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\begin{document}
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\section{最小方差无偏估计}
\begin{enumerate}
    \questionandanswerProof[1]{
        设总体概率函数是$p(x;\theta), x_1,x_2, \cdots ,x_n$ 是其样本，$T=T(x_1,x_2, \cdots ,x_n)$是$\theta$的充分统计量，则对$g(\theta)$的任一估计$\hat{g}$，令$\tilde{g}=E(\hat{g}|T)$，证明：$MSE(\tilde{g})\leqslant MSE(\hat{g})$。这说明，在均方误差准则下，人们只需要考虑基于充分统计量的估计。
    }{
        $$
        \begin{aligned}
            \operatorname{MSE}(\hat{g})&=E((\hat{g} - g(\theta))^{2})= E\left( (\hat{g} - \tilde{g} + \tilde{g} -g(\theta))^{2} \right)  \\
            &=E(\hat{g}-\tilde{g})^{2} + 2E(\hat{g}-\tilde{g})(\tilde{g} - g(\theta) ) + E(\tilde{g}-g(\theta))^{2} \\
            &=E(\hat{g} - \tilde{g})^{2}+2E(\hat{g}-\tilde{g})(\tilde{g}-g(\theta))+\operatorname{MSE}(\tilde{g}) \\
        \end{aligned}
        $$
        其中
        $$
        E(\hat{g}-\tilde{g})(\tilde{g}-g(\theta))=E(E((\hat{g}-\tilde{g})(\tilde{g}-g(\theta))|T))
        $$
        由于$T$是充分统计量，所以$E(\tilde{g}-g(\theta))$与$T$无关，所以
        $$
        \begin{aligned}
            E (\hat{g}-\tilde{g})(\tilde{g}-g(\theta))&=E(\tilde{g}-g(\theta)) E(E(\hat{g}-\tilde{g}|T)) \\
            &=[E(\tilde{g}) -E(g(\theta))]E(E(\hat{g}-\tilde{g}|T)) = 0 \\
        \end{aligned}
        $$
        所以
        $$
        \operatorname{MSE}(\hat{g}) = E(\hat{g}-\tilde{g})^{2}+\operatorname{MSE}(\tilde{g})\geqslant \operatorname{MSE}(\tilde{g})
        $$
    }
    \questionandanswerProof[3]{
        设$T$是$g(\theta)$的UMVUE，$\hat{g}$是$g(\theta)$的无偏估计，证明：若$\operatorname{Var}(\hat{g})<\infty$，则$\operatorname{Cov}(T,\hat{g})\geqslant 0$。
    }{
        由于$T$是$g(\theta)$的UMVUE，所以$E(T)=g(\theta), \operatorname{Var}(T)<\infty$；由于$\hat{g}$是$g(\theta)$的无偏估计，所以$E(\hat{g})=0$。从而$E(T-\hat{g})=E(T)-E(\hat{g})=0$，且$\operatorname{Var}(T-\hat{g})=\operatorname{Var}(T)+\operatorname{Var}(\hat{g})+\operatorname{Cov}(T,\hat{g})<\infty$（应该不存在方差有限但协方差无限的情况吧），所以根据判断准则，
        $$
        0=\operatorname{Cov}(T,T-\hat{g})=\operatorname{Var}(T)-\operatorname{Cov}(T,\hat{g})
        $$
        所以$\operatorname{Cov}(T,\hat{g})=\operatorname{Var}(T)>0$。
    }
    \questionandanswerProof[5]{
        设总体$p(x;\theta)$的费希尔信息量存在，若二阶导数$\displaystyle \frac{\partial ^{2}}{\partial \theta^{2}}p(x;\theta)$对一切的$\theta \in \Theta$存在，证明费希尔信息量
        $$
        I(\theta)=-E\left( \frac{\partial ^{2}}{\partial \theta^{2}}\ln p(x;\theta) \right) 
        $$
    }{
        $$
        \begin{aligned}
            -E\left( \frac{\partial ^{2}}{\partial \theta^{2}} \ln p(x;\theta) \right) =&-\int_{-\infty}^{+\infty} \frac{\partial ^{2} \ln p(x;\theta)}{\partial \theta^{2}} p(x;\theta)\mathrm{d}x \\
            =&\int_{-\infty}^{+\infty} \left( \frac{\partial \ln p(x;\theta)}{\partial \theta} \right) ^{2} p(x;\theta) \mathrm{d}x \\
            =&\int_{-\infty}^{+\infty} \frac{\partial \ln p(x;\theta)}{\partial \theta} \frac{\partial \ln p(x;\theta)}{\partial p(x;\theta)} \frac{\partial p(x;\theta)}{\partial \theta} p(x;\theta) \mathrm{d}x \\
            =&\int_{-\infty}^{+\infty} \frac{\partial \ln p(x;\theta)}{\partial \theta} \frac{1}{p(x;\theta)} \frac{\partial p(x;\theta)}{\partial \theta} p(x;\theta) \mathrm{d}x \\
            =&\int_{-\infty}^{+\infty} \frac{\partial \ln p(x;\theta)}{\partial \theta} \frac{\partial p(x;\theta)}{\partial \theta} \mathrm{d}x \\
            =& E_{x} \left( \frac{\partial \ln p(x;\theta)}{\partial \theta} \right) ^{2} = I(\theta) \\
        \end{aligned}
        $$
    }
    \questionandanswer[6]{
        设总体密度函数为$p(x;\theta)=\theta x^{\theta-1}, 0<x<1, \theta>0$, $x_1,x_2, \cdots ,x_n$是样本。
    }{}

