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)
Thank you. That is what I deserve for cursory reading.
He specifically specifies that A is true as well as A ⇒ B
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)
Thank you. That is what I deserve for cursory reading.