Index: The Book of Statistical Proofs ▷ General Theorems ▷ Probability theory ▷ Probability axioms ▷ Bonferroni's inequality

Theorem: The probability of the intersection of $A$ and $B$ is larger than or equal to the sum of the probabilities of $A$ and $B$ minus one:

\[\label{eq:bool-ineq} P(A \cap B) \geq P(A) + P(B) - 1 \; .\]

Proof: The addition law of probability states that, for two events $A$ and $B$, it holds true that

\[\label{eq:prob-add} P(A \cup B) = P(A) + P(B) - P(A \cap B) \; .\]

Rearranging for $P(A \cap B)$, we have:

\[\label{eq:bool-ineq-s1} P(A \cap B) = P(A) + P(B) - P(A \cup B) \; .\]

The range of probability is

\[\label{eq:prob-range} 0 \leq P(E) \leq 1 \; .\]

Thus, $P(A \cup B)$ is at most one. (This is the case, if $A$ and $B$ are collectively exhaustive and their union is thus equal to the sample space $\Omega$.) With this, we are able to derive a lower bound for $P(A \cap B)$, as given by the theorem:

\[\label{eq:bool-ineq-qed} P(A \cap B) \geq P(A) + P(B) - 1 \; .\]
∎
Sources:

Metadata: ID: P484 | shortcut: bonf-ineq | author: JoramSoch | date: 2025-01-10, 14:56.