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?

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    $\begingroup$ $A$ 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) $\endgroup$ – JMoravitz Jul 6 at 19:04
  • $\begingroup$ Maybe this comment should be an answer so that the OP can mark it as accepted. $\endgroup$ – 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.

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  • $\begingroup$ Thanks, I understand that they are equivalent, but intuitively one might ask "why should $P(A \mid B) = P(A)$ be equivalent to $P(B \mid A) = P(B)$" $\endgroup$ – twosigma Aug 1 at 16:00

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