# 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. This inequality provides an upper bound on the probability of occurrence of at least one of a countable number of events in terms of the individual probabilities of the events. Boole's inequality is named for its discoverer, George Boole.

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

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

${\mathbb {P} }\left(\bigcup _{i=1}^{n}A_{i}\cap A_{n+1}\right)\geq 0,$ 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 we modify the sets $A_{i}$ , so they become disjoint,

$B_{i}=A_{i}-\bigcup _{j=1}^{i-1}A_{j}$ we can show that

$\bigcup _{i=1}^{\infty }B_{i}=\bigcup _{i=1}^{\infty }A_{i}.$ by proving both directions of inclusion.

Suppose $x\in \bigcup _{i=1}^{\infty }A_{i}$ . Then $x\in A_{k}$ for some minimum $k$ such that $i . Therefore $x\in B_{k}=A_{k}-\bigcup _{j=1}^{k-1}A_{j}$ . So the first inclusion is true: $\bigcup _{i=1}^{\infty }A_{i}\subset \bigcup _{i=1}^{\infty }B_{i}$ .

Next suppose that $x\in \bigcup _{i=1}^{\infty }B_{i}$ . It follows that $x\in B_{k}$ for some $k$ . And $B_{k}=A_{k}-\bigcup _{j=1}^{k-1}A_{j}$ so $x\in A_{k}$ , and we have the other inclusion: $\bigcup _{i=1}^{\infty }B_{i}\subset \bigcup _{i=1}^{\infty }A_{i}$ .

By construction of each $B_{i}$ , $B_{i}\subset A_{i}$ . For $B\subset A,$ it is the case that $\mathbb {P} (B)\leq \mathbb {P} (A).$ So, we can conclude that the desired inequality is true:

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

Let

$S_{1}:=\sum _{i=1}^{n}{\mathbb {P} }(A_{i}),\quad S_{2}:=\sum _{1\leq i for all integers k in {1, ..., n}.

When n is odd, the sequence of inequalities:

${\mathbb {P} }\left(\bigcup _{i=1}^{n}A_{i}\right)=\sum _{j=1}^{n}(-1)^{j-1}S_{j}\leq \sum _{j=1}^{n-2}(-1)^{j-1}S_{j}\leq \sum _{j=1}^{n-4}(-1)^{j-1}S_{j}\leq \ldots \leq \sum _{j=1}^{1}(-1)^{j-1}S_{j}$ and

$\sum _{j=1}^{2}(-1)^{j-1}S_{j}\leq \ldots \leq \sum _{j=1}^{n-3}(-1)^{j-1}S_{j}\leq \sum _{j=1}^{n-1}(-1)^{j-1}S_{j}\leq \sum _{j=1}^{n}(-1)^{j-1}S_{j}={\mathbb {P} }\left(\bigcup _{i=1}^{n}A_{i}\right)$ both hold. When n is even, then:

${\mathbb {P} }\left(\bigcup _{i=1}^{n}A_{i}\right)=\sum _{j=1}^{n}(-1)^{j-1}S_{j}\leq \sum _{j=1}^{n-1}(-1)^{j-1}S_{j}\leq \sum _{j=1}^{n-3}(-1)^{j-1}S_{j}\leq \ldots \leq \sum _{j=1}^{1}(-1)^{j-1}S_{j}$ and

$\sum _{j=1}^{2}(-1)^{j-1}S_{j}\leq \ldots \leq \sum _{j=1}^{n-4}(-1)^{j-1}S_{j}\leq \sum _{j=1}^{n-2}(-1)^{j-1}S_{j}\leq \sum _{j=1}^{n}(-1)^{j-1}S_{j}={\mathbb {P} }\left(\bigcup _{i=1}^{n}A_{i}\right)$ both hold. The chains of inequalities among the partial sums follow from the observation that the events Sk form a decreasing sequence of sets. The equalities follow from the inclusion–exclusion principle, and Boole's inequality is the special case of the extremal upper bounds.

### Example

Suppose that you are estimating 5 parameters based on a random sample, and you can control each parameter separately. If you want your estimations of all five parameters to be good with a chance 95%, what should you do to each parameter?

Tuning each parameter's chance to be good to within 95% is not enough because "all are good" is a subset of each event "Estimate i is good". We can use Boole's Inequality to solve this problem. By finding the complement of event "all fives are good", we can change this question into another condition:

P( at least one estimation is bad) = 0.05 ≤ P( A1 is bad) + P( A2 is bad) + P( A3 is bad) + P( A4 is bad) + P( A5 is bad)

One way is to make each of them equal to 0.05/5 = 0.01, that is 1%. In another word, you have to guarantee each estimate good to 99%( for example, by constructing a 99% confidence interval) to make sure the total estimation to be good with a chance 95%. This is called Bonferroni Method of simultaneous inference.