Why does $A$ independent from $B$ imply $B$ independent from $A$

Suppose $$A, B$$ are events with $$P(A), P(B) > 0$$. If $$P(A \mid B) = P(A)$$, can I have an intuitive explanation of why $$P(B \mid A) = P(B)$$? And if $$P(A \mid B) \neq P(A)$$, then $$P(B \mid A) \neq P(B)$$? Preferably without any formulas, just simple examples or intuitive reasoning.

I can prove it using formulas:

$$P(A \mid B) = P(A) \iff \displaystyle \frac{P(A \cap B)}{P(B)} = P(A) \iff \frac{P(A \cap B)}{P(A)} = P(B) \iff P(B \mid A) = P(B)$$. However I am not sure how to picture this. For example if $$P(A \mid B) > P(A)$$ then what is an intuitive explanation for why $$P(B \mid A) > P(B)$$ as well?

• $A$ independent from $B$ if and only if $P(A\cap B) = P(A)\times P(B)$. Now.. recall that intersection of events is commutative and multiplication of real numbers is commutative as well, so the above further can be rewritten as $P(B\cap A) = P(B)\times P(A)$. It seems perfectly intuitive to me... and using that $P(A\cap B) = P(A)\times P(B)$ as the definition for independence I find to be easier in general (especially as it pertains to null events as conditioning on an impossible event is often tricky to explain or define properly) – JMoravitz Jul 6 at 19:04
• Maybe this comment should be an answer so that the OP can mark it as accepted. – Stelios Kounis Jul 8 at 10:35

More generally, given a probability space $$(\Omega,\mathcal F,\mathbb P)$$ and two sub-$$\sigma$$-algebras $$\mathcal G$$, $$\mathcal H\subset\mathcal F$$, we say that $$\mathcal G$$ and $$\mathcal H$$ are independent iff for all $$G\in\mathcal G$$ and $$H\in\mathcal H$$, $$\mathbb P(G\cap H) = \mathbb P(G)\mathbb P(H).$$ In the case where we are considering individual events, the definition is still the same - $$A,B\in\mathcal F$$ are independent iff $$\mathbb P(A\cap B) = \mathbb P(A)\mathbb P(B).$$ There aren't separate concepts like "$$A$$ independent from $$B$$" and "$$B$$ independent from $$A$$" - just the concept of the two events being independent.
• Thanks, I understand that they are equivalent, but intuitively one might ask "why should $P(A \mid B) = P(A)$ be equivalent to $P(B \mid A) = P(B)$" – twosigma Aug 1 at 16:00