# Bonferroni inequality

The Bonferroni inequalities are formulas used to estimate the probability of the average or union of events .

## Named after Bonferroni

The Bonferroni inequalities are not necessarily rightly named after Carlo Emilio Bonferroni .

Bonferroni was probably not the originator of these inequalities, but used them to define a statistical estimator ( Bonferroni method ). Naming it after him is therefore particularly popular in statistical circles. Because of their simplicity, the inequalities were most likely known before him.

The first of the following inequalities is more commonly referred to as the Boolean inequality after George Boole ; however, the inequalities are often mentioned without reference to a name.

## First inequality

In the following, arbitrary events are assumed in a probability space . Let it denote the probability of the event and the union of the events . Then: ${\ displaystyle E_ {i}}$ ${\ displaystyle (\ Omega, {\ mathcal {A}}, \ mathbb {P})}$${\ displaystyle \ mathbb {P} (E_ {i})}$${\ displaystyle E_ {i}}$${\ displaystyle \ bigcup _ {i = 1} ^ {n} E_ {i}}$${\ displaystyle E_ {1}, \ dots, E_ {n}}$

${\ displaystyle \ mathbb {P} \ left (\ bigcup _ {i = 1} ^ {n} E_ {i} \ right) \ leq \ sum _ {i = 1} ^ {n} \ mathbb {P} \ left (E_ {i} \ right)}$.

It also applies more generally:

${\ displaystyle \ mathbb {P} \ left (\ bigcup _ {i = 1} ^ {\ infty} E_ {i} \ right) \ leq \ sum _ {i = 1} ^ {\ infty} \ mathbb {P } \ left (E_ {i} \ right)}$

These inequalities are also called Boolean inequalities.

### proof

If you set

${\ displaystyle A_ {i} = E_ {i} \ setminus \ left (\ bigcup _ {j = 1} ^ {i-1} E_ {j} \ right),}$

then the pairs are disjoint and it applies ${\ displaystyle A_ {i}}$

${\ displaystyle \ bigcup _ {i} A_ {i} = \ bigcup _ {i} E_ {i}.}$

So follows

${\ displaystyle \ mathbb {P} \ left (\ bigcup _ {i} E_ {i} \ right) = \ mathbb {P} \ left (\ bigcup _ {i} A_ {i} \ right) = \ sum _ {i} \ mathbb {P} (A_ {i}) \ leq \ sum _ {i} \ mathbb {P} (E_ {i}).}$

The second equality applies because of the σ-additivity and the inequality because of the monotony of the probability measure. ${\ displaystyle A_ {i} \ subseteq E_ {i}}$

## Second inequality

In the following, let us again assume any events in a probability space . Also denote the complement of . Then follows: ${\ displaystyle E_ {i}}$ ${\ displaystyle (\ Omega, {\ mathcal {A}}, \ mathbb {P})}$${\ displaystyle {\ overline {E_ {i}}} = \ Omega \ setminus E_ {i}}$${\ displaystyle E_ {i}}$

${\ displaystyle \ mathbb {P} \ left (\ bigcap _ {i = 1} ^ {n} E_ {i} \ right) \ geq 1- \ sum _ {i = 1} ^ {n} \ mathbb {P } \ left ({\ overline {E_ {i}}} \ right) = \ sum _ {i = 1} ^ {n} \ mathbb {P} \ left (E_ {i} \ right) - (n-1 )}$

## Third inequality

Closely connected to the two above inequalities is the following, which by some authors as bonferronische inequality ( English Bonferroni's Inequality is called). It states (under the conditions mentioned):

${\ displaystyle \ mathbb {P} \ left (\ bigcup _ {i = 1} ^ {n} {E_ {i}} \ right) \ geq \ sum _ {i = 1} ^ {n} \ mathbb {P } \ left (E_ {i} \ right) - \ sum _ {i, j = 1, \ ldots, n \ atop \; {\ text {with}} \; i

## Examples

• Let be the set of results of a die roll. Designate the event of rolling an even number and the event of rolling at least a 5. Obviously, and . After the first Bonferroni's inequality applies to the event just a number , or 5 to dice at least one, that is ,${\ displaystyle \ Omega = \ {1,2,3,4,5,6 \}}$${\ displaystyle E_ {1} = \ {2,4,6 \}}$${\ displaystyle E_ {2} = \ {5.6 \}}$${\ displaystyle \ mathbb {P} (E_ {1}) = {\ frac {1} {2}}}$${\ displaystyle \ mathbb {P} (E_ {2}) = {\ frac {1} {3}}}$${\ displaystyle E = \ {2,4,5,6 \}}$
${\ displaystyle \ mathbb {P} \ left (E_ {1} \ cup E_ {2} \ right) \ leq \ mathbb {P} \ left (E_ {1} \ right) + \ mathbb {P} \ left ( E_ {2} \ right) = {\ frac {1} {2}} + {\ frac {1} {3}} = {\ frac {5} {6}}.}$
• Let the scenario be as in the previous example. After the second Bonferroni inequality applies for the event, just a number and 5 to roll at least one, ie ,${\ displaystyle E = \ {6 \}}$
${\ displaystyle \ mathbb {P} \ left (E_ {1} \ cap E_ {2} \ right) \ geq 1- \ mathbb {P} \ left ({\ overline {E_ {1}}} \ right) - \ mathbb {P} \ left ({\ overline {E_ {2}}} \ right) = 1- \ left (1 - {\ frac {1} {2}} \ right) - \ left (1 - {\ frac {1} {3}} \ right) = - {\ frac {1} {6}}}$
The result does not provide a useful statement, since every probability is greater than or equal to zero anyway.
However, it follows for the event to roll an even number and less than a 5, so ,${\ displaystyle E = \ {2,4 \}}$
${\ displaystyle \ mathbb {P} \ left (E_ {1} \ cap {\ overline {E_ {2}}} \ right) \ geq 1- \ mathbb {P} \ left ({\ overline {E_ {1} }} \ right) - \ mathbb {P} \ left (E_ {2} \ right) = 1- \ left (1 - {\ frac {1} {2}} \ right) - \ left (1 - {\ frac {2} {3}} \ right) = {\ frac {1} {6}}.}$

## Individual evidence

1. ^ Jürgen Bortz: Statistics for human and social scientists . 6th edition. Springer, 2005, p. 129 .
2. ^ J. Galambos: Bonferroni inequalities . In: Michiel Hazewinkel (Ed.): Encyclopaedia of Mathematics . Springer-Verlag , Berlin 2002, ISBN 978-1-55608-010-4 (English, online ).
3. Hans-Otto Georgii: Stochastics: Introduction to Probability Theory and Statistics. 4th edition. de Gruyter textbook, Berlin 2009, ISBN 978-3-11-021526-7 . P. 15.
4. Rosen et al: Handbook ... p. 433 .