Am i correct in presuming that
probability function satisfying point reflection symmetry at all points would be linear, ie F-1(x)=F(x);
where F(x) is a bi-jective strictly increasing function continuous function from a bounded closed interval F [0,1] to [0,1]; and hence (f(0)=0, F(1)=1)
satisfying merely at all points f-1(x)=f(x);
On the other hand, I can only presume that dual functions that are mirror symmetric,
-F(x)=G(1-x) for all x, F(x)+G(x)=1;
-Both satisfying -Absolute continuity and Equicontinuos
- Midpoint Convexity
_injective,
strictly montonely increasing/decreasing for G.
-x itself being linear probability function,
- Both F, and G, Strictly quasi Concave, strictly quasi convex,
-this condition expressed, as quasi equispaced at on all variables F(
- weird bisymmetry for all x F(x)=G(1-x); F-1(1-x)=G-1(x)
-for all x, standard,additivity F(x)+G(x)=1;
-quasi linear vertical additivity symmetry with regard F and F's inverse and G and G's inverse;
Symmetry
- F(x)+F(x1)=1 iff x+x1=1 ,F-1(x)+F-1(x1)=1 iff x+x1=1 (so it applies to the inverse function due to the iff claim);Likewise
-G(x)+G(x1)=1 iff x+y=1 ;G-1(x)+G-1(x1) iff x+y=1, and
-F(x)+G(y)=1 iff x+y=1, and F-1(x)+G-1(y)=1 iff x+y=1,
vertical additivity to a degree; otherwise F-1(x)+G-1(y)>1 x+y=1,F(x)+G(y)>1iff x+y>1, and likewise if F(x)+G(y)x3, F(x)+F(x1)>F(x2)+F(x3), x4>x1,x5>x3;F(x4)+F(x5)>F(x)+F(x1);F(x4)+F(x5)>>F(x2)+F(x3
and for both functions and their inverses.
-f F(x)+F(x1)>1 iff x+x1>1 and G(y)+G(y1)>1 iff x+x1>1 ;
F(x)+F(x1)x+x1
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