SchoolWork-LaTeX/概率论/平时作业/单元作业3.tex

\documentclass[全部作业]{subfiles}

\begin{document}
\chapter{单元作业3}
\begin{enumerate}
    \item 试用随机变量$X$的分布函数$F_{X}(x)$表示随机变量$-\min(X,0)$的分布函数。
    \begin{proof}[解]
    \begin{zhongwen}
        $$
        \begin{aligned}
        % F_{-\min(X,0)}(x)=-\min(F_{X}(x),0)
        F_{-\min(X,0)}(x) & = P(-\min(X,0)\leqslant x)  \\
        & = P(\min(X,0)\geqslant -x)  \\
        & = P(X\geqslant -x, 0\geqslant -x)  \\
        & = P(X\geqslant -x, x\geqslant 0)  \\
        &=\begin{cases} P(X\geqslant -x), & x\geqslant 0 \\ 0, & x<0 \end{cases} \\
        &=\begin{cases} 1-P(X<-x), & x\geqslant 0 \\ 0, & x<0 \end{cases} \\
        &=\begin{cases} 1-F_{X}(-x-0), & x\geqslant 0 \\ 0, & x<0 \end{cases} \\
        \end{aligned}
        $$
    \end{zhongwen}
    \end{proof}
    \item 设随机变量$X$等可能地取值0和1, 求$X$的分布函数。
    \begin{proof}[解]
    \begin{zhongwen}
        $$
        \begin{aligned}
        F_{X}(x)=\begin{cases} 0, & x<0 \\ \frac{1}{2}, & 0\leqslant x<1 \\ 1, & x\geqslant 1 \end{cases}
        \end{aligned}
        $$
    \end{zhongwen}
    \end{proof}
    \item 设离散型随机变量X的分布列为$\displaystyle P(X=k)=\frac{c}{2^{k}}, \quad k=0, 1, 2, \ldots $，求常数$c$的值。
    \begin{proof}[解]
    \begin{zhongwen}
        $$
        \begin{aligned}
        &\because \sum_{k = 0}^{\infty}\frac{c}{2^{k}}=1 \\
        &\therefore c \sum_{k=0}^{\infty}\frac{1}{2^{k}}=c \frac{1}{1-\frac{1}{2}}=2c=1 \\
        &\therefore c=\frac{1}{2} \\
        \end{aligned}
        $$
    \end{zhongwen}
    \end{proof}
    \item 设随机变量$X$的概率密度函数为$p(x)=\begin{cases} ce^{-\sqrt{x}}, & x>0 \\ 0, & x\leqslant 0 \end{cases}$，求常数$c$。
    \begin{proof}[解]
    \begin{zhongwen}
        $$
        \begin{aligned}
        &\because \int_{-\infty}^{+\infty} p(x) \mathrm{d}x=1 \\
        &\therefore \int_{0}^{+\infty} ce^{-\sqrt{x}} \mathrm{d}x=c \int_{0}^{+\infty} e^{-\sqrt{x}} \mathrm{d}x = 2c=1 \\
        &\therefore c=\frac{1}{2} \\
        \end{aligned}
        $$
    \end{zhongwen}
    \end{proof}
    \item 设$X\sim N(10,4)$，求概率$P(6<X\leqslant 9)$和$P(7\leqslant X<12)$。
    \begin{proof}[解]
    \begin{zhongwen}
        $$
        \begin{aligned}
        &\mu =10,\sigma ^{2}=4,\sigma =2 \\
        &\therefore \frac{x-10}{2}\sim N(0,1) \\
        P(6<X\leqslant 9) & = P(\frac{6-10}{2}<\frac{X-10}{2}\leqslant \frac{9-10}{2})  \\
        & = \Phi (\frac{9-10}{2})-\Phi (\frac{6-10}{2})  \\
        & = \Phi (-0.5)-\Phi (-2)  \\
        & = \Phi (2)-\Phi (0.5)  \\
        & \approx  0.9772-0.6915  \\
        & = 0.2857  \\
        P(7\leqslant X<12) & = P(\frac{7-10}{2}\leqslant \frac{X-10}{2}< \frac{12-10}{2})  \\
        & = \Phi (\frac{12-10}{2}-0)-\Phi (\frac{7-10}{2}-0)  \\
        & = \Phi (1)-\Phi (-1.5)  \\
        & = \Phi (1)-(1-\Phi (1.5))  \\
        & \approx  0.8413-(1-0.9332)  \\
        & = 0.7745  \\
        \end{aligned}
        $$
    \end{zhongwen}
    \end{proof}
    \item 设$X\sim \operatorname{Exp} (\lambda )$，求$\lambda $使得$P(X>1)=2P(X>2)$。
    \begin{proof}[解]
    \begin{zhongwen}
        $$
        \begin{aligned}
        由P(X>1)&=2P(X>2)可知： \\
        \int_{1}^{+\infty} \lambda e^{-\lambda x} \mathrm{d}x&=2\int_{2}^{+\infty} \lambda e^{-\lambda x} \mathrm{d}x \\
        \left. - e^{- \lambda x} \right|_{1}^{+\infty}&=-2\left. e^{- \lambda x} \right|_{2}^{+\infty} \\
        e^{-\lambda }&=2e^{-2\lambda } \\
        2(e^{-\lambda })^{2}-e^{-\lambda }&=0 \\
