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Prove the following version of the Bonferroni inequality- $$P\left(\bigcap_{i=1}^kA_i\right)\ge1-\sum_{i=1}^kP(A_i^c)$$

When creating simultaneous confidence interval, what are $A_i$ and $A_i^c$?

Would appreciate any help.

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Just note that by De Morgan's law, $\overline{\bigcap A_i} = \bigcup \overline{A_i} $ where $\overline{A_i}$ means "not $A_i$ or equivalently the complement $A_i^c$".

Now note that $P\left(A\cup B\right) = P(A)+P(B)-P(A\cap B) \le P(A)+P(B)$

So finally

$$ P\left(\bigcap A_i\right) = 1-P\left(\overline{\bigcap A_i}\right) = 1-P\left(\bigcup A_i^c\right) \ge 1-\sum P\left(A_i^c\right) $$

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