2
$\begingroup$

In a famous paper from 1974, Hal Varian proved (in Theorem 2.3) that:

In an economy with homogeneous divisible googs, if all agents have monotonic and convex preferences, then there exists an allocation which is both Pareto-efficient and envy-free.

He said (after Theorem 2.8) that, if preferences are not convex, then Pareto-efficient envy-free allocations might not exist. As an example, he brings the following economy: there are two agents with the same preferences:

$$u_1(x,y) = u_2(x,y) = \min(x,y)$$

and the total bundle is $(2,1)$.

I do not see how this is a negative example. If we give each agent a bundle of $(1,0.5)$, then the division is envy-free, since the value of every agent is exactly 0.5. It is also Pareto-efficient, since to increase the utility of one agent, it is necessary to give him some y from the other agent, but this hurts the utility of the other agent.

Am I missing something?

$\endgroup$
  • $\begingroup$ The preferences $$u_1(x,y) = u_2(x,y) = \min(x,y)$$ are convex. They are not strictly convex. I also don't see how a counterexample could be constructed using them. $\endgroup$ – Giskard Nov 29 '15 at 21:12
  • $\begingroup$ Also, given these utility functions not any bundle is prefered to the $0$ bundle, so this seems quite strange. $\endgroup$ – Giskard Nov 29 '15 at 21:15
  • 1
    $\begingroup$ Maybe there is a typo in the paper, and the function should be $\max$ instead of $\min$. Does it make sense? $\endgroup$ – Erel Segal-Halevi Nov 30 '15 at 2:26
  • 1
    $\begingroup$ That makes sense. Equitable allocations for that function provide 1 utility for each player, but that cannot be efficient as $(2,0)$, $(0,1)$ is feasible. And the utility function is not convex. I think you are right. $\endgroup$ – Giskard Nov 30 '15 at 7:45
  • $\begingroup$ @denesp OK, good to know! $\endgroup$ – Erel Segal-Halevi Nov 30 '15 at 7:48

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.