Wednesday, June 17, 2015


This result is probably known, and probably not optimal. A conditional probability function P is conglomerable provided that for any partition {Hi} (perhaps infinite and maybe even uncountable) of the state space if P(A|Hi)≥r for all i, then P(A)≥r.

Theorem. Assume the Axiom of Choice. Suppose P is a full conditional probability function (i.e., Popper function) P on an uncountable space such that:

  1. all singletons are measurable
  2. the function satisfies this regularity condition for all elements x and y: P({x}|{x,y})>0
  3. there is a partition of the probability space into two disjoint subsets A and B with the same cardinality such that P(A)>0 and P(B)>0
Then P is not conglomerable.

Conditions (2) and (3) are going to be intuitively satisfied for plausible continuous probabilities, like uniform and Gaussian ones. So in those cases there is no hope for a conglomerable conditional probability.

Sketch of proof: Let Q be a hyperreal-valued unconditional probability corresponding to P, so that P(X|Y)=Q(XY)/Q(Y). The regularity condition (2) implies that that there is a hyperreal α such that Q(F)/α is finite, non-zero and non-infinitesimal for each finite set F. (Just let α=Q({x0}) for any fixed x0.) Let R(F) be the standard part of Q(F)/α for any finite set α. Then P(F|G)=R(FG)/R(G) for any finite sets F and G. Moreover, R is finitely additive and non-zero on every singleton.

Since A2 has the same cardinality as A, there is a function f from B to the subsets of A with the property that f(b) and f(c) are disjoint if b and c are distinct and every f(b) is uncountable. Choose a finite number c such that P(A)<c/(1+c). For each b in B, choose a finite subset Fb of f(b) such that R(Fb)>cR({b}). Such a finite subset exists as R is finitely additive and the sum of an uncountable number of non-zero positive numbers is always infinity. Let H be the union of the Fb as b ranges over B. Then AH has at most the cardinality of B. Let h be a one-to-one function from AH to B. For each b in B, let Gb=Fb if there is no a in AH such that h(a)=b; otherwise, let Gb=Fb∪{a} for such an a. Let Hb={b}∪Gb. Then R(Gb)>cR({b}) and so R(Gb)/R(Hb)>c/(1+c). Then P(A|Hb)=P(Gb|Hb)=R(Gb)/R(Hb)>c/(1+c). But the Hb are a partition of our probability space, and P(A)<c/(1+c), so we have a violation of conglomerability.

1 comment:

IM2L844 said...

I don't know what any of that means, but what I think you're saying is that Ray Kurzweil may be too optimistic.