Boole's inequality

Proofs

Proof by induction

For finite collections of events, Boole's inequality can be proved by induction. First, for any event

A

1

{\displaystyle A_{1}}

, one trivially has

P

(

A

1

)

≤

P

(

A

1

)

.

{\displaystyle \mathbb {P} (A_{1})\leq \mathbb {P} (A_{1}).}

Next, assume that the inequality holds for any collection of

n

{\displaystyle n}

events, and consider n+1 events

A

1

,

…

,

A

n

.

{\displaystyle A_{1},\ldots ,A_{n}.}

Proof without using induction

Let events

A

1

,

A

2

,

A

3

,

…

{\displaystyle A_{1},A_{2},A_{3},\dots }

in our probability space be given. The countable additivity of the measure

P

{\displaystyle \mathbb {P} }

states that if

B

1

,

B

2

,

B

3

,

…

{\displaystyle B_{1},B_{2},B_{3},\dots }

are pairwise disjoint events, then

P

(

⋃

i

B

Bonferroni inequalities

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

Let

S

1

:=

∑

i

=

1

n

P

(

A

i

)

,

S

2

:=

∑

1

≤

i

1

<

i

2

≤

n

P

(

A

i

1

Proof for odd K

Let

E

=

⋂

i

=

1

n

B

i

{\displaystyle E=\bigcap _{i=1}^{n}B_{i}}

, where

B

i

∈

{

A

i

,

A

i

c

}

{\displaystyle B_{i}\in \{A_{i},A_{i}^{c}\}}

for each

i

=

1

,

…

,

n

{\displaystyle i=1,\dots ,n}

. These such

E

{\displaystyle E}

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.