Now we have:
- Axiom1: Completeness of $\succeq$.
- Axiom2: Transitivity of $\succeq$.
- Axiom3: Independence: For any $N$ and $p\in (0,1]$, if $L\succ M$, then $pL+(1-p)N\succ pM+(1-p)N$.
- Axiom4: Archimedean: if $L\succ M \succ N$, then there exists $\alpha, \beta\in(0,1)$ such that $\alpha L+(1-\alpha)N \succ M \succ \beta L+(1-\beta)N$.
Given these four axioms, how to prove that $\alpha>\beta$? By independence axiom we can only get preference relationship $\succ$ between lotteries, but not any numerical relationship $>$ between coefficients. Then how to get this between $\alpha, \beta$?
A dual problem would be, given coefficients $\alpha, \beta\in[0,1]$, $\alpha>\beta$, and preference $L\succ M$, how to prove that $\alpha L+(1-\alpha)M \succ \beta L+(1-\beta)M$?
I tried to construct intermediate lotteries to leverage Independence Axiom, but always failed at the gap between $\succ$ and $>$. Any suggestion on how to use coefficient relationship $>$ would be appreciated!
For the dual problem:
If $L\succ M$ and $\alpha\in (0,1)$ then by Ax3, $$L\,=\,\alpha L+(1-\alpha)L\,\succ\, \alpha L+(1-\alpha)M\,\succ\,\alpha M+(1-\alpha)M\,=\,M \,.$$
If also $\beta\in (0,\alpha)$ is given, then applying the same once again to $N:=\alpha L+(1-\alpha)M\,\succ M$ and $\gamma:=\frac\beta\alpha\in (0,1)$ we get $$N\,\succ\,\gamma N+(1-\gamma)M\,=\,\beta L+(1-\beta)M\,.$$ Swapping the role of $\alpha,\beta$ shows that if $\alpha<\beta$ then $\alpha L+(1-\alpha)M \prec \beta L+(1-\beta)M$, so it ensures $\alpha>\beta$ in Ax4.