<p>I can try and explain how the core shrinks using an example:</p>
<p>Consider a pure exchange economy with two consumers <span class="math-container">$A$</span> and <span class="math-container">$B$</span> and two goods <span class="math-container">$X$</span> and <span class="math-container">$Y$</span>. <span class="math-container">$A$</span> has an endowment of <span class="math-container">$1$</span> unit of <span class="math-container">$X$</span> and <span class="math-container">$B$</span>'s endowment is <span class="math-container">$1$</span> unit of <span class="math-container">$Y$</span>. So, <span class="math-container">$\omega_A = (1,0)$</span> and <span class="math-container">$\omega_B = (0,1)$</span>. Suppose utility functions of <span class="math-container">$A$</span> and <span class="math-container">$B$</span> are identical and of this form:
<span class="math-container">$u_A(x_A, y_A) = x_Ay_A$</span> and <span class="math-container">$u_B(x_B, y_B) = x_By_B$</span>.
In this set up, an allocation where <span class="math-container">$A$</span> consumes everything and <span class="math-container">$B$</span> consumes nothing is a core allocation.
Now let us consider a 2-replica of the above economy. We now have two agents of type-A and 2 agents of type-B. If we now consider the allocation where type-A consumes <span class="math-container">$(1,1)$</span> each and type-B consumes nothing. This allocation will no longer be in the core, because one type-A agent and two type-B agents can redivide their total endowment of 1 unit of X and 2 units of Y to make everyone in the group better off. We can consider this division: type-A gets <span class="math-container">$(1-2\epsilon, 2- 2\epsilon)$</span> and each of the type-B agent gets <span class="math-container">$(\epsilon, \epsilon)$</span>. For <span class="math-container">$\epsilon$</span> small enough this works.</p>

