Where F:[0,1] to [0,1]
(forall x,y z,m)\in dom (F)=[0,1] x+y=z+m, iff F(x)+F(y)=F(z)+F(m)
This is the identity, am I correct that this is roughly equivalent to Jensen's equality over the restricted interval. It says that the function is sum conserving, F(0.4)+F(0.3)=F(0.2)+F(0.5) for example as 0.4+ 0.3 =0.2=0.5=0.7
and that if F(0)=0 and F(1)=1 such a function will be F(x)=x? due to non-negativity of the domain.
I note that: F(1-x)+F(x)=1 and x+y =1 iff F(x)+F(y)=1
F(1-x-y)+F(y)+F(x)=1x+y+z=1 iff F(x)+F(z)+(y)=1
See the last section of the P/ Fishburn paper
'on Handas new theory of cardinal utility' and the maximization of expected utility', which is the first attachment
or 'Gleason explained' 2006 where Cauchy additivity is derived from constraints like these (4th attachment).
. It seems to have well known since the time of Wald Edwards 1953, that if one ranks subjective probability by objective probability over a homogeneous objective probability simplex (ie no triples go missing), and the rank is global, then even if additivity of subjective probability is only local (ie within a vector), the ability to put equalities from one vector (ie equal objective probabilities give equal subjective probabilities by the global rank ) will lead to global additivity, of subjective probability between vectors of subjective probability and function of objective probability. And give rise to of the above two global equations so long as each vector has at least three outcomes. One can get the first equation for n=2 and one can it for n=3 from closure under union or due to degenerate spaces, so that one only needs credal- normalization (local) to get the above two equations which are global (ie they hold whether the events are disjoint or not, or are on the same vector or not),
Th is is due to the fact that objective probabilities are also locally additive, for any three events with the same objective chances that sum to one, there will be a space where three events lie together and are disjoint and thus must sum to one in both objective probability and subjective probability., where the first event has the same objective chance for instance as the first event on the orthogonal column , the second= the second, the third = third, Due to the global rank objective PR(x) =objective PRr(k) iff Subjective PR(x)=subjective PR(K), this implies the same for the subjective probabilities, and as each of those events on row-same vector, have the same subjective probabilities now now as the three non disjoint events, in the column and thus must be assigned the same subjective credence.s not possible for two sums of three events, x1+x2 +x3=1, y1+y2+y3=1, x1=y1 where x1=y1,=0.6 for example
x2=y2=0.1, x3=y3=0.3, and so by subjective to objective probability (same objective probability x1=x2 iff same same subjective probability F (x1)=F(y1)) global rank F(x1)=F(y1), F(x2)=F(y2), F(x3)=F(y3) so F(x1)+F(x2)+F(x3)=F(y1)+F(y2)+F(y3
and thus they will have the same subjective probability sum, as its not really possible for F(x1)+F(x2)+F(x3)\neq F(y1)+F(y2)+F(y3) yet
F(x1)=F(y1), F(x2)=F(y2), F(x3)=F(y3) as
[x1=y1 and x1=y1 iff F(x1)=F(y1)], x2=y2 and x2=y2 iff F(x2)=F(y2),x3=x3 and x3=x3 iff F(x3)=F(y3)
and as the first three events lie on the same vector
F(y1), F(y2), F(y3) are disjoint and mutually exclusive as
are events on the same vector it is required that the subjective probabilities are locally additive and normalized, F(y1)+F(y2)+F(y3)= 1 but as F(x1)+F(x2)+F(x3)=F(y1)+F(y2)+F(y3) this enforces that F(x1)+F(x2)+F(x3)=1, despite these events not being disjoint at all. (and thus its a global condition.
