概率论中常见的不等式概率论中常见的等式与不等式；标红的是我认为有难度的题目 1不等式 (茆1.4.32)设证明: 证

概率论中常见的等式与不等式；

标红的是我认为有难度的题目

不等式

(茆1.4.32)设 $P(A)=p,P(B)=1-\alpha>0$ 证明: $\frac{p-\alpha}{1-\alpha}\le P\left( A|B \right) \le \frac{p}{1-\alpha}$

证明:

\begin{align*} &\because P\left( AB \right) \le P\left( A \right) =p\\ &1\ge P\left( A\cup B \right) =P\left( A \right) +P\left( B \right) -P\left( AB \right) \Longrightarrow P\left( AB \right) \ge P\left( A \right) +P\left( B \right) -1=p-\alpha \\ &\therefore P\left( A|B \right) =\frac{P\left( AB \right)}{P\left( B \right)}\in \left[ \frac{p-\alpha}{1-\alpha},\frac{p}{1-\alpha} \right] \end{align*}

（张宇1000题114面） ${\color[RGB]{240, 0, 0} \text{对任意事件}A,B,P\left( A \right) P\left( B \right) \ge P\left( AB \right) P\left( A\cup B \right) }$

证明:

\begin{align*} &P\left( A \right) =P\left( A\cap \left( B\cup \bar{B} \right) \right) =P\left( AB \right) +P\left( A\bar{B} \right)& \\ &\text{同理}P\left( B \right) =P\left( AB \right) +P\left( \bar{A}B \right) \end{align*}

\begin{align*} \text{故}P\left( A \right) P\left( B \right)& =\left( P\left( AB \right) +P\left( A\bar{B} \right) \right) \left( P\left( AB \right) +P\left( \bar{A}B \right) \right)& \\ &\,\, =P^2\left( AB \right) +P\left( AB \right) P\left( \bar{A}B \right) +P\left( AB \right) P\left( A\bar{B} \right) +P\left( \bar{A}B \right) P\left( A\bar{B} \right) \\ &\,\, =P\left( AB \right) \left[ P\left( AB \right) +P\left( \bar{A}B \right) +P\left( A\bar{B} \right) \right] +P\left( \bar{A}B \right) P\left( A\bar{B} \right) \end{align*}

\begin{align*} &\because P\left( AB \right) +P\left( \bar{A}B \right) +P\left( A\bar{B} \right) =P\left( A \right) +P\left( B \right) -P\left( AB \right) =P\left( A\cup B \right)& \\ &P\left( \bar{A}B \right) P\left( A\bar{B} \right) \ge 0\\ &\therefore P\left( A \right) P\left( B \right) \\ &=P\left( AB \right) \left[ P\left( AB \right) +P\left( \bar{A}B \right) +P\left( A\bar{B} \right) \right] +P\left( \bar{A}B \right) P\left( A\bar{B} \right) \\ &\ge P\left( AB \right) P\left( A\cup B \right) +0\\ &=P\left( AB \right) P\left( A\cup B \right) \\ &\text{当且仅当}P\left( \bar{A}B \right) P\left( A\bar{B} \right) =0,\text{即}P\left( AB \right) =P\left( A \right) \text{或}P\left( B \right) \text{时等号成立}\\ &注:P\left( AB \right) =P\left( A \right) \text{或}P\left( B \right) {\color[RGB]{179, 0, 179} \text{推不出}\left( A\in B\text{或}B\in A \right) } \end{align*}

（茆1.3.19(1)） $\text{对}\forall \text{事件}A,B,C.\text{证明}:P\left( AB \right) +P\left( AC \right) -P\left( BC \right) \le P\left( A \right)$

\begin{align*} &\text{证法}1,\text{从左往右证明:}&:\\ &P\left( AB \right) +P\left( AC \right) -P\left( BC \right) \\ &\le P\left( AB \right) +P\left( AC \right) -P\left( ABC \right) \\ &=P\left( AB \right) +P\left( AC\bar{B} \right) \\ &\le P\left( AB \right) +P\left( A\bar{B} \right) \\ &=P\left( A \right) \\ &\text{证法二}:\text{从右边往左证明}:\\ &P\left( A \right) \\ &\ge P\left( A\cap \left( B\cup C \right) \right) \\ &=P\left( AB \right) +P\left( AC \right) -P\left( ABC \right) \\ &\ge P\left( AB \right) +P\left( AC \right) -P\left( BC \right) \end{align*}

