Ordered Chain Rule of Probability

Question

For the chain rule of probability the order of variables does not matter and sets are not ordered as of themselves anyway. So therefore $P(A | B) P(B) = P(A \cap B) = P(B|A) P(A)$.

However, I cannot reconcile this with a scenario like in the following tree of (conditional) probabilities.

-| -- (.1) -- A -| | -- (.7) -- A | | | -- (.3) -- B | -- (.9) -- B -| -- (.4) -- A | -- (.6) -- B

This actually does not satisfy the symmetry from above, as the order does matter:

$P((B,A)) = P(B) P(A|B) = .9 \times .4$

and

$P((A,B)) = P(A) P(B|A) = .1 \times .3$

When I search for "ordered joint distribution" or "ordered chain rule of probability" I don't find anything useful.

So, my suspicion is that I am fundamentally confused about something here. If that is the case, can you point out my misconception?

Alternatively I am just using the wrong search terms...

Answer 1

The conditional proability $\mathsf P(A\mid B)$ is defined as $\mathsf P(A\cap B)\div\mathsf P(B)$, when $\mathsf P(B)\neq 0$.

Likewise $\mathsf P(B\mid A):=\mathsf P(A\cap B)\div\mathsf P(A)$ .   (Since $A\cap B = B\cap A)$ .

So it is definitely the case that $\mathsf P(A)~\mathsf P(B\mid A)~{=\mathsf P(A\cap B)\\=\mathsf P(B)~\mathsf P(A\mid B)}$

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However, I cannot reconcile this with a scenario like in the following tree of (conditional) probabilities.

The tree just does not make any sense.   Check the labels.