264 lines
14 KiB
TeX
264 lines
14 KiB
TeX
\documentclass[全部作业]{subfiles}
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\input{mysubpreamble}
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\begin{document}
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\renewcommand{\labelenumii}{(\alph{enumii})}
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\setcounter{chapter}{6}
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\setcounter{section}{5}
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\section*{期中考试}
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\begin{enumerate}
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\questionandanswer[]{
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(10') 设$X_1, \cdots ,X_n (n>6)$是来自指数分布$p(x,\lambda)=\lambda e^{-\lambda x} I(x\geqslant 0)$的样本，其中$\lambda>0$。用$X_{(i)}$表示样本的第$i$个次序统计量。求解：
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}{}
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\begin{enumerate}
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\questionandanswerSolution[]{
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$\lambda$的充分统计量$T$；
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}{
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样本联合密度为
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$$
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p(x_1,x_2, \cdots ;\lambda)=\lambda^{n} e^{-n \bar{x} \lambda} I(x_{(1)}\geqslant 0)
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$$
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由由因子分解定理知，$T=\bar{x}$为$\lambda$的充分统计量。
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}
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\questionandanswerSolution[]{
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$\operatorname{Cov}(X_{(3)}, X_{(6)})$。
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}{
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设$Y=\lambda X$，则$Y_1,Y_2, \cdots ,Y_n$为来自$\operatorname{Exp}(1)$的样本，所以$\operatorname{Cov}(X_{(3)},X_{(6)})=\frac{1}{\lambda^{2}}\operatorname{Cov}(Y_{(3)},Y_{(6)})$，下面只需求三个量：$E(Y_{(3)}Y_{(6)}),\ E(Y_{(3)}),\ E(Y_{(6)})$。
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$$
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\begin{aligned}
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\because p_{3}(u)&=\frac{n!}{2!(n-3)!} F^{2}(u) [1-F(u)]^{n-3} p(u) \\
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&=\frac{n!}{2!(n-3)!} (1-e^{-u})^{2} e^{-(n-2)u} I(u\geqslant 0) \\
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\therefore E(Y_{(3)}) &= \int_{0}^{+\infty} u p_3(u) \mathrm{d}u=\int_{0}^{+\infty} \frac{n!}{2!(n-3)!} u(1-e^{-u})^{2} e^{-(n-2)u} \mathrm{d}u \\
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&= \frac{n^{4} -3n^{2}+6n -2}{n(n-1)(n-2)} \\
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\end{aligned}
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$$
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同理算出$E(Y_{(6)})$与$E(Y_{(3)}Y_{(6)})$后可计算出
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$$
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\operatorname{Cov}(X_{(3)},X_{(6)})=\frac{1}{\lambda^{2}} \operatorname{Cov}(Y_{(3)},Y_{6})=\frac{1}{\lambda^{2}}\left[ E(Y_{(3)}Y_{(6)}) - E(Y_{(3)})E(Y_{(6)}) \right]
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$$
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}
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\end{enumerate}
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\questionandanswer[]{
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(15') 设$X_i, \ i=1,2,3$独立且都分别服从$N(i, i^{2})$，利用$X_i$构造下面分布的统计量：
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}{}
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\begin{enumerate}
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\questionandanswer[]{
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自由度为3的$\chi^{2}$分布；
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}{
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$$
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\left( \frac{X_1-1}{1} \right) ^{2}+\left( \frac{X_2-2}{2} \right) ^{2}+\left( \frac{X_3-3}{3} \right) ^{2} = \chi^{2}(3)
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$$
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}
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\questionandanswer[]{
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自由度为2的$t$分布；
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}{
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$$
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\frac{(X_1-1)\sqrt{2}}{\sqrt{\left( \frac{X_2-2}{2} \right) ^{2}+\left( \frac{X_3-3}{3} \right) ^{2}}} \sim t(2)
