Event Independence is Symmetric
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Theorem
Let $A$ and $B$ be events in a probability space.
Let $A$ be independent of $B$.
Then $B$ is independent of $A$.
That is, is independent of is a symmetric relation.
Proof
We assume throughout that $\map \Pr A > 0$ and $\map \Pr B > 0$.
Let $A$ be independent of $B$.
Then by definition:
- $\condprob A B = \map \Pr A$
From the definition of conditional probabilities, we have:
- $\condprob A B = \dfrac {\map \Pr {A \cap B} } {\map \Pr B}$
and also:
- $\condprob B A = \dfrac {\map \Pr {A \cap B} } {\map \Pr A}$
So if $\condprob A B = \map \Pr A$ we have:
\(\ds \map \Pr A\) | \(=\) | \(\ds \frac {\map \Pr {A \cap B} } {\map \Pr B}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map \Pr B\) | \(=\) | \(\ds \frac {\map \Pr {A \cap B} } {\map \Pr A}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map \Pr B\) | \(=\) | \(\ds \condprob B A\) |
So by definition, $B$ is independent of $A$.
$\blacksquare$
Sources
- 1986: Geoffrey Grimmett and Dominic Welsh: Probability: An Introduction ... (previous) ... (next): $\S 1.7$: Independent Events: Exercise $20$