Quasilinear Utility: Pareto Optimality Implies Total Utility Maximization?

Question

I read that if we have quasilinear utility for all consumers, then any pareto optimal allocation maximizes the sum of utility levels of all consumers. That is:

$\textbf{What we know:}$ $$1)\quad u^i(m^i,x^i)=m^i+\phi^i(x^i)\; \quad \forall i=1,...,I$$ $$2)\quad\phi^i(\;)\;\text{is continous and strictly increasing (but not necessarily differentiable)}$$ $$3)\quad \text{An allocation,}\,x\, \text{satisfies}\;\neg\,\exists\,\hat{x}\; s.t. \;\hat{m}^i+\phi^i(\hat{x}^i)\geq m^i+\phi(x^i)\;\forall i$$ $$\text{and} \quad \hat{m}^i+\phi^i(\hat{x}^i)> m^i+\phi(x^i)\,\text{for some}\,i$$

$\textbf{What to show:}$ $$x\;\text{solves}\;max\sum_{i=1}^Im^i+\phi^i(x^i)$$

Can anyone provide a proof of this? Any help would be greatly appreciated!

$\textbf{Edit:}\,$I don't know if this is the right path, but by the strict increasing property of $\phi(\,)$, preferences satisfy local non-satiation, which implies they satisfy the first welfare theorem. Now, If I could figure out whether all pareto optimal allocations are competitive equilibria with quasilinear utility, I may be on to something!

Answer 1

Edit: Edge cases suck; see comments. See also MWG Chapter 10 section C, D.

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Suppose $(\vec x^*, \vec m^*)$ solves

$$\max \sum^I_{i=1} m_i + \phi_i(x_i)$$

but is not Pareto optimal.

$$\begin{align} \implies \exists \ (x_i', m_i') \quad \text{s.t.} \quad & u_i(x_i', m_i') \geq u_i(x_i^*, m_i^*) \quad \forall \ i = 1,\cdots,I \\ & u_i(x_i', m_i') > u_i(x_i^*, m_i^*) \quad \text{for some} \ i \end{align}$$

$$\implies \sum^I_{i=1} m'_i + \phi_i(x'_i) > \sum^I_{i=1} m^*_i + \phi_i(x^*_i)$$

which is a contradiction. If we have a solution to the utility maximization problem, it must be Pareto optimal.

(Note that this comes form continuous and increasing properties of $\phi(\cdot)$)

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Suppose $(\vec x^*, \vec m^*)$ is a feasible Pareto optimal allocation, but does not solve

$$\max \sum^I_{i=1} m_i + \phi_i(x_i)$$

Because we treat $m_i$ as numeraire and $\phi_i(\cdot)$ is strictly increasing, we know $u_i(\cdot)$ is locally non-satiated. The Pareto allocation should be just feasible.

$$\exists \ (x_i', m_i') \quad \text{s.t.} \quad \sum^I_{i=1} m'_i + \phi_i(x'_i) > \sum^I_{i=1} m^*_i + \phi_i(x^*_i)\\ \implies \boxed{ \sum^I_{i=1} \phi_i(x'_i) > \sum^I_{i=1} \phi_i(x^*_i)}$$

If this is true because this alternative allocation simply gives an individual more of $x$, for all else equal, then the alternative allocation is infeasible. So we'd have a contradiction.

If this is true because in the alternative allocation, someone else is allocated more $x$ and just one other person is allocated less, then the original allocation would not be Pareto optimal. Suppose it was. If you took the original allocation and shifted $x$ in the way of the new allocation, then you would need a corresponding trade in the numeraire good, $m$, to keep whoever is losing $x$ at least at the same utility level. But trades in just the numeraire good can never change summed aggregate utility. From the original allocation, if you can trade $m$ for $x$ and make someone better off without hurting anyone, you weren't at a Pareto optimum, and if you can't trade $m$ for $x$ to make someone better off, you can't increase summed aggregate utility, which means the original allocation was a solution to the maximization problem.

