Re: Finsler and Sub-Finsler Geometry via PI

From: Osher Doctorow (mdoctorow_at_comcast.net)
Date: 07/24/04 

Date: Sat, 24 Jul 2004 02:21:48 +0000 (UTC)

On 23 Jul 04 01:37:58 -0400 (EDT), Osher Doctorow wrote:
>Much of the structure common to all of these things is contained in
>the expression:
>
>3) (A-->B) --> (C-->D)
>for set/events A, B, C, D, and its probability P{(A-->B) --> (C-->B)}
>which is easily seen to be:
>
>4) P{(A-->B)-->(C-->D)} = 1 - P(A-->B) + P{(A-->B)(C-->D)}

The last expression gives us considerable insight into the logical
implication (a-->b)-->(c-->d) for a, b, c, d propositions and
(a-->b) defined as "not 'a and not b'," which using negation ~ and
conjunction ^ is ~(a ^ ~b) and the latter reduces to ~a V b ("either
not a or b,") with V "or" (disjunction) including either-or.

As we would expect intuitively in (4) of last time (quoted above),
the left-hand-side probability increases when both (A-->B) and
(C-->D) increase together in terms of the probability of their
intersection. Before looking at 1 - P(A-->B) which on the surface
seems to go in the opposite direction to A-->B in a probabilistic
sense, let's notice that P(E-->F) for arbitrary E, F set/events can
also be written as P(E' U F) = P(E') + P(F) - P(E'F) and that the
subtraction of P(E'F) is due to its overlap (being counted twice),
that is to say it is already included in P(E') + P(F) implicitly in
both terms. So the entire three-term expression P(E') + P(F) -
P(E'F) really contains P(E'F'), P(E'F), P(FE), P(FE') added with
no subtraction, and so P(E' U F) increases with both FE and with
both FE', which means that it increases with F. Thus, in (4) from
last time, P{A-->B)-->(C-->D)} increases with P(C-->D).

Finally, going back to 1 - P(A-->B) in (4), the only "non-intuitive"
result is that P{(A-->B)-->(C-->D)} decreases as P(A-->B) increases,
but those of us already familiar with Rare Events or Probable Influ-
ence know that influencing events increase their probable influence
as their probabilities decrease toward 0 (from the right).

Osher Doctorow

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