# 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 $20$