\documentclass[全部作业]{subfiles}
\input{mysubpreamble}
\begin{document}
\setcounter{chapter}{5}
\setcounter{section}{3}
\section{三大抽样分布}
\begin{enumerate}
    \questionandanswerSolution[2]{
        设$x_1,x_2, \cdots ,x_n$是来自$N(\mu,16)$的样本，问$n$多大时才能使得$P(\left\vert \bar{x}-\mu \right\vert<1 )\geqslant 0.95$成立？
    }{
        由于$\bar{x}\sim N(\mu,\frac{16}{n})$，根据切比雪夫不等式，
        $$
        P(\left\vert \bar{x}-E \bar{x} \right\vert <\varepsilon)\geqslant 1-\frac{\operatorname{Var}\bar{x}}{\varepsilon}
        $$
        $$
        P(\left\vert \bar{x}-\mu \right\vert <1)\geqslant 1-\frac{16}{n}
        $$
        $$
        \frac{16}{n}=0.95 \Rightarrow n=\frac{16}{0.95} = \frac{320}{19} \approx  16.8421052631579
        $$
        因为$n$为整数，所以$n$至少为17时才能使得$P(\left\vert \bar{x}-\mu \right\vert<1 )\geqslant 0.95$成立。
    }
    \questionandanswerSolution[4]{
        由正态总体$N(\mu,\sigma^{2})$抽取容量为20的样本，试求$P\left( 10\sigma^{2}\leqslant \displaystyle \sum_{i=1}^{20} (x_i-\mu)^{2}\leqslant 30\sigma^{2} \right) $。
    }{
        由于$\displaystyle s^{2}=\frac{1}{19}\sum_{i=1}^{20} (x_i-\mu)^{2}$, $\displaystyle \frac{19s^{2}}{\sigma^{2}} \sim \chi^{2}(n-1)$，
        所以$\displaystyle \frac{1}{\sigma^{2}}\sum_{i=1}^{20} (x_i-\mu)^{2}\sim \chi^{2}(n-1)$。
        所以
        $$
        \begin{aligned}
            &P\left( 10\sigma^{2}\leqslant \sum_{i=1}^{20} (x_i-\mu)^{2}\leqslant 30\sigma^{2} \right) = P\left( 10\leqslant \frac{1}{\sigma^{2}}\sum_{i=1}^{20} (x_i-\mu)^{2}\leqslant 30 \right)  \\
            &=\int_{10}^{30} \frac{\left( \frac{1}{2} \right) ^{\frac{20}{2}}}{\Gamma\left( \frac{20}{2}-1 \right) } y^{\frac{20}{2}}e^{-\frac{y}{2}} \mathrm{d}y = \int_{10}^{30} \frac{\left( \frac{1}{2} \right) ^{10}}{9!} y^{9}e^{-\frac{y}{2}} \mathrm{d}y \approx  0.898318281994385 \\
        \end{aligned}
        $$
        \begin{center}
            \includegraphics[width=0.2\linewidth]{imgs/2024-03-20-14-05-40.png}
        \end{center}
    }
    \questionandanswerSolution[6]{
        设$x_1,x_2, \cdots ,x_n$是来自$N(\mu,1)$的样本，试确定最小的常数$c$，使得对任意的$\mu\geqslant 0$，有$P(\left\vert \bar{x} \right\vert <c)\leqslant \alpha$。
    }{
        这题什么意思？当$\mu=0$时，当$n \to \infty$时$\bar{x} \to 0$, $P(\left\vert \bar{x} \right\vert <c) \to 1$，怎么可能$P(\left\vert \bar{x} \right\vert <c)\leqslant \alpha$呢？
    }
    \questionandanswerProof[8]{
        设随机变量$X\sim F(n,m)$，证明$\displaystyle Z=\frac{n}{m}X \left\slash \left( 1+\frac{n}{m}X \right) \right.$服从贝塔分布，并指出其参数。
    }{
        设$Y=\dfrac{n}{m}X$，则
        $$
        p_{Y}(y)=\frac{\Gamma\left( \frac{m+n}{2} \right) }{\Gamma\left( \frac{m}{2} \right) \Gamma\left( \frac{n}{2} \right) } y^{\frac{m}{2}-1}(1+y)^{-\frac{m+n}{2}}
