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$\succsim$ 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.


The 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?

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    $\begingroup$ That's maybe not the most elegant way to write it down, but yes, I think it's correct. $\endgroup$
    – VARulle
    Mar 30, 2020 at 12:40

1 Answer 1

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If I'm reading right, you define a preference $\succsim$ on $\mathbb{R}^N$ to be linear if, for all $x,y$ and all $\alpha \in \mathbb{R}^N$, $$ x \succsim y \implies x + \alpha \succsim y + \alpha. $$ and $\succsim$ satisfies Condition 1 if, only for $\alpha \ge 0$, we have: $$ x \succsim y \implies x + \alpha \succsim y + \alpha. $$ Suppose we both require that Condition 1 also hold for strict preference, and ask that linearity also hold for strict preferences. Then I claim these properties are equivalent, without any further assumptions (such as local nonsatiation).

Proof: Suppose $\succsim$ (and $\succ$) satisfy condition 1, and suppose $x \succsim y$. Fix $\alpha \in \mathbb{R}^N$ arbitrarily. We want to show that $x + \alpha \succsim y+ \alpha$. Define: $$ \alpha^+ = \big(\max\{\alpha_1, 0\}, \ldots, \max\{\alpha_N, 0\}\big), $$ and $$ \alpha^- = \big(\min\{\alpha_1, 0\}, \ldots, \min\{\alpha_N, 0\}\big), $$ and let $x' = x + \alpha^-$, $y' = y + \alpha^-$. By condition 1 applied to $\succ$, and the assumption that $x \succsim y$, we have that $x' \succsim y'$. But then by condition 1 for $\succsim$, we have that $x' + \alpha^+ \succsim y'+ \alpha^+$. However, $x'+ \alpha^+ = x + \alpha$, and $y' + \alpha^+ = y + \alpha$. If instead $x \succ y$ an analogous argument holds. QED

A cautionary note on terminology, condition 1 (and hence equivalently what you're calling linear preferences) do not suffice, even with local nonsatiation, to guarantee existence of a linear utility representation, i.e. of the form $U(x) = \langle x ,\lambda \rangle$. You also need either continuity or a stronger monotonicity assumption. For a counterexample, consider $N=1$ and a preference represented by any discontinuous solution to the Cauchy functional equation.

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  • $\begingroup$ Nice answer! I am surprised that non-satiation isn't needed. Let me double check. $\endgroup$
    – High GPA
    Aug 18, 2022 at 13:25
  • $\begingroup$ $\alpha^-\leq 0$, so we don't have this: "$x\succsim y\implies x'\succsim y'$". Perhaps I missed some of your points. I think, let $a\geq 0$, we have: $[x\succsim y\implies x+a\succsim y+a]\iff[ x+a\prec y+a \implies x\prec y]$. $\endgroup$
    – High GPA
    Aug 18, 2022 at 13:38
  • $\begingroup$ But $\vert \alpha^-\vert$ (i.e. the component-wise absolute values of the vector $\alpha^-$) is component-wise non-negative. Now, we know $x' + \vert \alpha^- \vert \succsim y' + \vert \alpha^- \vert$ must be true (this is just $x \succsim y$). Suppose then that $x' \not \succsim y'$. Then as $\succsim$ is complete, $y' \succ x'$. Then by condition 1, it follows that $y' + \vert \alpha^- \vert\succ x' + \vert \alpha^- \vert$ which we know to be false by hypothesis. Thus $x' \succsim y'$. (Implicitly I'm also assuming the strict analogue of Condition 1 holds, which seems natural) $\endgroup$ Aug 18, 2022 at 14:31
  • $\begingroup$ I've made an edit to show that my answer both assumes that an analogue of condition 1 holds for strict preferences, but that it shows that under these hypotheses, both linearity and its strict analogue hold. The inclusion of the strict analogue is pretty standard for these kind of invariance axioms in the literature; case in point they let you get what you were after without continuity or LNS. That said if the intent behind your question was to look at whether CONT & LNS are substitutes for this condition, this may be less helpful. $\endgroup$ Aug 18, 2022 at 15:10
  • $\begingroup$ There is a slight difference, though. Linearity implies the strict analogue of linearity, so it is natural to also assume the strict analogue. But, the Condition 1 does not implies the strict analogue of Condition 1 without additional assumptions (I guess). $\endgroup$
    – High GPA
    Aug 19, 2022 at 9:10

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