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 A1, A2, A3, ..., 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.
case, it follows that For the case
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 we modify the sets
, so they become disjoint, we can show that by proving both directions of inclusion.
So the first inclusion is true:
So, we can conclude that the desired inequality is true: 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).
is odd: holds, and when
The equalities follow from the inclusion–exclusion principle, and Boole's inequality is the special case of
partition the sample space, and for each
contributes 0 to both sides of the inequality.
to the right side of the inequality, while it contributes to the left side of the inequality.
However, by Pascal's rule, this is equal to which telescopes to Thus, the inequality holds for all events
, we obtain the desired inequality: The proof for even
[3] 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 five are good", we can change this question into another condition: One way is to make each of them equal to 0.05/5 = 0.01, that is 1%.
In other words, 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 the Bonferroni Method of simultaneous inference.
This article incorporates material from Bonferroni inequalities on PlanetMath, which is licensed under the Creative Commons Attribution/Share-Alike License.