CC BY-SA 4.0
2022-03-02T03:43:43.867Z
2022-03-02T03:43:43.867Z
Amit
<p>I'm not sure this counts as intuition but is the logic of the proof and related proofs. I will purposefully confuse agents and types in what follows, since the logic works also in a limit models with a continuum of agents or for sequences of not necessarily replicas.</p>
<p>A competitive equilibrium is a feasible allocation together with a price system such that the consumption of an agent does not cost more than their endowment and everything better costs more. We will focus on the latter part first.</p>
<p>Suppose you have an allocation <span class="math-container">$x$</span> that is in the core of every replica of the economy. Clearly, it is feasible. We have to find a price system that supports it. That trick is that we want to find a price system <span class="math-container">$p$</span>, such that for every agent <span class="math-container">$i$</span> every commodity bundle <span class="math-container">$y$</span> such that <span class="math-container">$y\succ_i x_i$</span> one has <span class="math-container">$p\cdot y&gt;p\cdot \omega_i$</span>. One can alsways write <span class="math-container">$y$</span> as <span class="math-container">$\omega_i+z$</span> with <span class="math-container">$z=y-\omega_i$</span>. Then the problem reduces to showing that for every <span class="math-container">$z$</span> such that <span class="math-container">$z+\omega_i\succ_i x_i$</span> one has <span class="math-container">$p\cdot z&gt;0$</span>. We want that all beneficial net trade vectors have a positive price. So let <span class="math-container">$$Z=\{z\in\mathbb{R}^l\mid z+\omega_i\succ_i x_i\text{ for some agent } i\}.$$</span></p>
<p>The tool of choice to make a set of commodity bundels have a positive price is to use the separating hyperplane theorem. For this we have to separate it from some other convex set, and the strictly negative orthant <span class="math-container">$-\mathbb{R}^l_{++}$</span> works well.   No element of the strictly negative orthant can be in <span class="math-container">$Z$</span>. For then there would be some <span class="math-container">$i$</span> such that <span class="math-container">$z+\omega_i\succ_i x_i$</span>. But since <span class="math-container">$z\in -\mathbb{R}^l_{++}$</span>, we have <span class="math-container">$z+\omega_i\ll\omega_i$</span>. By strict monotonicity, <span class="math-container">$\omega_i\succ_i z+\omega_i$</span>. By transitivity <span class="math-container">$\omega_i\succ_i x_i$</span>. But then the coalition consisting of <span class="math-container">$i$</span> alone could block the allocation <span class="math-container">$x$</span>, in contradicition to <span class="math-container">$x$</span> being in the core of every replica. So <span class="math-container">$Z\cap -\mathbb{R}^l_{++}=\emptyset$</span>.</p>
<p>However, the  set <span class="math-container">$Z$</span> need not be convex, so we try to separate the convex hull <span class="math-container">$\text{con(Z)}$</span> from <span class="math-container">$-\mathbb{R}^l_{++}$</span>. This need not work in general, the difference between <span class="math-container">$Z$</span> and <span class="math-container">$\text{con(Z)}$</span> can be quite large. But for large replicas, an element of <span class="math-container">$-\mathbb{R}^l_{++}$</span> cannot be in the convex hull of <span class="math-container">$Z$</span> either. Otherwise, there would be a strictly negative vector that is a convex combination of elements with rational coefficients. So suppose <span class="math-container">$z,z'\in Z$</span>, <span class="math-container">$\alpha$</span> is a rational number between <span class="math-container">$0$</span> and <span class="math-container">$1$</span> and <span class="math-container">$\alpha  z + (1-\alpha)z'\in -\mathbb{R}^l_{++}$</span>. So write <span class="math-container">$\alpha$</span> as <span class="math-container">$p/q$</span>.
Now pick <span class="math-container">$q$</span> agents, <span class="math-container">$p$</span> of which would prefer <span class="math-container">$z$</span> plus their endowment to what they get, and <span class="math-container">$q-p$</span> who would prefer <span class="math-container">$z'$</span>. If one gives them exactly that, they would receive on average <span class="math-container">$\alpha  z + (1-\alpha)z'\in -\mathbb{R}^l_{++}$</span>. So they would be better and one could even dipose of something left. Or one distributes what is left among the agents, who prefer it even more because of monotonicity.</p>
<p>Now, the argument has not used that preferences are convex. The convexity assumption is only needed to prove the equal treatment property one needs to phrase everything in terms of replicas. There are versions of the argument not based on exact replicas that do without any convexity assumption. One can also make the argument work when preferences are not transitive, but this requires a bit more effort.</p>

CC BY-SA 4.0
2022-03-02T00:07:44.610Z
2022-03-02T00:07:44.610Z
Michael Greinecker
<p>The theorem states that if preferences are strictly convex and strongly monotone, then the Core eventually shrinks to the Competitive equilibrium as you replicate the economy enough times. How is this happening intuitively? Is it that you can form more richer coalitions and are able to block more and more core allocations?</p>

<microeconomics>
Core-Equivalence Theorem (Debreu-Scarf Theorem) Intuition
Rumi
# Core-Equivalence Theorem (Debreu-Scarf Theorem) Intuition

(Asked by: username_0 on 2022-02-27)

The theorem states that if preferences are strictly convex and strongly monotone, then the Core eventually shrinks to the Competitive equilibrium as you replicate the economy enough times. How is this happening intuitively? Is it that you can form more richer coalitions and are able to block more and more core allocations?

**Tags:** microeconomics


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## 2 answers

### Answer by username_2 on 2022-03-02. Score: 3 | Accepted answer

I'm not sure this counts as intuition but is the logic of the proof and related proofs. I will purposefully confuse agents and types in what follows, since the logic works also in a limit models with a continuum of agents or for sequences of not necessarily replicas.

A competitive equilibrium is a feasible allocation together with a price system such that the consumption of an agent does not cost more than their endowment and everything better costs more. We will focus on the latter part first.

