重构目录层次
0-课程笔记 1-平时作业 2-实验报告 3-期末大作业
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概率论/平时作业/mypreamble.tex
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概率论/平时作业/mypreamble.tex
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\usepackage[margin=1in]{geometry}
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\usepackage{environ} % 加了这个再\def\myitem就不报错了
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\usepackage{extarrows}
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\usepackage{amssymb, amsfonts, amstext, amsmath, amsopn, amsthm}
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% \usepackage{mathrsfs} % \mathscr
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\usepackage{enumitem}
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\usepackage{setspace}
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\usepackage{color}
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\usepackage{mylatex}
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\usepackage{diagbox}
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\usepackage{makecell}
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\usepackage{mathtools} % \coloneqq 在好几个包里都出现了，不知道引入哪个最好
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% \usepackage{floatflt}
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% \usepackage{wrapfig}
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\usepackage{picinpar}
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% \usepackage{cutwin}
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% https://www.zhihu.com/question/26837705 试了好几个，发现能在列表和证明环境中完美使用的只有picinpar
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\usepackage{amsrefs}
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\usepackage{hyperref}
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\usepackage{subfiles}
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\setlist[1]{label=\arabic{enumi}., listparindent=\parindent}
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\setlist[2]{label=(\arabic{enumii}), listparindent=\parindent}
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\definecolor{shadecolor}{RGB}{204,232,207}
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\def\myitem#1#2{
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\item \textbf{#1}
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\begin{enumerate}
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#2
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\end{enumerate}
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}
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\ExplSyntaxOn
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\cs_set:Nn \rawquestionandanswer:Nnnn {%
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\begin{shaded}%
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\ifstrequal{#2}{-}{}{\format_item:Nn #1{#2}} #3%
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\end{shaded}%
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\begin{zhongwen}%
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#4%
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\end{zhongwen}%
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}
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\cs_set:Nn \format_item:Nn {
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\IfBlankTF{#2}{
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\item
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}{
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\item[#1{#2}]
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}
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}
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\cs_set:Nn \simple_format:n {R#1.}
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\newcommand{\questionandanswer}[3][]{%
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\rawquestionandanswer:Nnnn \simple_format:n {#1}{#2}{#3}
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}
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\ExplSyntaxOff
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% 成功实现了，而且下划线也没问题！！！似乎是函数式程序设计，或者装饰器模式？
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18
概率论/平时作业/全部作业.tex
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概率论/平时作业/全部作业.tex
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\documentclass[a4paper]{ctexbook}
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\input{mypreamble}
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\title{\heiti\zihao{2} 《概率论》作业}
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\author{\songti 岳锦鹏}
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\newcommand{\mysignature}{10213903403 岳锦鹏}
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\date{2023年9月30日 —— 2024年1月4日}
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\begin{document}
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\renewcommand{\bar}{\xoverline} % 这一行在编译时可以取消注释，注意一定要放在\begin{document}下面才有用
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% \renewcommand{\overline}{\xoverline} % 这样会导致递归展开错误，暂时未解决
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\maketitle
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\tableofcontents
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\subfile{单元作业1}
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\subfile{单元作业2}
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\subfile{单元作业3}
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\subfile{单元作业4}
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\subfile{单元作业5}
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\subfile{单元作业6}
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\end{document}
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159
概率论/平时作业/单元作业1.tex
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概率论/平时作业/单元作业1.tex
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\documentclass[全部作业]{subfiles}
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\begin{document}
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\chapter{单元作业1}
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\begin{enumerate}
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\item 证明$P(AB)\ge P(A)+P(B)-1$。
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\begin{proof}
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\begin{zhongwen}
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$$
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\begin{aligned}
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&设\Omega 表示全集，则A \in \Omega, B \in \Omega \\
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&\therefore A\cup B\in \Omega \\
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&\therefore P(A\cup B)\le P(\Omega) \\
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&\therefore P(A)+P(B)-P(AB)\le 1 \\
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&\therefore P(AB)\ge P(A)+P(B)-1 \\
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\end{aligned}
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$$
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\end{zhongwen}
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\end{proof}
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\item 一副标准扑克牌52张一张一张轮流分发给给4名游戏者，求每人恰好得到1张A的概率。
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\begin{proof}[解]
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\begin{zhongwen}
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$$
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\begin{aligned}
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&设A表示事件“每人恰好得到一张A”，由盒子模型可知： \\
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&P(A) = \frac{\mathrm{P}_{4}^{4}}{4^{4}} = \frac{4!}{256} = \frac{3}{32} \\
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\end{aligned}
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$$
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\end{zhongwen}
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\end{proof}
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\item 设$P(A)=0.7,P(A-B)=0.3$，求概率$P(\overline{AB})$。
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\begin{proof}[解]
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\begin{zhongwen}
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$$
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\begin{aligned}
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P(\overline{AB})&=1-P(AB)=1-P(A-(A-B)) \\
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&=1-(P(A)-P(A-B))=1-(0.7-0.3)=0.6 \\
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\end{aligned}
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$$
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\end{zhongwen}
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\end{proof}
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\item 设$P(A)=a,P(B)=b,P(A\cup B)=c$，求概率$P(\overline{\bar{A}\cup \bar{B}})$。
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\begin{proof}[解]
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\begin{zhongwen}
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$$
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\begin{aligned}
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P(\overline{\bar{A}\cup \bar{B}})=P(A\cap B)=P(A)+P(B)-P(A\cup B)=a+b-c
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\end{aligned}
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$$
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\end{zhongwen}
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\end{proof}
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\item 设$A,B\in \mathcal{F}$，证明$P(A)=P(AB)+P(A \bar{B}), \quad P(A\triangle B)=P(A)+P(B)-2P(AB)$。
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\begin{proof}
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\begin{zhongwen}
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$$
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\begin{aligned}
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&设\Omega 表示全集 \\
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P(A)&=P(A\cap \Omega)=P(A\cap (B\cup \bar{B})) \\
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&=P(A(B+\bar{B}))=P(AB+A \bar{B})=P(AB)+P(A \bar{B}) \\
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P(A \triangle B)&=P((A\backslash B)\cup (B\backslash A))=P((A-(AB))+(B-(AB))) \\
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&=P(A)-P(AB)+P(B)-P(AB)=P(A)+P(B)-2P(AB) \\
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\end{aligned}
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$$
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\end{zhongwen}
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\end{proof}
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\item 设$\Omega =(-\infty, \infty), A=\{ x\in \Omega: 1\le x\le 5 \}, B=\{ x\in \Omega:3<x<7 \},C=\{ x\in \Omega:x<0 \}$，求下列事件$\bar{A}, A\cup B,B \bar{C}, \bar{A}\cap \bar{B}\cap \bar{C}, (A\cup B)C$。
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\begin{proof}[解]
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\begin{zhongwen}
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$$
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\begin{aligned}
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&\bar{A}=\{ x\in \Omega: x < 1 或x>5 \} \\
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&A\cup B=\{ x\in \Omega: 1\le x<7 \} \\
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&B \bar{C}=\{ x\in \Omega:3<x<7 \} \\
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&\bar{A}\cap \bar{B}\cap \bar{C}=\{ x\in \Omega:0\le x<1或x\ge 7 \} \\
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&(A\cup B)C=\varnothing \\
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\end{aligned}
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$$
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\end{zhongwen}
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\end{proof}
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\item 设\textit{I}是任意指标集，$\{ A_i, i\in I \}$是一事件类，证明$\overline{\displaystyle \bigcup_{i \in I} A_i}=\displaystyle \bigcap_{i \in I} \overline{A_i}$, \quad $\overline{\displaystyle \bigcap_{i \in I} A_i}=\displaystyle \bigcup_{i \in I} \overline{A_i}$。
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\begin{proof}
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\begin{zhongwen}
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$$
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\begin{aligned}
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&\forall x\in \overline{\bigcup_{i \in I} A_i},\ 即x \notin \bigcup_{i \in I} A_i, 即 \\
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&\forall i \in I,x \notin A_i, \\
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&\therefore \forall i \in I, x\in \overline{A_i} \\
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&\therefore x\in \bigcap_{i \in I} \overline{A_i} \\
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\end{aligned}
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$$
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\begin{equation}
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\therefore \overline{\bigcup_{i \in I} A_i} \subseteq \bigcap_{i \in I} \overline{A_i} \tag{1}\label{eq:7.1}