x1 might be in vector
xi ,
x2 in ,
x3 in whilst y1, y2, y3 are in the same vector and thus where , F(y1)+F(y2)+F(y3)=1, as they lie on the same vector
but as the as the ranking gives that as y1=x1 so F(x1)=F(y1) , x2=y2 so F(x2)=F(y2), and x3=y3 so F(x3)=F(y3) which entails, F(x1)+F(x2)+F(x3)=F(y1)+F(y2)+F(y3)and as
F(y1)+F(y2)+F(y3)=1, as they must have the same sum and the latter is one so is the former, vertical function sum F(x1)+F(x2)+F(x3)=1, yet x1, x2, x3 may be on totally different vectors on the simplex. So long as there there exist for any possible set of three events whose objective probabilities sum to one, (that are not on the same vector ) and are not presumed to add in credence, there is some vector where the three events have the same objective probabilities lie, and so by rank the subjective (x1)= Subjective pr(y2) iff x1=y2 (objective probabilities are equal) ; the entries x1, x2, x3, have the same subjective probabilities as n the row vector
the first entry in the row, has the same objective probabilities, as the first entry in the column , second = second , third =thrid, and the subjective probabilities are ranked globally,
so that if they have the same subjective probabilities they must have the same objective probabilties so this also applies, to the subjective probabilities, and as the subjective probabilities of the each entry taken one at a time are the same, the total subjective probability sum is the same on the rows and column, but the subjective probability sum on the row vector is 1 (the events are disjoint and mutually exclusive on rows and so are expected to add to one) which implies that it also on the column, despite these events being on distinct worlds
=0
seems to give F(x)=x as they entails Cauchy equations over 0
il'l ad some links.
One can do the same thing with the later equation above given F(0)=0 for Cauchy additive and F(0)=0, F(1) =1 for the Cauchy equation with F(1)=1
roughtly but given F:[0,1] to [0,1], and given non negativity,
or monotonicity equivalent F(x)=x over that resitricted domain,
F(0)=0, also gives F(1)=1 as well, in addition to F(1/3)=1/3 and oddness about 1/3
It seems obvious and it appears to be shown just then, but I want to be sure.
I have my proof that it entails cauchy equations given F(0)=0, one can get F(1)=1 from (eq1)x+y +z=1 then F(x)+F(y)+F(z)=1,
and derive eq(2) x+y =1 then F(x)+F(y)=1,
from F(0)=0 from from eq (1) x+y +z=1 then F(x)+F(y)+F(z)=1,
and from eq2 and eq 1 one can derive
cauchy additivity over the unit triangle
forall x,y, 0
Jensens equality (convexity and concavity) or jensen convexity?;
Ie the first I presume. in the equality case
ie its a saying events any two events with the same domain sum have the same function sum . with the same sum have the same function sum
(1)\forall x y, z m in dom F() x+y =z+m then F(x)+F(y)=F(z)+F(m), ie
by taking z=x/2+y/2, m=x/2+y/2 z+m=(x/2+y/2)+(x/2+y/2)=x+y, so long as (x/2+y/2), x,y, in dom(F), there are in the domain then (x/2+y/2)+(x/2+y/2)=x+y, implies from (1)
(x/2+y/2)+(x/2+y/2)=(x)+(y) implies F(x/2+y/2)+F(x/2+y/2)= F(x)+F(y)
ie 2F(x/2+y/2)=F(x)+F(y) and thus F(x/2+y/2)=F(x)/2+F(y)/2
\forall x y, z m in dom F() x+y =z+m iff F(x)+F(y)=F(z)+F(m) it seems to be jensen's equality case I presume
as jensens equation implies F(x)/2+F(y)/2=F(x/2+y/2), F(x)+F(y)=2F(x/2+y/2) which implies \forall (x,y) F(x)+F(y)=F(x/2+y/2)+F(x/2+y/2)
I am wondering whether they are literally equivalent. as the
jesnsens equality effectively relates this in terms of means ie F(x)+F(y)=2F(x/2+y/2).
Note that whilst jensen equation is also bounded for cauchy equation on [0,1] to [0,1] for cauchy additivity, no greater than one as well on [0,1] ie 0=2
so 0
I presume that with F(0)=0 and F(1) =1 with sum symmetry
x+y =z+m iff F(x)+F(y)=F(z)+F(m)
though which puts a direction on F relates it to a domain elements
on it it reduces to F(x)=x, ie with x=0, F(0)=0 and F(1)=1 and Cauchy's equation over the unit triangle F(x+y)=F(x)+F(y) for 0
@ peter I presume that it entails that the function is satisfies jensens equality on [0,1] on half of its domain ie F(x/2+y/2)=F(x)+F(y) where 0
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