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$$
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}
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\questionandanswer[]{
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自由度为1, 2的$F$分布。
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}{
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$$
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\frac{\left( \frac{X_1-1}{1} \right) ^{2}}{\left( \frac{X_2-2}{2} \right) ^{2}+\left( \frac{X_3-3}{3} \right) ^{2}} \sim F(1,2)
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$$
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}
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\end{enumerate}
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\questionandanswer[3]{
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(10') 设$X_1, \cdots ,X_n$是来自均值为$\mu$，方差为$\sigma^{2}$的总体的简单样本。
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}{}
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\begin{enumerate}
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\questionandanswerProof[]{
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证明：如果$\sum_{i=1}^{n} a_i=1$，则估计量$\sum_{i=1}^{n} a_i X_i$是$\mu$的一个无偏估计量；
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}{
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$$
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E\left( \sum_{i=1}^{n} a_i X_i \right) =\sum_{i=1}^{n} a_i (EX_{i})=\left( \sum_{i=1}^{n} a_i \right) \mu = 1 \cdot \mu = \mu
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$$
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所以估计量$\sum_{i=1}^{n} a_i X_i$是$\mu$的一个无偏估计量。
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}
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\questionandanswerSolution[]{
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在所有这类形式的估计量中求一个最小方差者，并计算其方差。
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}{
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$$
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\operatorname{Var}\left( \sum_{i=1}^{n} a_i X_i \right) =\left( \sum_{i=1}^{n} a_i^{2} \right) \sigma^{2} = \left( \sum_{i=1}^{n} a_i^{2} \cdot \sum_{i=1}^{n} 1^{2} \right) \sigma^{2} \frac{1}{n} \geqslant \left( \sum_{i=1}^{n} a_i \right) ^{2} \sigma^{2}
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$$
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所以当$\sum_{i=1}^{n} a_i=1$时方差最小，为$\sigma^{2}$。
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}
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\end{enumerate}
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\questionandanswer[4]{
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(15') 设简单样本$X_1, \cdots ,X_n \sim F$。对给定常数$x_0$，令$F_n(x_0)=\frac{1}{n} \sum_{i=1}^{n} I(X_i \leqslant x_0)$。回答以下问题：
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}{}
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\begin{enumerate}
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\questionandanswerProof[]{
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证明$F_n(x_0)$是$F(x_0)$的无偏估计；
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}{
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$$
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\begin{aligned}
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&\because \sum_{i=1}^{n} I(x_i \leqslant x_0)\text{为} n \text{重伯努利的独立和形式} \\
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&\therefore E\left( \sum_{i=1}^{n} I(x_i \leqslant x_0) \right) =n E(I (x_1\leqslant x_0)) = nF(x_0) \\
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\end{aligned}
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$$
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}
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\questionandanswerProof[]{
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证明$F_n(x_0)$是$F(x_0)$的相合估计；
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}{
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由于$E(F_n(x_0))=F(x_0)$, $\operatorname{Var}(F_n(x_0))=\frac{1}{n} F_n(x_0) (1\cdot F(x_0)) \to 0$ $(n \to \infty)$，因此$F_n(x_0)$是$F(x_0)$的相合估计。
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}
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\questionandanswerProof[]{
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证明$F_n(x_0)$的渐近正态性。
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}{
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$$
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\begin{aligned}
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&\because F_n(x_0) \text{为独立和} \\
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&\therefore \text{由中心极限定理知} \sqrt{n} (F_n(x_0)-F(x_0)) \xrightarrow{L} N(0, F(x_0) (1-F(x_0))) \\