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\end{equation}
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$$
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\begin{aligned}
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&\forall x\in \bigcap_{i \in I} \overline{A_i}, 即 \\
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&\forall i \in I,x\in \overline{A_i} \\
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&\therefore \forall i\in I,x \notin A_i \\
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&\therefore x\notin \bigcup_{i \in I} A_i \\
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&\therefore x\in \overline{\bigcup_{i \in I} A_i} \\
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\end{aligned}
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$$
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\begin{equation}
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\therefore \bigcap_{i \in I} \overline{A_i} \subseteq \overline{\bigcup_{i \in I} A_i} \tag{2}\label{eq:7.2}
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\end{equation}
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$$
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\begin{aligned}
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&由\eqref{eq:7.1}\eqref{eq:7.2}可知，\overline{\bigcup_{i \in I} A_i}=\bigcap_{i \in I} \overline{A_i} \\
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&由对偶公式可知, \overline{\overline{\bigcup_{i \in I} A_i}}=\overline{\bigcap_{i \in I} \overline{A_i}} \\
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&\therefore \overline{\bigcap_{i \in I} \overline{A_i}}=\bigcup_{i \in I} A_i \\
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&\because \{ \overline{A_i}|i\in I \}也是一事件类 \\
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&\therefore 将\overline{A_i}替换为A_i后可得 \\
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&\overline{\bigcap_{i \in I} A_i}=\bigcup_{i \in I} \overline{A_i} \\
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\end{aligned}
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$$
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\end{zhongwen}
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\end{proof}
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\item 从装有10双不同尺码或不同样式的皮鞋的箱子中, 任取4只, 求恰能成1双的概率。
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\begin{proof}[解]
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\begin{zhongwen}
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$$
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\begin{aligned}
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&设A表示事件“恰能成1双”，B表示事件“恰能成2双”，C表示事件“不能成双”， \\
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&则由不返回抽样模型可知： \\
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P(A)&=1-P(B)-P(C)=1-\frac{\mathrm{C}_{10}^{2}}{\mathrm{C}_{20}^{4}}-\frac{\mathrm{C}_{20}^{1}\mathrm{C}_{18}^{1}\mathrm{C}_{16}^{1}\mathrm{C}_{14}^{1}}{\mathrm{P}_{20}^{4}} \\
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&=1-\frac{\binom{10}{2}}{\binom{20}{4}}-\frac{\binom{20}{1}\binom{18}{1}\binom{16}{1}\binom{14}{1}}{\binom{20}{4}\times 4!} = \frac{96}{323} \approx 0.297213622291022 \\
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\end{aligned}
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$$
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\end{zhongwen}
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\end{proof}
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\item 现从有15名男生和30名女生的班级中随机挑选10名同学参加某项课外活动, 求在被挑选的同学中恰好有3名男生的概率。
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\begin{proof}[解]
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\begin{zhongwen}
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$$
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\begin{aligned}
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&设A表示事件“在被挑选的同学中恰好有3名男生”，则 \\
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&P(A)=\frac{\mathrm{C}_{15}^{3}\mathrm{C}_{30}^{7}}{\mathrm{C}_{45}^{10}}=\frac{\binom{15}{3}\binom{30}{7}}{\binom{45}{10}} = \frac{3958500}{13633279} \approx 0.290355680390609 \\
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\end{aligned}
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$$
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\end{zhongwen}
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\end{proof}
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\item 设$\mathcal{F}$是$\Omega$上的事件域，$A,B\in \mathcal{F}$，证明$A\cup B,AB\in \mathcal{F}$。
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\begin{proof}
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\begin{zhongwen}
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$$
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\begin{aligned}
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&设A_1=A,A_2=B,\forall n>2,A_n=\varnothing \\
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&\because \mathcal{F}是\Omega上的事件域,\forall n\geqslant 1,A_n\in \mathcal{F} \\
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&\therefore \bigcup_{n=1}^{\infty}A_n=A\cup B\in \mathcal{F} \\
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&又\because A,B\in \mathcal{F} \\
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&\therefore \bar{A},\bar{B}\in \mathcal{F} \\
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&同理，\bar{A}\cup \bar{B}\in \mathcal{F} \\
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&\therefore \overline{\bar{A}\cup \bar{B}}=AB\in \mathcal{F} \\
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\end{aligned}
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$$
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\end{zhongwen}
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\end{proof}
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\end{enumerate}
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\end{document}
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198
概率论/平时作业/单元作业2.tex
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概率论/平时作业/单元作业2.tex
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\documentclass[全部作业]{subfiles}
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\begin{document}
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\renewcommand{\bar}{\xoverline}
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\chapter{单元作业2}
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\begin{enumerate}
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\item 三人独立地对同一目标进行射击, 各人击中目标的概率分别是0.7, 0.8, 0.6, 求目标被击中的概率.
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\begin{proof}[解]
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\begin{zhongwen}
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$$
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\begin{aligned}
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&将三人击中目标的事件分别表示为A、B、C， \\
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&则P(A)=0.7, P(B)=0.8, P(C)=0.6, \\
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&P(目标被击中) = P(A\cup B\cup C) \\
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&\because 三人独立射击 \\
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&\therefore A、B、C相互独立 \\
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&\therefore \bar{A}、\bar{B}、\bar{C}相互独立 \\
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\end{aligned}
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$$ $$
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\begin{aligned}
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\therefore P(A\cup B\cup C)&=P(\overline{\bar{A} \bar{B} \bar{C}}) = 1 - P(\bar{A} \bar{B} \bar{C}) \\
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&=1 - P(\bar{A})P(\bar{B})P(\bar{C}) \\
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&=1 - (1-P(A))(1-P(B))(1-P(C)) \\
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&=1-(1-0.7)*(1-0.8)*(1-0.6) \\
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&= 0.976 \\
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\end{aligned}
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$$
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\end{zhongwen}
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\end{proof}
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\item 制作某个产品有两个关键工序, 第一道和第二道工序的不合格品的概率分别为3\%和5\%, 假定两道工序互不影响, 试问该产品为不合格品的概率(答案保留至小数点后4位).
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\begin{proof}[解]
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\begin{zhongwen}
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$$
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\begin{aligned}
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&令A：第一道工序不合格，B：第二道工序不合格 \\
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&则P(A)=0.03，P(B)=0.05 \\
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&则P(该产品为不合格品)=P(A\cup B) \\
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&\because 两道工序互不影响 \\
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&\therefore A与B相互独立 \\
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&\therefore \bar{A}与\bar{B}相互独立 \\
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\end{aligned}
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$$
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$$
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\begin{aligned}
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\therefore P(A\cup B)&=P(\overline{\bar{A}\bar{B}})=1-P(\bar{A}\bar{B}) \\
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&=1-P(\bar{A})P(\bar{B}) \\
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&=1- (1-P(A))(1-P(B)) \\
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&=1-(1-0.03)(1-0.05) \\
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&= 0.0785 \\
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\end{aligned}
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$$
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\end{zhongwen}
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\end{proof}
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\myitem{甲乙丙三个同学同时独立参加考试, 不及格的概率分别为: 0.2, 0.3, 0.4,}{
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\item 求恰有2位同学不及格的概率；\label{1}
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\item 若已知3位同学中有2位不及格,求其中1位是同学乙的概率. \label{2}
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\begin{proof}[解]
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\begin{zhongwen}
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$$
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\begin{aligned}
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&设A、B、C分别表示事件甲不合格、乙不合格、丙不合格 \\
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||||
&则P(A)=0.2, P(B)=0.3, P(C)=0.4 \\
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&P(\bar{A})=0.8, P(\bar{B})=0.7, P(\bar{C})=0.6 \\
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&且A、B、C相互独立 \\
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\end{aligned}
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$$
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\ref{1}
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$$
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\begin{aligned}
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P(恰有2位同学不合格)&=P(AB \bar{C}+A \bar{B} C+\bar{A}BC) \\
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||||
&=P(A)P(B)P(\bar{C})+P(A)P(\bar{B})P(C)+P(\bar{A})P(B)P(C) \\
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&=0.2\times 0.3\times 0.6+0.2\times 0.7\times 0.4+0.8\times 0.3\times 0.4 \\
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&= 0.188 \\
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||||
\end{aligned}
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$$
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\ref{2}
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||||
\setlength{\lineskip}{2.5pt}
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\setlength{\lineskiplimit}{2.5pt}
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$$
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||||
\begin{aligned}
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||||
% \setstretch{1.5}
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P(其中1位是同学乙)&=P(\left. B\ \right|\ 恰有2位同学不合格) \\
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||||
&=\frac{P(B\cap 恰有2位同学不合格)}{P(恰有2位同学不合格)} \\
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||||
&=\frac{P(AB \bar{C} + \bar{A} BC)}{P(AB \bar{C}+A \bar{B}C+\bar{A}BC)} \\
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||||
&=\frac{P(A)P(B)P(\bar{C})+P(\bar{A})P(B)P(C)}{P(A)P(B)P(\bar{C})+P(A)P(\bar{B})P(C)+P(\bar{A})P(B)P(C)} \\
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||||
&=\frac{0.2\times 0.3\times 0.6+0.8\times 0.3\times 0.4}{0.2\times 0.3\times 0.6+0.2\times 0.7\times 0.4+0.8\times 0.3\times 0.4} \\
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||||
&= \frac{33}{47} \\
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||||
&\approx 0.702127659574468 \\
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||||
\end{aligned}
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$$
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||||
\end{zhongwen}
|
||||
\end{proof}
|
||||
}
|
||||
\item 甲袋中装有2个白球和4个黑球, 乙袋中装有3个白球和2个黑球, 现随机地从乙袋中取出一球放入甲袋, 然后从甲袋中随机取出一球, 试求从甲袋中取得的球是白球的概率.
