# Boole's inequality

In probability theory, Boole's inequality, also known as the union bound, says that for any finite or countable set of events, the probability that at least one of the events happens is no greater than the sum of the probabilities of the individual events. Boole's inequality is named after George Boole.

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

$\mathbb {P} \left(\bigcup _{i}A_{i}\right)\leq \sum _{i}{\mathbb {P} }(A_{i}).$ In measure-theoretic terms, Boole's inequality follows from the fact that a measure (and certainly any probability measure) is σ-sub-additive.

## Proof

### Proof using induction

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

For the $n=1$ case, it follows that

$\mathbb {P} (A_{1})\leq \mathbb {P} (A_{1}).$ For the case $n$ , we have

${\mathbb {P} }\left(\bigcup _{i=1}^{n}A_{i}\right)\leq \sum _{i=1}^{n}{\mathbb {P} }(A_{i}).$ 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

$\mathbb {P} \left(\bigcup _{i=1}^{n+1}A_{i}\right)=\mathbb {P} \left(\bigcup _{i=1}^{n}A_{i}\right)+\mathbb {P} (A_{n+1})-\mathbb {P} \left(\bigcup _{i=1}^{n}A_{i}\cap A_{n+1}\right).$ Since

${\mathbb {P} }{\biggl (}\bigcup _{i=1}^{n}A_{i}\cap A_{n+1}{\biggr )}\geq 0,$ by the first axiom of probability, we have

$\mathbb {P} \left(\bigcup _{i=1}^{n+1}A_{i}\right)\leq \mathbb {P} \left(\bigcup _{i=1}^{n}A_{i}\right)+\mathbb {P} (A_{n+1}),$ and therefore

$\mathbb {P} \left(\bigcup _{i=1}^{n+1}A_{i}\right)\leq \sum _{i=1}^{n}\mathbb {P} (A_{i})+\mathbb {P} (A_{n+1})=\sum _{i=1}^{n+1}\mathbb {P} (A_{i}).$ ### Proof without using induction

For any events in $A_{1},A_{2},A_{3},\dots$ in our probability space we have

$\mathbb {P} \left(\bigcup _{i}A_{i}\right)\leq \sum _{i}\mathbb {P} (A_{i})$ One of the axioms of a probability space is that if $B_{1},B_{2},B_{3},\dots$ are disjoint subsets of the probability space then

$\mathbb {P} \left(\bigcup _{i}B_{i}\right)=\sum _{i}\mathbb {P} (B_{i})$ If $B\subset A$ then $\mathbb {P} (B)\leq \mathbb {P} (A)$ Indeed, from the axioms of a probability distribution,

$\mathbb {P} (A)=\mathbb {P} (B)+\mathbb {P} (A-B)$ Note that both terms on the right are nonnegative.

Now we have to modify the sets $A_{i}$ , so they become disjoint.

$B_{i}=A_{i}-\bigcup _{j=1}^{i-1}A_{j}$ So if $B_{i}\subset A_{i}$ , then we know

$\bigcup _{i=1}^{\infty }B_{i}=\bigcup _{i=1}^{\infty }A_{i}$ Therefore, we can deduce the following equation

$\mathbb {P} \left(\bigcup _{i}A_{i}\right)=\mathbb {P} \left(\bigcup _{i}B_{i}\right)=\sum _{i}\mathbb {P} (B_{i})\leq \sum _{i}\mathbb {P} (A_{i})$ ## Bonferroni inequalities

Boole's inequality may be generalized 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\leq i as well as

$S_{k}:=\sum _{1\leq i_{1}<\cdots for all integers k in {3, ..., n}.

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

${\mathbb {P} }\left(\bigcup _{i=1}^{n}A_{i}\right)\leq \sum _{j=1}^{k}(-1)^{j-1}S_{j},$ and for even k in {2, ..., n},

${\mathbb {P} }\left(\bigcup _{i=1}^{n}A_{i}\right)\geq \sum _{j=1}^{k}(-1)^{j-1}S_{j}.$ Boole's inequality is recovered by setting k = 1. When k = n, then equality holds and the resulting identity is the inclusion–exclusion principle.