If for argument sake one modeled another probability function over entire the canonical 2 probability simplex, (set of all vectors of 3 non negative numbers that sum to 1 triples as conceived to be an indidividual probability space, but over all of them, that is the entire canonical simplex);
as a probability function of said simplex values interpreted as probability values; and had the function globally ranked as a frame probability like function globally ranked by itself the values in and between each event, and global context free (E) below with the dimensions or cardinality of each vector itself being 3
Although there are un-countably many elements in the order (the entire canonical probability simplex, closed under all and only convex combinations of three non negative values that sum to one) ; enough to force both function to be equal will F just re-produce the simplex
on the condition that one cannot (mess around with the unions, or faces, the vectors with a single zero entry, to get a 4th fixed pt, ie in addition to F(1)=1 F(0)=0 and F(1/3)
(regardless of whether these are the same vector or not) or an x, or y or z event).
And one had the both functions effectively working on the same axioms, (A)non-negativity, (B)
regular /positivity 0F(xj|yj|zj 2) (everywhere)
\forall vectors (i, j, )forall events): xi|yi|zi0 \iff 1>F(x|y|z)>0
Then is the fusion of vector dependent normalization, within each vector (same cartesian point in R^3) triple of three values of both entities function and domain, Where each vector is probabilty space ; 1>=F(x), F(z), F(y)>=0;
On each individual vector F(x)+F(y)+F(z)=1 ; x+y+z=1 , 1>=x,y,z>=0;
regularity, 1 >x|y|z>0 \iff 1>F(x|y|z)>0
and F(x,y,z,)=1 iff x|y|z=1
F(x|y|z|)=0 iff x|y|z=0 else
and global continnum dense numerical (Context free rank vector independent,, equalities between distinct events or not; distinct vector , x versus y versus z or one distinct vectors x1 versus x2 versus z3 versus y2 are all assigned the same numerical function value iff their domain value (their x value or z value etc in [0,1] are the same)
(1/3, 1/3, 1/3)
The answer is yes; the 1- simplex is not.
By the above, i mean if another probability function is defined over the scalar coordinate equivalence classes in the canonical probability simplex (every coordinate is non-negative, and at any point the sum of the coordinates is 1)
(ie the equivalence class 1/2, here is any x,y,z coordinate in any vector/point in the simplex, , with a value of 1/2; for example x=1/2 at , y=1/2 0,(1-x-y)>0: F(1-x-y)+F(x)+F(y)=1. This Initial positivity restriction holds if one can only make use of the vertices and the interior of the simplex (it is lifted and the simplex is uniquely represented by itself; ie F just is the simplex, if one uses only one (appropriate_ non-vertex edge point, or, if one imposes continuity (I think)
(E) Where I (I abuse notation a little) if two coordinates, have the same value, then so long as they are elements of distinct vectors, they are considered distinct scalars for the purposes of the above (so x=1/2 @ , is a distinct 'scalar' in this order, from x=1/2 @ but is in the same equivalence class as it.
To get the point, it can be seen as a connected set of every possible distinct probability function with the same sample space, but defined only over sample space, (not a boolean algebra as such, as the sample space is not an element of itself, and so it technically does not meet the definition).
This is what I mean by a probability function being ordered by the scalars in the elements of simplex, or the simplex being ranked by itself (its a set of all probability functions, defined over the atomic events, and by my question I mean whether another function,f, defined over the simplex coordinates, and which preserves the order of these coordinate scalars, and is probabilistic in the same manner (as above), would simply spit out/reproduce the simplex k out).
That is over these, rather than the points in the plane/vectors, themselves. In fact, one can get by the restriction (EDGE):
(EDGE): One only needs the (3) scalars in ONE (rather than all) non-vertex point along the edges of the simplex (scalars in a vectors, with precisely one zero entry with the remaining two being positive), in addition to those (9) scalars/coordinates (in {0,1}) which are coordinates of the 3 vertices (no other edge points, or rather no other scalars, that are coordinates of an edge point, need be used in addition to these.
The edge points, which are a permutations of , work, one can get to (F):F(x)+F(1-x), (I wont mention how here)
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