|
||||
\begin{proof}[解]
|
||||
\begin{zhongwen}
|
||||
$$
|
||||
\begin{aligned}
|
||||
&设A表示从乙袋中取出的是白球，B表示从甲袋中取出的是白球 \\
|
||||
则P(B) &= P(A)P(B|A)+P(\bar{A})P(B|\bar{A}) \\
|
||||
&=\frac{3}{3+2}\times \frac{2+1}{2+4+1}+\frac{2}{3+2}\times \frac{2}{2+4+1} \\
|
||||
&= \frac{13}{35} \\
|
||||
&\approx 0.371428571428571 \\
|
||||
\end{aligned}
|
||||
$$
|
||||
\end{zhongwen}
|
||||
\end{proof}
|
||||
\item 设n只罐子的每一只中装有4个白球和6个黑球, 另有一只罐子中装有5个白球和5个黑球.从这n+1个罐子中随机地选择一只罐子, 从中任取两个球, 结果发现两个都是黑球. 已知在此条件下, 有5个白球和3个黑球留在选出的罐子中的条件概率是1/7, 求n的值.
|
||||
\begin{proof}[解]
|
||||
\begin{zhongwen}
|
||||
$$
|
||||
\begin{aligned}
|
||||
&设A表示事件：选中的是第n+1个罐子 \\
|
||||
&设B表示事件：取出的两个都是黑球 \\
|
||||
&则事件：有5个白球和3个黑球留在选出的罐子中，\\
|
||||
&即选中的是第n+1个罐子，并且取出的两个都是黑球， 可表示为AB\\
|
||||
&\therefore P(A|B)=\frac{1}{7}, P(A)=\frac{1}{n+1} \\
|
||||
& P(B|A)=\frac{\mathrm{C}_{5}^{2}}{\mathrm{C}_{10}^{2}}=\frac{2}{9} \\
|
||||
\therefore P(B)&=P(A)P(B|A)+P(\bar{A})P(B|\bar{A})\\
|
||||
&=\frac{1}{n+1}\times \frac{\mathrm{C}_{5}^{2}}{\mathrm{C}_{10}^{2}}+\frac{n}{n+1}\times \frac{\mathrm{C}_{6}^{2}}{\mathrm{C}_{10}^{2}} \\
|
||||
&=\frac{3 n + 2}{9 (n + 1)} \\
|
||||
% &\because A与B相互独立 \\
|
||||
\therefore P(A|B)&=\frac{P(A)P(B|A)}{P(B)} \\
|
||||
&=\frac{\frac{1}{n+1}\times \frac{2}{9}}{\frac{3n+2}{9(n+1)}} \\
|
||||
&= \frac{2}{3 n + 2} = \frac{1}{7} \\
|
||||
&\therefore n=4 \\
|
||||
% solve(latex2sympy(r"\frac{2}{3 n + 2} = \frac{1}{7}")) = {n: 4} = []
|
||||
% solve(latex2sympy(r"\frac{1}{2} = n"))
|
||||
% var["n"]
|
||||
\end{aligned}
|
||||
$$
|
||||
\end{zhongwen}
|
||||
\end{proof}
|
||||
\item 设有三张卡片, 第一张两面皆为红色, 第二张两面皆为黄色, 第三张一面是红色一面是黄色. 随机地选择一张卡片并随机地选择其中一面. 如果已知此面是红色, 求另一面也是红色的概率.
|
||||
\begin{proof}[解]
|
||||
\begin{zhongwen}
|
||||
$$
|
||||
\begin{aligned}
|
||||
&设A、B、C分别表示选到了第一、二、三张卡片，设R表示选择的一面是红色。 \\
|
||||
&则P(A) = P(B) = P(C) = \frac{1}{3} \\
|
||||
&P(R|A)=1, P(R|B)=0, P(R|C)=\frac{1}{2} \\
|
||||
\end{aligned}
|
||||
$$
|
||||
$$
|
||||
\begin{aligned}
|
||||
则P(另一面也是红色) &= P(A|R) \\
|
||||
&=\frac{P(A)P(R|A)}{P(A)P(R|A)+P(B)P(R|B)+P(C)P(R|C)} \\
|
||||
&=\frac{\frac{1}{3}\times 1}{\frac{1}{3}\times 1+\frac{1}{3}\times 0+\frac{1}{3}\times \frac{1}{2}} \\
|
||||
&= \frac{2}{3} \\
|
||||
\end{aligned}
|
||||
$$
|
||||
\end{zhongwen}
|
||||
\end{proof}
|
||||
\item 从装有r个红球和w个白球的盒子中不返回的取出两只, 求事件“第一只为红球, 第二只为白球”的概率.
|
||||
\begin{proof}[解]
|
||||
\begin{zhongwen}
|
||||
$$
|
||||
\begin{aligned}
|
||||
&设A表示取出的两只球一红一白 \\
|
||||
&则P(第一只为红球，第二只为白球) = \frac{P(A)}{\mathrm{P}_{2}^{2}} \\
|
||||
&=\frac{\frac{\mathrm{C}_{r}^{1}\mathrm{C}_{w}^{1}}{\mathrm{C}_{r+w}^{2}}}{\mathrm{P}_{2}^{2}} \\
|
||||
&=\frac{\frac{\binom{r}{1}\binom{w}{1}}{\binom{r+w}{2}}}{2!} \\
|
||||
&= \frac{r w}{(r + w) (r + w - 1)} \\
|
||||
\end{aligned}
|
||||
$$
|
||||
\end{zhongwen}
|
||||
\end{proof}
|
||||
\item 已知$P(A)=\frac{1}{4}, P(B|A)=\frac{1}{3},P(A|B)=\frac{1}{2}$,求概率$P(A\cup B)$。
|
||||
\begin{proof}[解]
|
||||
\begin{zhongwen}
|
||||
$$
|
||||
\begin{aligned}
|
||||
&P(AB) = P(B|A)P(A) = \frac{1}{3}\times \frac{1}{4} = \frac{1}{12} \\
|
||||
&P(B)=\frac{P(AB)}{P(A|B)}=\frac{\frac{1}{12}}{\frac{1}{2}} = \frac{1}{6} \\
|
||||
&\therefore P(A\cup B)=P(A)+P(B)-P(AB) \\
|
||||
&=\frac{1}{4}+\frac{1}{6} - \frac{1}{12} \\
|
||||
&= \frac{1}{3} \\
|
||||
\end{aligned}
|
||||
$$
|
||||
\end{zhongwen}
|
||||
\end{proof}
|
||||
\item 设$P(A)=P(B)=\frac{1}{3},P(A|B)=\frac{1}{6}$，求概率$P(\bar{A}|\bar{B})$。
|
||||
\begin{proof}[解]
|
||||
\begin{zhongwen}
|
||||
$$
|
||||
\begin{aligned}
|
||||
&P(AB)=P(B)P(A|B)=\frac{1}{3}\times \frac{1}{6} = \frac{1}{18} \\
|
||||
&P(\bar{A} \bar{B})=P(\overline{A\cup B}) = 1 - P(A\cup B)=1 - (P(A)+P(B)-P(AB))\\
|
||||
&=1 - (\frac{1}{3}+\frac{1}{3}-\frac{1}{18}) = \frac{7}{18}\\
|
||||
&P(\bar{B}) = 1-P(B) = 1 - \frac{1}{3} = \frac{2}{3} \\
|
||||
&\therefore P(\bar{A}|\bar{B})=\frac{P(\bar{A} \bar{B})}{P(\bar{B})}=\frac{\frac{7}{18}}{\frac{2}{3}} = \frac{7}{12} \\
|
||||
\end{aligned}
|
||||
$$
|
||||
\end{zhongwen}
|
||||
\end{proof}
|
||||
\end{enumerate}
|
||||
\end{document}
|
||||
168
概率论/平时作业/单元作业3.tex
Normal file
168
概率论/平时作业/单元作业3.tex
Normal file
@@ -0,0 +1,168 @@
|
||||
\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}
|
||||
228
概率论/平时作业/单元作业4.tex
Normal file
228
概率论/平时作业/单元作业4.tex
Normal file
@@ -0,0 +1,228 @@
|
||||
\documentclass[全部作业]{subfiles}
|
||||
|
||||
\begin{document}
|
||||
\chapter{单元作业4}
|
||||
\begin{enumerate}
|
||||
\item 若二维离散型随机变量$(X,Y)$有联合分布列如下：
|
||||
|
||||
\begin{center}
|
||||
% \renewcommand\arraystretch{1.5}
|
||||
\begin{tabular}{c|ccc}
|
||||
% \hline
|
||||
\diagbox{$X$}{$Y$} & $0$ & $1$ & $2$ \\
|
||||
\hline
|
||||
$0$ & $\frac{1}{2}$ & $\frac{1}{8}$ & $\frac{1}{4}$ \\
|
||||
% \hline
|
||||
$1$ & $\frac{1}{16}$ & $\frac{1}{16}$ & $0$ \\
|
||||
% \hline
