|||
|
|
$\displaystyle
|
|||
|
|
EX=\sum_{k=2}^{\infty} k(k-1)\theta^{2}(1-\theta)^{k-2} = \frac{2}{\theta}
|
|||
|
|
$
|
|||
|
|
，所以$\theta=\dfrac{2}{EX}$，所以$\theta$的矩估计为
|
|||
|
|
$\displaystyle
|
|||
|
|
\hat{\theta}=\frac{2}{\bar{x}}
|
|||
|
|
$。
|
|||
|
|
}
|
|||
|
|
\end{enumerate}
|
|||
|
|
\questionandanswer[4]{
|
|||
|
|
设总体密度函数如下，$x_1,x_2, \cdots ,x_n$是样本，试求未知参数的矩估计：
|
|||
|
|
}{}
|
|||
|
|
\begin{enumerate}
|
|||
|
|
\questionandanswerSolution[]{
|
|||
|
|
$p(x;\theta)=\frac{2}{\theta^{2}}(\theta-x),\quad 0<x<\theta, \quad \theta>0$；
|
|||
|
|
}{
|
|||
|
|
$\displaystyle
|
|||
|
|
EX=\int_{0}^{\theta} x\frac{2}{\theta^{2}}(\theta-x) \mathrm{d}x = \frac{\theta}{3}
|
|||
|
|
$
|
|||
|
|
，所以$\theta=3EX$，所以$\theta$的矩估计是$\hat{\theta}=3 \bar{x}$。
|
|||
|
|
}
|
|||
|
|
\questionandanswerSolution[]{
|
|||
|
|
$p(x;\theta)=(\theta+1)x^{\theta},\quad 0<x<1,\quad \theta>0$；
|
|||
|
|
}{
|
|||
|
|
$\displaystyle EX=\int_{0}^{1} x(\theta+1)x^{\theta} \mathrm{d}x = \frac{\theta + 1}{\theta + 2} $，所以$\theta$的矩估计是$\displaystyle \hat{\theta}=\frac{1}{1-\bar{x}}-2$。
|
|||
|
|
}
|
|||
|
|
\questionandanswerSolution[]{
|
|||
|
|
$p(x;\theta)=\sqrt{\theta}x^{\sqrt{\theta}-1},\quad 0<x<1, \quad ,\theta>0$；
|
|||
|
|
}{
|
|||
|
|
$\displaystyle
|
|||
|
|
EX=\int_{0}^{1} x\sqrt{\theta}x^{\sqrt{\theta}-1} \mathrm{d}x = \frac{\sqrt{\theta}}{\sqrt{\theta} + 1}
|
|||
|
|
$
|
|||
|
|
，所以$\theta$的矩估计是$\displaystyle \hat{\theta}=\left( \frac{\bar{x}}{1-\bar{x}} \right) ^{2}$。
|
|||
|
|
}
|
|||
|
|
\questionandanswerSolution[]{
|
|||
|
|
$\displaystyle p(x;\theta,\mu)=\frac{1}{\theta}e^{-\frac{x-\mu}{\theta}}, \quad x>\mu,\quad \theta>0$。
|
|||
|
|
}{
|
|||
|
|
$$
|
|||
|
|
EX=\int_{\mu}^{+\infty} x \cdot \frac{1}{\theta}e^{-\frac{x-\mu}{\theta}} \mathrm{d}x=\theta+\mu
|
|||
|
|
$$
|
|||
|
|
$$
|
|||
|
|
EX^{2}=\int_{\mu}^{+\infty} x^{2}\cdot \frac{1}{\theta}e^{-\frac{x-\mu}{\theta}} \mathrm{d}x = 2\theta^{2}+2\mu \theta+\mu^{2}
|
|||
|
|
$$
|
|||
|
|
$$
|
|||
|
|
\operatorname{Var}X=EX^{2}-(EX)^{2}=2\theta^{2}+2\mu \theta+\mu^{2}-(\theta+\mu)^{2} = \theta^{2}
|
|||
|
|
$$
|
|||
|
|
所以$\theta$和$\mu$的矩估计是
|
|||
|
|
$$
|
|||
|
|
\hat{\theta}=s, \quad \hat{\mu}=\bar{x}-s
|
|||
|
|
$$
|
|||
|
|
}
|
|||
|
|
\end{enumerate}
|
|||
|
|
\questionandanswerSolution[5]{
|
|||
|
|
设总体为$N(\mu,1)$，现对该总体观测$n$次，发现有$k$次观测值为正，使用频率替换方法求$\mu$的估计。
|
|||
|
|
}{
|
|||
|
|
设总体为$X$，则根据频率替换方法，$P(X>0)=\dfrac{k}{n}$。设标准正态分布的累积分布函数为$\Phi(x)$，则
|
|||
|
|
$$
|
|||
|
|
\frac{k}{n}=P(X>0)=P\left( \frac{x-\mu}{1}>\frac{0-\mu}{1} \right) =1-P\left( \frac{x-\mu}{1}\leqslant -\mu \right) =1-\Phi(-\mu)
|
|||
|
|
$$
|
|||
|
|
所以$\mu$的估计为
|
|||
|
|
$$
|
|||
|
|
\hat{\mu}=-\Phi^{-1}(1-\frac{k}{n})
|
|||
|
|
$$
|
|||
|
|
}
|
|||
|
|
\questionandanswerSolution[7]{
|
|||
|
|
设总体$X$服从二项分布$b(m,p)$，其中$m,p$为未知参数，$x_1,x_2, \cdots ,x_n$为$X$的一个样本，求$m$与$p$的矩估计。
|
|||
|
|
}{
|
|||
|
|
因为
|
|||
|
|
$\displaystyle
|
|||
|
|
EX=mp,\ \operatorname{Var}X=mp(1-p)
|
|||
|
|
$
|
|||
|
|
，所以$\displaystyle p=1-\frac{\operatorname{Var}X}{EX}$，$\displaystyle m=\frac{EX}{p}=\frac{(EX)^{2}}{EX-\operatorname{Var}X}$，所以$m$与$p$的矩估计为
|
|||
|
|
$$
|
|||
|
|
m=1-\frac{s}{\bar{x}},\qquad p=\frac{\bar{x}^{2}}{\bar{x}-s}
|
|||
|
|
$$
|
|||
|
|
}
|
|||
|
|
\end{enumerate}
|
|||
|
|
\end{document}
|