183 lines
11 KiB
TeX
183 lines
11 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}{4}
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\section{充分统计量}
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\begin{enumerate}
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\questionandanswerProof[1]{
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设$x_1,x_2, \cdots ,x_n$是来自几何分布
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$$
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P(X=x)=\theta(1-\theta)^{x},\quad x=0,1,2, \cdots
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$$
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的样本，证明 $\displaystyle T=\sum_{i=1}^{n} x_i$是充分统计量。
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}{
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$$
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p(x_1,x_2, \cdots ,x_n;\theta)=\prod_{i=1}^{n} \theta(1-\theta)^{x_i}=\theta^{n}(1-\theta)^{\sum_{i=1}^{n} x_i}=\theta^{n}(1-\theta)^{T}
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$$
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取$g(T,\theta)=\theta^{n}(1-\theta)^{T}, h(X)=1$，
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由因子分解定理可知 $\displaystyle T=\sum_{i=1}^{n} x_i$是$\theta$的充分统计量。
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}
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\questionandanswer[3]{
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设总体为如下离散分布：
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\begin{tabular}{c|cccc}
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$x$ & $a_1$ & $a_2$ & $\cdots$ & $a_k$ \\
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\hline
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$p$ & $p_1$ & $p_2$ & $\cdots$ & $p_k$ \\
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\end{tabular}。
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$x_1,x_2, \cdots ,x_n$是来自该总体的样本，
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}{}
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\begin{enumerate}
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\questionandanswerProof[]{
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证明次序统计量$(x_{(1)},x_{(2)}, \cdots , x_{(n)})$是充分统计量；
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}{
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设$T=(x_{(1)},x_{(2)}, \cdots , x_{(n)})$，$X$表示一次取样。则
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$$
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\begin{aligned}
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P(X=(x_1,x_2, \cdots ,x_n)|T=t) &= \frac{P(X=(x_1,x_2, \cdots ,x_n), T=t)}{P(T=t)} \\
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&=\frac{\prod_{i=1}^{n} p_{i}}{\mathrm{P}_{n}^{n}\prod_{i=1}^{n} p_{i}}=\frac{1}{\mathrm{P}_{n}^{n}}=\frac{1}{n!} \\
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\end{aligned}
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$$
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可见与$T$无关，所以次序统计量$(x_{(1)},x_{(2)}, \cdots , x_{(n)})$是充分统计量。
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}
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\questionandanswer[]{
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以$n_j$表示$x_1,x_2, \cdots ,x_n$中等于$a_j$的个数，证明$(n_1,n_2, \cdots ,n_k)$是充分统计量。
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}{
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设$T=(n_1,n_2, \cdots , n_k)$，$X$表示一次取样。则
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$$
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\begin{aligned}
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P(X=(x_1,x_2, \cdots ,x_n)|T=t) &= \frac{P(X=(x_1,x_2, \cdots ,x_n), T=t)}{P(T=t)} \\
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&=\frac{\prod_{j=1}^{n} p_j}{\mathrm{P}_{n}^{n} \prod_{j=1}^{n} p_j^{n_j}} \\
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\end{aligned}
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$$
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应该与$T$无关，所以$(n_1,n_2, \cdots ,n_k)$是充分统计量。
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}
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\end{enumerate}
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\questionandanswerSolution[8]{
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设$x_1,x_2, \cdots ,x_n$是来自拉普拉斯（Laplace）分布
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$$