|
||||
\end{tabular}
|
||||
\end{center}
|
||||
求$X$与$Y$的边际分布列。
|
||||
|
||||
\begin{proof}[解]
|
||||
\begin{zhongwen}
|
||||
|
||||
$X$和$Y$的边际分布列分别为
|
||||
|
||||
\begin{center}
|
||||
\renewcommand\arraystretch{1.5}
|
||||
\begin{tabular}{c|cc}
|
||||
$X$ & $0$ & $1$ \\
|
||||
\hline
|
||||
$P$ & \makecell{$\dfrac{7}{8}$} & $\dfrac{1}{8}$ \\
|
||||
\end{tabular} \qquad
|
||||
\begin{tabular}{c|ccc}
|
||||
$Y$ & $0$ & $1$ & $2$ \\
|
||||
\hline
|
||||
$P$ & $\dfrac{9}{16}$ & $\dfrac{3}{16}$ & $\dfrac{1}{4}$ \\
|
||||
\end{tabular}
|
||||
\end{center}
|
||||
\end{zhongwen}
|
||||
\end{proof}
|
||||
|
||||
\item 设随机变量$(X,Y)$的概率密度函数为
|
||||
$$
|
||||
p(x,y)=\begin{cases}
|
||||
3x,\quad & \text{若}0<y<x<1; \\
|
||||
0,\quad & \text{其他}. \\
|
||||
\end{cases}
|
||||
$$
|
||||
求$X$与$Y$的边际概率密度函数，并判断$X$与$Y$是否互相独立。
|
||||
|
||||
\begin{proof}[解]
|
||||
\begin{zhongwen}
|
||||
|
||||
\begin{window}[0,r,
|
||||
{\includexopp[0.3]{4.2.1}},
|
||||
{}]
|
||||
$$
|
||||
\begin{aligned}
|
||||
p_{X}(x)&=\int_{-\infty}^{+\infty} p(x,y) \mathrm{d}y=\begin{cases}
|
||||
\int_{0}^{x} 3x \mathrm{d}y,\quad & 0\leqslant x\leqslant 1 \\
|
||||
0,\quad & x<0 \text{或} x>1 \\
|
||||
\end{cases}\\
|
||||
&=\begin{cases}
|
||||
3x^{2},\quad & 0\leqslant x\leqslant 1 \\
|
||||
0,\quad & x<0 或x>1 \\
|
||||
\end{cases}
|
||||
\end{aligned}
|
||||
$$
|
||||
|
||||
$$
|
||||
\begin{aligned}
|
||||
p_{Y}(y)&=\int_{-\infty}^{+\infty} p(x,y) \mathrm{d}x=\begin{cases}
|
||||
\int_{y}^{1} 3x \mathrm{d}x, \quad & 0\leqslant y\leqslant 1 \\
|
||||
0,\quad & x<0 或 x>1 \\
|
||||
\end{cases}\\
|
||||
&=\begin{cases}
|
||||
\frac{3}{2}-\frac{3}{2}y^{2},\quad & 0\leqslant y\leqslant 1 \\
|
||||
0,\quad & y<0 或 y>1 \\
|
||||
\end{cases}
|
||||
\end{aligned}
|
||||
$$
|
||||
\end{window}
|
||||
|
||||
\noindent$p(1,1)=3,p_{X}(1)=3,p_{Y}(1)=0$，于是$p(1,1)\neq p_{X}(1)\cdot p_{Y}(1)$，所以$X$和$Y$互相不独立。
|
||||
|
||||
\end{zhongwen}
|
||||
\end{proof}
|
||||
|
||||
\item 设随机变量$X$与$Y$相互独立，且$X$服从均匀分布$U(0,1)$，$Y$服从指数分布$\operatorname{Exp}(1)$。求
|
||||
|
||||
\begin{enumerate}
|
||||
\item $(X,Y)$的联合概率密度函数$p(x,y)$；
|
||||
\begin{proof}[解]
|
||||
\begin{zhongwen}
|
||||
|
||||
$$
|
||||
p_{X}(x)=\begin{cases}
|
||||
1,\quad & x\in [0,1] \\
|
||||
0,\quad & x<0 或 x>1 \\
|
||||
\end{cases}\quad p_{Y}(y)=\begin{cases}
|
||||
e^{-y},\quad & y\geqslant 0 \\
|
||||
0,\quad & y< 0 \\
|
||||
\end{cases}
|
||||
$$
|
||||
|
||||
$$
|
||||
p(x,y)=\begin{cases}
|
||||
e^{-y},\quad & x\in [0,1], y\geqslant 0 \\
|
||||
0,\quad & 其他 \\
|
||||
\end{cases}
|
||||
$$
|
||||
\end{zhongwen}
|
||||
\end{proof}
|
||||
\item 概率$P(X+Y\leqslant 1)$；
|
||||
|
||||
\begin{proof}[解]
|
||||
\begin{zhongwen}
|
||||
|
||||
令$Z=X+Y$，则随机变量$Z$的概率密度函数为
|
||||
$$
|
||||
p_{Z}(z)=\int_{-\infty}^{+\infty} p_{X}(x)p_{Y}(z-x) \mathrm{d}x=\begin{cases}
|
||||
\int_{0}^{1} e^{x-z} \mathrm{d}x ,\quad & z>1\\
|
||||
\int_{0}^{z} e^{x-z} \mathrm{d}x,\quad & 0\leqslant z\leqslant 1 \\
|
||||
0,\quad & z<0 \\
|
||||
\end{cases}=\begin{cases}
|
||||
e^{1-z}-e^{-z},\quad & z>1 \\
|
||||
1-e^{-z},\quad & 0\leqslant z\leqslant 1 \\
|
||||
0,\quad & z<0 \\
|
||||
\end{cases}
|
||||
$$
|
||||
所以
|
||||
$$
|
||||
P(X+Y\leqslant 1)=\int_{-\infty}^{1} p_{Z}(z) \mathrm{d}z=\int_{0}^{1} (1-e^{-z}) \mathrm{d}z=1+(\frac{1}{e}-1)=\frac{1}{e}
|
||||
$$
|
||||
\end{zhongwen}
|
||||
\end{proof}
|
||||
\item 概率$P(X\leqslant Y)$。
|
||||
|
||||
\begin{proof}[解]
|
||||
\begin{zhongwen}
|
||||
令随机变量$Z=X-Y$，则$Z$的概率密度函数为
|
||||
$$
|
||||
p_{Z}(z)=\int_{-\infty}^{+\infty} p(x,x-z) \mathrm{d}x=\begin{cases}
|
||||
0,\quad & z>1 \\
|
||||
\int_{z}^{1} e^{z-x} \mathrm{d}x,\quad & 0\leqslant z\leqslant 1 \\
|
||||
\int_{0}^{1} e^{z-x} \mathrm{d}x,\quad & z<0 \\
|
||||
\end{cases}=\begin{cases}
|
||||
0,\quad & z>1 \\
|
||||
1-e^{z-1},\quad & 0\leqslant z\leqslant 1 \\
|
||||
e^{z}-e^{z-1},\quad & z<0 \\
|
||||
\end{cases}
|
||||
$$
|
||||
所以
|
||||
$$
|
||||
P(X\leqslant Y)=P(X-Y\leqslant 0)=\int_{-\infty}^{0} (e^{z}-e^{z-1}) \mathrm{d}z=1-\frac{1}{e}
|
||||
$$
|
||||
\end{zhongwen}
|
||||
\end{proof}
|
||||
\end{enumerate}
|
||||
|
||||
\item 设随机变量$(X,Y)$的概率密度函数为
|
||||
$$
|
||||
p(x,y)=\begin{cases}
|
||||
Ae^{-(3x+2y)},\quad & x>0,y>0; \\
|
||||
0,\quad & \text{otherwise}. \\
|
||||
\end{cases}
|
||||
$$
|
||||
求：
|
||||
\begin{enumerate}
|
||||
\item 常数$A$的值；
|
||||
|
||||
\begin{proof}[解]
|
||||
\begin{zhongwen}
|
||||
$$
|
||||
\begin{aligned}
|
||||
&\int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} p(x,y) \mathrm{d}x \mathrm{d}y=\int_{0}^{+\infty} \int_{0}^{+\infty} Ae^{-(3x+2y)} \mathrm{d}x \mathrm{d}y=\int_{0}^{+\infty} \left. -\frac{1}{3}Ae^{-(3x+2y)} \right|_{0}^{+\infty} \mathrm{d}y\\
|
||||
=&\int_{0}^{+\infty} \frac{1}{3}Ae^{-2y} \mathrm{d}y=\left. -\frac{1}{6}Ae^{-2y} \right|_{0}^{+\infty}=\frac{1}{6}A=1
