Does anyone know of the technical (functional equations) conditions that are required for a function to satisfy mirror/reflection/square symmetry or square symmetry let alone point reflection symmetry;
Segal defines standard symmetric probability functions (of other probabilistic entities) as f(1-p)+ f(p)=1; although my function also appears to satisfy this, beforehand mentioned definitions (around the centre and origin) as well as, and f-1(p) +f-1(1-p)=1; and a few other conditions, pertaining to partial sums.
However, Famlgane uses the notions of being 'centred', anchored, uncrossed, and balanced, and I am wondering whether my balanced he means segals condition F(1-p)+F-1(p)=1; or something stronger, I am wondering what conditions are in ordinary terms to render curves parallel.
Is is some isometry or point reflection symmetry, a triangle inequality,
I
(and correspondingly F(0)=0, f(1) by bijectivity and strict increasingness,) would be a linear; f(x)=xF(x+y+z...=F(x)+f(z)...F(deltaz), delta F(z). ie F(x)
It appears to weaker than point reflection symmetry which f-1(x)=f(x); but stronger then standard symmetry or central symmetry; as exhibited which is
s in https://en.wikipedia.org/wiki/Symmetric_probability_distribution ie in my F'(0.5+delta)=F(0.5-delta), and F'(0+delta)=F'(delta)=F'(1-delta)
Segal defines standard symmetric probability functions (of other probabilistic entities) as f(1-p)+ f(p)=1; although my function also appears to satisfy this, as well as the beforehand mentioned definitions (around the centre and origin) as well as, and f-1(p) +f-1(1-p)=1;
I presume they are quite similar. and a few other conditions, pertaining to partial sums. I know what would render them parallel but I am wondering if weakening, midpoint convexity at the expense something stronger then strict quasi convexity strict quasi concavity might work or whether there are some other symmetry conditions that entail point reflection symmetry; other then something like F-1(x)+F(1-x)=1; or the dual functions being inverses on each others or using signed charges. Its a lot simpler with outcomes;
However, McFalmgane (see attached) uses the notions of being centred, anchored, uncrossed, and balanced, and I am wondering whether my balanced he means segals condition F(1-p)+F-1(p)=1 or something else such as the centred mean symmetry of wikipedia. Which amount to similar things. Please see attached
and by 'symmetric' does McFalmgane's means something like the central symmetry condition as in the wikipedia article above (but for fechner functions) F'(o.5+delta)=F'(0.5delta)=F'(1-delta); and not F(y)+F(x)iff x+y=1 F-1(y)+F-1(x) iff x+y=1? Any help is would be much appreciated, as these terms appear to apply to different things. I dont just mean standard addivitity by symmetry of course, as this condition applies betwixt different probability spaces; ie a quasi homogeity requirement or quasi linearity.
I am also interested in whether there are strongly bi-symmetric probability functions; I presume they are functions of two variables, and that this is generally a weaker property than symmetry.?
as one doesnt need to use a distance metric, or a weird, additive difference domination conditional,that is, to order the sums (for clear reasons), to induce equi-spacing along the adjacents. F(0.4)-F(0.3)=F(0.4)-F(0.5) in the infinite limit.And not just within the origin.
The other interesting condition is also that for all x; f(x)+F(0.5)-F(0.5-x)=f(x) +F(0.5+x)-F(0.5)=F(0.5)=F(1)-F(0.5+x)+F(0.5)-F(0.5-x)=F(1)-F(0.5+delta) +F(0.5+delta)-F(0.5)=0.5;F(x+delta)-f(x-delta)=2[F(x+delta)-F(0.5)], and likewise, F where F(1)-F(0.5+delta)=F(x)-F(0)=F(x);0.5+delta=1-x, although this hardly seems independent of the rest, and likewise; as you can imagineF(1)- F(1-x)-=F(x)-F(0)=F(x) and is >0.5 0.5 and hence F(x)>0.5 and G(x)>0.5 otherwise smaller and consequently
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