If we have $\mathbb{P}(A|B) = \mathbb{P}(B|A)$ then what is this special case called, and are there special properties? I'm interested in a simpler way of computing one of them and would like to take advantage of such properties.
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3This is true if and only if $P(A) = P(B)$. Is this the kind of relationship you are looking for or...? – Jun 04 '14 at 02:27
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12@Matthew Just to nitpick a little, the result $P(A\mid B) = P(B\mid A)$ does not hold only if $P(A) = P(B)$ since it also holds when $P(A\cap B) = 0$ in which case whether $P(A)$ equals $P(B)$ or not is not germane. One of my favorite True/False questions is the pair "If $P(A) = P(B)$, then $P(A\mid B) = P(B\mid A)$" followed immediately by "If $P(A\mid B) = P(B\mid A)$, then $P(A) = P(B)$" – Dilip Sarwate Jun 04 '14 at 02:34
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1@DilipSarwate good catch! – Jun 04 '14 at 02:44
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Just to formalise this answer (from the comments), note that your condition entails the requirement that:
$$\mathbb{P}(A|B) = \frac{\mathbb{P}(A \cap B)}{\mathbb{P}(B)} = \frac{\mathbb{P}(A \cap B)}{\mathbb{P}(A)} = \mathbb{P}(B|A).$$
This equation holds either if $\mathbb{P}(A \cap B) = 0$ (the events are mutually exclusive) or if $\mathbb{P}(A)=\mathbb{P}(B)$ (the events have the same marginal probability). There is no special name for this that I am aware of, other than naming it directly by description ---i.e., it is the case where the conditional probability is equal to its "inverse". Although there is no special name for this condition, there is a name for the fallacy of equating these two things when the conditions for their equality do not hold --- this is called the conditional probability fallacy (also known as the "confusion of the inverse").
Ben
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