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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.^{[1]}

Formally, for a countable set of events *A*_{1}, *A*_{2}, *A*_{3}, ..., we have

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

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

For the case, it follows that

For the case , we have

Since and because the union operation is associative, we have

Since

by the first axiom of probability, we have

and therefore

For any events in in our probability space we have

One of the axioms of a probability space is that if are *disjoint* subsets of the probability space then

this is called *countable additivity.*

If then

Indeed, from the axioms of a probability distribution,

Note that both terms on the right are nonnegative.

Now we have to modify the sets , so they become disjoint.

So if , then we know

Therefore, we can deduce the following equation

Boole's inequality may be generalized to find upper and lower bounds on the probability of finite unions of events.^{[2]} These bounds are known as **Bonferroni inequalities**, after Carlo Emilio Bonferroni; see Bonferroni (1936).

Define

and

as well as

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

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

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

Boole's inequality is the initial case, *k* = 1. When *k* = *n*, then equality holds and the resulting identity is the inclusion–exclusion principle.

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%, how should you do to each parameter?

Obviously, controlling each parameter good with a chance 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( A _{1} is bad) + P( A_{2} is bad) + P( A_{3} is bad) + P( A_{4} is bad) + P( A_{5} 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.

- Bonferroni, Carlo E. (1936), "Teoria statistica delle classi e calcolo delle probabilità",
*Pubbl. D. R. Ist. Super. Di Sci. Econom. E Commerciali di Firenze*(in Italian),**8**: 1–62, Zbl 0016.41103 - Dohmen, Klaus (2003),
*Improved Bonferroni Inequalities via Abstract Tubes. Inequalities and Identities of Inclusion–Exclusion Type*, Lecture Notes in Mathematics, vol. 1826, Berlin: Springer-Verlag, pp. viii+113, ISBN 3-540-20025-8, MR 2019293, Zbl 1026.05009 - Galambos, János; Simonelli, Italo (1996),
*Bonferroni-Type Inequalities with Applications*, Probability and Its Applications, New York: Springer-Verlag, pp. x+269, ISBN 0-387-94776-0, MR 1402242, Zbl 0869.60014 - Galambos, János (1977), "Bonferroni inequalities",
*Annals of Probability*,**5**(4): 577–581, doi:10.1214/aop/1176995765, JSTOR 2243081, MR 0448478, Zbl 0369.60018 - Galambos, János (2001) [1994], "Bonferroni inequalities",
*Encyclopedia of Mathematics*, EMS Press

*This article incorporates material from Bonferroni inequalities on PlanetMath, which is licensed under the Creative Commons Attribution/Share-Alike License.*