    \begin{enumerate}
        \questionandanswerSolution[]{
            求$g(\theta)=\dfrac{1}{\theta}$的最大似然估计；
        }{
            对数似然函数为
            $$
            \ln L(\theta)=\sum_{i=1}^{n} \ln \theta x^{\theta-1}=\sum_{i=1}^{n} \left( \ln \theta+(\theta-1) \ln x_i \right) =n \ln \theta+(\theta-1) \sum_{i=1}^{n} \ln x_i
            $$
            对$\theta$求导并令其为0，
            $$
            \frac{\partial L(\theta)}{\partial \theta} = \frac{n}{\theta}+\sum_{i=1}^{n} x_i = 0
            $$
            则$\theta$的最大似然估计为$\hat{\theta} = \dfrac{n}{\sum_{i=1}^{n} x_i}$，根据最大似然估计的不变性，$g(\theta)=\dfrac{1}{\theta}$的最大似然估计为
            $$
            \widehat{g(\theta)}=\frac{1}{\hat{\theta}}=\frac{1}{n} \sum_{i=1}^{n} x_i
            $$
        }
        \questionandanswerSolution[]{
            求$g(\theta)$的有效估计。
        }{
            可以猜测上一小题中的
            $
            \widehat{g(\theta)}=\frac{1}{n} \sum_{i=1}^{n} x_i
            $
            为有效估计，接下来验证一下。

            % $$
            % \frac{\theta}{\theta+2}-\left( \frac{\theta}{\theta+1} \right) ^{2} = \frac{\theta}{\theta^{3} + 4 \theta^{2} + 5 \theta + 2}
            % $$
            可以计算得到总体的方差为$\dfrac{1}{\theta^{2}}$，因此 $g(\hat{\theta})=\bar{x}$的方差为$\dfrac{1}{n \theta^{2}} $。

            由于$\ln p(x;\theta) = \ln \theta +(\theta-1)\ln x$, 
            $$
            \frac{\partial \ln p(x;\theta)}{\partial \theta}=\frac{1}{\theta}+\ln x, \quad \frac{\partial^{2} \ln p(x;\theta)}{\partial \theta^{2}}=-\frac{1}{\theta^{2}}, \quad I(\theta)=-E\left( \frac{\partial ^{2}\ln p(x;\theta)}{\partial \theta^{2}} \right) =\frac{1}{\theta^{2}}
            $$
            所以$I(\frac{1}{\theta})=\theta^{2}$，所以
            $$
            \operatorname{Var}(\widehat{g(\theta)})= \frac{1}{n \theta^{2}}=\frac{1}{I(\frac{1}{\theta})}=\frac{1}{I(g(\theta))}
            $$
            因此$\widehat{g(\theta)}=\frac{1}{n} \sum_{i=1}^{n} x_i$为$g(\theta)$的有效估计。
        }
    \end{enumerate}
    \questionandanswerSolution[7]{
        设总体密度函数为$\displaystyle p(x;\theta)=\frac{2\theta}{x^{3}} e^{-\frac{\theta}{x^{2}}},x>0,\theta>0$，求$\theta$的费希尔信息量$I(\theta)$。
    }{
        $$
        \ln p(x;\theta)=\ln 2+\ln \theta-3\ln x -\frac{\theta}{x^{2}}
        $$
        $$
        \frac{\partial \ln p(x;\theta)}{\partial \theta}=\frac{1}{\theta} - \frac{1}{x^{2}}, \quad \frac{\partial ^{2}\ln p(x;\theta)}{\partial \theta^{2}} = -\frac{1}{\theta^{2}}
        $$
        所以
        $$
        I(\theta)=-E\left( \frac{\partial ^{2}\ln p(x;\theta)}{\partial \theta^{2}} \right) =\frac{1}{\theta^{2}}
        $$
    }
    \questionandanswerProof[10]{
        设$x_1,x_2, \cdots ,x_n$是来自$\operatorname{Ga}(\alpha,\lambda)$的样本，$\alpha>0$已知，试证明$\dfrac{\bar{x}}{\alpha}$是$g(\lambda)=\dfrac{1}{\lambda}$的有效估计，从而也是UMVUE。
    }{
        $$
        p(x;\lambda) = \frac{\lambda^{\alpha}}{\Gamma(\alpha)} x^{\alpha-1} e^{-\lambda x}, x>0; \quad \ln p(x;\lambda)=\alpha\ln \lambda-\ln \Gamma(\alpha)+(\alpha-1)\ln x-\lambda x
        $$
        $$
        \frac{\partial \ln p(x;\lambda)}{\partial \lambda}=\frac{\alpha}{\lambda}-x; \quad \frac{\partial ^{2} \ln p(x;\lambda)}{\partial \lambda^{2}}=-\frac{\alpha}{\lambda^{2}}
        $$
        所以
        $$
        I(\lambda)=-E\left( \frac{\partial ^{2}\ln p(x;\lambda)}{\partial \lambda^{2}} \right) =\frac{\alpha}{\lambda^{2}}; \quad \text{C-R下界}=\frac{(g'(\lambda))^{2}}{n I(\lambda)}=\frac{(-\frac{1}{\lambda^{2}})^{2}}{n \frac{\alpha}{\lambda^{2}}} = \frac{1}{\alpha \lambda^{2} n}
        $$
        由于总体的方差为$\dfrac{\alpha}{\lambda^{2}}$，所以$\bar{x}$的方差为$\dfrac{\alpha}{n \lambda^{2}}$，所以$\dfrac{\bar{x}}{\alpha}$的方差为$\dfrac{1}{n \alpha \lambda^{2}}$，等于C-R下界。
    }
\end{enumerate}
\end{document}