<p>If you are happy to restrict attention to the interior, then in a two-good-two-agent economy - given differentiable, strictly concave, and increasing utility functions - the Pareto curve must be increasing. Then, <span class="math-container">$\textit{a fortiori}$</span>, the contract curve is also increasing.</p>
<p>To see the Pareto set is representable by an increasing function in this case, first recall that given differentiability, strict concavity, and monotonicity, an interior allocation is Paretian if and only if <span class="math-container">$MRS_1(x_{11},x_{12})=MRS_2(x_{21},x_{22})$</span>. Therefore, given an economy endowment of <span class="math-container">$(\omega_1,\omega_2)$</span>, the Pareto set is implicitly defined by the following relation (preferences are locally non-satiated so the endowment constraints bind), <span class="math-container">$$P(x_{11},x_{12}):=MRS_1(x_{11},x_{12})-MRS_2(\omega_1-x_{11},\omega_2-x_{12})=0$$</span></p>
<p>Applying the implicit function theorem, we obtain <span class="math-container">$$\frac{d x_{12}}{d x_{11}}=-\frac{\partial_{x_{11}}P}{\partial_{x_{12}}P}=-\frac{\partial_{x_{11}}MRS_1+\partial_{x_{21}}MRS_2}{\partial_{x_{12}}MRS_1+\partial_{x_{22}}MRS_2}$$</span></p>
<p>Given strict concavity (which implies decreasing <span class="math-container">$MRS$</span>), we have <span class="math-container">$$\begin{align}\partial_{x_{11}}MRS_1&lt;0 \\ \partial_{x_{21}}MRS_2&lt;0 \\ \partial_{x_{12}}MRS_1&gt;0 \\ \partial_{x_{22}}MRS_2&gt;0 \end{align}$$</span> Therefore, over the Pareto set <span class="math-container">$$\frac{d x_{12}}{d x_{11}}&gt;0$$</span> Whence, as the contract curve is a subset of the Pareto set, it is also an increasing curve in the Edgeworth box. Additionally, since the curve is strictly increasing, this also proves the Pareto set relation cannot be many-to-one.</p>
<p>As for your question about whether the Pareto set relation can be one-to-many, the implicit function theorem proves the existence of a unique <span class="math-container">$\textbf{function}$</span> (so not a one-to-many relation) characterising the Pareto set. More explicitly, decreasing <span class="math-container">$MRS$</span> removes one-to-many behaviour as a possibility. Suppose there is some <span class="math-container">$(\tilde{x}_{11},\tilde{x}_{12})$</span> which is Paretian. Fix some <span class="math-container">$\hat{x}_{12}&gt; \tilde{x}_{12}$</span> which is feasible. By decreasing <span class="math-container">$MRS$</span>, <span class="math-container">$$ \begin{align} MRS_1(\tilde{x}_{11},\hat{x}_{12})&amp;&gt;MRS_1(\tilde{x}_{11},\tilde{x}_{12}) \\
MRS_2(\omega_1-\tilde{x}_{11},\omega_2-\hat{x}_{12})&amp;&lt;MRS_2(\omega_1-\tilde{x}_{11},\omega_2-\hat{x}_{12})
\end{align}$$</span> So then <span class="math-container">$P(\tilde{x}_{11},\hat{x}_{12})&gt;P(\tilde{x}_{11},\tilde{x}_{12})=0$</span> meaning <span class="math-container">$(\tilde{x}_{11},\hat{x}_{12})$</span> cannot be Paretian as <span class="math-container">$MRS_1&gt;MRS_2$</span>. An analogous argument works, <span class="math-container">$\textit{mutatis mutandis}$</span>, to show no <span class="math-container">$\hat{x}_{12}&lt;\tilde{x}_{12}$</span> can result in a Pareto allocation either. Hence, the Pareto set correspondence cannot be one-to-many.</p>

CC BY-SA 4.0
2024-06-28T17:25:45.750Z
I see! My intuition is: the slope / direction of the pareto set has to be perpendicular to the tangent of the indifference curves. Since the slope of the tangent has to negative, the pareto set has to be increasing.
Vibius
CC BY-SA 4.0
2024-06-28T08:35:49.510Z
2024-06-28T08:35:49.510Z
Joseph Basford
<p>In the edgeworth box model, is the pareto set / contract curve necessarily shaped like a monotonically increasing function? This seems to be stated / implied in various places (such as <a href="https://en.wikipedia.org/wiki/Contract_curve" rel="nofollow noreferrer">Wikipedia</a>), but I can't seem to find a rigorous argument / proof for it.</p>
<p>Given the standard caboodle of consumer preference assumptions (completeness, transitivity, continuity, non-satiation, diminishing marginal rate of substitution), is it possible to get, say, both <span class="math-container">$(4, 3)$</span> and <span class="math-container">$(5, 1)$</span> in the pareto set, like in my hand-drawn &quot;counterexample&quot; below? Can the pareto-set relation be one-to-many or many-to-one (perhaps barring corner solutions)?</p>
<p>(Let's assume the marginal rate of substitution is strictly diminishing. <em>I.e.</em>, no straight-line indifference curves)</p>
<p>Thanks!</p>
<p><a href="https://i.sstatic.net/UmftsXTE.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/UmftsXTE.png" alt="Hand-drawn &quot;counterexample&quot;" /></a></p>

