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From: Osher Doctorow on 4 Jul 2010 16:39
From Osher Doctorow

In Elementary Particle Physics, "Interaction" is roughly speaking
viewed as objects (for example, fermions) "passing" other objects
(bosons) to third objects (for example fermions). The wave or field
pictures suggest that we look into representing this in terms of sets
and set intersections, such as how waves or wave/particles or fields
or field/particles intersect. So we look for "Interaction" in terms
of intersection.

We know from Conditional Probability (CP for short) that:

1) P(B|A) = P(B) for P(A) not 0, iff P(AB) = P(A)P(B), that is to say
A and B are (Statistically, Probabilistically) Independent.

The PI-analog of (1), a type of PI-Independence, is:

2) P(A-->B) = P(B) (a type of PI-Independence)

which can be proven to be equivalent to each of the following:

3) P(A ' --> B) = 1
4) P(AB) = P(A) + P(B) - 1

Notice the analogy between (4) and (1). (1) says P(AB) = P(A)P(B),
while (4) says P(AB) = P(A) + P(B) - 1. In both cases, P(AB) is
determined not just by P(A) or by P(B) but by both, to degree 1
(exponent 1) of each.

If we take the partial derivative D or D_P(A) of P(AB) with respect to
P(A) in (4), assuming that P(B) is not mathematically influenced by
P(A), then we get:

5) D[P(AB)] = 1 in (4), or D_P(A)[P(AB)] = 1 for A, B PI-Independent.

Is there some scenario in which D_P(A)[P(AB)] is greater than 1?
This would conjecturally involve something more than PI-Independence,
namely PI-Dependence. The answer is "yes".

Consider the equation:

6) P(A-->B) = P(A)

This intuitively says that the Probable Causation/Influence of A on B
is maximally dependent on A, unlike (2). (6) can be proven to be
equivalent to:

7) P(AB) = 2P(A) - 1 (equivalent to P(A-->B) = P(A) of (6)).

Now the partial derivative D_P(A)[P(AB)] is just the ordinary
derivative dP(AB)/dP(A):

8) D_P(A)[P(AB)] = dP(AB)/dP(A) = 2 (when (7) holds, equivalent to
P(A-->B) = P(A)).

Comparing (8) and (5), D_P(A)[P(AB)] jumps from 1 to 2 when P(A-->B)
goes from P(B) to P(A), that is from PI-Independence to what is
undoubtedly a form of PI-Dependence of P(AB) on P(A).

Readers can work out the proofs of the above, and also study the
equations:

9) P(A-->B) = P(A ' ) = 1 - P(A)
10) P ' (A-->B) = P(A)
11) P ' (A-->B) = P(B)

Try to solve (9), (10), (11) for P(AB) or if not involved for P(B) or
if not for P(A).

Osher Doctorow



From: Osher Doctorow on 4 Jul 2010 17:45
From Osher Doctorow

We get:

1) P(A-->B) = P(A ' ) = 1 - P(A) iff P(AB) = 0.

2) P ' (A-->B) = P(A) iff P(B) = 2P(A) - 1

3) P ' (A-->B) = P(B) iff P(A) = 1.

Readers can see that (2) is similar to (7) of the previous subsection,
while (3) is basically impossible unless A is the Universe (up to sets/
events of probability 0).

Equation (1) tells us that P(A-->B) = P(A ' ) is even more
"independent" than P(A-->B) = P(B) in terms of P(AB), since then
partial derivatives of P(AB) with respect to P(A) or P(B) (regarded as
varying without influencing each other on the constraint space) are 0
rather than 1. This is intuitively what would be expected roughly
since P(A-->B) = P(A ' ) is even more "opposed" to A than P(A-->B) =
P(B).

Notice that (2) and (7) reduce to each other when AB = B almost
everywhere (with probability 1, or except for sets of probability 0).

Osher Doctorow
From: Osher Doctorow on 4 Jul 2010 17:56
From Osher Doctorow

The proof of P(A-->B) = P(A ' ) = 1 - P(A) iff P(AB) = 0 is simply
obtained by evaluating P(A-->B) - 1 + P(A) which is 1 + P(AB) - P(A) -
1 + P(A) = P(AB), so P(A-->B) - 1 + P(A) = 0 iff P(AB) = 0.

For P(A-->B) = P(A) iff P(AB) = 2P(A) - 1, we evaluate P(A-->B) - P(A)
= 1 + P(AB) - P(A) - P(A) = 1 + P(AB) - 2P(A), which is 0 iff P(AB) =
2P(A) - 1.

For P(A-->B) = P(B) iff P(AB) = P(A) + P(B) - 1, we evaluate P(A-->B)
- P(B) = 1 + P(AB) - P(A) - P(B) which is 0 iff P(A) = P(A) + P(B) -
1.

For P ' (A-->B) = P(A) iff P(B) = 2P(A) - 1, we evaluate P ' (A-->B) -
P(A) which is 1 + P(B) - P(A) - P(A) = 1 + P(B) - 2P(A) which is 0 iff
P(B) = 2P(A) - 1 provided that P(B) < = P(A).

For P ' (A-->B) = P(B), we evaluate P ' (A-->B) - P(B) = 1 + P(B) -
P(A) - P(B) = 1 - P(A) = 0 iff P(A) = 1 for P(B) < = P(A).

Osher Doctorow
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