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\end{aligned}
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$$
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}
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\end{enumerate}
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\questionandanswer[5]{
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(15') 设随机变量$Y_1, \cdots ,Y_n$满足
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$$
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Y_i = x_i \beta + \varepsilon_i, \ i=1, \cdots ,n,
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$$
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其中$x_1, \cdots ,x_n$是固定常数，$\varepsilon_1, \cdots , \varepsilon_n$独立同分布于$N(0, \sigma^{2})$，其中$\sigma^{2}$未知。回答以下问题：
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}{}
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\begin{enumerate}
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\questionandanswerSolution[]{
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求关于$(\beta,\sigma^{2})$的一个2维充分统计量。
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}{
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由于$Y_1,Y_2, \cdots ,Y_n$的联合密度函数为$Y_1 \sim N(x_i\beta, \alpha^{2})$，
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$$
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\begin{aligned}
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&L= \prod_{i=1}^{n} \frac{1}{\sqrt{2\pi} \sigma} e^{-\frac{1}{2\sigma^{2}}(Y_i-x_i \beta)^{2}} \\
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&=\left( \frac{1}{\sqrt{2\pi} \sigma} \right) ^{n} \exp \left\{ -\frac{1}{2\sigma^{2}} \sum_{i=1}^{n} (Y_i-x_i\beta)^{2} \right\} \\
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&=\left( \frac{1}{\sqrt{2\pi} \sigma} \right) ^{n} \cdot e^{-\frac{1}{2\sigma^{2}} \sum_{i=1}^{n} Y_i^{2}} \cdot e^{\frac{1}{2\sigma^{2}} \sum_{i=1}^{n} 2 x_i Y_i \beta} e^{-\frac{1}{2\sigma^{2}} \sum_{i=1}^{n} x_i^{2} \beta^{2}} \\
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\end{aligned}
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$$
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设$T_1=\sum_{i=1}^{n} Y_i^{2}$, $T_2=\sum_{i=1}^{n} x_i Y_i$，有因子分解定理知$(T_1, T_2)$为$(\beta, \alpha^{2})$的二维充分统计量。
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}
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\questionandanswer[]{
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求$\beta$的MLE并证明它是$\beta$的一个无偏估计；
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}{
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令$\frac{\mathrm{d}L}{\mathrm{d}\beta}=0$，得$-\frac{1}{\sigma^{2}} \sum_{i=1}^{n} (Y_i - x_i \beta) x_i = 0 \implies \hat{\beta}_{ML}=\frac{\sum_{i=1}^{n} Y_i x_i}{\sum_{i=1}^{x} x_i^{2}}=\frac{\overline{Y x}}{ \overline{x ^{2}}}$。
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$$
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\begin{aligned}
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E\left( \hat{\beta}_{ML} \right) &=E\left( \frac{1}{\sum_{i=1}^{n} x_i^{2}} \sum_{i=1}^{n} x_i Y_i \right) =\frac{1}{\sum_{i=1}^{n} x_i^{2}} \sum_{i=1}^{n} E(x_i Y_i) = \frac{1}{\sum_{i=1}^{n} x_i^{2}} \sum_{i=1}^{n} x_i (EY_{i}) \\
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&=\frac{1}{\sum_{i=1}^{n} x_i^{2}} \sum_{i=1}^{n} x_i(x_i \beta) = \beta \\
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\end{aligned}
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$$
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}
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\questionandanswerSolution[]{
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求$\beta$的MLE的分布。
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}{
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$$
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\hat{\beta}_{ML} = \frac{\sum_{i=1}^{n} x_i Y_i}{\sum_{i=1}^{n} x_i^{2}}
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$$
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}
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\end{enumerate}
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\questionandanswer[6]{
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(15') 设$X_1, X_2, \cdots , X_n$是来自$N(\mu,1)$的简单随机样本，其中$\mu$未知。用$\bar{X}$表示样本均值。令$p=P(X_1\geqslant 0)$，回答以下问题：
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}{}
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\begin{enumerate}
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\questionandanswerSolution[]{
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求$p$的极大似然估计$\hat{p}$；
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}{
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因为$\mu$的MLE为$\bar{x}$，而$p = P(X_1\geqslant 0)=P\left( \frac{x_1-\mu}{1} \geqslant -\mu \right) =1-\Phi(-\mu)=\Phi(\mu)$，
