Question

Use induction to generalize Bonferroni’s inequality to $n$events. That is, show that

$P(E_1{E}_2···E_n)\ge P\left(E_1\right)+···+P\left(E_n\right)-(n-1)$.

Short answer

proven by the principle of mathematical induction.

Step-by-step solution

Given Information.

Given, generalize Bonferroni’s inequality to $n$events.

Explanation.

For events$E_1,E_2,\dots E_n$.

$P\left(E_1{E}_2{E}_3·\dots ·E_n\right)=P\left(E_1\right)+P\left(E_2\right)+\dots +P\left(E_n\right)-(n-1)$

Proof by mathematical induction

For $n=2$

$P\left(E_1{E}_2\right)=P\left(E_1\right)+P\left(E_2\right)-1$

For proof of this refer to the Theoretical exerciserole="math" localid="1649259597999" $11$.

If this is true for some $n\in \mathrm{\mathbb{N}}$. i.e.

$P\left(E_1{E}_2{E}_3·\dots ·E_n\right)=P\left(E_1\right)+P\left(E_2\right)+\dots +P\left(E_n\right)-(n-1)$

For events$E_1,E_2,\dots E_n,E_{n+1}$

$P\left(E_1{E}_2{E}_3·\dots ·E_n{E}_{n+1}\right)=P\left[\left(E_1{E}_2{E}_3·\dots ·E_n\right)E_{n+1}\right]$

$=P\left(E_1{E}_2{E}_3·\dots ·E_n\right)+P\left(E_{n+1}\right)-1$

$$\begin{gathered} \stackrel{\left(1\right)}{=}P\left(E_1\right)+P\left(E_2\right)+\dots +P\left(E_n\right)-(n-1)+P\left(E_{n+1}\right)-1 \\ \stackrel{\left(2\right)}{=}P\left(E_1\right)+P\left(E_2\right)+\dots +P\left(E_n\right)+P\left(E_{n+1}\right)-(n+1-1) \end{gathered}$$

The statement holds for $n+1$thus by the principle of mathematical induction it holds for every$n\in \mathrm{\mathbb{N}}$.