# 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:

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

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

$\blacksquare$