This logic applies no matter how you rearrange $x$ between multiple people.

$\square$

Answer 2

I don't think it is true in a standard pure exchange economy the question is referring to. Consider the following counterexample: Suppose

$I = \{1,2\}$ and $u_1(x_1, m_1) = \sqrt{x_1} + m_1$ and $u_2(x_2, m_2) = \sqrt{x_2} + m_2$.

and let the set of feasible allocations be

$\{((x_1, m_1), (x_2, m_2))\in\mathbb{R}^2_+\times\mathbb{R}^2_+: x_1+x_2 = 2, m_1+m_2 = 2 \}$.

Notice that the allocation $a_1 = ((x_1, m_1), (x_2, m_2)) = ((2,2),(0,0))$ is Pareto efficient, but does not maximize the sum of utilities. The reason is that allocation $a_2 = ((1,1),(1,1))$ yields the higher sum.

$u_1(2,2) + u_2(0,0) = \sqrt{2} + 2 < 2 + 2 = u_1(1,1) + u_2(1,1)$.

Answer 3

I believe you are referring to the following result: Any PE allocation maximizes $\sum_{i=1}^{I}\phi_{i}(x_{i})$, but it is hard to know precisely since you are not specific about feasibility.

Let me be more specific. For each $i\in\{1,\ldots,I\}$, $(x_{i},m_{i})\in\mathbb{R}_{+}\times\mathbb{R}$. An allocation is $a=(x_{i},m_{i})_{i=1}^{I}$. The set of feasible allocations is $F=\{(x_{i},m_{i})_{i=1}^{I}|(x_{i},m_{i})\in\mathbb{R}_{+}\times\mathbb{R}\forall i\in\{1,\ldots,I\},\sum_{i=1}^{I}x_{i}\leq c_{x},\sum_{i=1}^{I}m_{i}\leq c_{m}\}$. Utility of $i\in\{1,\ldots,I\}$ from $a\in F$ is $u_{i}(a)=m_{i}+\phi_{i}(x_{i})$, where $\phi_{i}$ is strictly increasing.

Definition of PE allocation is standard: $a\in F$ is PE if $\nexists a'\in F$ such that $u_{i}(a')\geq u_{i}(a)$ for all $i$ and $u_{i}(a')>u_{i}(a)$ for some $i$.

Now I claim that if $a$ is PE then $a$ is a solution to $\displaystyle\max_{a\in F}\sum_{i=1}^{I}\phi_{i}(x_{i})$, or, making the maximization with respect to $x_{i}$s explicit, $\displaystyle\max_{(x_{i})_{i=1}^{I}\in\mathbb{R}_{+}^{I}}\sum_{i=1}^{I}\phi_{i}(x_{i})$ s.t. $\sum_{i=1}^{I}x_{i}\leq c_{x}$.

I am not going to prove the claim here, but the key idea is simple and is as follows. Suppose $a^{*}$ is PE but does not solve the maximization problem. Then we can find another feasible $a'$ such that $\sum_{i=1}^{I}\phi_{i}(x_{i}')>\sum_{i=1}^{I}\phi_{i}(x_{i}^{*})$. True, in $a'$, relative to $a^{*}$, come agents are worse off, but we can use money, $m_{i}$s, to make them equally well of as under $a^{*}$, and still be left with some money since we increased the sum of utility coming from $x_{i}$s.

Another way to say this is that the sum of utility from $a\in F$ is $\sum_{i=1}^{I}m_{i}+\sum_{i=1}^{I}\phi_{i}(x_{i})$. Now any non-wasteful allocation $a\in F$ will have the first term identical.

Yet another way to think about this is that $x_{i}$s determine the size of the pie and money, $m_{i}$s, determine redistribution. By quasi-linearity, decreasing $m_{i}$ by one unit and increasing $m_{j}$ by one unit leaves leaves $m_{i}+m_{j}$ unchanged. This is not true for $x_{i}$ and $x_{j}$.

This also implies that any $a\in F$ that solves the maximization problem is PE.