Independence of composite events
Dependencies:
- Probability
- Independence of events
- /sets-and-relations/de-morgan-laws
Let events $A_1$ and $A_2$ be independent of $B$. Suppose $A_1 \subseteq A_2$. Then $A_2 - A_1$ is independent of $B$.
$\Omega$ is independent of $B$, so by replacing $A_2$ by $\Omega$ we get that $\overline{A}$ is independent of $B$.
Let $S$ be a countable set of pairwise-disjoint events such that $\forall A \in S, A$ is independent of $B$. Then $\bigcup_{A \in S} A$ is independent of $B$. Furthermore, this theorem has a counterexample when sets in $S$ are not pairwise-disjoint.
Proof
\begin{align} & \Pr((A_2 - A_1) \cap B) \\ &= \Pr((A_2 \cap B) - (A_1 \cap B)) \\ &= \Pr(A_2 \cap B) - \Pr(A_1 \cap B) \tag{$A_1 \cap B \subseteq A_2 \cap B$} \\ &= \Pr(A_2)\Pr(B) - \Pr(A_1)\Pr(B) \tag{by independence} \\ &= \Pr(B)(\Pr(A_2) - \Pr(A_1)) \\ &= \Pr(B)\Pr(A_2 - A_1) \tag{$A_1 \subseteq A_2$} \end{align} Therefore, $A_2 - A_1$ is independent of $B$.
\begin{align} & \Pr\left(\left(\bigcup_{A \in S} A\right) \cap B\right) \\ &= \Pr\left(\bigcup_{A \in S} (A \cap B)\right) \tag{De Morgan's law} \\ &= \sum_{A \in S} \Pr(A \cap B) \tag{events in $\{A \cap B: A \in S\}$ are disjoint} \\ &= \sum_{A \in S} \Pr(A)\Pr(B) \tag{by independence} \\ &= \Pr(B)\Pr\left(\bigcup_{A \in S} A\right) \tag{events in $S$ are disjoint} \end{align} Therefore, $\bigcup_{A \in S} A$ is independent of $B$.
Counterexample
Let $\Omega = \{1, 2, 3, 4\}$, $\Pr(X) = |X|/4$, $B = \{2, 4\}$, $A_1 = \{1, 4\}$, $A_2 = \{3, 4\}$ and $S = \{A_1, A_2\}$. Then $A_1$ and $A_2$ are not disjoint, $A_1$ is independent of $B$, $A_2$ is independent of $B$ and $A_1 \cup A_2$ is not independent of $B$.
Dependency for:
Info:
- Depth: 6
- Number of transitive dependencies: 8
Transitive dependencies:
- /sets-and-relations/countable-set
- /sets-and-relations/de-morgan-laws
- σ-algebra
- Measure
- σ-algebra is closed under countable intersections
- Probability
- Conditional probability (incomplete)
- Independence of events