0
$\begingroup$

This question is from Harvard seminar problem set (Q-3 part b) https://www.studocu.com/en-us/document/harvard-university/economics/mandatory-assignments/econ2020a-14-ps01-please-give-as-much-additional-information-as-possible/3513583/view

Show that it holds continuous, monotonic, LNS and transversality. $$|x_1-x_2 -y_1-y_2| \le 1$$ if x~y

I can just show transversality condition

first, for x~y $$-1\le x_1-x_2-y_1-y_2\le 1 $$ for y~z $$-1\le y_1-y_2-z_1-z_2\le 1 $$

When I sum these in equalities, I will obtain

$$-2\le x_1-x_2-z_1-z_2\le 2 $$

So, At the same time, I can say that

$$-1\le x_1-x_2-z_1-z_2\le 1$$

This implies that x~z. Thus, transversality holds. But I cannot prove for others even though I know their definitions. Please help me to do this.

$\endgroup$
1
  • $\begingroup$ There's a typo, it should be $|x_1+x_2 -y_1-y_2|$. Then it should be "transitivity" instead of "transversality". And your proof is wrong: $-2\le a \le 2$ it doesn't imply $-1\le a \le 1$, as $a=1.5$ shows. $\endgroup$
    – VARulle
    Commented Jan 25, 2021 at 10:18

1 Answer 1

1
$\begingroup$

The preference relation is not transitive. As a counterexample take $x=(0,0), y=(0,1), z=(1,1)$.

For monotonicity let $x=(x_1,x_2)=(y_1+1, y_2)$ so that the $x$ bundle contains more of some commodity but no less of any, than $y=(y_1,y_2)$. Then $$ (y_1+1+y_2-y_1-y_2) = 1 \leq 1 $$

so that the preference relation is not monotonic.

so that the preference relation is strictly monotonic.

The preference relation is continuous since $$ \{x| x\succeq y\} = \{x_1, x_2 | x_1 + x_2 - y_1 - y_2 \geq -1\} $$ is closed and $$ \{x| x\preceq y\} = \{x_1, x_2 | x_1 + x_2 - y_1 - y_2 \leq 1\} $$ is also closed

Finally, it is not locally non satiated. Take any $\epsilon\leq 1$ as an example.

EDIT: It is not strictly monotonic, see the comment below

$\endgroup$
1
  • 1
    $\begingroup$ Let $x=(x_1,x_2)=(y_1+0.1, y_2+0.1)$ so that the $x$ bundle contains more of all commodities than $y=(y_1,y_2)$. Then $ x_1+x_2-y_1-y_2 = 0.2 < 1 $, so it is not strictly monotonic. $\endgroup$
    – VARulle
    Commented Jan 25, 2021 at 10:25

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.