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p(x;\theta)=\frac{1}{2\theta} e^{-\frac{\left\vert x \right\vert }{\theta}}, \theta>0
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$$
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的样本，试给出一个充分统计量。
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}{
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设$X$表示一次取样，则
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$$
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\begin{aligned}
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P(X=(x_1,x_2, \cdots ,x_n);\theta)&=\prod_{i=1}^{n} p(x_i;\theta)=\prod_{i=1}^{n} \frac{1}{2\theta} e^{-\frac{\left\vert x \right\vert }{\theta}} = \left( \frac{1}{2\theta} \right) ^{n} e^{-\frac{1}{\theta}\sum_{i=1}^{n} \left\vert x_i \right\vert }\\
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% =\left( \frac{1}{2\theta} \right) ^{n} \left( e^{\sum_{i=1}^{n} \left\vert x_i \right\vert } \right) ^{-\frac{1}{\theta}} \\
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\end{aligned}
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$$
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令$T=\displaystyle \sum_{i=1}^{n} \left\vert x_i \right\vert $，则上式$=\displaystyle \left( \frac{1}{2\theta} \right) ^{n} \left( e^{-\frac{T}{\theta}} \right) $。则可以令$g(T,\theta)=\displaystyle \left( \frac{1}{2\theta} \right) ^{n} \left( e^{-\frac{T}{\theta}} \right)$, $h(X)=1$，由因子分解定理可知$T=\displaystyle \sum_{i=1}^{n} \left\vert x_i \right\vert $是$\theta$的充分统计量。
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}
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\questionandanswer[10]{
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设$x_1,x_2, \cdots ,x_n$是来自正态分布$N(\mu,\sigma^{2})$的样本。
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}{}
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\begin{enumerate}
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\questionandanswerSolution[]{
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在$\mu$已知时给出$\sigma^{2}$的一个充分统计量。
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}{
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$$
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p(x_1,x_2, \cdots ,x_n; \sigma^{2})=(2\pi\sigma^{2})^{-\frac{n}{2}} \exp \left\{ -\frac{1}{2\sigma^{2}} \sum_{i=1}^{n} (x_i-\mu)^{2}\right\}
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$$
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所以可以令$\displaystyle T=\sum_{i=1}^{n} (x_i-\mu)^{2}$，则$T$是$\sigma^{2}$的一个充分统计量。
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}
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\questionandanswerSolution[]{
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在$\sigma^{2}$已知时给出$\mu$的一个充分统计量。
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}{
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$$
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\begin{aligned}
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p(x_1,x_2, \cdots ,x_n; \sigma^{2})&=(2\pi\sigma^{2})^{-\frac{n}{2}} \exp \left\{ -\frac{1}{2\sigma^{2}} \sum_{i=1}^{n} (x_i-\mu)^{2}\right\} \\
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&=(2\pi \sigma^{2})^{-\frac{n}{2}} \exp \left\{ -\frac{n\mu^{2}}{2\sigma^{2}} \right\} \exp \left\{ -\frac{1}{2\sigma^{2}}\sum_{i=1}^{n} x_i^{2} \right\} \exp \left\{ \frac{\mu}{\sigma^{2}}\sum_{i=1}^{n} x_i \right\} \\
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\end{aligned}
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$$
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% 理论上来说，对于正态分布的参数$\mu$，可以使用样本均值$\displaystyle \bar{x}= \sum_{i=1}^{n} x_i$来估计，但无法使用因子分解定理证明，那只能认为$\bar{x}$是$\mu$的一个充分统计量了。
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令$\displaystyle T=\sum_{i=1}^{n} x_i$，则$\displaystyle g(\mu, T)=(2\pi \sigma^{2})^{-\frac{n}{2}} \exp \left\{ -\frac{n\mu^{2}}{2\sigma^{2}} \right\}\exp \left\{ \frac{\mu}{\sigma^{2}}T \right\}$，$\displaystyle h(\overrightarrow{x})=\exp \left\{ -\frac{1}{2\sigma^{2}}\sum_{i=1}^{n} x_i^{2} \right\} $。
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所以$T$是$\mu$的一个充分统计量。
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}
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\end{enumerate}
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\questionandanswerSolution[11]{
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设$x_1,x_2, \cdots ,x_n$是来自均匀分布$U(\theta_1, \theta_2)$的样本，试给出一个充分统计量。
