Von Neumann–Morgenstern: compare coefficients in Archimedean axiom

axiomsdecision-theoryeconomicsutility

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!

Best Answer

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.

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