|
||||
\end{aligned}
|
||||
$$
|
||||
所以常数$A$的值为$6$。
|
||||
\end{zhongwen}
|
||||
\end{proof}
|
||||
\item 分布函数$F(x,y)$；
|
||||
|
||||
\begin{proof}[解]
|
||||
\begin{zhongwen}
|
||||
|
||||
由题意可知，当$x\leqslant 0$ 或$y\leqslant 0$时，$F(x,y)=0$。
|
||||
|
||||
当$x>0,y>0$时，$
|
||||
\displaystyle F(x,y)
|
||||
=\int_{0}^{y} \int_{0}^{x} 6e^{-(3u+2v)} \mathrm{d}u \mathrm{d}v
|
||||
=\int_{0}^{y} \left. -2e^{-(3u+2v)} \right|_{0}^{x} \mathrm{d}v
|
||||
=\int_{0}^{y} \left(2e^{-2v}-2e^{-(3x+2v)}\right) \mathrm{d}v
|
||||
=\left. -e^{-2v} \right|_{0}^{y}+\left. e^{-(3x+2v)} \right|_{0}^{y}
|
||||
=-e^{-2y}+1+e^{-(3x+2y)}-e^{-3x}
|
||||
=e^{-(3x+2y)}-e^{-3x}-e^{-2y}+1
|
||||
$
|
||||
|
||||
所以
|
||||
$$
|
||||
\begin{aligned}
|
||||
F(x,y)=\begin{cases}
|
||||
e^{-(3x+2y)}-e^{-3x}-e^{-2y}+1,\quad & x>0,y>0 \\
|
||||
0,\quad & \text{otherwise} \\
|
||||
\end{cases}
|
||||
\end{aligned}
|
||||
$$
|
||||
\end{zhongwen}
|
||||
\end{proof}
|
||||
\item 概率$P(-2<X\leqslant 2,-3<Y\leqslant 3)$。
|
||||
|
||||
\begin{proof}[解]
|
||||
\begin{zhongwen}
|
||||
$$
|
||||
\begin{aligned}
|
||||
&P(-2<X\leqslant 2,-3<Y\leqslant 3)=P(X\leqslant 2,Y\leqslant 3)=F(2,3)\\
|
||||
=&e^{-(3\times 2+2\times 3)}-e^{-3\times 2}-e^{-2\times 3}+1 \\
|
||||
=&1 +e^{-12} - 2e^{-6}
|
||||
\end{aligned}
|
||||
$$
|
||||
\end{zhongwen}
|
||||
\end{proof}
|
||||
\end{enumerate}
|
||||
\end{enumerate}
|
||||
|
||||
\end{document}
|
||||
239
概率论/平时作业/单元作业5.tex
Normal file
239
概率论/平时作业/单元作业5.tex
Normal file
@@ -0,0 +1,239 @@
|
||||
\documentclass[全部作业]{subfiles}
|
||||
\setlength{\textheight}{300em}
|
||||
% \PassOptionsToPackage{papersize={170mm, 1000em}}{geometry}
|
||||
% \usepackage[textheight=1em]{geometry}
|
||||
\setlength{\paperheight}{300em}
|
||||
|
||||
\begin{document}
|
||||
\chapter{单元作业5}
|
||||
\begin{enumerate}
|
||||
\questionandanswer[]{设随机变量$X$与$Y$满足$EX=EY=0$, $\operatorname{Var}X=\operatorname{Var}Y=1$, $\operatorname{Cov}(X,Y)=\rho $，证明$E\max(X^{2},Y^{2})\leqslant 1+\sqrt{1-\rho ^{2}}$}{
|
||||
\begin{proof}
|
||||
|
||||
因为$\operatorname{Cov}(X,Y)=EXY-EXEY=\rho $，而$EX=EY=0$，所以$EXY=\rho $。\\
|
||||
而$\operatorname{Var}X=EX^{2}-(EX)^{2}=EX^{2}=1$，$\operatorname{Var}Y=EY^{2}-(EY)^{2}=EY^{2}=1$。
|
||||
|
||||
根据$\max(a,b)=\frac{\left\vert a+b \right\vert +\left\vert a-b \right\vert }{2}$和期望的线性性质，
|
||||
\begin{equation}\label{eq:1}\tag{1}
|
||||
E \max(X^{2},Y^{2})=E \frac{\left\vert X^{2}+Y^{2} \right\vert +\left\vert X^{2}-Y^{2} \right\vert }{2}=\frac{1}{2}E\left\vert X^{2}+Y^{2} \right\vert +\frac{1}{2}E\left\vert X^{2}-Y^{2} \right\vert
|
||||
\end{equation}
|
||||
|
||||
其中，$E\left\vert X^{2}+Y^{2} \right\vert =E(X^{2}+Y^{2})=EX^{2}+EY^{2}=2$，
|
||||
$E\left\vert X^{2}-Y^{2} \right\vert =E\left\vert X+Y \right\vert \left\vert X-Y \right\vert $。
|
||||
根据Cauchy-Schwarz不等式，$\left(E\left\vert X+Y \right\vert \left\vert X-Y \right\vert\right) ^{2}\leqslant E\left\vert X+Y \right\vert ^{2}E\left\vert X-Y \right\vert ^{2}$。
|
||||
|
||||
其中$E\left\vert X+Y \right\vert ^{2}=E(X^{2}+2XY+Y^{2})=EX^{2}+2EXY+EY^{2}=2+2\rho $，\\
|
||||
$E\left\vert X-Y \right\vert ^{2}=E(X^{2}-2XY+Y^{2})=EX^{2}-2XY+EY^{2}=2-2\rho $。
|
||||
|
||||
因此$\left( E\left\vert X+Y \right\vert \left\vert X-Y \right\vert \right) ^{2}\leqslant (2+2\rho )(2-2\rho )=4(1-\rho ^{2})$，两边同取根号可得
|
||||
$$E\left\vert X^{2}-Y^{2} \right\vert =E\left\vert X+Y \right\vert \left\vert X-Y \right\vert \leqslant 2\sqrt{1-\rho ^{2}}$$
|
||||
|
||||
再代入回 \eqref{eq:1} ，即可得到
|
||||
$$
|
||||
E\max(X^{2},Y^{2})\leqslant \frac{1}{2}\times 2+\frac{1}{2}\times 2\sqrt{1-\rho ^{2}}=1+\sqrt{1-\rho ^{2}}
|
||||
$$
|
||||
|
||||
\end{proof}
|
||||
}
|
||||
\questionandanswer[]{设随机变量$(X,Y)$服从均匀分布$U(D)$，其中$D=\{ (x,y)\ |\ x^{2}+y^{2}\leqslant 1 \}$，求$X$与$Y$的协方差。}{
|
||||
\begin{proof}[解]
|
||||
|
||||
设$p(x,y)$表示随机变量$(X,Y)$的联合概率密度函数，则可以得到
|
||||
$$
|
||||
p(x,y)=\begin{cases}
|
||||
\frac{1}{\pi },\quad & (x,y)\in D \\
|
||||
0,\quad & (x,y)\not \in D \\
|
||||
\end{cases}
|
||||
$$
|
||||
$$
|
||||
EX=\iint xp(x,y)\mathrm{d}x\mathrm{d}y=\iint_{D}x \cdot \frac{1}{\pi }\mathrm{d}x\mathrm{d}y\xlongequal{\textit{对称性}} 0
|
||||
$$
|
||||
$$
|
||||
EY=\iint yp(x,y)\mathrm{d}x\mathrm{d}y=\iint_{D}y \cdot \frac{1}{\pi }\mathrm{d}x\mathrm{d}y\xlongequal{\textit{对称性}} 0
|
||||
$$
|
||||
$$
|
||||
EXY=\iint xyp(x,y)\mathrm{d}x\mathrm{d}y=\iint_{D}xy \cdot \frac{1}{\pi }\mathrm{d}x\mathrm{d}y\xlongequal{\textit{对称性}} 0
|
||||
$$
|
||||
所以$X$与$Y$的协方差为
|
||||
$$
|
||||
\operatorname{Cov}(X,Y)=EXY-EXEY=0
|
||||
$$
|
||||
|
||||
\end{proof}
|
||||
}
|
||||
\questionandanswer[]{设随机变量$(X,Y)$服从均匀分布$U(D)$，其中$D=\{ (x,y)\ |\ 0<x<y<1 \}$，求相关系数$\operatorname{Corr}(X,Y)$。}{
|
||||
\begin{proof}[解]
|
||||
|
||||
设$p(x,y)$表示随机变量$(X,Y)$的联合概率密度函数，则可以得到
|
||||
$$
|
||||
