K K K ∑a4)=∑a∑mn-∑∑a6|m k=1n=1 n=1k=1 K ∑ akLk k=1 so linearity holds More surprisingly, if K K ∑ akLk ∑aU(LA) then we can write N Lk=(,N)=∑pnLn Where In places a probability of 1 on state n and zero on all other states
k1 K kULk k1 K k n1 N unpn k n1 N k1 K kpn k un U k1 K kLk so linearity holds. More surprisingly, if U k1 K kLk k1 K kULk then we can write Lk p1 k ,...pN k n1 N pnLn where Ln places a probability of 1 on state n and zero on all other states
We have then UL)=∑pln=∑p(n)=∑ nun which has the expected utility form
We have then ULk U n1 N pnLn n1 N pnULn n1 N pnun which has the expected utility form