Quantum Gravity 344.8: P(A ' --> B ' ), P(A ' --> B), P(A --> B ' ) in Terms of P(A-->B)
OsherD
We can easily prove:
1) P(A ' --> B ' ) = P(A U B ' )
2) P(A ' --> B ) = P(A U B)
3) P(A --> B) = P(A ' U B)
4) P(A --> B ' ) = P(A ' U B ' )
5) P{(A ' --> B ' )(A --> B)} = P(A<-->B)
6) P{(A ' --> B)(A-->B)} = P(B)
7) P{(A --> B)(A -->B)} = P(A --> B)
8) P{(A --> B ' )(A --> B)} = P(A ' )
To get any of the above left hand sides, for example P(A ' --> B ' ),
in terms of P(A-->B), we proceed as follows (I will only give the
proof of this example here):
9) P{(A ' --> B ' ) U (A --> B)} = P(A U B ' U A ' U B) = 1 = P(A ' --
> B ' ) + P(A --> B) - P{(A ' --> B ' )(A --> B)} = P(A ' --> B ' ) + P
(A --> B) - P(A<-->B)
and therefore:
10) P(A ' --> B ' ) = 1 + P(A<-->B) - P(A --> B)
For the remaining cases we get in a similar way:
11) P(A ' --> B) = 1 + P(B) - P(A-->B)
12) P(A --> B ' ) = 1 + P(A ' ) - P(A-->B)
So the left hand side of (10), (11), (12) are all equal to 1 - P(A--
>B), but a positive term is added to the latter which is respectively P
(A<-->B), P(B), or P(A ' ).
Is there a pattern as to what is added to the latter in the last
sentence? Yes. If the quantity on the left hand side is symmetric
in "primes", that is to say P(A ' --> B ' ) (or trivially P(A-->B)
which needn't be written in terms of itself), then a symmetric term P
(A<-->B) is added or 0. If the quantity on the left hand side is
asymmetric, as for example P(A ' --> B), then an asymmetric term is
added, that is to say only a term with one of A, B, A ' , or B ' .
If the "prime" is on the Cause, as with P(A ' --> B), then the
unprimed Effect P(B) is added. If the "prime" is on the Effect, as
with P(A --> B ' ), then the primed Cause P(A ' ) is added.
The most interesting cases to compare are P(A ' --> B ' ) and P(A ' --
> B) since we are trying to consider a Repulsive Cause when the
original Cause (A) is Attractive. Then, rewriting (10) and (11) to
compare them, we have:
13) P(A ' --> B ' ) = 1 - P(A-->B) + P(A<-->B)
14) P(A ' --> B ) = 1 - P(A-->B) + P(B)
Notice that while the intersection probability P(AB) is explicit in P
(A-->B) = 1 - P(A) + P(AB), it is not explicit in (13) and (14), but
the latter two in a sense have clearer relationships to the right hand
side probabilities and sets that they involve. In fact, the right
hand side of (14) has a pattern somewhat similar to conditional
probability's "multiplicative law" P(B|A)P(A) = P(AB) except that the
former is additive and B rather than A "passes through".
Recall that in 344.7 we were concerned mostly with P(A ' --> B ' ) in
terms of P ' (A-->B). In this Section, we see somewhat greater
clarity in comparing both P(A ' --> B ' ) and P(A ' --> B) with P(A --
> B).
Osher Doctorow
OsherD
The previous post's results are especially interesting when we take A,
B as bounded sets and A ' , B ' as unbounded or infinite in the sense
of unbounded. This corresponds to the intuitive idea of Attraction
(which we can represent by A) as "directed toward the center of an
object", in this case directed toward the bounded set A, while
Repulsion (which we can represent by A ' ) is "directed away from (the
center) of A".
Now when A ' acts on B ', which is unbounded, where B is any bounded
object or Effect, then P(A ' --> B ' ) is the opposite of P(A-->B),
namely 1 - P(A --> B), plus a term P(A<-->B) that increases the more
Probably Correlated A, B are.
When A ' acts on B, which is bounded, then P(A ' --> B) is again 1 - P
(A-->B), but this time plus a term P(B) that increases with the
probability of bounded Effect B. Since I typically take P(C) for any
object C to be proportional to the geometric size of C if C is a
geometric object (size such as volume, area, length, etc.) in physics
applications, we get the remarkable result that "the bigger they are,
the harder they Repel", provided that "they" is bounded. For A '
acting on the unbounded object B ', this is no longer the case.
Rather, it is how strong the Probable Correlation P(A<-->B) of A and B
is that increases the Repulsion over and above 1 - P(A-->B). This is
suggestive of an "Entanglement" in this latter case.
Osher Doctorow