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所以由MLE的不变性知 $\hat{P}_{MLE} = \Phi(\bar{x})$
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}
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\questionandanswerSolution[]{
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求$\sqrt{n}(\hat{p}-p)$的极限分布；
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}{
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因为$\hat{\mu} = \bar{x} \sim N(\mu, \frac{1}{n})$，设$g(x)= \Phi(x)$，由Delta方法知
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$$
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\sqrt{n} \left( \hat{p} - p \right) =\sqrt{n} \left( \Phi(\bar{x}) -\Phi(\mu) \right) = \xrightarrow{L} N(0, \phi^{2}(\mu) \frac{1}{n})
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$$
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其中$\phi$为标准正态分布的概率密度函数。
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}
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\questionandanswerSolution[]{
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求$p$的UMVUE。
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}{
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$\Phi\left( \sqrt{\frac{n}{n-1}} \cdot \bar{x} \right) $为$p$的UMVME。
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}
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\end{enumerate}
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\questionandanswerSolution[]{
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(10') 已知总体密度为$p(x;\theta) = \theta^{2} x^{\theta^{2}-1} (\theta>0, 0<x<1)$。若有容量为$n$的样本，求参数$\theta$无偏估计的C-R下界。
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}{
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$$
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\begin{aligned}
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&\because &\ln p&=2 \ln \theta + (\theta^{2}-1) \ln x \\
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&&\frac{\partial \ln p}{\partial \theta}& = \frac{2}{\theta} + 2\theta \ln x \\
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&&\frac{\partial ^{2} \ln p}{\partial \theta^{2}}& = -\frac{2}{\theta^{2}} + 2\ln x \\
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&\therefore& I(\theta) &= - E\left( \frac{\partial ^{2}\ln p}{\partial \theta^{2}} \right) =\frac{2}{\theta^{2}} - 2E(\ln x) \\
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&&&=\frac{2}{\theta^{2}} - 2\int_{0}^{1} \ln x \cdot \theta^{2} x^{\theta^{2}-1} \mathrm{d}x = \frac{2}{\theta^{2}} + \frac{2}{\theta^{2}} = \frac{4}{\theta^{2}} \\
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&\therefore&& \text{C-R下界为} \frac{1}{nI(\theta)} = \frac{\theta^{2}}{4n} \\
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\end{aligned}
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$$
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}
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\questionandanswerSolution[8]{
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设$X_1, \cdots ,X_n$是来自泊松分布Poisson($\lambda$)（其中$\lambda >0$）的一个样本。参数$\lambda$的先验分布是$\phi(\lambda)=e^{-\lambda}$（其中$\lambda>0$），求$\lambda$的后验密度和贝叶斯估计。
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}{
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后验分布为
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$$
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\begin{aligned}
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h(\theta |X_1, \cdots ,X_n) &= c \prod_{i=1}^{n} f(x_i |\lambda) \cdot \pi(\lambda) \\
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&=\frac{\lambda^{X_1+X_2+ \cdots +X_n}}{x_1! x_2!\cdots X_n!} e^{-n\lambda} e^{-\lambda} \\
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&=c \lambda^{\sum_{i=1}^{n} x_i} e^{-(n+1) \lambda} \sim \operatorname{Ga}\left( \sum_{i=1}^{n} x_i +1, n+1 \right) \\
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\end{aligned}
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$$
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后验均值为贝叶斯估计 $\displaystyle \hat{\lambda} = \frac{\sum_{i=1}^{n} x_i + 1}{n+1}$。
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}
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\end{enumerate}
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\subsection*{附加题}
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\begin{enumerate}
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\questionandanswer[]{
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(10') 设$X_1, \cdots X_n, X_{n+1}$是来自$N(\mu,\sigma^{2})$的样本，记$\bar{X}_n = \frac{1}{n} \sum_{i=1}^{n} X_i, \ S_n^{2} = \frac{1}{n-1} \sum_{i=1}^{n} (X_i-\bar{X}_n)^{2}$。求解：
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}{}
