# Boole's inequality

Short description: Inequality applying to probability spaces

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

$\displaystyle{ {\mathbb P}\left(\bigcup_{i=1}^{\infty} A_i \right) \le \sum_{i=1}^{\infty} {\mathbb P}(A_i). }$

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 $\displaystyle{ n }$ events using the method of induction.

For the $\displaystyle{ n=1 }$ case, it follows that

$\displaystyle{ \mathbb P(A_1) \le \mathbb P(A_1). }$

For the case $\displaystyle{ n }$, we have

$\displaystyle{ {\mathbb P}\left(\bigcup_{i=1}^{n} A_i \right) \le \sum_{i=1}^{n} {\mathbb P}(A_i). }$

Since $\displaystyle{ \mathbb P(A \cup B) = \mathbb P(A) + \mathbb{P}(B) - \mathbb{P}(A \cap B), }$ and because the union operation is associative, we have

$\displaystyle{ \mathbb{P}\left(\bigcup_{i=1}^{n+1}A_i\right) = \mathbb{P}\left(\bigcup_{i=1}^n A_i\right) + \mathbb{P}(A_{n+1}) -\mathbb{P}\left(\bigcup_{i=1}^n A_i \cap A_{n+1}\right). }$

Since

$\displaystyle{ {\mathbb P}\left(\bigcup_{i=1}^n A_i \cap A_{n+1}\right) \ge 0, }$

by the first axiom of probability, we have

$\displaystyle{ \mathbb{P}\left(\bigcup_{i=1}^{n+1} A_i \right) \le \mathbb{P} \left(\bigcup_{i=1}^n A_i\right) + \mathbb{P}(A_{n+1}), }$

and therefore

$\displaystyle{ \mathbb{P}\left(\bigcup_{i=1}^{n+1} A_i \right) \le \sum_{i=1}^{n} \mathbb{P}(A_i) + \mathbb{P}(A_{n+1}) = \sum_{i=1}^{n+1} \mathbb{P}(A_i). }$

### Proof without using induction

For any events in $\displaystyle{ A_1, A_2, A_3, \dots }$in our probability space we have

$\displaystyle{ \mathbb{P}\left(\bigcup_{i} A_i\right) \leq \sum_i \mathbb P(A_i). }$

One of the axioms of a probability space is that if $\displaystyle{ B_1, B_2, B_3, \dots }$ are disjoint subsets of the probability space then

$\displaystyle{ \mathbb{P}\left(\bigcup_{i} B_i\right) = \sum_i \mathbb P(B_i); }$

If we modify the sets $\displaystyle{ A_i }$, so they become disjoint,

$\displaystyle{ B_i = A_i - \bigcup^{i-1}_{j=1} A_j }$

we can show that

$\displaystyle{ \bigcup^{\infty}_{i=1} B_i = \bigcup^{\infty}_{i=1} A_i. }$

by proving both directions of inclusion.

Suppose $\displaystyle{ x \in \bigcup^{\infty}_{i=1} A_i }$. Then $\displaystyle{ x \in A_k }$ for some minimum $\displaystyle{ k }$ such that $\displaystyle{ i \lt k \implies x \notin A_i }$. Therefore $\displaystyle{ x \in B_k = A_k - \bigcup^{k-1}_{j=1} A_j }$. So the first inclusion is true: $\displaystyle{ \bigcup^{\infty}_{i=1} A_i \subset \bigcup^{\infty}_{i=1} B_i }$.

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

By construction of each $\displaystyle{ B_i }$, $\displaystyle{ B_i \subset A_i }$. For $\displaystyle{ B \subset A, }$ it is the case that $\displaystyle{ \mathbb P (B) \leq \mathbb P(A). }$

So, we can conclude that the desired inequality is true:

$\displaystyle{ \mathbb P\left(\bigcup_iA_i\right) = \mathbb P\left(\bigcup_iB_i\right) = \sum_i \mathbb P (B_i) \leq \sum_i \mathbb P(A_i). }$

## Bonferroni inequalities

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

Let

$\displaystyle{ S_1 := \sum_{i=1}^n {\mathbb P}(A_i), \quad S_2 := \sum_{1\le i_1 \lt i_2\le n} {\mathbb P}(A_{i_1} \cap A_{i_2} ),\quad \ldots,\quad S_k := \sum_{1\le i_1\lt \cdots\lt i_k\le n} {\mathbb P}(A_{i_1}\cap \cdots \cap A_{i_k} ) }$

for all integers k in {1, ..., n}.

Then, when $\displaystyle{ K \leq n }$ is odd:

$\displaystyle{ \sum_{j=1}^K (-1)^{j-1} S_j \geq \mathbb{P}\left(\bigcup_{i=1}^n A_i\right) = \sum_{j=1}^n (-1)^{j-1} S_j }$

holds, and when $\displaystyle{ K \leq n }$ is even:

$\displaystyle{ \sum_{j=1}^K (-1)^{j-1} S_j \leq \mathbb{P}\left(\bigcup_{i=1}^n A_i\right) = \sum_{j=1}^n (-1)^{j-1} S_j }$

holds.

The equalities follow from the inclusion–exclusion principle, and Boole's inequality is the special case of $\displaystyle{ K=1 }$.

### Proof for odd K

Let $\displaystyle{ E = \bigcap_{i=1}^n B_i }$, where $\displaystyle{ B_i \in \{A_i, A_i^c\} }$ for each $\displaystyle{ i = 1, \dots, n }$. These such $\displaystyle{ E }$ partition the sample space, and for each $\displaystyle{ E }$ and every $\displaystyle{ i }$, $\displaystyle{ E }$ is either contained in $\displaystyle{ A_i }$ or disjoint from it.

If $\displaystyle{ E = \bigcap_{i=1}^n A_i^c }$, then $\displaystyle{ E }$ contributes 0 to both sides of the inequality.

Otherwise, assume $\displaystyle{ E }$ is contained in exactly $\displaystyle{ L }$ of the $\displaystyle{ A_i }$. Then $\displaystyle{ E }$ contributes exactly $\displaystyle{ \mathbb{P}(E) }$ to the right side of the inequality, while it contributes

$\displaystyle{ \sum_{j=1}^K (-1)^{j-1} {L \choose j} \mathbb{P}(E) }$

to the left side of the inequality. However, by Pascal's rule, this is equal to

$\displaystyle{ \sum_{j=1}^K (-1)^{j-1} \left({L-1 \choose j-1} + {L-1 \choose j} \right)\mathbb{P}(E) }$

which telescopes to

$\displaystyle{ \left( 1 + {L-1 \choose K}\right) \mathbb{P}(E) \geq \mathbb{P}(E) }$

Thus, the inequality holds for all events $\displaystyle{ E }$, and so by summing over $\displaystyle{ E }$, we obtain the desired inequality:

$\displaystyle{ \sum_{j=1}^K (-1)^{j-1} S_j \geq \mathbb{P}\left(\bigcup_{i=1}^n A_i\right) }$

The proof for even $\displaystyle{ K }$ is nearly identical.[3]

### 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 the Bonferroni Method of simultaneous inference.