p(x,y)=\begin{cases}
|
||||
2,\quad & 0<x<y<1 \\
|
||||
0,\quad & \textit{其他} \\
|
||||
\end{cases}
|
||||
$$
|
||||
$$
|
||||
EX=\iint xp(x,y)\mathrm{d}x\mathrm{d}y=\iint_{D}x\cdot 2\mathrm{d}x\mathrm{d}y=\int_0^{1}\int_0^{y}2x\mathrm{d}x\mathrm{d}y = \frac{1}{3}
|
||||
$$
|
||||
$$
|
||||
EY=\iint yp(x,y)\mathrm{d}x\mathrm{d}y=\iint_{D}y\cdot 2\mathrm{d}x\mathrm{d}y=\int_0^{1}\int_0^{y}2y\mathrm{d}x\mathrm{d}y = \frac{2}{3}
|
||||
$$
|
||||
$$
|
||||
EXY=\iint xyp(x,y)\mathrm{d}x\mathrm{d}y=\iint_{D}xy\cdot 2\mathrm{d}x\mathrm{d}y=\int_0^{1}\int_0^{y}2xy\mathrm{d}x\mathrm{d}y = \frac{1}{4}
|
||||
$$
|
||||
所以$X$与$Y$的协方差为
|
||||
$$
|
||||
\operatorname{Cov}(X,Y)=EXY-EXEY=\frac{1}{4}-\frac{1}{3}\times \frac{2}{3} = \frac{1}{36}
|
||||
$$
|
||||
还需要计算
|
||||
$$
|
||||
EX^{2}=\iint x^{2}p(x,y)\mathrm{d}x\mathrm{d}y=\iint_{D} x^{2}\cdot 2\mathrm{d}x\mathrm{d}y=\int_0^{1}\int_0^{y}2x^{2}\mathrm{d}x\mathrm{d}y = \frac{1}{6}
|
||||
$$
|
||||
$$
|
||||
EY^{2}=\iint y^{2}p(x,y)\mathrm{d}x\mathrm{d}y=\iint_{D}y^{2}\cdot 2\mathrm{d}x\mathrm{d}y=\int_0^{1}\int_0^{y}2y^{2}\mathrm{d}x\mathrm{d}y = \frac{1}{2}
|
||||
$$
|
||||
所以$X$与$Y$的相关系数为
|
||||
$$
|
||||
\begin{aligned}
|
||||
\operatorname{Corr}(X,Y)&=\frac{\operatorname{Cov}(X,Y)}{\sqrt{\operatorname{Var}X\cdot \operatorname{Var}Y}}=\frac{\operatorname{Cov}(X,Y)}{\sqrt{(EX^{2}-(EX)^{2})\cdot (EY^{2}-(EY)^{2})}}\\
|
||||
&=\frac{\dfrac{1}{36}}{\sqrt{\left( \dfrac{1}{6}-\left( \dfrac{1}{3} \right) ^{2} \right) \left( \dfrac{1}{2}-\left( \dfrac{2}{3} \right) ^{2} \right) }} = \dfrac{1}{2}\\
|
||||
\end{aligned}
|
||||
$$
|
||||
|
||||
\end{proof}
|
||||
}
|
||||
\questionandanswer[]{设随机变量$(X,Y)$的联合概率密度函数为$p(x,y)=\begin{cases}
|
||||
e^{-y},\quad & 0<x<y \\
|
||||
0,\quad & \text{otherwise} \\
|
||||
\end{cases}$, 设$y>0$，求在$Y=y$时$X$的条件概率密度函数$p_{X|Y}(x|y)$，条件数学期望$E(X|Y=y)$。进一步地，利用重期望公式求$EX$。}{
|
||||
\begin{proof}[解]
|
||||
|
||||
先计算边际概率密度函数
|
||||
$$
|
||||
p_{Y}(y)=\int_{-\infty}^{+\infty} p(x,y) \mathrm{d}x=\begin{cases}
|
||||
\int_{0}^{y} e^{-y} \mathrm{d}x,\quad & y>0 \\
|
||||
0,\quad & y\leqslant 0 \\
|
||||
\end{cases}=\begin{cases}
|
||||
ye^{-y},\quad & y>0 \\
|
||||
0,\quad & y\leqslant 0 \\
|
||||
\end{cases}
|
||||
$$
|
||||
之后计算条件概率密度函数
|
||||
$$
|
||||
p_{X|Y}(x|y)=\frac{p(x,y)}{p_{Y}(y)}=\begin{cases}
|
||||
\frac{1}{y},\quad & 0<x<y \\
|
||||
0,\quad & 0< y\leqslant x \textit{或}y>0,x\leqslant 0 \\
|
||||
\end{cases}
|
||||
$$
|
||||
根据数学期望的定义，
|
||||
$$
|
||||
E(X|Y=y)=\int_{-\infty}^{+\infty} x p_{X|Y}(x|y) \mathrm{d}x = \int_{0}^{y} x\cdot \frac{1}{y} \mathrm{d}x = \frac{y}{2}
|
||||
$$
|
||||
再利用重期望公式，
|
||||
$$
|
||||
EX=\int_{-\infty}^{+\infty} E(X|Y=y)p_{Y}(y) \mathrm{d}y=\int_{0}^{+\infty} \frac{y}{2}\cdot ye^{-y} \mathrm{d}y = 1
|
||||
$$
|
||||
|
||||
\end{proof}
|
||||
}
|
||||
\questionandanswer[]{设$X$服从指数分布$\operatorname{Exp}(\lambda )$，求$Y=[X]$的分布。（这里符号$[a]$表示不超过$a$的最大整数。}{
|
||||
\begin{proof}[解]
|
||||
|
||||
由于$Y=[X]$是离散型分布，所以求出$Y$的分布列即可
|
||||
$$
|
||||
p_i=\int_{i}^{i+1} \lambda e^{-\lambda x} \mathrm{d}x = (e^{\lambda} - 1) e^{- \lambda (i + 1)}
|
||||
$$
|
||||
|
||||
\end{proof}
|
||||
}
|
||||
\questionandanswer[]{设随机变量$X$服从标准正态分布$N(0,1)$, $a>0$，记$Y=\begin{cases}
|
||||
X,\quad & \left\vert X \right\vert <a \\
|
||||
-X,\quad & \left\vert X \right\vert \geqslant a \\
|
||||
\end{cases}$，求随机变量$Y$的分布。}{
|
||||
\begin{proof}[解]
|
||||
|
||||
因为$X\sim N(0,1)$，所以$-X\sim N(0,1)$，所以随机变量$Y$服从标准正态分布$N(0,1)$，即
|
||||
$$
|
||||
Y\sim N(0,1)
|
||||
$$
|
||||
|
||||
\end{proof}
|
||||
}
|
||||
\questionandanswer[]{设二维随机变量$(X,Y)$的联合概率密度函数为 $$
|
||||
p(x,y)=\begin{cases}
|
||||
2,\quad & \textit{若}0<x<y<1; \\
|
||||
0,\quad & \textit{其他.} \\
|
||||
\end{cases}
|
||||
$$ 求
|
||||
\begin{enumerate}
|
||||
\item 随机变量$T=X-Y$的概率密度函数$p_{T}(t)$；
|
||||
\item 概率$P(Y-X\leqslant \frac{1}{2})$。
|
||||
\end{enumerate}
|
||||
}{
|
||||
\begin{proof}[解]
|
||||
|
||||
\begin{enumerate}
|
||||
\item
|
||||
根据连续情形的卷积公式，
|
||||
$$
|
||||
p_{T}(t)=\int_{-\infty}^{+\infty} p(x,x-t) \mathrm{d}x=\begin{cases}
|
||||
\int_{0}^{t+1} 2 \mathrm{d}x,\quad & -1<t<0 \\
|
||||
0,\quad & t\leqslant -1 \textit{或}t\geqslant 0 \\
|
||||
\end{cases}=\begin{cases}
|
||||
2t+2,\quad & -1<t<0 \\
|
||||
0,\quad & t\leqslant -1\textit{或}t\geqslant 0 \\
|
||||
\end{cases}
|
||||
$$
|
||||
|
||||
\item
|
||||
$$
|
||||
P(Y-X\leqslant \frac{1}{2})=P(X-Y\geqslant \frac{1}{2})=P(T\geqslant \frac{1}{2})=0
|
||||
$$
|
||||
\end{enumerate}
|
||||
|
||||
\end{proof}
|
||||
}
|
||||
\questionandanswer[]{设$X$与$Y$独立同分布，共同分布为$N(0,1)$，求概率$P(\left\vert X+Y \right\vert \leqslant \left\vert X-Y \right\vert )$。}{
|
||||
\begin{proof}[解]
|
||||
|
||||
% 根据正态分布的可加性，
|
||||
% $$
|
||||
% X+Y\sim N(0,2),\quad X-Y\sim N(0,2)
|
||||
% $$
|
||||
% 又因为
|
||||
% $$
|
||||
