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

$\blacksquare$


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