Boole's inequality on arvau

Wikipedia / Boole's inequality

In probability theory, Boole's inequality, also known as the union bound, says that for any finite or countable set of eventss, 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 is replaced by any measure.

Proof

Proof using induction

events using the method of induction.

case, it follows that

, we have

and because the union operation is associative, we have

Since

by the first axiom of probability, we have

and therefore

Proof without using induction

in our probability space are pairwise disjoint events, then

Set

are pairwise disjoint. We claim that:

for the minimum positive integer.

Therefore

for all i.

Bonferroni inequalities

Boole's inequality for a finite number of events may be generalized to certain upper and lower boundss 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}.

is odd:

is even:

holds.

The inequalities follow from the inclusion–exclusion principle is replaced by any finite content, in the sense of measure theory.

Proof for odd K

partition the sample space or disjoint from it.

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

, we obtain the desired inequality:

is nearly identical.

Example

Suppose that you are estimating five 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.