（茆1.3.23） ${\color[RGB]{240, 0, 0} \text{对任意事件}A,B,\text{证明}:\left| P\left( AB \right) -P\left( A \right) P\left( B \right) \right|\le \frac{1}{4}}$

\begin{align*} &\text{证明}:\text{不妨设}P\left( A \right) \ge P\left( B \right)& \\ &\text{故}P\left( AB \right) -P\left( A \right) P\left( B \right) \le P\left( B \right) -P^2\left( B \right) \le \frac{1}{4}\\ &\text{另一方面}P\left( B \right) ={\color[RGB]{0, 0, 240} P\left( AB \right) +P\left( \bar{A}B \right) }\\ &P\left( AB \right) -P\left( A \right) P\left( B \right) \\ &=P\left( AB \right) -P\left( A \right) \left( {\color[RGB]{0, 0, 240} P\left( AB \right) +P\left( \bar{A}B \right) } \right) \\ &=P\left( AB \right) -P\left( A \right) P\left( AB \right) -P\left( A \right) {\color[RGB]{0, 0, 240} P\left( \bar{A}B \right) }\\ &{\color[RGB]{0, 0, 240} =}P\left( AB \right) \left( 1-P\left( A \right) \right) -P\left( A \right) {\color[RGB]{0, 0, 240} P\left( \bar{A}B \right) }\\ &{\color[RGB]{0, 0, 240} \ge }-P\left( A \right) {\color[RGB]{0, 0, 240} P\left( \bar{A}B \right) }\\ &{\color[RGB]{0, 0, 240} \ge }-P\left( A \right) {\color[RGB]{0, 0, 240} P\left( \bar{A} \right) }\\ &{\color[RGB]{0, 0, 240} =-P\left( A \right) \left( 1-P\left( A \right) \right) }\\ &{\color[RGB]{0, 0, 240} \ge -\frac{1}{4}}\\ &{\color[RGB]{0, 0, 240} \text{即}-}\frac{1}{4}{\color[RGB]{0, 0, 240} \le }P\left( AB \right) -P\left( A \right) P\left( B \right) \le \frac{1}{4},\text{即}\left| P\left( AB \right) -P\left( A \right) P\left( B \right) \right|\le \frac{1}{4}\\ &\text{若}P\left( A \right) \le P\left( B \right) ,\text{同理可得}\left| P\left( AB \right) -P\left( A \right) P\left( B \right) \right|\le \frac{1}{4}\\ &\text{综上可得}\left| P\left( AB \right) -P\left( A \right) P\left( B \right) \right|\le \frac{1}{4} \end{align*}

(茆1.3.22)

$P\left( AB \right) \ge P\left( A \right) +P\left( B \right) -1$

$\displaystyle P\left( A_1A_2\cdots A_n \right) \ge \sum_{i=1}^n{A_i}-\left( n-1 \right)$

证明:

在第一题中已经给出证明过程
$\begin{align*} &\text{当}n=2\text{时显然}P\left( AB \right) \ge P\left( A \right) +P\left( B \right) -1\text{成立}\\ &\text{设}n=k\text{时，}P\left( A_1A_2\cdots A_k \right) \ge \sum_{i=1}^k{P\left( A_i \right)}-\left( k-1 \right) \\ &\text{则当}n=k+1\text{时，记}C=A_1A_2\cdots A_k,P\left( C \right) =P\left( A_1A_2\cdots A_k \right) \ge \sum_{i=1}^k{P\left( A_i \right)}-\left( k-1 \right) \end{align*}$

\begin{align*} P\left( A_1A_2\cdots A_{k+1} \right) =P\left( CA_{k+1} \right)&\ge P\left( C \right) +P\left( A_{k+1} \right) -1&\\ &\,\, \ge \sum_{i=1}^k{P\left( A_i \right)}-\left( k-1 \right) +P\left( A_{k+1} \right) -1\\ &\,\, =\sum_{i=1}^{k+1}{P\left( A_i \right)}-\left( k+1-1 \right) \end{align*}

\begin{align*} &\text{则由数学归纳法可知}P\left( A_1A_2\cdots A_n \right) \ge \sum_{i=1}^n{A_i}-\left( n-1 \right) & \end{align*}

本文由mdnice多平台发布