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\begin{enumerate}
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\questionandanswerSolution[]{
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常数$a$和$b$使得$T=(a X_{n+1} + b \bar{X}_n)/S_n$服从$t$分布，并指出$t$分布的自由度；
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}{
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若$T=(a X_{n+1} + b \bar{X}_n)/S_n$服从$t$分布，则因为$X_{n+1}, \bar{X}_n$都与$S_n$独立，所以
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$$
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\begin{aligned}
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ET=0 &\implies a E(X_{n+1}) + b(\bar{X}_{n}) = 0 \\
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&\implies a \mu + b \mu =0 \implies a = -b \\
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\end{aligned}
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$$
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又因为$\displaystyle \frac{(\bar{X}_n - X_{n+1})}{\sqrt{\frac{n+1}{n} }\sigma}=\frac{\bar{X}_n - \mu-(X_{n+1}-\mu)}{\sqrt{\frac{n+1}{n}} \sigma}\sim N(0,1)$，$\displaystyle \frac{(n-1) S_n^{2}}{\sigma^{2}} \sim \chi^{2}(n-1)$，所以
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$$
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T=\left.\frac{(\bar{X}_n-X_{n+1})}{\sqrt{\frac{n+1}{n}}\sigma} \middle/ \sqrt{\frac{(n-1)S_n^{2}}{\sigma^{2}(n-1)}}\right. \sim t(n-1)
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$$
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整理得
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$
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\displaystyle \sqrt{\frac{n}{n+1}} (\bar{X}_n-X_{n+1}) /S_n \sim t(n-1)
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$\\
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即$a=\pm \sqrt{\frac{n}{n+1}}$, $b=\mp \sqrt{\frac{n}{n+1}}$，自由度为$n-1$。
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}
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\questionandanswerSolution[]{
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计算$\mathbb{E}(S_n^{3})$。
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}{
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$$
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\begin{aligned}
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&\because X = \frac{(n-1)S_n^{2}}{\sigma^{2}} \sim \chi^{2}(n-1) \\
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&\therefore X\text{的密度函数为} p(x)= \frac{\left( \frac{1}{2} \right) ^{\frac{n-1}{2}}}{\Gamma\left( \frac{n-1}{2} \right) }x^{\frac{n}{2}-1} e^{-\frac{x}{2}}\ (x>0) \\
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\end{aligned}
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$$
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取$y=\sqrt{x}$，则$Y=\sqrt{X}$的密度函数为
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$$
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g(y)= p(x) \frac{\mathrm{d}x}{\mathrm{d}y} =p(x) \cdot 2y = p(y^{2})\cdot 2y = \frac{\left( \frac{1}{2} \right) ^{\frac{n-1}{2}}}{\Gamma\left( \frac{n-1}{2} \right) } \cdot 2y^{n-2} \cdot y \cdot e^{-\frac{y^{2}}{2}}
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$$
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$$
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\begin{aligned}
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E(Y^{3}) &= \int_{0}^{+\infty} y^{3}g(y) \mathrm{d}y = \int_{0}^{+\infty} \frac{2\left( \frac{1}{2} \right) ^{\frac{n-1}{2}}}{\Gamma\left( \frac{n-1}{2} \right) } y^{n+1} \cdot e^{-\frac{y^{2}}{2}} \mathrm{d}y \\
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&=\frac{2^{\frac{n+1}{2}}}{\Gamma\left( \frac{n-1}{2} \right) } \int_{0}^{+\infty} y^{n+1} e^{-\frac{y^{2}}{2}} \mathrm{d}y = \begin{cases}
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\frac{2^{\frac{n+1}{2}}}{\Gamma\left( \frac{n-1}{2} \right) } (2k-1)!!,\quad & n=2k-1 \\
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\frac{2^{\frac{n+1}{2}}}{\Gamma\left( \frac{n-1}{2} \right) } (2k)!!,\quad & n=2k \\
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\end{cases}
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\end{aligned}
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$$
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由$\displaystyle \sqrt{\frac{(n-1)S_n^{2}}{\sigma^{2}}}=Y \implies E S_n^{3}=\left( \frac{\sigma}{\sqrt{n-1}} \right) ^{3} \cdot E(Y^{3})$代入即得。
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}
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\end{enumerate}
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\end{enumerate}
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\textbf{注：}正常题总分为100分，附加题总分为10分。总评分为两者之和，但总评分不超过100分。
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\end{document} |