布尔不等式

${\displaystyle P(\bigcup _{i}A_{i})\leq \sum _{i}P(A_{i})}$

证明

${\displaystyle P(A_{1})\leq P(A_{1})}$

${\displaystyle P(\bigcup _{i=_{1}}^{n}A_{i})\leq \sum _{i=_{1}}^{n}P(A_{i})}$
${\displaystyle P(A\cup B)=P(A)+P(B)-P(A\cap B)}$
${\displaystyle P(\bigcup _{i=_{1}}^{n+1}A_{i})=P(\bigcup _{i=_{1}}^{n}A_{i})+P(A_{n+1})-P(\bigcup _{i=_{1}}^{n}A_{i}\cap A_{n+1})}$
${\displaystyle P(\bigcup _{i=_{1}}^{n}A_{i}\cap A_{n+1})\geq 0,}$
${\displaystyle P(\bigcup _{i=_{1}}^{n+1}A_{i})\leq P(\bigcup _{i=_{1}}^{n}A_{i})+P(A_{n+1})}$
${\displaystyle P(\bigcup _{i=_{1}}^{n+1}A_{i})\leq \sum _{i=_{1}}^{n}P(A_{i})+P(A_{n+1})=\sum _{i=_{1}}^{n+1}P(A_{i})}$.

使用马尔可夫不等式的证明

${\displaystyle A_{1},A_{2},\cdots ,A_{n}}$是任意概率事件${\displaystyle X}$是各种事件${\displaystyle A_{i}}$的发生次数的随机变量。显然有：

${\displaystyle E(X)=P(A_{1})+P(A_{2})+\cdots +P(A_{n})=\sum _{i=1}^{n}P(A_{i})}$

${\displaystyle P(X\geqslant 1)\leqslant E(X)=\sum _{i=1}^{n}P(A_{i})}$

Bonferroni不等式

${\displaystyle S_{1}=\sum _{i=1}^{n}P(A_{i}),}$
${\displaystyle S_{2}=\sum _{1\leq i
${\displaystyle S_{k}=\sum _{1\leq i_{1}<\cdots

${\displaystyle P(\bigcup _{i=1}^{n}A_{i})\leq \sum _{j=1}^{k}(-1)^{j-1}S_{j}}$

${\displaystyle P(\bigcup _{i=1}^{n}A_{i})\geq \sum _{j=1}^{k}(-1)^{j-1}S_{j}}$