<p>Finding Pareto efficient allocations via social planner's problems is a special case of <em>scalarization</em> in convex optimization.</p>
<p>Suppose agents' utility functions are  <span class="math-container">$u_i:\mathbb{R}^n \rightarrow \mathbb{R}$</span>, for <span class="math-container">$i = 1, 2, \cdots, m$</span>, and the feasible allocations are given by <span class="math-container">$g(x) \geq 0$</span> for some <span class="math-container">$g:\mathbb{R}^n \rightarrow \mathbb{R}^p$</span>.
The general result is as follows.</p>
<p><strong>Sufficiency under general conditions</strong></p>
<p>If an allocation <span class="math-container">$x$</span> solves the social planner's problem with strictly positive social weights, then it is Pareto. This is true with no assumptions on utility functions and feasibility constraint.
In other words, if
<span class="math-container">$$
x \in \arg\max_{g(x') \geq 0} \sum_{i = 1}^m \lambda_i u_i(x'), \mbox{ for some } \lambda \in \mathbb{R}^m_{++}, \quad (*)
$$</span>
then it is Pareto optimal.</p>
<p>This is easy to see. If <span class="math-container">$y$</span> Pareto-improves <span class="math-container">$x$</span>, then
<span class="math-container">$$
\sum_{i = 1}^m \lambda_i u_i(y) &gt; \sum_{i = 1}^m \lambda_i u_i(x)
$$</span>
for any  <span class="math-container">$\lambda \in \mathbb{R}^m_{++}$</span>. It is also easy to see that zero entry in <span class="math-container">$\lambda$</span> cannot be allowed---it would make the statement false.</p>
<p>The converse is not true in general---this sufficient condition is not necessary. There are Pareto allocations that do not solve <span class="math-container">$(*)$</span>. This is because the set of achievable utility values
<span class="math-container">$$
\mathcal{U} = \{ (u_1 (x), \cdots, u_m(x)); g(x) \geq 0 \}
$$</span>
is in general not a convex subset of <span class="math-container">$\mathbb{R}^m$</span>. The social planner's weights <span class="math-container">$\lambda$</span> corresponds to certain supporting hyperplanes of <span class="math-container">$\mathcal{U}$</span>. If <span class="math-container">$\mathcal{U}$</span> is not convex, its Pareto frontier cannot be recovered by varying <span class="math-container">$\lambda$</span>.</p>
<p>However, there is a <em>partial</em> converse.</p>
<p><strong>Necessity under concavity</strong></p>
<p>Assume <span class="math-container">$u_i$</span>'s and <span class="math-container">$g$</span> are concave. If an allocation <span class="math-container">$x$</span> is Pareto, then it solves a social planner's problem with nonnegative social weights.
In other words, if <span class="math-container">$x$</span> is Pareto then
<span class="math-container">$$
x \in \arg\max_{g(x') \geq 0} \sum_{i = 1}^m \lambda_i u_i(x') \mbox{ for some } \lambda \in \mathbb{R}^m_{+}. \quad (**)
$$</span>
This is true because of convex sets and separating hyperplanes.</p>
<p>This is a partial converse because it's necessary but not sufficient. There are allocations that solve <span class="math-container">$(**)$</span> but not Pareto.</p>
<p><strong>Comment</strong></p>
<p>Even under concavity, <span class="math-container">$(**)$</span> is a only a partial converse of <span class="math-container">$(*)$</span>. Notice the gap between strictly positive and nonnegative social weights.</p>
<p>To recover the Pareto frontier, the general procedure is to first find solutions to <span class="math-container">$(*)$</span>. Then check the solutions to <span class="math-container">$(**)$</span> case-by-case for Pareto optimality. Alternatively, one can also take the closure (the limit points) of solutions to <span class="math-container">$(*)$</span>.</p>

