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 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 For the case n, we have Since and because the union operation is associative, we have Since by the first axiom of probability, we have and therefore

Proof without using induction

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 A_i, so they become disjoint, we can show that by proving both directions of inclusion. Suppose. Then x \in A_k for some minimum k such that. Therefore. So the first inclusion is true:. Next suppose that. It follows that x \in B_k for some k. And so x \in A_k, and we have the other inclusion:. By construction of each B_i,. For it is the case that So, we can conclude that the desired inequality is true:

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. Let for all integers k in {1, ..., n}. Then, when K \leq n is odd: holds, and when K \leq n is even: holds. The equalities follow from the inclusion–exclusion principle, and Boole's inequality is the special case of K=1.

Proof for odd K

Let, where for each. These such E partition the sample space, and for each E and every i, E is either contained in A_i or disjoint from it. If, then E contributes 0 to both sides of the inequality. Otherwise, assume E is contained in exactly L of the A_i. Then E contributes exactly 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 E, and so by summing over E, we obtain the desired inequality: The proof for even K is nearly identical.

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 five 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 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.

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