Quantum Gravity 347.4: If P(A) < 1/2 and P(A<-->B) = 1 + DEP(A, B), then we have P ' {(A --> B)

Subject:	Quantum Gravity 347.4: If P(A) < 1/2 and P(A<-->B) = 1 + DEP(A, B), then we have P ' {(A --> B) \| A)} > P(B)
Date:	Sun, 6 Dec 2009 20:14:30 -0800 (PST)

>From Osher Doctorow

The proof of:

1) If P(A) < 1/2 and P(A<-->B) = 1 + DEP(A, B), then P ' {(A --> B) |
A)} > P(B)

is direct from QG 347.3 by simply interchanging A ' and A in that
theorem. Notice that P(A ' ) > 1/2 is equivalent to P(A) < 1/2 since P
(A ' ) = 1 - P(A) > 1/2 iff P(A) < 1/2. Also recall that we have P
(A<-->B) = P(A ' <--> B ' ), proven a few posts back.

Notice that (1) says that under the conditions of P(A) < 1/2 ("low
probability events) and with the interpretation of P(A<-->B) as 1 + DEP
(A, B), then the Probable Causation (P ' type) of B by A with A
"given" or fixed is higher than P(B), which intuitively corresponds to
positive Dependence since Statistical Independence is equivalent to P
(B|A) = P(B) when P(A) is not 0. The latater equation P(B | A) = P(B)
when written out is P(AB)/P(A) = P(B), or P(AB) = P(A)P(B). In terms
of the Statistical Dependence DEP(A, B) = P(AB) - P(A)P(B), the latter
statement says DEP(A, B) = 0, while DEP(A, B) > 0 is equivalent to P
(AB) - P(A)P(B) > 0 which is equivalent to P(AB)/P(A) > P(B) or P(AB)
| A) > P(B).

The implications appear to be considerable. Dependence of (random)
variables cannot be understood in general without Probable Causation/
Influence (PI) and its linkage to mainstream Statistical (Quadrant)
Dependence. Foundations of Quantum Probability/Statistics that do not
account for this, whether as special cases or limiting cases, become
questionable. Information Theory, which makes extensive uses of
Conditional Information, Mutual Information, etc., is also restricted
by the same requirements, and again Quantum Information is affected.

Osher Doctorow

Quantum Gravity 347.4: If P(A) < 1/2 and P(A<-->B) = 1 + DEP(A, B), then we have P ' {(A --> B) | A)} > P(B)