        e^{-\lambda }(2e^{-\lambda }-1)&=0 \\
        \end{aligned} \hspace{5em}
        \begin{aligned}
        &\because e^{-\lambda }\neq 0 \\
        &\therefore 2e^{-\lambda }-1=0 \\
        &\therefore e^{-\lambda }=\frac{1}{2} \\
        &\therefore \lambda =-\ln (\frac{1}{2})=\ln 2 \\
        % &\left. \frac{(- \lambda x - 1) e^{- \lambda x}}{\lambda} \right|_{1}^{+\infty} =2\left. \frac{(- \lambda x - 1) e^{- \lambda x}}{\lambda} \right|_{2}^{+\infty} \\
        % &0-\frac{(-\lambda -1)e^{-\lambda }}{\lambda }=2 (0-\frac{(-2\lambda -1)e^{-2\lambda }}{\lambda }) \\
        % &(\lambda +1)e^{-\lambda }=2(2\lambda +1)e^{-2\lambda } \\
        % &\lambda e^{-\lambda }+e^{-\lambda }=4\lambda e^{-2\lambda }+2e^{-2\lambda } \\
        % &使用fx-991 \mathrm{CN}\ \mathrm{X}解得\lambda =2857100 \\
        % solve(latex2sympy(r"0-\frac{(-\lambda -1)e^{-\lambda }}{\lambda }=2 (0-\frac{(-2\lambda -1)e^{-2\lambda }}{\lambda })"))
        % solve(0 - (-y - 1)*exp(-y)/y - 2*(0 - (-1*2*y - 1)*exp(-1*2*y)/y))
        \end{aligned}
        $$
        \begin{center}
            % \includegraphics[width=0.5\linewidth]{imgs/2023-11-08-13-38-39.png}
        \end{center}
    \end{zhongwen}
    \end{proof}
    \item 设随机变量$X$服从几何分布$\operatorname{Ge}(p)$，求$X$的数学期望$EX$。
    \begin{proof}[解]
    \begin{zhongwen}
        $$
        \begin{aligned}
        &\because  0<p<1 \\
        &\therefore EX=\sum_{k=1}^{\infty}kp(1-p)^{k-1} = \frac{1}{p} \\
        \end{aligned}
        $$
    \end{zhongwen}
    \end{proof}
    \item 设随机变量$X$服从几何分布$\operatorname{Ge}(p)$，求$X$的方差$\operatorname{Var}X$。
    \begin{proof}[解]
    \begin{zhongwen}
        $$
        \begin{aligned}
        &\because 0<p<1 \\
        \therefore \operatorname{Var}X & = EX^{2}-(EX)^{2}  \\
        & = \sum_{k=1}^{\infty}k^{2}p(1-p)^{k-1} - \left( \frac{1}{p} \right) ^{2}  \\
        & = \frac{2 - p}{p^{2}} - \frac{1}{p^{2}}  \\
        & = \frac{1-p}{p^{2}}  \\
        \end{aligned}
        $$
    \end{zhongwen}
    \end{proof}
    \item 设随机变量$X$服从Gamma分布$\operatorname{Ga}(\alpha ,\lambda )$，求数学期望$EX^{n}$。
    \begin{proof}[解]
    \begin{zhongwen}
        $$
        \begin{aligned}
        EX^{n} & = \int_{-\infty}^{+\infty} x^{n}p(x) \mathrm{d}x  \\
        &=\int_{0}^{+\infty} x^{n} \cdot \frac{\lambda ^{\alpha }}{\Gamma (\alpha )}x^{\alpha -1}e^{-\lambda x} \mathrm{d}x \\
        &=\frac{1}{\lambda ^{n}\Gamma(\alpha )}\int_{0}^{+\infty} (\lambda x)^{n+\alpha -1}e^{-\lambda x} \mathrm{d}(\lambda x) \\
        &=\frac{\Gamma(n+\alpha )}{\lambda ^{n}\Gamma(\alpha )} \\
        &=\frac{\alpha (\alpha +1)\cdots(\alpha +n-1)}{\lambda ^{n}} \\
        \end{aligned}
        $$
    \end{zhongwen}
    \end{proof}
    \item 设随机变量$X$服从均匀分布$U(0,1)$，计算随机变量$X^{3}$的方差。
    \begin{proof}[解]
    \begin{zhongwen}
        $$
        由题意可知，X的概率密度函数为p(x)=\begin{cases} 0, & x<0 或x>1\\ 1, & 0\leqslant x\leqslant 1 \end{cases}
        $$
        $$
        \begin{aligned}
        \operatorname{Var}X^{3} & = E(X^{3})^{2}-(EX^{3})^{2}  \\
        &=\int_{-\infty}^{+\infty} x^{6}p(x) \mathrm{d}x-\left( \int_{-\infty}^{+\infty} x^{3}p(x) \mathrm{d}x \right) ^{2} \\
        &=\int_{0}^{1} x^{6} \mathrm{d}x-\left( \int_{0}^{1} x^{3} \mathrm{d}x \right) ^{2} \\
        & = \frac{9}{112}  \\
        & \approx  0.0803571428571429 \\
        \end{aligned}
        $$
    \end{zhongwen}
    \end{proof}
\end{enumerate}
\end{document}