$A,B_1,B_2$ stochastically independent $\Rightarrow$ $A$ independent of $B_1 \cup B_2$?

Question

Let $A,B_1,B_2$ be stochastically independent. Does this mean that $A$ is stochastically independent of $B_1 \cup B_2$? My gut feeling says no but how can I prove this?

Thanks in advance!

Answer 1

If $B_1\cup B_2$ were to be independent of $A$ then we should have $P(B_1 \cup B_2|A)=P(B_1 \cup B_2)$. This is consequence of the definition of independence between the events. So lets check that,

$P(B_1 \cup B_2|A) =\frac{P(A \cap (B_1 \cup B_2))}{P(A)}$

$\frac{P(A \cap (B_1 \cup B_2))}{P(A)} =\frac{P((A\cap B_1)\cup(A \cap B_2))}{P(A)} =\frac{P(A\cap B_1)+P((A\cap B_2)-P(A \cap B_1)P(A \cap B_2)}{P(A)}$

$$ =\frac{P(A)(P(B_1)+P(B_2)-P(B_1)P(B_2))}{P(A)} =P(B_1 \cup B_2)$$

Answer 2

You just have to show that $P((B_1 \cup B_2) \cap A) = P(B_1 \cup B_2)P(A)$.

First, note that by distributivity $(B_1 \cup B_2) \cap A = (B_1 \cap A) \cup (B_2 \cap A)$.

By inclusion-exclusion, $$P((B_1 \cap A) \cup (B_2 \cap A)) = P(B_1 \cap A) + P(B_2 \cap A) - P((B_1 \cap A) \cap (B_2 \cap A)). \tag{1}$$

Using our assumption about independence in (1) and the fact that $(B_1 \cap A) \cap (B_2 \cap A) = B_1 \cap B_2 \cap A$ gets us \begin{align} P((B_1 \cap A) \cup (B_2 \cap A)) &= P(B_1)P(A) + P(B_2)P(A) - P(B_1)P(B_2)P(A)\\ &= P(A) \big[ P(B_1) + P(B_2) - P(B_1)P(B_2) \big] \tag{2} \end{align}

Using inclusion-exclusion and our independence assumption again, we see that $P(B_1) + P(B_2) - P(B_1)P(B_2) = P(B_1 \cup B_2)$. Plugging this into (2) we get $$P((B_1 \cap A) \cup (B_2 \cap A)) = P(B_1 \cup B_2)P(A),$$ which is the desired result.