<consumer-theory><proof><edgeworth-box>
Shape of the contract "curve"
Vibius
# Shape of the contract "curve"

(Asked by: username_0 on 2024-06-27)

In the edgeworth box model, is the pareto set / contract curve necessarily shaped like a monotonically increasing function? This seems to be stated / implied in various places (such as [Wikipedia](https://en.wikipedia.org/wiki/Contract_curve)), but I can't seem to find a rigorous argument / proof for it.

Given the standard caboodle of consumer preference assumptions (completeness, transitivity, continuity, non-satiation, diminishing marginal rate of substitution), is it possible to get, say, both $(4, 3)$ and $(5, 1)$ in the pareto set, like in my hand-drawn "counterexample" below? Can the pareto-set relation be one-to-many or many-to-one (perhaps barring corner solutions)?

(Let's assume the marginal rate of substitution is strictly diminishing. *I.e.*, no straight-line indifference curves)

Thanks!

[![Hand-drawn ](https://i.sstatic.net/UmftsXTE.png)](https://i.sstatic.net/UmftsXTE.png)

**Tags:** consumer-theory, proof, edgeworth-box

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## 1 answers

### Answer by username_1 on 2024-06-28. Score: 2 | Accepted answer

If you are happy to restrict attention to the interior, then in a two-good-two-agent economy - given differentiable, strictly concave, and increasing utility functions - the Pareto curve must be increasing. Then, $\textit{a fortiori}$, the contract curve is also increasing.

To see the Pareto set is representable by an increasing function in this case, first recall that given differentiability, strict concavity, and monotonicity, an interior allocation is Paretian if and only if $MRS\_1(x\_{11},x\_{12})=MRS\_2(x\_{21},x\_{22})$. Therefore, given an economy endowment of $(\omega\_1,\omega\_2)$, the Pareto set is implicitly defined by the following relation (preferences are locally non-satiated so the endowment constraints bind), $$P(x\_{11},x\_{12}):=MRS\_1(x\_{11},x\_{12})-MRS\_2(\omega\_1-x\_{11},\omega\_2-x\_{12})=0$$

Applying the implicit function theorem, we obtain $$\frac{d x\_{12}}{d x\_{11}}=-\frac{\partial\_{x\_{11}}P}{\partial\_{x\_{12}}P}=-\frac{\partial\_{x\_{11}}MRS\_1+\partial\_{x\_{21}}MRS\_2}{\partial\_{x\_{12}}MRS\_1+\partial\_{x\_{22}}MRS\_2}$$

Given strict concavity (which implies decreasing $MRS$), we have $$\begin{align}\partial\_{x\_{11}}MRS\_1<0 \\ \partial\_{x\_{21}}MRS\_2<0 \\ \partial\_{x\_{12}}MRS\_1>0 \\ \partial\_{x\_{22}}MRS\_2>0 \end{align}$$ Therefore, over the Pareto set $$\frac{d x\_{12}}{d x\_{11}}>0$$ Whence, as the contract curve is a subset of the Pareto set, it is also an increasing curve in the Edgeworth box. Additionally, since the curve is strictly increasing, this also proves the Pareto set relation cannot be many-to-one.

As for your question about whether the Pareto set relation can be one-to-many, the implicit function theorem proves the existence of a unique $\textbf{function}$ (so not a one-to-many relation) characterising the Pareto set. More explicitly, decreasing $MRS$ removes one-to-many behaviour as a possibility. Suppose there is some $(\tilde{x}\_{11},\tilde{x}\_{12})$ which is Paretian. Fix some $\hat{x}\_{12}> \tilde{x}\_{12}$ which is feasible. By decreasing $MRS$, $$ \begin{align} MRS\_1(\tilde{x}\_{11},\hat{x}\_{12})&>MRS\_1(\tilde{x}\_{11},\tilde{x}\_{12}) \\
MRS\_2(\omega\_1-\tilde{x}\_{11},\omega\_2-\hat{x}\_{12})& So then $P(\tilde{x}\_{11},\hat{x}\_{12})>P(\tilde{x}\_{11},\tilde{x}\_{12})=0$ meaning $(\tilde{x}\_{11},\hat{x}\_{12})$ cannot be Paretian as $MRS\_1>MRS\_2$. An analogous argument works, $\textit{mutatis mutandis}$, to show no $\hat{x}\_{12}<\tilde{x}\_{12}$ can result in a Pareto allocation either. Hence, the Pareto set correspondence cannot be one-to-many.

> **Comments:**
> **username_0**: I see! My intuition is: the slope / direction of the pareto set has to be perpendicular to the tangent of the indifference curves. Since the slope of the tangent has to negative, the pareto set has to be increasing.  

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