The product rule and Bayes rule

[aaaa202]

Is the rule:

P(AB I) = P(BA I)

(which is used to derive Bayes rule) an axiom for probability? And if so, do you guys find it intuitive that it should hold. For instance consider a box with green and red beads. Do you think it is strictly obvious that the probability of getting red-green is the same as the probability for getting green-red?

My question is because I do not think that Bayes rule is intuitive yet it is derived from exactly this principle.

 

[Fredrik]

I don't understand your notation. Regarding your example, I don't think it's at all obvious. I didn't even know that it was true. (I compared the probability of "red, then green" to the probability of "green, then red" in a few simple examples, and they were equal every time, so it seems to be true).

But I don't see what it has to do with Bayes theorem. The following example seems more appropriate: Suppose that the total number of beads in your example is n, and that each of them has a unique integer between 1 and n written on it. Now the probability that the first bead you draw "is red and has an odd number" is the same as the probability that the first bead you draw "has an odd number and is red". That is intuitively obvious.

I have never taken a course on probability, so maybe I'm missing something simple, but isn't the derivation of Bayes theorem just what I'm doing below?

We define P(A|B) by
$$P(A|B)=\frac{P(A\cap B)}{P(B)}.$$ This equality implies that $$P(A\cap B)=P(A|B)P(B).$$ Since A and B are arbitrary here, we also have $$P(B\cap A)=P(B|A)P(A).$$ Since the two left-hand sides are equal (by definition of $\cap$), the two right-hand sides must be equal. This implies that
$$P(A|B)=\frac{P(B|A)P(A)}{P(B)}.$$

 

[NegativeDept]

I have never taken a course on probability, so maybe I'm missing something simple, but isn't the derivation of Bayes theorem just what I'm doing below?

That looks right to me. In practice, conditional probability can become really confusing, especially when $A$ and $B$ are uncountable and $P$ is some kind of integral. But the main ideas follow from the set-theory argument you wrote. (Also, it only works if $P(B)\neq0$, otherwise we just divided by zero.)