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!

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    $\begingroup$ Are you sure that $m^i$ under $\hat x^i$ is the same as $m^i$ under $x^i$? A budget/resource constraint seems to be missing. And with that, you should be able to get what you want by summing inequalities in (3) over $i$. $\endgroup$ – Herr K. Mar 9 '16 at 18:04
  • $\begingroup$ @HerrK. That's an excellent point and a rather embarassing blunder by me, I will change that $\endgroup$ – DornerA Mar 10 '16 at 14:24
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    $\begingroup$ Are there any properties for the function of X? For example, if it is strictly increasing but concave then the PO allocation where one agent takes the total endowment should yield less total utility that splitting that allocation evenly between two agents. $\endgroup$ – 123 Jun 8 '16 at 19:59
  • $\begingroup$ @123 there is no other assumptions made about $\phi^i(\frac{}{})$ than the ones listed above unfortunately $\endgroup$ – DornerA Jun 8 '16 at 20:46

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

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 $\phi(\cdot)$)

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.


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    $\begingroup$ I see that the OP accepted this answer but this doesn't prove his actual proposition. OP claims that any PO allocation solves the given maximization problem. This proof shows that a solution to the maximization problem is PO. However, this result follows immediately from the fact that the utility function makes clear that preferences satisfy local non-satiation. And we know that there doesn't necessarily exist a bijection between CE and PO points The original proposition is likely false, depending on the restrictions placed on the function of X. (Using phone so hard to use LaTex - sorry.) $\endgroup$ – 123 Jun 11 '16 at 10:56
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    $\begingroup$ I don't think the proposition is true in standard pure exchange economy environments. Here is the counter example: economics.stackexchange.com/a/15146/11824 $\endgroup$ – Amit Jan 24 '17 at 7:35
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    $\begingroup$ @Amit I think you are right. However the statement seems to hold with the added condition that the PO allocation $(x,m)$ is such that for all consumers $i$: $m_i>0$. Or alternatively if the problem allows for negative values for $m_i$. In this case your counterexample would not be PO. $\endgroup$ – Giskard Jan 24 '17 at 9:18
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    $\begingroup$ @KitsuneCavalry Here lies the mistake: "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..." or you can't make the trade because it would violate a non-negativity constraint. Boo, swindler! :D Give back the 50 points :D $\endgroup$ – Giskard Jan 24 '17 at 9:24
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    $\begingroup$ @denesp I agree that the result holds if we either allow $m_i$ to be any real number, or only strictly positive real number, for all $i$. $\endgroup$ – Amit Jan 24 '17 at 9:29

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.

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

  • $\begingroup$ @DornerA your thoughts on this? $\endgroup$ – Giskard Jan 24 '17 at 9:15

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.

  • $\begingroup$ Have you read the other two answers? One basically states the same. The other provides a counterexample. $\endgroup$ – Giskard Jan 24 '17 at 21:16
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    $\begingroup$ @denesp Yes I read the answers and I am saying different thing. The two answers are talking about maximization of sum of utilities, I am talking about maximization of sum from the $x_{i}$s. In the counter-example, the critical assumption is that $m_{i}\geq0$ $\forall i\in\{1,2\}$. If $m_{i}\in\mathbb{R}$ for $i\in\{1,2\}$, then what I am saying applies. Which assumption is 'standard' is arguable. I've been brought up by MWG. $\endgroup$ – Jan Jan 24 '17 at 21:25
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    $\begingroup$ One more comment, Mas-Colell, Whinston, Green chapter 10, especially parts C and even more especially part D, are good textbook treatment of the issue OP asks about. $\endgroup$ – Jan Jan 24 '17 at 21:30

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