0
$\begingroup$

$x,y,a$ 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 $x\sim y$ implies $x+a\sim y+a$ for any $a\geq0$ and $x,y\in\mathbb R^n$,

Then the preference is linear.


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

$\endgroup$
2
  • $\begingroup$ Isn't the proof just restating the definition? $\endgroup$
    – Herr K.
    Mar 21 '20 at 21:24
  • 1
    $\begingroup$ @HerrK.: In the definition of linearity the indifference has to hold for all $a$, by assumption we only know that it holds for $a\ge 0$. $\endgroup$
    – VARulle
    Mar 23 '20 at 15:54
1
$\begingroup$

It is not true. Let us consider $\mathbb{R}^2$ so bundles are $x = (x_1,x_2)$.

Consider the preference:

(i) If $x_1 \leq 0$, preferences are lexicographic, i.e. $$ x \succ y \Leftrightarrow \begin{cases} x_1 > y_1 \\ \text{ or } \\ x_1 = y_1 \text{ and } x_2 > y_2 \end{cases} $$ (ii) If $x_1 \geq 0$, $u(x_1,x_2)=x_1+x_2$.

Notice that no indifference occurs on $\mathbb{R}_{<0} \times \mathbb{R}$, and the required condition holds on $\mathbb{R}_{\geq 0}\times \mathbb{R}$ since it is linear there.

However, $(0,4) \sim (2,2)$, choose $a = (-1,-1)$, we get $$(2-a,2-a) = (1,1) \sim (0,2) \succ (-1,3) = (0-a,4-a) $$ hence the second condition fails.

$\endgroup$
1
  • $\begingroup$ Hi Walrasian I am sorry that I forgot to mention the preference has to be continuous. I will open another question! Thank you for your smart counterexample by the way! $\endgroup$
    – High GPA
    Mar 21 '20 at 9:33

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.