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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?

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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.

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  • $\begingroup$ "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, 2015 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, 2015 at 20:34
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    $\begingroup$ 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, 2015 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, 2015 at 21:08

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