From Osher Doctorow

COPYRIGHT NOTICE
x + y + z - t as Partial Inverse Causal Metric
Copyright By Owner Osher Doctorow Ph.D.
First Published 2005

The metric:

1) dx^2 + dy^2 + dz^2 - dt^2

also known as the proper time interval squared (e.g., in reverse order
of (1), that is to say, multiplying (1) by -1), could never be
transformed to x + y + z - t, right? Well, maybe the word
"transformed" should be replaced by "represented by" in a generalized
sense, but in fact there's a way to do it.

I've proven in earlier postings to sci.stat.math and geometry.research
and elsewhere that:

2) 1 + y - x

is a partial one-sided inverse to Euclidean (and Euclidean-like)
distance of type:

3) sqrt(x^2 + y^2)

where for convenience I've left the second point (besides (x, y)) as
(0, 0). However, readers may recognize the expression above (2) as
P(A--B) or P' (A--B) = 1 + P(B) - P(A) with y = P(B), x = P(A).

Let's consider the expression:

4) P(A--B) = r P(A)

Is there anything curious about this, with r a positive real number?
Well, those who know conditional probability can look at the analog
there with P(A) replaced by P(B)

5) P(B|A) = rP(B)

With r = 1, (5) is equivalent to A and B being statistically
independent. A simply has "no effect" on B in (5). Arguably, if A and
B were very dependent, then A would "come through" instead of B on the
right side of (5) to yield:

6) P(B|A)= rP(A)

with r = 1 presumably. The Probable Influence (PI) analog of this
argument is (4) above with r to be solved for.

Here's a theorem regarding (4) which readers can try to prove as
homework.

Theorem. The following holds:

7) [P'A--B) + P'C--B) + P'D--B) - P'T--B)]/20 = P(A) + P(C) + P(D)
- P(T)

holds if P(A), P(C), P(D), P(T) for sets A, B, C, D are each between
..01 and .05 and P(A) = P(B), P(C) = P(B), P(D) = P(B), P(T) =
P(B), and P(T) = P(A), P(T) = P(C), P(T) = P(D). (I've typed
P'A--B) for P'(A--B) because my keyboard is having backward erasing
problems.)

Hint: Use the same r in (4) for P'(A--B), P'(C--B), P'(D--B), and
P'(T--B), and see what conditions are required for the remaining
inequalities to hold. Notice that P(A) .05 is a basic definition of
a Rare Event A, although one could also choose P(A) .01 for example.

Osher Doctorow

From Osher Doctorow

Maybe I should illustrate part of the calculation for the proof of the
last theorem.

Consider:

1) P ' (A--B) = 1 + P(B) - P(A) = r P(A)

for r 0 a real number and P(B) = P(A).

We get:

2) 1 + P(B) = (r + 1)P(A)

and solving for P(A):

3) P(A) = 1/(r+1) + P(B)/(r + 1)

Now, P(A) = P(B) and P(A) .05 and P(A) .01, so since P(B) = 0
from the definition of probability, choosing r to be something like 1
or 2 would yield P(A) 1/2 or P(A) 1/3 respectively, which fails to
make P(A) .05. If r were 20 or larger, however, things are O.K. For
example, with r = 20, (3) becomes:

4) P(A) = (1/21) + P(B)/21

Now, 1/21 1/20 = .05, so this doesn't violate P(A) .05 provided
P(B) is small enough and provided that P(A) 1/21. Actually, if we
had chosen r = 19, things would have just failed because:

5) P(A) = 1/20 + P(B)/20

is violated by P(A) .05 = 1/20. But anything greater than 19, such
as 19.1, would work with appropriately shorter intervals similar to the
above.

Now carry on (as the British say). In other words, continue.

Osher Doctorow

From Osher Doctorow

Attentive mathematical readers (or maybe others) might notice that in
the last posting example, x .05 = .01 had to change to x 1/21 when
r = 20 was chosen. This could be incorporated into the premise or
hypothesis of the theorem by having x 1/(r + 1), and similarly for
other similar modifications.

Osher Doctorow

From Osher Doctorow

This thread also relates to my previous thread on spheres and curvature
via the Pythagorean Theorem and so on.

Osher Doctorow