Does every allocation have a maximal Pareto-improvement?

Question

Consider an economy with a finite number of goods and a finite quantity of each good. Each agent $i$ has a preference-relation $\succeq_i$ which is a total, reflexive and transitive relation over the set of bundles. Given an allocation $x$ of the goods among the agents, say that $x^*$ is a maximal Pareto-improvement of $x$ if:

• $x^*$ is Pareto-optimal,
• Every agent weakly prefers the allocation $x^*$ over the allocation $x$.

What is the weakest condition on the preference-relations $\succeq_i$ that guarantees that every allocation has a maximal Pareto-improvement?

I would guess that it is somehow related to compactness, however, I do not see how to define this condition on preference-relations.

Answer 1

I think there is a short proof if you also assume that the number of agents $n$ is finite and that the preferences are continuous.

Given the second assumption Debreu's theorem (1954, "Representation of a preference ordering by a numerical function") states that a continuous utility function exists which represents the preferences. I will denote the utility function representing the preference of agent $i$ by $U_i$.

Though this is somewhat uncommon, I will denote the utility agent $i$ gets from an allocation $x$ by $U_i(x)$.

An algorithm resulting in a maximum Pareto-improvement of $x$:
Step 1. Initially set $i=1$ and set $y = x$.
Step 2. Maximize the utility of agent $i$ over the set of allocation where all other agents $j$ have at least utility $U_j(y)$. As all functions $U_j$ are continuous, there is a finite number of goods and a finite quantity of each good, for all $j$ the upper set of $U_j$ is a closed set. A finite intersection of closed sets is also a closed set. As $U_i$ is also continuous, it will have a maximum over this set. Select an allocation $y'$ which maximizes $U_i$ and let this be the our allocation $y$ from now on, so set $y = y'$.
Step 3. If $i = n$, stop. Otherwise set $i = i+1$ and proceed to step 2.

Notice the following:

• Any move from $y$ to $y'$ in Step 2. is a Pareto-improvement.
• Pareto-improvements are transitive, therefore the set of Pareto-improvements is narrowing, i.e. if $y'$ is a Pareto-improvement to $y$ then the set of Pareto-improvements to $y'$ is included in the set of Pareto-improvements to $y$.
• As the utility of agent $i$ is maximized in Step 2. over the set of Pareto-improvements to $y$, it is impossible that his utility can be further increased by a Pareto-improvement. Therefore once we reach Step 3. for $i = n$ we have a Pareto-optimal allocation $y$, that is either a Pareto-improvement of $x$ or where $\forall i$: $x\sim_i y$. Hence $y$ is a maximal Pareto-improvement of $x$.

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A comment: It seems to me you could replace continuity with monotonicity, but that would require another proof. Hopefully someone else is up to it!

Answer 2

This is essentially a variation on the answer of denesp that requires slightly fewer assumptions.

Assume there are $l$ commodities and $m$ agents. An allocation is then a point in $\mathbb{R}^{lm}_+$. If the aggregate endowment is $e\in\mathbb{R^l}_+$, an allocation is a point in $\sum^{-1}(\{e\})$, where $\sum:\mathbb{R}^{lm}_+\to\mathbb{R}^l$ is the continuous "summation function". Since this function is continuous and the set $\{e\}\subseteq\mathbb{R}^l$ closed, the space of allocations is closed too. It is also clearly bounded, so the space of allocation is compact. Let $A\subseteq\mathbb{R}^{lm}_+$ be the nonempty compact space of feasible allocations.

Define the relation $\succeq$ on $A$ such that for allocations $x=(x_1,\ldots,x_m)$ and $y=(y_1,\ldots,y_m)$, we have $x\succeq y$ if and only if $x_i\succeq_i y_i$ for every agent $i$. Now $x^*\in A$ is a maximal Pareto improvement over $x$ exactly if $x^*\succeq x$ and there is no $y\in A$ such that $y\succeq x^*$ but not $x^*\succeq y$.

Assume now that for all $a\in\mathbb{R}^l_+$ and every agent $i$, the "weakly-better-set" $\{b\in\mathbb{R}^l\mid b\succeq_i a\}$ is closed. Then the set $A_x=\{y\in A\mid y\succeq x\}$ is closed and, as the closed subset of a compact set, compact. Our problem reduces to showing that there exists a $\succeq$-maximal element $x^*\in A_x$.

Let $\succ$ be the asymmetric part of $\succeq$. It is transitive and irreflexive and therefore acyclic. Also, the "upper sections" of $\succeq$ are closed and therefore the lower sections of $\succ$ open. The existence of a $\succ$-maximal element follows then from what is sometimes referred to as the Walker-Bergstrom theorem (first proven by Sloss....). For the sake of completeness, I give the easy proof here.

Let $L_z=\{y\in A_y\mid y\prec z\}$ be the lower set of $\succ$ at $z$. Assume for the sake of contradiction that there is no $\succ$-maximal element in $A_x$. Then every point in $A_x$ lies in some $L_z$ with $z\in A_x$. Also, the $L_z$ are relatively open in the compact space $A_x$. So $\{L_z\mid z\in A_x\}$ is an open cover of $A_x$ and, by compactness, there is a finite set $F\subseteq A_x$ such that $\{L_z\mid z\in F\}$ is still an open cover of $A_x$. In particular, for each $z\in F$, there is some $z'\in F$ such that $z\in L_{z'}$ or, equivalently, $z'\succ z$. So the relation $\succ$ has no maximal element on the finite set $F$. This means there exists an infinite sequence $\langle z_n\rangle$ such that $z_{n+1}\succ z_n$ for all $n$. Since $\succ$ is acyclic, the sequence consists of infinitely many distinct elements. Since $F$ is finite, this is impossible.