CC BY-SA 4.0
2021-02-02T11:38:19.677Z
A quick question about this (very helpful!) answer. In the usual framework, utility is taken to be an ordinal concept; which means that properties like the concavity of utility are not very meaningful. Indeed, I can always rescale utility functions so that they are not concave. But here, concavity (not quasi-concavity!) appears to play a special role. What is going on?
afreelunch
CC BY-SA 4.0
2021-02-02T11:41:42.370Z
And to further clarify: is the converse meant to hold for *all* concave utility functions, not just concave, quasi-linear utility functions?
afreelunch
CC BY-SA 4.0
2021-04-09T08:40:51.453Z
@afreelunch Well, what is going on is with the concavity restriction is that it allows one to consider linear SWF's. That is, a natural separation of convex sets arise under the concavity assumption and the separating hyperplane is a SWF. Maximizing a (strictly increasing) SWF implies Pareto (without assuming concavity) but it's not clear whether one obtain a characterization all SWF's as in the concave case.
Michael
CC BY-SA 4.0
2020-08-13T23:13:07.750Z
2020-08-13T23:29:30.583Z
2020-08-13T23:29:30.583Z
Michael
<p>I'm trying to understand the relation between maximizing sum of utilities and finding Pareto efficient allocations. According to <a href="https://econ.ucsb.edu/%7Etedb/Courses/UCSBpf/pflectures/chap2.pdf" rel="nofollow noreferrer">https://econ.ucsb.edu/~tedb/Courses/UCSBpf/pflectures/chap2.pdf</a> (page 13), concavity plays an important role in whether maximizing the sum of utilities will help obtain the efficient allocations or not. Could someone explain this further?</p>
<p>A similar question is posted at <a href="https://economics.stackexchange.com/questions/11033/quasilinear-utility-pareto-optimality-implies-total-utility-maximization">Quasilinear Utility: Pareto Optimality Implies Total Utility Maximization?</a>, but the counterexample in <a href="https://economics.stackexchange.com/a/15146/18507">this answer</a> seems contradictory to the result stated in the above document.</p>
<p>Specifically, my questions are:</p>
<ol>
<li>When can we maximize the sum of utilities to find all the efficient allocations?</li>
<li>What criteria can be used to locate all the efficient allocations in corner/edge cases?</li>
</ol>

CC BY-SA 4.0
2020-07-30T09:04:48.260Z
What do you mean when you write "*the efficient allocation*"? There are usually several Pareto-efficient allocations. Do you just want to find one?
Giskard
CC BY-SA 4.0
2020-07-30T09:58:10.737Z
Your first link says on another page "it is true that maximizing the sum of utilities is a *sufficient* condition for Pareto optimality.  But maximizing the sum of utilities is not a *necessary* condition for Pareto optimality."
Henry
CC BY-SA 4.0
2020-07-30T11:42:21.713Z
@Giskard Have edited my question. I am interested in finding the Pareto set, all efficient allocations.
PGupta
CC BY-SA 4.0
2020-07-30T11:46:48.250Z
@Henry Yes, so my understanding is that if the sum of utilities is maximized at an allocation, it is Pareto efficient. But the discussion at the same link seems to suggest that this can lead to wrong answers, hence the confusion.
PGupta
<pareto-efficiency>
Pareto efficiency and maximization of sum of utilities
PGupta
# Pareto efficiency and maximization of sum of utilities

(Asked by: username_0 on 2020-07-30)

I'm trying to understand the relation between maximizing sum of utilities and finding Pareto efficient allocations. According to [https://econ.ucsb.edu/~tedb/Courses/UCSBpf/pflectures/chap2.pdf](https://econ.ucsb.edu/%7Etedb/Courses/UCSBpf/pflectures/chap2.pdf) (page 13), concavity plays an important role in whether maximizing the sum of utilities will help obtain the efficient allocations or not. Could someone explain this further?

A similar question is posted at [Quasilinear Utility: Pareto Optimality Implies Total Utility Maximization?](https://economics.stackexchange.com/questions/11033/quasilinear-utility-pareto-optimality-implies-total-utility-maximization), but the counterexample in [this answer](https://economics.stackexchange.com/a/15146/18507) seems contradictory to the result stated in the above document.

Specifically, my questions are:

1. When can we maximize the sum of utilities to find all the efficient allocations?
2. What criteria can be used to locate all the efficient allocations in corner/edge cases?

