2
$\begingroup$

Consider a preference relation $\succeq$ on $X=\mathbb{R^2_{+}}$. If $\succeq$ satisifies: $$ \begin{align} &1.\mbox{ }(a_1,a_2)\succeq (b_1,b_2)\implies(a_1+t,a_2+s)\succeq (b_1+t,b_2+s),\forall t,s\\ &2.\mbox{ }a_1\geq b_1 \mbox{ and } a_2\geq b_2 \implies (a_1,a_2)\succeq (b_1,b_2)\mbox{ (and the analogous for }\succ\mbox{)}\\ &3.\mbox{ Continuity } \end{align} $$ Then: exists a linear representation for $\succeq$.

Could anyone give me some hints on how to prove this?

Thanks for helping! :D

$\endgroup$
5
  • $\begingroup$ You can look at Koopman's very famous Econometrica paper to have some insight on these kind of stuff. $\endgroup$ Aug 31 '15 at 0:10
  • $\begingroup$ I remember having seen this proof somewhere before, I tried finding it on Mas-colell but without success. Maybe your suggestion can help me somehow. Thanks! :D BTW, which paper are you talking about? $\endgroup$ Aug 31 '15 at 0:11
  • $\begingroup$ Here it is ; mikael.cozic.free.fr/koopmans60.pdf $\endgroup$ Aug 31 '15 at 10:28
  • $\begingroup$ You could make this into an answer. $\endgroup$
    – BKay
    Aug 31 '15 at 22:53
  • $\begingroup$ Who proposed this independence axiom? I cannot find it on Koopmans 1960 $\endgroup$
    – High GPA
    Jun 5 '19 at 22:05
2
$\begingroup$

How about this: For each vector $(x,y)$, there is a unique $z\in \mathbb R$ such that $(x,y) \sim (z,z)$. WLOG assume $x \geq y$. Then to see this claim, first notice by A2 that $(x,x) \succeq (x,y) \succeq (y,x)$. Then traveling along the $45^\circ$ from $(y,y)$ to $(x,x)$, A3 ensures the existence of our $z$. (Strict) Monotonicity assures uniqueness in the obvious way. Let $u: (x,y) \mapsto z$ where $z$ is defined in this way.

Now let $(x,y) \sim (z,z)$ and $(x',y') \sim (z',z')$. Then by A1 we have \begin{align} (x + x',y+y') &\sim (z + x',z + y') \\ (x' + z,y' +z) &\sim (z'+z,z'+z) \end{align} so by transitivity, $(x+x',y+y') \sim (z+z',z+z')$. Verifying that $u$ is continuous is trivial.

So $u$ is additive and continuous. As this Math SE post explains if $\mathbb R^2 \to \mathbb{R}$ is additive and continuous then it is linear.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.