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# Boole's inequality

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 Title: Boole's inequality Author: World Heritage Encyclopedia Language: English Subject: Collection: Publisher: World Heritage Encyclopedia Publication Date:

### Boole's inequality

In George Boole.

Formally, for a countable set of events A1, A2, A3, ..., we have

{\mathbb P}\biggl(\bigcup_{i} A_i\biggr) \le \sum_i {\mathbb P}(A_i).

In measure-theoretic terms, Boole's inequality follows from the fact that a measure (and certainly any probability measure) is σ-sub-additive.

## Contents

• Proof 1
• Bonferroni inequalities 2
• References 4

## Proof

Boole's inequality may be proved for finite collections of events using the method of induction.

For the n=1 case, it follows that

\mathbb P(A_1) \le \mathbb P(A_1).

For the case n, we have

{\mathbb P}\biggl(\bigcup_{i=_1}^{n} A_i\biggr) \le \sum_{i=_1}^{n} {\mathbb P}(A_i).

Since \mathbb P(A \cup B) = \mathbb P(A) + \mathbb P(B) - \mathbb P(A \cap B), and because the union operation is associative, we have

{\mathbb P}\biggl(\bigcup_{i=_1}^{n+1} A_i\biggr) = {\mathbb P}\biggl(\bigcup_{i=_1}^n A_i\biggr) + \mathbb P(A_{n+1}) - {\mathbb P}\biggl(\bigcup_{i=_1}^n A_i \cap A_{n+1}\biggr).

Since

{\mathbb P}\biggl(\bigcup_{i=_1}^n A_i \cap A_{n+1}\biggr) \ge 0,

by the first axiom of probability, we have

{\mathbb P}\biggl(\bigcup_{i=_1}^{n+1} A_i\biggr) \le {\mathbb P}\biggl(\bigcup_{i=_1}^n A_i\biggr) + \mathbb P(A_{n+1}),

and therefore

{\mathbb P}\biggl(\bigcup_{i=_1}^{n+1} A_i\biggr) \le \sum_{i=_1}^{n} {\mathbb P}(A_i) + \mathbb P(A_{n+1}) = \sum_{i=_1}^{n+1} {\mathbb P}(A_i).

## Bonferroni inequalities

Boole's inequality may be generalised to find upper and lower bounds on the probability of finite unions of events. These bounds are known as Bonferroni inequalities, after Carlo Emilio Bonferroni, see Bonferroni (1936).

Define

S_1 := \sum_{i=1}^n {\mathbb P}(A_i),

and

S_2 := \sum_{1\le i

as well as

S_k := \sum_{1\le i_1<\cdots

for all integers k in {3, ..., n}.

Then, for odd k in {1, ..., n},

{\mathbb P}\biggl( \bigcup_{i=1}^n A_i \biggr) \le \sum_{j=1}^k (-1)^{j-1} S_j,

and for even k in {2, ..., n},

{\mathbb P}\biggl( \bigcup_{i=1}^n A_i \biggr) \ge \sum_{j=1}^k (-1)^{j-1} S_j.

Boole's inequality is recovered by setting k = 1. When k = n, then equality holds and the resulting identity is the inclusion–exclusion principle.