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}{
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$$
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p(x_1,x_2, \cdots ,x_n; \theta_1, \theta_2)= \prod_{i=1}^{n} \frac{1}{\theta_2-\theta_1} 1_{[\theta_1, \theta_2]}(x_i)=\left( \frac{1}{\theta_2-\theta_1} \right) ^{n} 1_{[\theta_1,\theta_2]}(x_{(1)}, x_{(n)})
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$$
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所以$(x_{(1)}, x_{(n)})$是一个充分统计量。
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}
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\questionandanswerSolution[12]{
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设$x_1,x_2, \cdots ,x_n$是来自均匀分布$U(\theta,2\theta), \theta>0$的样本，试给出充分统计量。
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}{
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$$
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p(x_1,x_2, \cdots ,x_n; \theta)=\prod_{i=1}^{n} \frac{1}{\theta} 1_{[\theta,2\theta]}(x_i)=\frac{1}{\theta^{n}} 1_{[\theta, 2\theta]}(x_{(1)}, x_{(n)})
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$$
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所以$(x_{(1)}, x_{(n)})$是一个充分统计量。
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}
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\questionandanswerSolution[17]{
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设$\displaystyle \binom{x_i}{y_i}, i=1,2, \cdots ,n$是来自正态分布族
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$$
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\left\{ N\left( \binom{\theta_1}{\theta_2}, \begin{pmatrix}
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\sigma_1^{2} & \rho\sigma_1\sigma_2 \\
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\rho\sigma_1\sigma_2 & \sigma_2^{2} \\
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\end{pmatrix} \right) \ ;\ -\infty<\theta_1,\theta_2<\infty, \sigma_1,\sigma_2>0,\left\vert \rho \right\vert \leqslant 1 \right\}
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$$
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的一个二维样本，寻求$(\theta_1,\sigma_1,\theta_2,\sigma_2,\rho)$的充分统计量。
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}{
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$$
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\begin{aligned}
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&p\left( \binom{x_i}{y_i};(\theta_1,\sigma_1,\theta_2,\sigma_2,\rho) \right) = \prod_{i=1}^{n} \frac{1}{2\pi\sigma_1\sigma_2\sqrt{1-\rho^{2}}} \exp \left\{ -\frac{1}{2(1-\rho^{2})}(a_i^{2}+b_i^{2}-2\rho a_i b_i) \right\} \\
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&=\left( \frac{1}{2\pi\sigma_1\sigma_2\sqrt{1-\rho^{2}}} \right) ^{n} \exp \left\{ -\frac{1}{2(1-\rho^{2})} \left( \sum_{i=1}^{n} a_i^{2}+\sum_{i=1}^{n} b_i^{2}-2\rho \sum_{i=1}^{n} a_i b_i \right)\right\} \\
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\end{aligned}
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$$
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其中
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$$
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\sum_{i=1}^{n} a_i^{2}=\sum_{i=1}^{n} \left( \frac{x_i-\theta_1}{\sigma_1} \right) ^{2}=\frac{1}{\sigma_1^{2}}\sum_{i=1}^{n} (x_i^{2}-2\theta_1 x_i+\theta_1^{2})=\frac{1}{\sigma_1^{2}}\sum_{i=1}^{n} x_i^{2}-\frac{2\theta_1}{\sigma_1^{2}}\sum_{i=1}^{n} x_i+ \frac{\theta_1^{2}}{\sigma_1^{2}}
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$$
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$$
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\sum_{i=1}^{n} b_i^{2}=\sum_{i=1}^{n} \left( \frac{y_i-\theta_2}{\sigma_2} \right) ^{2}=\frac{1}{\sigma_2^{2}}\sum_{i=1}^{n} (y_i^{2}-2\theta_2 y_i+\theta_2^{2})=\frac{1}{\sigma_2^{2}}\sum_{i=1}^{n} y_i^{2}-\frac{2\theta_2}{\sigma_2^{2}}\sum_{i=1}^{n} y_i+\frac{\theta_2^{2}}{\sigma_2^{2}}
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$$
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$$
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\begin{aligned}
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&\sum_{i=1}^{n} a_i b_i =\sum_{i=1}^{n} \left( \frac{x_i-\theta_1}{\sigma_1} \right) \left( \frac{y_i-\theta_2}{\sigma_2} \right) =\frac{1}{\sigma_1\sigma_2}\sum_{i=1}^{n} (x_i y_i- \theta_1 y_i - \theta_2 x_i+\theta_1 \theta_2) \\