% \begin{aligned}
|
||||
% p_{(X+Y,X-Y)}(u,v) & = p_{(X,Y)} \left(\frac{u+v}{2},\frac{u-v}{2}\right) \left\vert J(u,v) \right\vert \\
|
||||
% & = p_{X} \left( \frac{u+v}{2} \right) p_{Y} \left( \frac{u-v}{2} \right) \begin{vmatrix}
|
||||
% \frac{1}{2} & \frac{1}{2} \\
|
||||
% \frac{1}{2} & -\frac{1}{2} \\
|
||||
% \end{vmatrix}\\
|
||||
% &=\frac{1}{2}\varphi\left(\frac{u+v}{2}\right)\varphi\left(\frac{u-v}{2}\right)\\
|
||||
% &=\frac{1}{2}\cdot \frac{1}{\sqrt{2\pi }}e^{-\frac{1}{2} \left( \frac{u+v}{2} \right) ^{2}}\cdot \frac{1}{\sqrt{2\pi }}e^{-\frac{1}{2}\left( \frac{u-v}{2} \right) ^{2}} \\
|
||||
% &=\frac{1}{\sqrt{2\pi }\cdot \sqrt{2}}\frac{1}{\sqrt{2\pi }\cdot \sqrt{2}}e^{-\frac{1}{8}\left( 2u^{2}+2v^{2} \right) } \\
|
||||
% &=\frac{1}{\sqrt{2\pi }\sqrt{2}}e^{-\frac{u^{2}}{2\cdot 2}}\cdot \frac{1}{\sqrt{2\pi }\sqrt{2}}e^{-\frac{v^{2}}{2\cdot 2}} \\
|
||||
% &=p_{X+Y}(u)\cdot p_{X-Y}(v) \\
|
||||
% \end{aligned}
|
||||
% $$
|
||||
% 所以$X+Y$与$X-Y$相互独立，从而独立同分布，所以$\left\vert X+Y \right\vert $与$\left\vert X-Y \right\vert $也独立同分布，
|
||||
% 设他们的概率密度函数为$p_1(x)$，则根据独立同分布的可加性，$\left\vert X+Y \right\vert -\left\vert X-Y \right\vert $的概率密度函数可表示为$p_2(x)=p_1(x)+p_1(-x)$，是偶函数，于是
|
||||
% $$
|
||||
% P(\left\vert X+Y \right\vert \leqslant \left\vert X-Y \right\vert )=P(\left\vert X+Y \right\vert -\left\vert X-Y \right\vert \leqslant 0) = \frac{1}{2}
|
||||
% $$
|
||||
|
||||
$$
|
||||
\begin{aligned}
|
||||
P(\left\vert X+Y \right\vert \leqslant \left\vert X-Y \right\vert ) & = P\left( (X+Y)^{2}\leqslant (X-Y)^{2} \right) \\
|
||||
& = P(X^{2}+2XY+Y^{2}\leqslant X^{2}-2XY+Y^{2}) \\
|
||||
& = P(4XY\leqslant 0) \\
|
||||
& = P(XY\leqslant 0) \\
|
||||
&=P(X\leqslant 0,Y>0) + P(X>0,Y\leqslant 0) \\
|
||||
&\xlongequal{X\textit{与}Y\textit{}独立}P(X\leqslant 0)P(Y>0)+P(X>0)P(Y\leqslant 0) \\
|
||||
&\xlongequal{X\sim N(0,1),Y\sim N(0,1)}\frac{1}{2}\times \frac{1}{2}+\frac{1}{2}\times \frac{1}{2} \\
|
||||
&=\frac{1}{2} \\
|
||||
\end{aligned}
|
||||
$$
|
||||
|
||||
\end{proof}
|
||||
}
|
||||
\end{enumerate}
|
||||
\end{document}
|
||||
156
概率论/平时作业/单元作业6.tex
Normal file
156
概率论/平时作业/单元作业6.tex
Normal file
@@ -0,0 +1,156 @@
|
||||
\documentclass[全部作业]{subfiles}
|
||||
\setlength{\textheight}{210em}
|
||||
\setlength{\paperheight}{210em}
|
||||
|
||||
\begin{document}
|
||||
\chapter{单元作业6}
|
||||
\begin{enumerate}
|
||||
\item 证明随机变量$X$的特征函数是实值函数当且仅当$X$与$-X$同分布。
|
||||
|
||||
\begin{proof}
|
||||
\begin{zhongwen}
|
||||
|
||||
设$X$的特征函数为$f_{X}(t)$，则
|
||||
$$
|
||||
\begin{aligned}
|
||||
X的特征函数是实值函数 \iff& f_{X}(t)=\overline{f_{X}(t)} \\
|
||||
\iff& f_{X}(t)=f_{X}(-t) \\
|
||||
\iff& Ee^{itX}=Ee^{i(-t)X} \\
|
||||
\iff& Ee^{itX}=Ee^{it(-X)} \\
|
||||
\iff& f_{X}(t)=f_{-X}(t) \\
|
||||
\iff& X与-X同分布 \\
|
||||
\end{aligned}
|
||||
$$
|
||||
|
||||
\end{zhongwen}
|
||||
\end{proof}
|
||||
|
||||
\item 设随机变量$X$的特征函数为$f(t)=\left( \frac{2-it}{2} \right) ^{-2}$，求$EX$和$\operatorname{Var}X$。
|
||||
|
||||
\begin{proof}[解]
|
||||
\begin{zhongwen}
|
||||
|
||||
% 根据唯一性定理的推论，可知$X$的概率密度函数为
|
||||
% $$
|
||||
% p(x)=\frac{1}{2\pi }\int_{-\infty}^{+\infty} e^{-itx}\left( \frac{2-it}{2} \right) ^{-2} \mathrm{d}t
|
||||
% $$
|
||||
根据特征函数的性质，可知
|
||||
$$
|
||||
iEX = \left. f'(t) \right|_{t=0}=\left. \left( -\frac{i}{2}\cdot (-2)\left( \frac{2-it}{2} \right) ^{-3} \right) \right|_{t=0}=i\cdot 1^{-3}=i
|
||||
$$
|
||||
$$
|
||||
-EX^{2}=i^{2}EX^{2}=\left. f''(t) \right|_{t=0}=\left.\left( i\cdot \left( -\frac{i}{2} \right) \cdot (-3)\left( \frac{2-it}{2} \right) ^{-4} \right) \right|_{t=0}=\left( -\frac{3}{2} \right) \cdot 1^{-4}=-\frac{3}{2}
|
||||
$$
|
||||
所以有
|
||||
$$
|
||||
EX=1, \qquad EX^{2}=\frac{3}{2}, \qquad \operatorname{Var}X=EX^{2}-(EX)^{2}=\frac{3}{2}-1^{2}=\frac{1}{2}
|
||||
$$
|
||||
|
||||
\end{zhongwen}
|
||||
\end{proof}
|
||||
|
||||
\item 设$X_i$独立同分布，且$X_i\sim \operatorname{Exp}(\lambda ),i=1,2, \ldots ,n$。试用特征函数的方法证明：
|
||||
$$
|
||||
Y_n=\sum_{i=1}^{n}X_i\sim \operatorname{Ga}(n,\lambda )
|
||||
$$
|
||||
\begin{proof}
|
||||
\begin{zhongwen}
|
||||
|
||||
根据题意可知$X_i$的概率密度函数为$p(x)=\begin{cases}
|
||||
\lambda e^{-\lambda x},\quad & x>0 \\
|
||||
0,\quad & x\leqslant 0 \\
|
||||
\end{cases}$。根据特征函数的定义，计算$X_i$的特征函数$f(t)$：
|
||||
$$
|
||||
\begin{aligned}
|
||||
f(t)&=Ee^{itX_{i}}=\int_{-\infty}^{+\infty} e^{itx}p(x) \mathrm{d}x=\int_{0}^{+\infty} e^{itx}\lambda e^{-\lambda x} \mathrm{d}x=\frac{\lambda}{it-\lambda }\int_{0}^{+\infty} e^{(it-\lambda )x} \mathrm{d}(it-\lambda )x \\
|
||||
&=\frac{\lambda}{it-\lambda }\cdot \left. e^{(it-\lambda )x} \right|_{0}^{+\infty}=\frac{\lambda }{it-\lambda } \cdot \left( \lim_{x \to +\infty}e^{itx-\lambda x}-1 \right)=\frac{\lambda}{it-\lambda }\cdot \left( \lim_{x \to +\infty}e^{-\lambda x}e^{itx} -1 \right) \\
|
||||
\end{aligned}
|
||||
$$
