| Topic: |
Science > Physics |
| User: |
"OsherD" |
| Date: |
15 Jun 2005 08:18:32 AM |
| Object: |
Probable Influence Type 2 in Quantum Theory |
From Osher Doctorow
COPYRIGHT NOTICE
Probable Influence Type 2 in Quantum Theory
Copyright By Owner Osher Doctorow Ph.D.
First Published 2005
Professor Peter Mittelstaedt of U. Cologne/Koln pointed out in the
early 1970s that quantum implication works with the modification:
1) (a-->b) = a' U ab
for lattice elements a, b which we can think of as propositions. My
type 2 PI, namely P', which I introduced on sci.stat.math some years
ago and even here quite a while back, is ideally suited for the
corresponding set situation:
2) P'(A-->B) = 1 + P(B) - P(A)
because the corresponding type 1 PI (or just PI without qualification
if no confusion occurs) is:
3) P(A-->B) = P(A' U AB) = 1 - P(A) + P(AB) = 1 iff P(A) = P(AB)
which is to say iff A is a subset of B with probability 1. Although
(3) is perfectly satisfactory (it yields a "deterministic" limit so to
speak of Probable Influence, namely Influence or Causation, up to sets
of probability 0), the advantage of using (2) in this quantum scenario
is that it still discriminates between A and B and varies from 1.
Osher Doctorow
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| User: "OsherD" |
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| Title: Re: Probable Influence Type 2 in Quantum Theory |
15 Jun 2005 08:30:03 AM |
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From Osher Doctorow
Here's an interesting theorem.
Theorem. P' (A-->B) and P(A-->B) are related by:
1) P' (A-->B) = P(A-->B) + [P(A U B) - P(A)]
and so P' (A-->B) = 1 iff P(A-->B) = 1.
I'll prove this next time hopefully.
Osher Doctorow
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| User: "OsherD" |
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| Title: Re: Probable Influence Type 2 in Quantum Theory |
15 Jun 2005 08:45:46 AM |
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From Osher Doctorow
Proof of last theorem.
Since P' (A-->B) = 1 + P(B) - P(A) and since P(A-->B) = 1 + P(AB) -
P(A), it follows by direct algebra that P' (A-->B) = P(A-->B) + P(B) -
P(AB). However, we have:
1) P(B) - P(AB) = P(A' B) = P(A U B) - P(A)
because P(A U B) - P(A) = P(A) + P(B) - P(AB) - P(A) = P(B) - P(AB)
which is precisely P(A' B) because P(B) = P(AB) + P(A' B) by additivity
of probability of disjoint (mutually exclusive) sets. Q.E.D.
So have I excluded P' (A-->B) for the same reason that P(A-->B) is
excluded? Yes, but now define P*(A-->B):
2) P*(A-->B) = P(B) - P(A) if P(B) > P(A)
3) P*(A-->B) = P' (A-->B) - 1 = 0 if P(AB) = P(A)
In a manner of speaking, we've managed to eat our cake and keep it too,
since we now have an extra quantum scenario, namely in which P(B) >
P(A) and A is not a subset of B, in which case P(AB) does not equal
P(A) up to sets of probability 0, and therefore (2) applies but not (3)
and P*(A-->B) is not 1.
Osher Doctorow
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| User: "OsherD" |
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| Title: Re: Probable Influence Type 2 in Quantum Theory |
15 Jun 2005 09:04:11 AM |
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From Osher Doctorow
Now watch this.
We know that:
1) P(A-->B) - P(B-->A) = P(B) - P(A)
from the definitions P(A-->B) = 1 + P(AB) - P(A), P(B-->A) = 1 + P(AB)
- P(B). Therefore:
2) P*(A-->B) = P(A-->B) - P(B-->A) = P(B) - P(A) when P(B) > = P(A)
So our new Type 3 PI, namely P*(A-->B), is the excess of Probable
Influence/Causation of A over the Probable Influence/Causation of B.
Notice that if P(B) < P(A), then P(B) - P(A) is negative and so can't
be a probability so that P*(A-->B) in that case becomes undefined.
Also in that case P' (A-->B) is undefined because P' (A-->B) = 1 + P(B)
- P(A) is defined only for P(B) < = P(A) since otherwise 1 + P(B) -
P(A) > 1, which is impossible for probability which is always between 0
and 1 inclusive.
Osher Doctorow
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| User: "OsherD" |
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| Title: Re: Probable Influence Type 2 in Quantum Theory |
15 Jun 2005 09:23:10 AM |
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From Osher Doctorow
The new Type 3 PI, P*(A-->B) = P(B) - P(A) for P(B) > = P(A), doesn't
have to be optimized or maximized at 1, but it is an interesting
exercise to maximize it since it turns out that it equals 1 iff P(B) =
1 and P(A) = 0, which is to say that A is an extremely Rare Event as a
cause and B is an extremely "Certain" (non-Rare) Event as an effect.
If one were to accept the typical Quantum Theory claim that the theory
is deterministic via the Schrodinger equation even though ww* is a
probability with w the wave function, then up to sets of probability 0
we would set P*(A-->B) = 1 which indicates "certainty" in a
probabilistic sense, in which case by the last paragraph quantum events
would be deterministic as effects but very Rare Event probabilistic as
causes! So you can see that the Stochastic (Probabilistic) School
wins.
Osher Doctorow
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