From Osher Doctorow
Here are some additional P(A-->B) vs P' (A-->B) theorems (see my
Quantum Gravity thread for more theorems and results, especially
section 100 and higher).
Theorem 1. P(A-->B) reduces to P' (A-->B) if B is a subset of A (up
to a set of probability 0), while if A is a subset of B then both are
equal and P(A-->B) = P' (A-->B) = 1.
Theorem 2. If P(A) is not 0, then we have for all events A, B:
1) P(B|A) < = P(A --> B) < = P' (A --> B)
where P(B|A) is conditional probability of B given A (defined as P(AB)
divided by P(A)) and the other quantities are respectively P(A-->B) =
1 + P(AB) - P(A) and P' (A-->B) = 1 + P(B) - P(A) with P(B) < = P(A)
for the latter.
Readers can easily prove Theorem 1, while I'll outline the proof of
Theorem 2 here.
P(A --> B) = 1 + P(AB) - P(A) < = P' (A-->B) = 1 + P(B) - P(A) for
P(B) < = P(A) because P(AB) < = P(B) by monotonicity of probability
(that is to say, AB is a subset of B, so P(AB) < = P(B)). As for P(B|
A) < = P(A-->B) = 1 + P(AB) - P(A), we have:
2) P(A-->B) - P(B|A) = 1 + y - x - y/x = 1 + y(1 - 1/x) - x = (1 - x)
+ y(1 - 1/x) = (1 - x) + (y/x)(x - 1) = (1 - x)(1 - y/x) > = 0 since x
< = 1 and y < = x for y = P(AB) and x = P(A)
Q.E.D.
Osher Doctorow
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