Is there a general name for these 'pair' of symmetic probability functional equations (1) and (2); see https://math.stackexchange.com/posts/2239756/edit
1. F(1-x)+F(x)=1 = F(1)=F-1(1)
2. F^(-1)(p)+F^(-1)(p)=1 =F^(-1)(1)=F(1);
See https://en.wikipedia.org/wiki/Minkowski%27s_question_mark_function, the second equation that is highlighted in the third attachment below. Its that property as bi-conditional or in other words, it applies to the inverse function as well. With these additional properties. What is the official (if at all ) name of this functional equation. and for its bi-conditioonal or inverse form?
where F:[0,1] to [0,1]; F is injective, ()uniformly continuity) and strictly monotonic increasing
F(1)=1
F(0)=0
F(1/2)=1/2 (yes these are redundant but I specify them any,way as these points were not derived via the symmetry as such but through standard addivity, which when combined with other feature of my model; induce this weird upward addivity). As F(x)+G(1-x)=1 is standard additivity, where G governs the behaviour of the other outcome on that probability space but its matched pair for pair so that -same chance same credence so G(0.75)=F(0.75) so that so that this forces A with 75 percent chance F(0.25)to function act like its negation with relata to F(0.25) (A with 0.25 chance) when its complement is actually (~A with 0.25 chance) G(0.25). Despite being on different spaces
That is x,y in dom(F) then x +y=1 iff F(x)+F(y)=1 and x,y in im(F) x+y=1 iff F^(-1)(x)+F-1(x)=1, where for every x there exists one and only one y such that x +y =1,where x~=y nless x=y=F(x)=F(y)=0.5, and for every F(u)\in co-domain, there exist one and one, F(u2)\ in cod-domain s.t F(u)+F(u2)=1, F(u)~=F(u2) unless F(u)=F(u2)=u=u2=0.5
and x
The function also satisfies these relations
$$∀(x1,x2)∈Dom(F)=[0,1]:[x_1+x_2]>1↔[F(x1)+F(x2)]>1$$
$$∀(x1,x2)∈Dom(F)=[0,1]:[x_1+x_2]1$$
$$\forall(p1,p2)∈Im(F);[p_1+p_2]>1↔[F^{−1}(p_1)+F^{−1}(p_2)]
IT is that is correct, @Raed; but in this case I express that as F(x)+G(1-x)=1 F guides the behaviour of the outcome A alone as its chance increasing between distinct probability space; so its say that the probabbility in A with chance 25 + the probability of A with chance 75 must sum to one.
I distinguish' probability 'from 'chance'; in this case, the chance,x, is one's interpretation of a physical probability (such as the amplitude moduli square intepretation in qm , or single case propensity in my case, the classical interpretation of probability, frequentism, bayesianism) for example),
and F(x) is probability is another probabilistic entitty that you want align with (such as the frame function probability in qm, for the born rule, or the objective rational credence, for the principle principal) that yo want to align with ones chance
So this is saying that between two different bases where A has chance, x1= 0.25, and the other distinct space, where A has chance, x2= 0.75 (so I do not mean ~A) must be sch that function values at those points F(x1)+F(x2)=1; F(0.25) + F(0.75)=1; as x2=1-x1 ie the credence or frame function vale at those if you will must sum to 1
The other condition; the inverse conditon express this as a biconditional only if the x1 +x2=1 does F(x1)+ F(x2)=1 or in other words if x1+x2 ~= 1 then F(x1)+ F(x2)~=1, where this '~=' means 'does not equall. So if and only if. Not all probability functions have the first property, Segal, and Quiggin call a probability function with the first property symmetric probability function. I also seen it referred in a property of dyadic monoids- such as cauchy;s function; or in minkowski's question mark function. I am not sure what its official functional name is.
It differs from normalization in the same sense that cauchy s equation differs F(x+y)=F(x)+F(y) differs from F(x or y)= F(x) + F(y) where x,y are disjoint. So normalizatoon wouild F(x or ~x)= F(x) + F(~x) =1,where this says F(x +y)=F(x)+F(y) iff x+y=1
@Peter, I have attempted to fix it; That should read as the inverse function
Well the function has the second property mentioned in the pasted wikipedia page on the minkowski ? mark property, but as a bi-conditional (or to put it another way it, applies to the inverse function as well, and as result both are bi-conditionals, '(iff;'
You are correct; I knew those contraints were superfluous; I just stressed to be sure, perhaps it was pedantic. Yes F(0.5)=0.5 follows ; I dont think that without a metric such as midpoint convexity that it would, Although i had an intuition that even without by symmetry constraints. That just from Proper Convexity rather then midpoint convexity, that if the function is strictly monotonically increasing, and agrees with F(x) on those three points that the function must be F(x)=x. I first required it to be twice differentiable with strictly positive first derivative (not always a conseqence of strict monotonicity) although some of the characters on maths stack exchange informed that they did not need need that. The intuited rested on the fact that the line must be on or below the tangents, but if a strictly monotonic function can have turning pts (local minima or maxima) how is going get from F(0.5) =0.5to F(1)=1. I still dont believe. Nonetheless I believe that given my symmetry constraints one can derive jensens equation from midpoint convexity; i should post the derivation in lyx to avoid the difficult in posting it here
I dont think that the symmetry flows up or between the equal intervals betwist whereever the functions sums to one; that is one cannot say that the adjacent intervals, for example either within a side (that is where both intervals are both below, or above, 0.5) are equal; [0.5,0.6] is equal to [0.4, 0.5] as in the function difference. The interval [0.3, 0.4] of corse and [0.7. 0.6 are are; but one cannot say that [0.6,0.7] is equal in function length to [0.6,0.7] if you impose midpoint convexity, the symmetry midpoint convexity will give you midpoint concavity (the other inequality) and thus, jensen equality. I might be able to use something weaker. Whatever comes up comes down in a symmetric probability function
@George do you know if all univariate strictly monotonic functions are trivially schur convex (and or schur concave). Ie there is nothing to permute; so that they permutation symmetric and always generally strictly quasi convex, quasi convex, strictly quasi concave and quasi concave- where quasi convexity and permutation symmetry generally gives schur convexity (although I am not exactly sure if what it is supposed to mean for standard function). Other wise one can appear to derive jensens equation from the symmetry constraints and midpoint convextiy- but I did not use all of them such as the inverse symmetry of the last four conjections. Or injectivity (well I may have)
2-power boundaries; how do you mean by this_ dont you mean, midpoints between two domain points x and y where y is some dyadic even multiple of x, ie y=2^K\timesx; i doubt you would be able to read any other number off other than 0.5
; or get to F(3/4+x) +3/4=3/4-F (3/4-x); et to F(3/4+x) +F(3/4)=F(3/4)-F (3/4-x);this is correct
(1/2+x)=1-F(1/2-x). Yes. That follows too. as a result of substitions One can simply use F(1-x)+F(x)=1 with x=0.5 and f a function to get F(0.5)=0.5
That' one can get F(3/4+x)= 1-F(1/4-x);
F(3/4+x)-3/4=1/4-F(1/4-x) so that F(3/4)-3/4=1/4-F(1/4) and likewise with the inverse so trhat F-1(3/4)+F-1(1/4) iso that t F(x)-F^(-1)(x)=F(1-x)-F(-1)(1-x) is parit wise constant; one cannot deducde that this is constant is zero. without further structure
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