P{D[Ai,Ai+1]} = 0 (all i) implies P(A1<-->...<-->An) = 1

From: Osher Doctorow (mdoctorow_at_comcast.net)
Date: 12/18/04 

Date: Sat, 18 Dec 2004 03:21:25 +0000 (UTC)

 From Osher Doctorow mdoctorow@comcast.net

COPYRIGHT NOTICE
P{D[Ai,Ai+1]} = 0 (all i) implies P(A1<-->...<-->An) = 1
Copyright By Owner Osher Doctorow Ph.D.
First Published 2004.

D[Ai,Ai+1] is defined as D[ ] for 2 sets A, B, which is AB' U A'B
when A = Ai, B = Ai+1, so it is AiAi+1' U Ai'Ai+1. Since its prob-
ability is 0, Ai = Ai+1 with probability 1, since if Ai = Ai+1 U E
where P(E) > 0 and E is disjoint from Ai+1, then P(Ai) = P(Ai+1) +
P(E) > P(Ai+1) and AiAi+1 = Ai+1Ai+1 U EAi+1 = Ai+1 and Ai'Ai+1 =
(Ai+1'E') = Ai+1' since Ai+1 doesn't meet E, so AiAi+1 U Ai'Ai+1'
= Ai+1 U Ai+1' = universe so its probability is 1.

Thus, if D[Ai,Ai+1] = 0, then P(Ai<-->Ai+1) = P(AiAi+1 U Ai'Ai+1') =
1.

Now let's compare:

1) P(A1<-->A2<-->...<-->An) and P(A1<-->A2)P(A2<-->A3)...P(An-1An)

If the left hand side is greater than or equal to the right hand
side, we'll call the situation nonnegative statistical PI dependence
since statistical dependence involves P(AB) > P(A)P(B) for positive
quadrant statistical dependence of Lehmann etc. We formulate the
theorem:

Theorem. Let A1, A2, ..., An be nonnegative statistically PI
dependent and D[Ai,Ai+1] = 0, i = 1 to n-1, then P(A1<-->A2<-->...
<-->An) = 1.

Proof. This was proven above since P(Ai<-->Ai+1) = 1 for i = 1 to
n-1 which forces the left hand side of (1) to be 1. Q.E.D.

Remark. The condition of nonnegative statistical PI dependence
couldn't be fit into the title of this posting/thread without an
over-long title, I thought at least.

Osher Doctorow

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