The only uniformly (equi-continuous, at least) functions that satisfy these constraints?
These are probability functions, roughly such that are,
which are
1.The domains equal the range (endo-morphic), domains and ranges are both [0,1] for f and G, for both such that
The only uniformly (equi-continuous, at least) functions that satisfy these constraints?
F(x);x\in [0,1] etc, such that F:[0,1]\to[0,1]
. G:[0,1]\to]0,1] y\in[0,1]
Each x,y, pair (such that x+y=1) denote a distinct hypothetical chance set up that is counter-factually/modally linked with all of the other pairs; These rise to a complete ordering over the entire set of pairs x,y, such ,forall x\in[0,1],for all y\in[0,1];that x+y=1
.Consider a pair of strictly monotonically increasing functions F(x) and G(y);F(x) the credal measure for outcome A, given that A's chance is x, and where G(y), gives the chance of B its mutually exclusive alternative given that its chance is y (y=1-x)
1. Strictly monotonically increasing
2. Uniformly equi-continuous
3.and sur-jective (if not bi-jective presumably as they are injective)
4. Such that G(1)=1,G(0.5)=0.5, G(0)=0, and F(0)=0, F(0.5)=0.5. F(1)=1.
5.And subject to: F(x) + F(1-x)=1, and, G(y)+G(1-y)=1; I am not sure what this, a strong symmetry condition?
5 . for any specific x =m, y=1-m, and thus F(x)+G(y)=G(x)+G(y+1-x)=1,where x+y=1 always equal 1 (this is any singular any single chance space; where F, and G are the subjective credence measures)
7. At any two distinct points, such that y=x\forall possible values in [0,1]; G(x)=F(y);if y=m (which will generally be different modal-ly connected set ups chance set ups, unless, x=y=0.5).
Some of these may be redundant as the earlier conditions imply some of the latter and vice versa.What kinds of functions are compatible with these. Is there a name for F(p) +F(1-p)=1 condition (remembering that this concerns the same outcome between spaces; its a structural, uniquenes condition that appears to fall out, not one required for the representation, as it does not concern additivity within a space which is condition 5.
I am not sure what kind of unique-ness and representations theorems are used for these types of qualitative probability structures (PCS concatenation, additive difference, lexicographic, multi-dimensional, or quantum like representations, but some of the earlier ones concerns products, and expected utility and not so much single case affairs).
It may be non-atomic (or strongly so), but not in the right way; unless an ordering on the differences is induced. One could add up the probability credence differences between events on spaces and not just the fact that it is greater, but one would have to justify a complete ordering on these as well .The symmetry condition places some structure to these differences.
The differences will trivially add (at least in finite and countably infinite cases); and presumably to one, with very strong archimedean constraints (via some kind of strong monotone continuity as well, countably additivity I presume), if conducted correctly, but whether they appropriately and uniquely converge to 1 (so as to yield the 'right' or a unique solution is another matter.
Would I have to presume that these differences, are themselves some kind of sigma algebra. ie P(x=0.1)\subsetP(x=0.2)-P(x-0.19999...... etc); for example in the finite case (f(1, f(1)-f(0.9)=P(0.1), f(0.9)-P(0.8), f(0.8)-f(0.7)..........) for each outcome or rather for each partiton [0,0.5] and [0.5,1];
I wonder if this will approach F(x)=x; as otherwise if more restrictions are placed on it such as;\ [x1+y1]>[x2+y2]then \sum(F(x1)+G(x2);where, one of y1, x1 might be smaller then x2, y2. Unless a stronger form of continuity is required.
There are other ways, to swap gaps using additive differences, as Id rather not go down that road for unique-ness.See
It appears not to be required for three outcomes or more, in my case. I do not want to employ splitting within a space, or random lotteries or extraneous chance devices I can help it, so whilst there are infinitely many events, there are only two atomic (or rather four in F with \omega, and the empty set) for each pair
;or y1+y2>y3+y4;such that [G(x1)+G(x2)]>G(y3)+G(y4)
, (or (x1 +x2)>x3+x4;then F((x1)+F(x2)>F(x5;x5=x3+x4) (Presumably it would just fall out then)++;where,x1>x1,or,x2, ).
Where cr(A or B)=1 on any pair, cr(emptyset)=0 , on any pair etc;
I doubt that unless certain constraints hold that the symmetry condition 4(balanced and centred or symmetric and centred? bi-symmetric, I am not sure the what the official name is for this)
entails this or that it generalised to three points which sum to 1
(then it would be linear); F(x1)+f(x2)+F(3) I think
8.