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&=\frac{1}{\sigma_1\sigma_2}\sum_{i=1}^{n} x_i y_i- \frac{\theta_1}{\sigma_1\sigma_2}\sum_{i=1}^{n} y_i - \frac{\theta_2}{\sigma_1\sigma_2}\sum_{i=1}^{n} x_i+\frac{n\theta_1\theta_2}{\sigma_1\sigma_2} \\
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\end{aligned}
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$$
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仔细观察即可发现
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$$
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\left( \sum_{i=1}^{n} x_i,\ \sum_{i=1}^{n} x_i^{2},\ \sum_{i=1}^{n} y_i,\ \sum_{i=1}^{n} y_i^{2},\ \sum_{i=1}^{n} x_i y_i \right)
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$$
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是此二维正态分布的充分统计量。
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}
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\questionandanswerProof[19]{
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设$x_1,x_2, \cdots ,x_n$是来自两参数指数分布
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$$
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p(x;\theta,\mu)=\frac{1}{\theta} e^{-\frac{x-\mu}{\theta}}, \quad x>\mu, \theta>0
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$$
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的样本，证明$(\bar{x},x_{(1)})$是充分统计量。
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}{
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$$
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\begin{aligned}
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&p(x_1,x_2, \cdots ,x_n; \theta,\mu)=\prod_{i=1}^{n} \frac{1}{\theta} e^{-\frac{x_i-\mu}{\theta}}=\frac{1}{\theta^{n}} \exp \left\{ -\frac{1}{\theta} \sum_{i=1}^{n} (x_i-\mu) \right\} \\
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=&\frac{1}{\theta^{n}} \exp \left\{ -\frac{1}{\theta}\sum_{i=1}^{n} x_i \right\} \exp \left\{ \frac{n\mu}{\theta} \right\} , \quad x_1,x_2, \cdots ,x_n > \mu \\
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\end{aligned}
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$$
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其中$x_1,x_2, \cdots ,x_n>\mu \iff x_{(1)} > \mu$，并且$\displaystyle \sum_{i=1}^{n} x_i=n \bar{x}$，
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所以$(\bar{x}, x_{(1)})$是充分统计量。
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}
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\questionandanswerSolution[20]{
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设随机变量$Y_i\sim N(\beta_0+\beta_1 x_i, \sigma^{2}), i=1,2, \cdots ,n$，诸$Y_i$独立，$x_1,x_2, \cdots ,x_n$是已知常数，证明$\displaystyle \left( \sum_{i=1}^{n} Y_i,\ \sum_{i=1}^{n} x_i Y_i,\ \sum_{i=1}^{n} Y_i^{2} \right) $是充分统计量。
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}{
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$$
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\begin{aligned}
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&p(Y_1,Y_2, \cdots ,Y_n; \beta_0, \beta_1, \sigma^{2})=\prod_{i=1}^{n} \frac{1}{\sqrt{2\pi}\sigma} \exp \left\{ -\frac{1}{2}\left( \frac{Y_i-(\beta_0+\beta_1 x)}{\sigma} \right) ^{2} \right\} \\
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=&\left( \frac{1}{\sqrt{2\pi}\sigma} \right) ^{n} \exp \left\{ -\frac{1}{2\sigma^{2}} \sum_{i=1}^{n} \left( Y_i-\beta_0-\beta_1 x_i \right) ^{2} \right\} \\
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\end{aligned}
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$$
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其中
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$$
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\sum_{i=1}^{n} (Y_i-\beta_0-\beta_1 x_i)^{2}=\sum_{i=1}^{n} Y_i^{2}+n \beta_0^{2}+n\beta_1^{2}\sum_{i=1}^{n} x_i^{2} - 2\beta_0\sum_{i=1}^{n} Y_i -2\beta_1 \sum_{i=1}^{n} x_i Y_i + \beta_0\beta_1 \sum_{i=1}^{n} x_i
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$$
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其中$\beta_0,\beta_1, \sigma$为参数，$x_1,x_2, \cdots ,x_n$已知，
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所以$\displaystyle \left( \sum_{i=1}^{n} Y_i,\ \sum_{i=1}^{n} x_i Y_i,\ \sum_{i=1}^{n} Y_i^{2} \right) $是充分统计量。
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}
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\end{enumerate}
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\end{document}
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