From: Osher Doctorow on 26 Jul 2010 10:37 From Osher Doctorow From the physical importance of A "enwrapping" B (embedding B in itself), which involves AB (the intersection of A and B), we are stimulated to return to the expression: 1) P ' (A-->B) - P(A-->B) since when B is embedded in A, that is to say P(B C A) = 1, then (1) is 0 because P(AB) = P(B). The complete expansion of (1) is: 2) P ' (A-->B) - P(A-->B)= P(B) - P(AB) = P(A ' B) and this leads us to evaluate the analogous: 3) P ' (A ' --> B) - P(A ' --> B) = P(AB) The "Embedding-Expansion-Repulsion Theorem" is then the following ((4) and (5) below): 4) P ' (A-->A ' ) - P(A-->A ' ) = P(A-->A ' ) if P(A ' ) < = P(A). 5) P ' (A ' --> A) - P(A ' --> A) = P(A ' --> A) if P(A) < = P(A ' ). The proof is easy, and readers can try to prove it before I give the proof. Notice that P(A-->A ' ) is Expansion/Repulsion from the previous posts, and P(A ' --> A) is Contraction/Attraction from the previous posts. So Embedding is clearly related to Expansion and Contraction by (4) and (5). The Embedding part is perhaps more obvious from (2) and (3), noting that (3) represents embedding and (2) represents "counter-embedding". Osher Doctorow
From: Osher Doctorow on 26 Jul 2010 10:50 From Osher Doctorow (4) is proven by (2), and (5) is proven analogously. Recall also that: 1) P(A-->B) = 1 + P(AB) - P(A) 2) P ' (A-->B) = 1 + P(B) - P(A) if P(B) < = P(A) 3) P(A-->A ' ) = P(A ' ) = 1 - P(A) 4) P(A ' --> A) = P(A) 5) P ' (A-->A ' ) = 1 + P(A ' ) - P(A) (if P(A ' ) < = P(A)) = 1 + (1 - P(A) - P(A)) = 2 - 2P(A) = 2(1 - P(A)) 6) P ' (A ' --> A) = 1 + P(A) - P(A ' ) = 1 + P(A) - [1 - P(A)] = 2P(A) Substituting these as appropriate into the equations to be proven yields the desired results. For example, 2(1 - P(A)) - (1 - P(A)) = 1 - P(A), and 2P(A) - P(A) = P(A). Osher Doctorow
|
Pages: 1 Prev: Geometrical-probability theory #4.23 & #239 Correcting Math & Atom Totality Next: Making Waves |