A mixture space is a set P and a binary operation on PxP that assigns for each a in [0,1] an element aP+(1-a)Q s.t.

• M1: 1P+0Q=P
• M2: aP+(1-a)Q = (1-a)Q+aP
• M3: a[bP+(1-b)Q]+(1-a)Q=abP+(1-ab)Q

M1,M2,M3 =>

• M4: aP+(1-a)P=P
• M5: a[bP+(1-b)Q]+(1-a)[gP+(1-g)Q]=[ab+(1-a)g]Q+[a(1-b)+(1-a)(1-g)]R

Example:
Let S,X finite, P =set of all probability charges on X and let F=P^S=set of all functions from S into P.

f,g in F /\ af+(1-a)g== forall s in S (af+(1-a)g)(s)=af(s)+(1-a)g(s)

=> F is a mixture space.

a representation theorem:
P a mixture space and "=>" a binary relation on P => "=>" satsifies J1,J2,J3 <=> there exists U:P->R s.t. U represents "=>":

Proof:
Lemma: P a mixture space and "=>" satisfies J1,J2,J3 =>

• P">"Q /\ 0<=a<b<=1 => bP+(1-b)Q"=>"aP+(1-a)Q
• P"=>"Q"=>"R /\ P">"R => there exists ! a in [0,1]:Q~aP+(1-a)R
• P">"Q /\ R">"S /\ a in [0,1] => aP+(1-a)R">"aQ+(1-a)S
• P~Q /\ a in [0,1] => aP + (1-a)Q ~P
• P~Q /\ a in [0,1] => aP + (1-a)Q ~ aQ + (1-a)P

Use the above lemma and the representation theorem on probability charges to prove this one.

source
jl@crush.caltech.edu index
probability_charge