Suppose you have an allocation $x$ that is in the core of every replica of the economy. Clearly, it is feasible. We have to find a price system that supports it. That trick is that we want to find a price system $p$, such that for every agent $i$ every commodity bundle $y$ such that $y\succ\_i x\_i$ one has $p\cdot y>p\cdot \omega\_i$. One can alsways write $y$ as $\omega\_i+z$ with $z=y-\omega\_i$. Then the problem reduces to showing that for every $z$ such that $z+\omega\_i\succ\_i x\_i$ one has $p\cdot z>0$. We want that all beneficial net trade vectors have a positive price. So let $$Z=\{z\in\mathbb{R}^l\mid z+\omega\_i\succ\_i x\_i\text{ for some agent } i\}.$$

The tool of choice to make a set of commodity bundels have a positive price is to use the separating hyperplane theorem. For this we have to separate it from some other convex set, and the strictly negative orthant $-\mathbb{R}^l\_{++}$ works well. No element of the strictly negative orthant can be in $Z$. For then there would be some $i$ such that $z+\omega\_i\succ\_i x\_i$. But since $z\in -\mathbb{R}^l\_{++}$, we have $z+\omega\_i\ll\omega\_i$. By strict monotonicity, $\omega\_i\succ\_i z+\omega\_i$. By transitivity $\omega\_i\succ\_i x\_i$. But then the coalition consisting of $i$ alone could block the allocation $x$, in contradicition to $x$ being in the core of every replica. So $Z\cap -\mathbb{R}^l\_{++}=\emptyset$.

However, the set $Z$ need not be convex, so we try to separate the convex hull $\text{con(Z)}$ from $-\mathbb{R}^l\_{++}$. This need not work in general, the difference between $Z$ and $\text{con(Z)}$ can be quite large. But for large replicas, an element of $-\mathbb{R}^l\_{++}$ cannot be in the convex hull of $Z$ either. Otherwise, there would be a strictly negative vector that is a convex combination of elements with rational coefficients. So suppose $z,z'\in Z$, $\alpha$ is a rational number between $0$ and $1$ and $\alpha z + (1-\alpha)z'\in -\mathbb{R}^l\_{++}$. So write $\alpha$ as $p/q$.
Now pick $q$ agents, $p$ of which would prefer $z$ plus their endowment to what they get, and $q-p$ who would prefer $z'$. If one gives them exactly that, they would receive on average $\alpha z + (1-\alpha)z'\in -\mathbb{R}^l\_{++}$. So they would be better and one could even dipose of something left. Or one distributes what is left among the agents, who prefer it even more because of monotonicity.

Now, the argument has not used that preferences are convex. The convexity assumption is only needed to prove the equal treatment property one needs to phrase everything in terms of replicas. There are versions of the argument not based on exact replicas that do without any convexity assumption. One can also make the argument work when preferences are not transitive, but this requires a bit more effort.


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### Answer by username_1 on 2022-03-02. Score: 1

I can try and explain how the core shrinks using an example:

Consider a pure exchange economy with two consumers $A$ and $B$ and two goods $X$ and $Y$. $A$ has an endowment of $1$ unit of $X$ and $B$'s endowment is $1$ unit of $Y$. So, $\omega\_A = (1,0)$ and $\omega\_B = (0,1)$. Suppose utility functions of $A$ and $B$ are identical and of this form:
$u\_A(x\_A, y\_A) = x\_Ay\_A$ and $u\_B(x\_B, y\_B) = x\_By\_B$.
In this set up, an allocation where $A$ consumes everything and $B$ consumes nothing is a core allocation.
Now let us consider a 2-replica of the above economy. We now have two agents of type-A and 2 agents of type-B. If we now consider the allocation where type-A consumes $(1,1)$ each and type-B consumes nothing. This allocation will no longer be in the core, because one type-A agent and two type-B agents can redivide their total endowment of 1 unit of X and 2 units of Y to make everyone in the group better off. We can consider this division: type-A gets $(1-2\epsilon, 2- 2\epsilon)$ and each of the type-B agent gets $(\epsilon, \epsilon)$. For $\epsilon$ small enough this works.


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