Event Independence is Symmetric

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 $\Pr \left({A}\right) > 0$ and $\Pr \left({B}\right) > 0$.

Let $A$ be independent of $B$.

Then by definition:

$\Pr \left({A \mid B}\right) = \Pr \left({A}\right)$

From the definition of conditional probabilities, we have:

$\Pr \left({A \mid B}\right) = \dfrac{\Pr \left({A \cap B}\right)} {\Pr \left({B}\right)}$

and also:

$\Pr \left({B \mid A}\right) = \dfrac{\Pr \left({A \cap B}\right)} {\Pr \left({A}\right)}$

So if $\Pr \left({A \mid B}\right) = \Pr \left({A}\right)$ we have:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \Pr \left({A}\right)\)	\(=\)	\(\displaystyle \frac{\Pr \left({A \cap B}\right)} {\Pr \left({B}\right)}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \Pr \left({B}\right)\)	\(=\)	\(\displaystyle \frac{\Pr \left({A \cap B}\right)} {\Pr \left({A}\right)}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \Pr \left({B}\right)\)	\(=\)	\(\displaystyle \Pr \left({B \mid A}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

So by definition, $B$ is independent of $A$.

$\blacksquare$

Sources

• Geoffrey Grimmett and Dominic Welsh: Probability: An Introduction (1986): $\S 1.7$: Exercise $20$