Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Define $$S_1 = \sum_{i=1}^n P(A_i)$$ and $$S_2 =\sum_{1 \le i < j \le n}^n P(A_i \cap A_j)$$ as well as $$S_k =\sum_{1 \le i_1 < \cdots < i_k \le n}^n P(A_{i_1} \cap \cdots \cap A_{i_k})$$ Then for odd $k$ in $\{1,\ldots,n\}$ $$P\left(\bigcup_{i=1}^n A_i\right) \le \sum_{j=1}^{k}(-1)^{j-1} S_j$$ For even $k$ in $\{2,\ldots,n\}$ $$P\left(\bigcup_{i=1}^n A_i\right) \ge \sum_{j=1}^{k}(-1)^{j-1}S_j$$

More details of Bonferroni inequalities or Boole's inequality is here.

share|improve this question
1  
It's better to write explicitly the inequality here. –  Patrick Li Oct 7 '12 at 14:38
    
I've added the description. –  LittleSweet Oct 7 '12 at 15:07
1  
isn't this inclusion exclusion principle? For odd number of terms you overcount the probability, for even number of terms you undercount. –  john mangual Oct 7 '12 at 15:24
add comment

1 Answer 1

up vote 2 down vote accepted

A proof is there. The main idea is that this is the integrated version of analogous pointwise inequalities and that, for every $k$, $$ S_k=\mathbb E\left({T\choose k}\right),\qquad T=\sum_{i=1}^n\mathbf 1_{A_i}. $$ Hence the result follows from the stronger inequalities asserting that, for every positive integer $N$, $$ \sum_{i=0}^k(-1)^ia_i,\qquad a_i={N\choose i}, $$ is nonnegative when $k$ is even and nonpositive when $k$ is odd. In turn, this fact follows from the properties that the sequence $(a_i)_{0\leqslant i\leqslant N}$ is unimodal and that $\sum\limits_{i=0}^N(-1)^ia_i=0$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.