When it was shown that de finetti's conjectures was false, by way of Krafts M=5 example; by which I mean that all finite qualitative orderings which meet his conditions are representatbe even if if the number of states are greater then 4. BY this do they mean the number of events in the 'boolean algebra of events, F; the number of events in the qualitative ordering/ (the measurable events') or just the atomic states of the world omega. ?
For example if I have three disjoint events in omega (the atomic sample state, {A, B, C} ) which could represented as the three possible truth valuations, where Fl contain 8 measurable/comparable propositions (the power set of omega) will something like scotts theorem of the other sufficient conditions for strong finite representation be necessary? I presume this is not the case if there are only two states in omega and thus that very limited conditions need to hold in addition to get a 'albeit' trivial strong representation
Ie if the only events in F are {{A},{B}, {empty set}, {A v B} where A and B are disjoint
is it sufficient for the existence of a strong representation to just show
asymmetry: If A >- B then not (B >- A) (and trivial total ordering between the four events in F); nontriviality: S >- 0; nonnegativity: A > 0; monotonicity:and the trivial cases of transitivity and addivitity?
I was wondering if one could then use the trivial strong representation onf distinct spaces, and use comparison relations between these distinct spaces to derive a unique solution when A>B; as if some kind of entangled product space; or else perhaps scotts theorem (if not scotts infintie theorem would be required)- i presume the latters conditions are almost impossible to derive (can one even use proof by induction for scotts infinite theorem; I presume its not even infinitely (countably) axiomatizable)
strong representation generally means that there exists a probability function that agrees with your ordering so it gets all of the strict ordering inequalities correctly (>, < ,=;
A>=B iff A>B, so it requires a total qualitative (comparative ordering) (A,B entails Not (A
a total qualitative credence ordering induced by numerical propensities in my case in my case.
. 'Agreement' just means that a strong representation probabilistic exists that is compatible with said ranking; that is your ordering is compatible with finite additivity, and there exists such a function that gets the inequalities and equalities correct; equals here means equi-probable. The ordering is just a set of events which are judged more, less, or equally probable- and for strong representation it generally is a total ordering, lest one has incomparable events which are given a numerical probability value in the representation whichould violate A=>B iff P(A)>P(B), );> here means more 'probable than' not implication (at least generally)
So I mean by this that the numerical representation does not get the numerical inequalities out of sync with the qualitative ones; presuming that its finitely additive function and the events are totally ordered. That is exists a probabilistic representatiion (finitely additive) such that for all of the events ordered; A=>B iff P(A)>P(B), all of the strict inequalities and equalities are preserved in the numerical representation in both directions.
Thus for example a simple ordering would be four events in F; one being the contradictionT_, the other being the sample space T (A or Bt)he other two being, A, B, which are disjoint ) such that T_
ok forget what I meant by propensity at the moment; what I am saying essentially is that I have a whole set of hypothetical qualitative probability spaces; i (for reals numbers, iin [0,1]; one of which might be actual the world or chance set up. There are two atomic events in sigma; A, B, and F={ A, B, A v B (the sample space), empty set)};ie the boolean algebra of events if you will, or generally speaking the the power set of sigma
In each of these hypothetical spaces i, the qualitative ranking denotes whether ones judges, or ranks, any given two events, A, B etc, in Fi, equally probable (A=B), more probable than (A>B ) or less probable than (AB denotes that a full belief in A, epistemically dominates a full belief B and so. How i justify these claims is a different story, which is through my propensity interpretation. The idea is see if only need a finite representation of each hypothetical space (to see if these ranking are compatible with a finitely additive probability representation which preserves these orderings; so that B>A, B is given a higher probability value then A), whilst using some other kind of representation,theorem, (as if its one product space) to
see if (1) these ranking are compatible with the existence of an infinite representation that preserves these ranking, or whether indeed an infinite representation is neededed at all,
(2) To then see if there that there a unique representation; (there is only one probability function, or set of values for each event in each space,that each such event in every one of these spaces can have) (3) such that it matches the propensities values from which this qualititave ordering is based
Technically speaking it is a total order or rather meets the condition for totality; which is that every two event on a given space, it iss such that A>B or A=B (again means judged equiprobable), or Bdoes not mean implication here; although it borders on it in my case, as they way I justify these claims, is 'as if equally probable events occurs in the cases as each other, 'more probable events occur in a superset of possible cases as events that they are less probable then'; So I am likewise unsure as to whether it would be consistent to just use the truth valuations 'A true, B false', 'A false, B true' (the sample space essentially) as the possible worlds on which one would use scotts theorem if I have to use it; ie where one uses the sum of the indicator function value of any two equations, A+B= B+ A,in each world, to show whether they are balanced; BY balanced i mean that 'in all possible worlds, or rather under all truth valuations, or atomic states in sigma', the sum of the truth values of all propositions mentioned in said state, LHS is equal to that of sum of truth values of (propositions) mentioned in the RHS, in said state.
So in between any two states (s1A true, B false, and s2B false, A true) , the sum for any given side (RHS for example, s1 sum=1+0=1, s2sum=0+1=1), may not need to be the same between both states , but it must nonetheless be equal to whatever is the sum of the values of the other side (LHS) on said state. Now of course in that trivial given case, it does so happen to be the case that for both two states, s2, s1, mentioned, the sums of the truth values of the propositions in LHS are the same; but need not be the case in general for scotts conditions to hold, it just required that whatever the sum happens to be in any given state, the sums (of the truth falues, the indicator function sum over the propositions mentioned) in the LHS and RHS must be equal
any other
However, there are relations between the events in F
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