3
$\begingroup$

So I'm looking at a 2-agent Arrow-Debreu economy with one good. Consumption and endowments are zero in t=0, and 2 states are possible in t=1 with aggregate endowment in both states equal to 1.

We assume utility is strictly increasing and strictly quasiconcave. My question is this:

My professor says by strict monotonicity

$\dfrac{v^{'}_{1}\left(x^1_1\right)}{v^{'}_{1}\left(x^2_1\right)} = \dfrac{v^{'}_{2}\left(1 - x^1_1\right)}{v^{'}_{2}\left(1 - x^2_1\right)} \Rightarrow x^1_1 = x^2_1$

I can see this is obviously true if $v(\cdot)$ is concave, but we only have strict quasi-concavity. For example $f(x) = x^2$ is strictly convex yet strictly quasiconcave. Since agent 1 and agent 2 are not required to have the same utility function, it is possible for agent 1 to have a convex utility and agent 2 to have a concave one. In short, we cannot say the second derivatives are the same sign without the concavity assumption.

Moreover, wouldn't a counter example be aif $v_i(x) = x$ for $i = 1,2$. Then $x_1^1 \neq x_1^2$ would still imply the ratio holds. Is it then something about an interior solution?

$\endgroup$
2
$\begingroup$

In two details you seem to be mistaken:

  1. You need strict concavity of $v(\cdot)$, not concavity.
  2. According to the definition of quasiconcavity, the function $x^2$ is not quasiconcave, it is quasiconvex.

Your main point is correct. If both functions are linear, then $x_i^1 = x_i^2$ no longer needs to hold, as the goods $x_i^1$ and $x_i^2$ are perfect substitutes for consumer $i$. Hence all consumption decisions are correct as long as $i$ trades at a price ratio of 1. This is indeed the equilibrium price ratio. The same can be said for the other consumer.

$\endgroup$
  • $\begingroup$ "According to the definition of quasiconcavity, the function $x^2$ is not quasiconcave, it is quasiconvex." Isn't it also quasiconcave ? It seems to me that it satisfies the definition of quasiconcavity when defined on $\mathbb{R}_+$. Going back to the original question, I think that the examples provided by @user176153 are correct and that strict quasiconcavity is not sufficient (but strict concavity is). $\endgroup$ – Oliv Jun 16 '15 at 18:46
  • $\begingroup$ @Oliv I also agree with user176153 's example. (I said as much in the answer.) $x^2$ is not quasiconcave. E.g.: There is no $\lambda \in (0,1)$ such that $$ (1 - \lambda)^2 = (\lambda \cdot 0 + (1 - \lambda) \cdot 1)^2 \geq \min (0^2,1^2) = 1. $$ $\endgroup$ – Giskard Jun 16 '15 at 20:34
  • 1
    $\begingroup$ Yes, my last sentence was about the original question and not about your answer. About your example, $\min(0^2,1^2)=0$, not $1$ and I maintain that $x \rightarrow x^2$ is quasiconcave on $\mathbb{R}_{+}$. More generally, any strictly increasing function is both quasiconcave and quasiconvex. $\endgroup$ – Oliv Jun 16 '15 at 20:53
  • $\begingroup$ You are right, my mistake. In the answer I meant that $x^2$ is not quasiconcave on $\mathbb{R}$ which is true. $$ 1 > (λ⋅1+(1−λ)⋅1)^2 \geq \min((-1)^2,1^2) = 1. $$ (You are right about it being quasiconcave on $\mathbb{R}_+$.) Your comment made me realise I have some false concepts about quasiconcavity, thank you for that! $\endgroup$ – Giskard Jun 16 '15 at 21:08

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.