<p>In two details you seem to be mistaken:</p>

<ol>
<li>You need strict concavity of $v(\cdot)$, not concavity.</li>
<li>According to the <a href="https://en.wikipedia.org/wiki/Quasiconvex_function" rel="nofollow">definition</a> of quasiconcavity, the function $x^2$ is not quasiconcave, it is quasiconvex.</li>
</ol>

<p>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.</p>

CC BY-SA 3.0
2015-06-16T18:46:30.877Z
"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).
Oliv
CC BY-SA 3.0
2015-06-16T20:34:18.763Z
@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
$$
(1 - \lambda)^2 = (\lambda \cdot 0 + (1 - \lambda) \cdot 1)^2 \geq \min (0^2,1^2) = 1.
$$
Giskard
CC BY-SA 3.0
2015-06-16T20:53:25.563Z
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.
Oliv
CC BY-SA 3.0
2015-06-16T21:08:13.040Z
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!
Giskard
CC BY-SA 3.0
2015-06-14T22:05:08.057Z
2015-06-16T21:10:55.583Z
2015-06-16T21:10:55.583Z
Giskard
<p>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. </p>

<p>We assume utility is strictly increasing and strictly quasiconcave. My question is this:</p>

<p>My professor says by strict monotonicity </p>

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

<p>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. </p>

<p>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?</p>

<general-equilibrium><risk>
Perfect risk sharing in Arrow-Debreu w/ the same subjective beliefs over states
hipHopMetropolisHastings
# Perfect risk sharing in Arrow-Debreu w/ the same subjective beliefs over states

(Asked by: username_0 on 2015-06-14)

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?

**Tags:** general-equilibrium, risk


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## 1 answers

### Answer by username_1 on 2015-06-14. Score: 2 | Accepted answer

In two details you seem to be mistaken:

1. You need strict concavity of $v(\cdot)$, not concavity.
2. According to the [definition](https://en.wikipedia.org/wiki/Quasiconvex_function) 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.

> **Comments:**
> **username_2**: "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).  
> **username_1**: @username_2 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 $$ (1 - \lambda)^2 = (\lambda \cdot 0 + (1 - \lambda) \cdot 1)^2 \geq \min (0^2,1^2) = 1. $$  
> **username_2**: 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.  
> **username_1**: 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!  

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