So technically the functions will be identical; except that G will, as a function of x,be actually strictly monotoncally decreasing as x increases; but will be functionally identical as a function of y. Or least will be functionally equivalent on these domains as G(0.1)=F(0.1)0.5 and to the same degree ie G(0.2)-G(0.1)=F(0.2)-G(0.2). G(0.6)=F(0.6)>5;
However, they the pair of functions only correpond to the same point in the joint domain, which is when x=0.5,y=0.5; and at this point they will be equal. I am not sure if these will correspond to something close to a linear function. Although notice that clearly F(0.6)-F(0.4)=2 \timesF(0.6-0.5)=2\timesG(0.6-0.5)=2times [F(0.5)-0.4]
;F(0.6)-F(0.5)=F(0.4)-f(0.5)=G(0.5)-G(0.4)=G(0.6)-G(0.5)_and likewise for for combinations of points F(0.4)-F(0.3)=F(0.7)-F(0.6)=G(0.4)-G(0.3)
There are other ways, to swap gaps using additive differences, as Id rather not go down that road for unique-ness.See
It appears not to be required for three outcomes or more, in my case. I do not want to employ splitting within a space, or random lotteries or extraneous chance devices I can help it, so whilst there are infinitely many events, there are only two atomic (or rather four in F with \omega, and the empty set) for each pair
;or y1+y2>y3+y4;such that [G(x1)+G(x2)]>G(y3)+G(y4)
, (or (x1 +x2)>x3+x4;then F((x1)+F(x2)>F(x5;x5=x3+x4) (Presumably it would just fall out then)++;where,x1>x1,or,x2, ).
Where cr(A or B)=1 on any pair, cr(emptyset)=0 , on any pair etc;
I doubt that unless certain constraints hold that the symmetry condition 4(balanced and centred or symmetric and centred? bi-symmetric, I am not sure the what the official name is for this)
entails this or that it generalised to three points which sum to 1
(then it would be linear); F(x1)+f(x2)+F(3) I think
8.
So technically the functions will be identical; except that G will, as a function of x,be actually strictly monotoncally decreasing as x increases; but will be functionally identical as a function of y. Or least will be functionally equivalent on these domains as G(0.1)=F(0.1)0.5 and to the same degree ie G(0.2)-G(0.1)=F(0.2)-G(0.2). G(0.6)=F(0.6)>5;
However, they the pair of functions only correpond to the same point in the joint domain, which is when x=0.5,y=0.5; and at this point they will be equal. I am not sure if these will correspond to something close to a linear function. Although notice that clearly F(0.6)-F(0.4)=2 \timesF(0.6-0.5)=2\timesG(0.6-0.5)=2times [F(0.5)-0.4]
;F(0.6)-F(0.5)=F(0.4)-f(0.5)=G(0.5)-G(0.4)=G(0.6)-G(0.5)_and likewise for for combinations of points F(0.4)-F(0.3)=F(0.7)-F(0.6)=G(0.4)-G(0.3)
I want to rule out functions such as chews; Mine is more then symmetric around its mean and it also symmetric to some degree equispaced wherever it sums to one. and Is symmeric around its origin and centre. Moreover there are other constraints
For any x or y, such that x+y~=1; or any x1+x2~=1, y1 or y2,, y1+y2notequal1 or any combinations of two factors; it cannot be that either f(x1)+f(x2)=1, or ; and or g(y1) + g(y)=1; where y~=1-x, or whether they are on different space or not.
If i could get that also for any three variables f(x1) + f(x2)+f(x3) =1 iff x1+x2+x3=1, and or as well, which would follow from my system given this f(x1) + f(x2)+g(x3)=1 presumably it would come out that they all would have to.
Nonetheless there are furrther restriction for any x1+x2>1;iff, f(x)+f(x2)>1; likewise for g(x1+x2>1; g(x)+g(x2)>1; and any combination of the two; and likewise x1+x20.5 is identical to f(1)-(0.2)=f(0.8)=f(0.8)-f(0)>0.5; as f(1)=1 f(1)-f(0.8)=f(0.2)=f(0.2)-f(0)0.5; and I think f(0.1)=f(0.6)=f(0.5) +f(0.6)-f(0.5)=f(0.5)-f(0.4)]+f(0.5)=f(0.6); so that wherever it over laps or meets in the middle or the origin or centre, it is symmetric, it may be that (f(0.6)=f(0.7)-f(0.1)=f(0.9)-f(0.3),in any case the difference beween f(0.1) and f(0.7)>0.5 and equals the difference between f(0.2) and f(0.8) between f(0.3) and f(0.9) and thus >0.5 I think ; where the difference between f(0.9) and 1 ie f(1) clearly equals f(0.10
f(0.9)-
and is
Likewise for any x1 and x2 where x and x2> are such that x1>x2, x2>x3; f(x1+x2)>f(x1+x2) and likewise for G and any combination of the two.
Inaddition, the function are bounded sums are bounded by the equalities f(0.4) + g(0.4) > f(0.3) +g(0.3) in that any x1,x2,,y1,y,2 that are in between them F(0.35) + f(0.34) or G or any combinaton of the two of them must be smaller then first and larger then the second, and I think this may hold even if is not the case that both of x1, x2 >0.3 or smaller then 0.4
There are two values of any chance, one A and one B, ie two 25 percent chances, G(0.25) and f(0.25) for all x, and these values are all equal
If only i could get this adjacents to line up to get a uniform structure; if got that f(0.75)-f(0.25)=0.5=f(0.5) ie that or that f(0.75-0.5)=f(1)-f(0.75)=f(0.5)=2\times f(0.25)
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