|
||||
对于$\displaystyle \lim_{x \to +\infty}e^{-\lambda x}e^{itx}$，设$a=e^{-\lambda x}e^{itx}$，将其看作复数的模长辐角表示法，则$a$的模长为$e^{-\lambda x}$，辐角为$tx$。由于$\lambda >0$，所以当$x \to +\infty$时，$a$的模长$e^{-\lambda x} \to 0$（辐角$tx \to +\infty$，但不重要），因此$\displaystyle \lim_{x \to +\infty}e^{-\lambda x}e^{itx}=\lim_{x \to +\infty}a=0$。
|
||||
|
||||
于是$\displaystyle f(t)=\frac{\lambda}{it-\lambda }\cdot (0-1)=\frac{\lambda }{\lambda -it}$。
|
||||
|
||||
由于$X_i$相互独立，所以$Y_n$的特征函数为：
|
||||
$$
|
||||
g(t)=\left[ f(t) \right] ^{n}=\left( \frac{\lambda }{\lambda -it} \right) ^{n}
|
||||
$$
|
||||
|
||||
再计算$Y\sim \operatorname{Ga}(n,\lambda )$的特征函数：
|
||||
$$
|
||||
\begin{aligned}
|
||||
h(t)&=Ee^{itY}=\int_{-\infty}^{+\infty} e^{itx}p(x) \mathrm{d}x=\int_{0}^{+\infty} e^{itx} \frac{\lambda ^{n}}{\Gamma (n)}x^{n-1}e^{-\lambda x} \mathrm{d}x \\
|
||||
&=\frac{\lambda ^{n}}{(\lambda -it)^{n}}\int_{0}^{+\infty} \frac{(\lambda -it)^{n}}{\Gamma (n)}x^{n-1}e^{(it-\lambda )x} \mathrm{d}x \\
|
||||
\end{aligned}
|
||||
$$
|
||||
注意到$\displaystyle \frac{(\lambda -it)^{n}}{\Gamma (n)}x^{n-1}e^{(it-\lambda )}$（零延拓后）为$\operatorname{Ga}(n, \lambda -it)$的概率密度函数，根据分布函数的正则性，$\displaystyle \int_{0}^{+\infty} \frac{(\lambda -it)^{n}}{\Gamma (n)}x^{n-1}e^{(it-\lambda )} \mathrm{d}x=1$，所以
|
||||
$$
|
||||
h(t)=\frac{\lambda ^{n}}{(\lambda -it)^{n}}=\left( \frac{\lambda }{\lambda -it} \right) ^{n}=g(t)
|
||||
$$
|
||||
根据特征函数的唯一性，$Y_n$与$Y$同分布，即
|
||||
$$
|
||||
Y_n=\sum_{i=1}^{n}X_i\sim \operatorname{Ga}(n,\lambda )
|
||||
$$
|
||||
|
||||
\end{zhongwen}
|
||||
\end{proof}
|
||||
|
||||
\item 设$X_n\xrightarrow{P}X,X_n\xrightarrow{P}Y$。证明$P(X=Y)=1$。
|
||||
|
||||
\begin{proof}
|
||||
\begin{zhongwen}
|
||||
|
||||
根据依概率收敛的运算性质，有
|
||||
$$
|
||||
0=X_n-X_n\xrightarrow{P}X-Y
|
||||
$$
|
||||
即对任意的$\varepsilon>0$，有
|
||||
$$
|
||||
\lim_{n \to \infty} P(\left\vert 0-(X-Y) \right\vert \geqslant \varepsilon)=\lim_{n \to \infty}P(\left\vert X-Y \right\vert \geqslant \varepsilon)=P(\left\vert X-Y \right\vert \geqslant \varepsilon)=0
|
||||
$$
|
||||
令$\varepsilon \to 0$，则根据概率的下连续性，可得
|
||||
$$
|
||||
P(\left\vert X-Y \right\vert >0)=0
|
||||
$$
|
||||
由于$\left\vert X-Y \right\vert \geqslant 0$，所以
|
||||
$$
|
||||
P(\left\vert X-Y \right\vert =0)=1-P(\left\vert X-Y \right\vert >0)=1
|
||||
$$
|
||||
即$P(X=Y)=1$。
|
||||
|
||||
\end{zhongwen}
|
||||
\end{proof}
|
||||
|
||||
\item 证明$\displaystyle \lim_{n \to \infty}E\left( \frac{\left\vert X_n \right\vert }{1+\left\vert X_n \right\vert } \right) =0$当且仅当$X_n\xrightarrow{P}0$。
|
||||
|
||||
\begin{proof}
|
||||
\begin{zhongwen}
|
||||
|
||||
对任意的$\varepsilon>0$，$1=1_{\{ \left\vert X_n \right\vert \leqslant \varepsilon \}}+1_{\{ \left\vert X_n \right\vert >\varepsilon \}}$。所以有如下不等式：
|
||||
|
||||
$$
|
||||
\begin{aligned}
|
||||
\frac{\left\vert X_n \right\vert }{1+\left\vert X_n \right\vert }&=\frac{\left\vert X_n \right\vert }{1+\left\vert X_n \right\vert }1_{\{ \left\vert X_n \right\vert \leqslant \varepsilon \}}+\frac{\left\vert X_n \right\vert }{1+\left\vert X_n \right\vert }1_{\{ \left\vert X_n \right\vert >\varepsilon \}} \\
|
||||
&\leqslant \left\vert X_n \right\vert 1_{\{ \left\vert X_n \right\vert \leqslant \varepsilon \}}+1_{\{ \left\vert X_n \right\vert >\varepsilon \}}\leqslant \varepsilon+1_{\{ \left\vert X_n \right\vert >\varepsilon \}} \\
|
||||
\end{aligned}
|
||||
$$
|
||||
等式两边同时取期望，则
|
||||
$$
|
||||
E\left( \frac{\left\vert X_n \right\vert }{1+\left\vert X_n \right\vert } \right) \leqslant E(\varepsilon+1_{\{ \left\vert X_n \right\vert >\varepsilon \}})=\varepsilon+E(1_{\{ \left\vert X_n \right\vert >\varepsilon \}})
|
||||
$$
|
||||
若$X_n\xrightarrow{P}0$，则$\displaystyle \lim_{n \to \infty}P(\left\vert X_n \right\vert >\varepsilon)=0$，所以$\displaystyle \lim_{n \to \infty}E(1_{\{ \left\vert X_n \right\vert >\varepsilon \}})=0$。
|
||||
于是上述不等式两边同时取$n \to 0$时的极限可得
|
||||
$$
|
||||
\lim_{n \to \infty}E\left( \frac{\left\vert X_n \right\vert }{1+\left\vert X_n \right\vert } \right) \leqslant \varepsilon
|
||||
$$
|
||||
所以$\displaystyle \lim_{n \to \infty}E\left( \frac{\left\vert X_n \right\vert }{1+\left\vert X_n \right\vert } \right) =0$。
|
||||
|
||||
若$\displaystyle \lim_{n \to \infty}E\left( \frac{\left\vert X_n \right\vert }{1+\left\vert X_n \right\vert } \right) =0$，这里使用和书上不同的方法。根据马尔可夫不等式的一般形式，有
|
||||
$$
|
||||
P(\left\vert X_n \right\vert >\varepsilon)\leqslant \frac{E\left( \frac{\left\vert X_n \right\vert }{1+\left\vert X_n \right\vert } \right) }{\frac{\varepsilon}{1+\varepsilon}}
|
||||
$$
|
||||
不等式两边同时取$n \to 0$时的极限可得$\displaystyle \lim_{n \to \infty}P(\left\vert X_n \right\vert >\varepsilon)=0$，即$X_n\xrightarrow{P}0$。
|
||||
|
||||
\end{zhongwen}
|
||||
\end{proof}
|
||||
\end{enumerate}
|
||||
\end{document}
|
||||
Reference in New Issue
Block a user