**Tags:** pareto-efficiency

> **Comments:**
> **username_1**: What do you mean when you write "*the efficient allocation*"? There are usually several Pareto-efficient allocations. Do you just want to find one?  
> **username_2**: Your first link says on another page "it is true that maximizing the sum of utilities is a *sufficient* condition for Pareto optimality.  But maximizing the sum of utilities is not a *necessary* condition for Pareto optimality."  
> **username_0**: @username_1 Have edited my question. I am interested in finding the Pareto set, all efficient allocations.  
> **username_0**: @username_2 Yes, so my understanding is that if the sum of utilities is maximized at an allocation, it is Pareto efficient. But the discussion at the same link seems to suggest that this can lead to wrong answers, hence the confusion.  

---

## 1 answers

### Answer by username_3 on 2020-08-13. Score: 2

Finding Pareto efficient allocations via social planner's problems is a special case of *scalarization* in convex optimization.

Suppose agents' utility functions are $u\_i:\mathbb{R}^n \rightarrow \mathbb{R}$, for $i = 1, 2, \cdots, m$, and the feasible allocations are given by $g(x) \geq 0$ for some $g:\mathbb{R}^n \rightarrow \mathbb{R}^p$.
The general result is as follows.

**Sufficiency under general conditions**

If an allocation $x$ solves the social planner's problem with strictly positive social weights, then it is Pareto. This is true with no assumptions on utility functions and feasibility constraint.
In other words, if
$$
x \in \arg\max\_{g(x') \geq 0} \sum\_{i = 1}^m \lambda\_i u\_i(x'), \mbox{ for some } \lambda \in \mathbb{R}^m\_{++}, \quad (\*)
$$
then it is Pareto optimal.

This is easy to see. If $y$ Pareto-improves $x$, then
$$
\sum\_{i = 1}^m \lambda\_i u\_i(y) > \sum\_{i = 1}^m \lambda\_i u\_i(x)
$$
for any $\lambda \in \mathbb{R}^m\_{++}$. It is also easy to see that zero entry in $\lambda$ cannot be allowed---it would make the statement false.

The converse is not true in general---this sufficient condition is not necessary. There are Pareto allocations that do not solve $(\*)$. This is because the set of achievable utility values
$$
\mathcal{U} = \{ (u\_1 (x), \cdots, u\_m(x)); g(x) \geq 0 \}
$$
is in general not a convex subset of $\mathbb{R}^m$. The social planner's weights $\lambda$ corresponds to certain supporting hyperplanes of $\mathcal{U}$. If $\mathcal{U}$ is not convex, its Pareto frontier cannot be recovered by varying $\lambda$.

However, there is a *partial* converse.

**Necessity under concavity**

Assume $u\_i$'s and $g$ are concave. If an allocation $x$ is Pareto, then it solves a social planner's problem with nonnegative social weights.
In other words, if $x$ is Pareto then
$$
x \in \arg\max\_{g(x') \geq 0} \sum\_{i = 1}^m \lambda\_i u\_i(x') \mbox{ for some } \lambda \in \mathbb{R}^m\_{+}. \quad (\*\*)
$$
This is true because of convex sets and separating hyperplanes.

This is a partial converse because it's necessary but not sufficient. There are allocations that solve $(\*\*)$ but not Pareto.

**Comment**

Even under concavity, $(\*\*)$ is a only a partial converse of $(\*)$. Notice the gap between strictly positive and nonnegative social weights.

To recover the Pareto frontier, the general procedure is to first find solutions to $(\*)$. Then check the solutions to $(\*\*)$ case-by-case for Pareto optimality. Alternatively, one can also take the closure (the limit points) of solutions to $(\*)$.

> **Comments:**
> **username_4**: A quick question about this (very helpful!) answer. In the usual framework, utility is taken to be an ordinal concept; which means that properties like the concavity of utility are not very meaningful. Indeed, I can always rescale utility functions so that they are not concave. But here, concavity (not quasi-concavity!) appears to play a special role. What is going on?  
> **username_4**: And to further clarify: is the converse meant to hold for *all* concave utility functions, not just concave, quasi-linear utility functions?  
> **username_3**: @username_4 Well, what is going on is with the concavity restriction is that it allows one to consider linear SWF's. That is, a natural separation of convex sets arise under the concavity assumption and the separating hyperplane is a SWF. Maximizing a (strictly increasing) SWF implies Pareto (without assuming concavity) but it's not clear whether one obtain a characterization all SWF's as in the concave case.  

---

