benelliott comments on Logic in the language of probability - Less Wrong

12 Post author: Stuart_Armstrong 26 April 2013 06:45PM

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Comment author: benelliott 27 April 2013 10:38:10AM 5 points [-]

Sure this is right? After all, the implication is also true in the case of A being false, the conjuntion certainly is not.

He specifically specifies that A is true as well as A => B

Intuitively I suggest there should be an inequality, too, seeing as B|A is not necessarily independent of A.

B|A is not an event, so it makes no sense to talk about whether or not it is independent of A.

To see why this is a valid theorem, break it up into three posibilities, P(A & B) = x, P(A & ~B) = y, P(~A) = 1 - x - y.

Then P(A) = P(A & B) + P(A & ~B) = x + y

For P(B|A), restrict to the space where A is true, this has size x + y of which B takes up x, so P(B|A) = x / (x+y)

Thus P(A)P(B|A) = x = P(A & B)

Comment author: Metus 27 April 2013 12:01:55PM 0 points [-]

Thank you. That is what I deserve for cursory reading.