Exercise 1.43

Question

Show that for any events $A$ and $B$,
$$P(A) + P(B) - 1 \leq P(A \cap B) \leq P(A \cap B) \leq P(A) + P(B).$$
For each of these three inequalities, give a simple criterion for when the inequality is
actually an equality (e.g., give a simple condition such that $P (A \cap B) = P (A \cup B)$ if and only if the condition holds).

fifthist x · Accepted Answer

\begin{enumerate}[label=(\alph*)]

\item Inequality can be demonstrated using the first property of probabilities,
$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$ and the first axiom of probabilities,
$$P(S) = 1.$$

$P(A) + P(B) - P(A \cap B) \leq 1 \implies P(A) + P(B) - 1 \leq P(A \cap B)$.

Strict equality holds if and only if $A \cup B = S$ where $S$ is the sample space.

\item Since $A \cap B \subseteq A \cup B$, $P(A \cap B) \leq P(A \cup B)$ by
 the second property of probabilites.

Strict equality holds if and only if $A = B.$

\item Inequality follows directly from the first property of probabilities
 with strict equality if and only if $P(A \cap B) = 0.$

\end{enumerate}

Exercise 1.43

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Exercise 1.43

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