# Proving Bonferroni's Inequality

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• Feb 28th 2008, 06:31 AM
janvdl
Proving Bonferroni's Inequality
Bonferroni's Inequality states that:

$\displaystyle P(A \cap B) \geq P(A) + P(B) - 1$

But the first axiom of probability states that:

$\displaystyle P(\Omega ) = 1$

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Am I allowed to do the following?

$\displaystyle P(A \cap B) \geq P(A \cup B) - 1$

$\displaystyle P(A \cap B) \geq P(A \cup B) - P(\Omega )$

$\displaystyle P(\Omega ) \geq P(A \cup B) - P(A \cap B)$

And the last line must be true because the entire sample size must be at least equal to the union of A and B minus their intersection.

OR

Should i rather try to prove it this way:

Rearrange so that the 1 is isolated...

$\displaystyle 1 \geq P(A) + P(B) - P(A \cap B)$

$\displaystyle 1 \geq P(A \cup B)$

$\displaystyle P(\Omega ) \geq P(A \cup B)$

Which is also true...

Anyone got an idea which one of these is correct?
Thanks in advance.

EDIT: The second one might be the correct one seeing as A and B are not disjoint. Am i right?
• Feb 28th 2008, 07:38 AM
tukeywilliams
$\displaystyle \Omega$ is the sample space. Then $\displaystyle P(A\cup B) \leq P(\Omega) = 1$.

And $\displaystyle P(A \cup B) = P(A) + P(B) - P(A \cap B)$.

Hence $\displaystyle P(A \cap B) \geq P(A) + P(B) -1$.
• Feb 28th 2008, 07:42 AM
janvdl
Quote:

Originally Posted by tukeywilliams
$\displaystyle \Omega$ is the sample space. Then $\displaystyle P(A\cup B) \leq P(\Omega) = 1$.

And $\displaystyle P(A \cup B) = P(A) + P(B) - P(A \cap B)$.

Hence $\displaystyle P(A \cap B) \geq P(A) + P(B) -1$.

Thanks a lot TukeyWilliams +rep+ (Handshake) :D