Is the canonical unit 2 standard probability simplex, the convex hull of the equilateral triangle in the three dimensional Cartesian plane whose vertices are (1,0,0) , (0,1,0) and (0,0,1) in euclidean coordinates, closed under all and only all convex combinations of probability vector
that is the set of all non negative triples/vectors of three real numbers that are non negative and sum to 1, ?
Do any unit probability vectors, set of three non negative three numbers at each pt, if conceveid as a probability vector space, go missing; for example may not be an element of the domain if the probability simplex in barry-centric /probabilty coordinate s a function of p1, p2, p3 .
where y denotes p2, and z denotes p3, is not constructed appropriately?
and the pi entries of each vector, p1, p2 p3 in p1+p2+p3=1 pi>=0
in the x,y,z plane where x =m=1/3 for example, denotes the set of probability vectors whose first entry is 1/3 ie < p1=1/3, p2, p3> p2+p3=2/3 p1, p2, p3>=0; p1=1/3 +p2+p3=1?
p1=1/3, the coordinates value of all vectors whose first entry is x=p1=m =1/3 ie
Does using absolute barry-centric coordinates rule out this possibility? That vector going missing?
where is the vector located at p1, p2 ,p3 in absolute barycentric coordinates.
Given that its convex hull, - it is the smallest such set such that inscribed in the equilateral such that any subset of its not closed under all convex combinations of the vertices (I presume that this means all and only triples of non negative pi that sum to 1 are included, and because any subset may not include the vertices etc). so that the there are no vectors with negative entries
every go missing in the domain in the , when its traditionally described in three coordinates, as the convex hull of three standard unit vectors (1,0 0) (0,0,1 and (0,1,0), the equilateral triangle in Cartesian coordinates x,y,z in three dimensional euclidean spaces whose vertices are Or can this only be guaranteed by representing in this fashion.
As apparently there can be issues when using a function of only two variables, such as with x,y Cartesian coordinates which correspond to this probability simplex in Barry-centric coordinates but have to be absolute Barry-centric coordinates for it to be isomorphic to to set of all convex combinations of probability vectors as the dimensions increase.
For example the volume within the solids (as one moves to higher dimensions, the unit standard probability 4 simplex expressed in absolute
As apparently there can be issues when using a function of only two variables, such as with x,y Cartesian coordinates which correspond to this probability simplex in bary-centric coordinates but have to be absolute bary-centric coordinates for it to be isomorphic to to set of all convex combinations of probability vectors as the dimensions increase.
For example the volume within the solids (as one moves to higher dimensions, the unit standard probability 4 simplex expressed in absolute barycentric coordinates, has a volume that can contract leaving most values in each vector for all but finitely many vectors most 0, except for finitely many points. I think this is what meant sometimes by the curse of dimensional and that sometimes the polygon needs to be literally convex, ie the norm of the p values needs >1.
Unless this is talking about the entries within each vector, as the dimensions increase to infinity, so that it becomes an essentially infinitely many dimensional space of non atomic probability vectors having uncountable many values, and thus nearly all values within each vector will be zero if they are going sum 'in any sense of the word' to one?
, has a volume that can contract leaving most values in each vector for all but finitely many vectors most 0, except for finitely many points. I think this is what meant sometimes by the curse of dimensional and that sometimes the polygon needs to be literally convex, ie the norm of the p values needs >1.
Unless this is talking about the entries within each vector, as the dimensions increase to infinity, so that it becomes an essentially infinitely many dimensional space of non atomic probability vectors having uncountable many values, and thus nearly all values within each vector will be zero if they are going sum 'in any sense of the word' to one?
Of course it is, its what it is by definition. I probably should have thought more about this, back when I originally posted. If it isnt, then nothing is. Its the "closed" convex hull of its vertices, or all points in [0,1]^3, that can be expressed by convex combinations of its vertices:, (1,0,0), (0,0,1), (0,1,0).
So that any vector (x,y,z), x+y+z=1 , or rather point, (x,y,z), in [0,1]^3, where x+y+z=1,and x >=0,y,>=0z>=0 can be simply expressed by x* (1,0,0)+(0,1,0)+z*(0,0,1)=(x,y,z), as closed under all convex combinations of the vertices convexity (of this form) simply means/ requires that its contains all points in [0,1]^3, that can be expressed by-negative c1, c2,c3 ; c1*(1,0,0)+c2*(0,1,0) +c3*(0,0,1) where c1+c2+c3 =1, it just is the set of 3 all and only non-negative coordinates, that sum to one . And clearly we can set c1=x, c2=y, and c3=y and these are non-negative and sum to one. So its all, and "only" probably triples (x,y,z),x+y+z=1 .
I should have thought this through.
The main questions are (1) and (2):
(1): does the canonical 2 probability simplex, contain the canonical 2 simplex of the sums at each point x+y, x+z, z+y, clearly at each point (x,y,z) in the canonical 2 probability simplex, x+y,in [0,1] x+z\in [0,1], z+yin [0,1] and their sum (x+y+x+z+z+y=2(x+y+z)=2*1=1, but does it contain at "every" point, every such convex combination of (x+y=l, x+z=g, z+y=h), 1>=h>=0,1>=g>=0, 1>=l>=0 & l+g, g+h, h+l in where h+g+l=2 at some point (x,y,z)?
Id say yes, because if there were a (l,g,h), l+g+h=2, 1>=h>=0,1>=g>=0, 1>=l>=0, but with no accompanying (x,y,z) x+y+z=1, (x,y,z)>=0, l= x+y, g=x+z, h=z+y
, g=x+h-y, y=x+h-g, so l=x+x+h-g, l=2x+h-: x=1/2*(l-h+g), which is uniquely determined where l-h+g=0, and clearly x+y+z=1/2(2x+2y+2x)1/2(l+g+h)=1, the only way that point could not be in the simplex is if, for example x
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