Quantum Gravity 347.2: A Relationship Between Probable Correlation P(A<-->B) of PI and Statistical Dependence (Lehmann) Via The Complex Field

Subject:	Quantum Gravity 347.2: A Relationship Between Probable Correlation P(A<-->B) of PI and Statistical Dependence (Lehmann) Via The Complex Field
Date:	Sat, 5 Dec 2009 08:06:38 -0800 (PST)

>From Osher Doctorow

Since Probable Correlation P(A<-->B) = P(AB) + P(A ' B ' ) in Probable
Causation/Influence (PI) and E. L. Lehmann's Statistical Quadrant
Dependence (Positive or Negative - "Some concepts of dependence," Ann.
Math. Stat. 37 (1966) 1137-1153) which can be expressed as DEP(A, B) =
P(AB) - P(A)P(B) both measure the Probabilistic/Statistical strength
of the relationship between sets/events and between random variables,
a key question is:

1) What is the mathematical relationship between P(A<-->B) and DEP(A,
B)?

If we try to equate them, we get:

2) P(A<-->B) = P(AB) + P(A ' B ' ) = DEP(A, B) = P(AB) - P(A)P(B) ?

which yields:

3) P(A ' B ' ) = - P(A)P(B) if (2) is correct.

Since probabilities are always nonnegative at least in this Universe,
(3) cannot hold, but there are some surprising things about (3). We
can easily prove that:

4) P(A ' B ' ) = P{(A U B) ' } = 1 - P(A U B)

and substituting from (4) for P(A ' B ' ) into (3) yields:

5) 1 - P(A U B) = - P(A)P(B) if (2) is correct.

Now, P(A U B) and P(A)P(B) both increase together, even though the
first is "set-additive" and the second is real-multiplicative. So (5)
is plausible under at least a wide range of conditions except for the
constant 1, or at least a modification of (5):

6) 1 - P(A U B) = -kP(A)P(B), k a positive real number.

One "natural" suggestion is to change to complex variables so that a 1
appears on the right hand side of (6) from -i^2 = 1 with a
modification in (2).

Another possibility is to change the definition of Statistical
Dependence by simply adding 1 to it and replacing -P(A)P(B) by -kP(A)P
(B) for k real positive number in its definition above (1). That is:

7) Modified DEP(A, B) = 1 + P(AB) - kP(A)P(B), k > 0

Then the same procedure as above with DEP(A, B) replaced by Modified
DEP(A, B) yields:

8) P(A U B) = kP(A)P(B), k > 0.

Although this may not always hold for all sets/events, it is plausible
under a wide variety of conditions.

Osher Doctorow