        $$
        $$
        Z=\frac{Y}{1+Y} = 1 - \frac{1}{1+Y} \Longrightarrow Y=\frac{1}{1-Z}-1=\frac{Z}{1-Z}
        $$
        所以
        $$
        \begin{aligned}
            &p_{Z}(z)=\frac{\Gamma(\frac{m+n}{2})}{\Gamma(\frac{m}{2})\Gamma(\frac{n}{2})}\left( \frac{z}{1-z} \right) ^{\frac{m}{2}-1} \left( \frac{1}{1-z} \right) ^{- \frac{m+n}{2}} \\
            &=\frac{\Gamma(\frac{m+n}{2})}{\Gamma(\frac{m}{2})\Gamma(\frac{n}{2})} \left( \frac{z}{1-z} \right) ^{\frac{m}{2}} \left( \frac{1-z}{z} \right) \left( 1-z \right) ^{\frac{m}{2}} \left( 1-z \right) ^{\frac{n}{2}} \\
            &=\frac{\Gamma(\frac{m+n}{2})}{\Gamma(\frac{m}{2})\Gamma(\frac{n}{2})}z^{\frac{m}{2}}(1-z)^{\frac{n}{2}}\left( \frac{1-z}{z} \right)  \\
            &=\frac{\Gamma(\frac{m+n}{2})}{\Gamma(\frac{m}{2})\Gamma(\frac{n}{2})} z^{\frac{m}{2}-1} (1-z)^{\frac{n}{2}+1} \\
        \end{aligned}
        $$

        所以$Z$服从贝塔分布，其参数为$\dfrac{m}{2}$和$\dfrac{n}{2}$。
    }
    \questionandanswerSolution[9]{
        设$x_1,x_2$是来自$N(0,\sigma^{2})$的样本，试求$\displaystyle Y=\left( \frac{x_1+x_2}{x_1-x_2} \right) ^{2}$的分布。
    }{
        $$
        Y=\left( \frac{x_1+x_2}{x_1-x_2} \right) ^{2}=\left( \frac{\frac{x_1}{x_2}+1}{\frac{x_1}{x_2}-1} \right) ^{2}=\left( 1+\frac{2}{\frac{x_1}{x_2}-1} \right) ^{2}
        $$
        其中$\dfrac{x_1}{x_2}$的概率密度函数为
        $$
        \begin{aligned}
        p_{\frac{x_1}{x_2}}(t)&=\int_{-\infty}^{+\infty} \left\vert x \right\vert p(x,tx) \mathrm{d}x=\int_{-\infty}^{+\infty} \left\vert x \right\vert \phi(x)\phi(tx) \mathrm{d}x =\int_{-\infty}^{+\infty} \left\vert x \right\vert \frac{1}{\sqrt{2\pi}}e^{\frac{-\sigma^{2}x^{2}}{2}}\cdot \frac{1}{\sqrt{2\pi}}e^{\frac{-\sigma^{2}t^{2}x^{2}}{2}} \mathrm{d}x \\
        &=\frac{1}{\sigma^{2}\pi}\int_{0}^{+\infty} x e^{\frac{-\sigma^{2}x^{2}}{2}(1+t^{2})} \mathrm{d}x = \frac{1}{\pi \sigma^{4} (t^{2} + 1)} \\
        \end{aligned}
        $$
        设随机变量$Z=1+\frac{2}{\frac{x_1}{x_2}-1}$，则$\frac{x_1}{x_2}=1+\frac{2}{Z-1}$, $Y=Z^{2}$，所以
        $$
        p_{Z}(z)=\frac{1}{\pi\sigma^{4}\left[ \left( 1+\frac{2}{z-1} \right) ^{2}+1 \right] } = \frac{(z - 1)^{2}}{\pi \sigma^{4} ((z - 1)^{2} + (z + 1)^{2})}
        $$
        $$
        \begin{aligned}
            p_{Y}(y)&=p_{Z}(\sqrt{y})+p_{Z}(-\sqrt{y})=\frac{(\sqrt{y} - 1)^{2}}{\pi \sigma^{4} ((\sqrt{y} - 1)^{2} + (\sqrt{y} + 1)^{2})}+\frac{(\sqrt{y} - 1)^{2}}{\pi \sigma^{4} ((\sqrt{y} - 1)^{2} + (\sqrt{y} + 1)^{2})}  \\
            &= \frac{- 2 \sqrt{y} + y + 1}{\pi \sigma^{4} (y + 1)} \\
        \end{aligned}
        $$
        此即为$Y$的概率密度函数。
    }
\end{enumerate}
\end{document}