1
$\begingroup$

$\succsim$ is a continuous and local non-satiate weak order.

$x,y,a$ are vectors in $\mathbb R^n$

We say $a\geq0$ if all directions of the vector $a$ is greater or equal to zero.

We want to prove (or disprove by counterexample) that:

Suppose $x\succsim y$ implies $x+a\succsim y+a$ for any $a\geq0$ and $x,y\in\mathbb R^n$ (condition 1),

Then the preference is linear.


One definition of linear preference is that $x\succsim y$ implies $x+a\succsim y+a$ for any $x,y,a$.

Proof by contradiction. Suppose $\succsim$ is not linear then there exists $x\succsim y$ but $x+a\prec y+a$.

By non-satiation and continuity, there exists $x+\epsilon\succ y$ and $x+a+\epsilon \prec y+a$

Denote $x'=x+\epsilon$

Here if $a_i\geq 0$ or $a_i\leq 0$ for all index $i\in\{1,..,n\}$ then the proof is done.

Now suppose that $a_i\geq0$ for some indexes but $a_j\leq 0$ for some other indexes.

Let $c_i=\min\{0,a_i\}$

$v:=x'+c$ is a point such that $v\leq x'$ and $v\leq x'+a$

$w:=y+c$ is a point such that $w\leq y$ and $w\leq y+a$

If $v\succsim w$, then by condition (1) we must have $x'\succsim y$ and $x'+a\succsim y+a$, contradition!

If $v\precsim w$, then by condition (1) we must have $x'\precsim y$ and $x'+a\precsim y+a$, contradition!

Is the proof sounds rigorous?

$\endgroup$
  • 1
    $\begingroup$ That's maybe not the most elegant way to write it down, but yes, I think it's correct. $\endgroup$ – VARulle Mar 30 at 12:40

Your Answer

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

Browse other questions tagged or ask your own question.