1
$\begingroup$

Given the following two conditions:

$x\succ y$ implies $x+a\succsim y+a$,

And,

$x\prec y$ implies $x+a\precsim y+a$

We want to prove that $\succsim$ is a linear preference.


One of the definition of linear preference is that: $x\succsim y \Leftrightarrow x+a\succsim y+a$

So I am trying to do this:

Since $x\succsim y$ means that $x\succ y$ or $x\sim y$

We already know that $x\succ y$ implies $x+a\succsim y+a$,

all things left is to prove that $x\sim y$ also implies that $x+a\succsim y+a$.

$\endgroup$
2
  • 1
    $\begingroup$ The two conditions you provide are really only one condition (by exchanging $x$ and $y$ in the first you get the second). And what is "left to prove" according to your partial answer is actually true by definition, so that's not what is left to prove. $\endgroup$
    – VARulle
    Mar 21, 2020 at 15:34
  • $\begingroup$ @VARulle I am truly sorry that I made a mistake in my partial "proof". We must show that $x\sim y$ implies $x+a\succsim y+a$ $\endgroup$
    – High GPA
    Mar 22, 2020 at 1:02

1 Answer 1

2
$\begingroup$

You cannot prove this. It is wrong.

Define $u(x)=\min{\{x,0\}}$. Let $\succsim$ be the preference relation represented by $u$. This preference relation satisfies $x\succ y \Longrightarrow x+a\succsim y+a$ for all $x,y,a\in\mathbb R$. But let $x=0$, $y=1$, and $a=-1$. Then $x\sim y$, but $y+a=0\succ -1=x+a$, thus $x+a\not\succsim y+a$ and $\succsim$ is not linear.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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