<p>Given a pure-exchange economy:</p>
<ul>
<li><span class="math-container">$u_1(x_1,y_1) = \min(3x_1+y_1,x_1+3y_1)$</span>, <span class="math-container">$u_2(x_2,y_2)= x_2^{0.5}y_2^{0.5}$</span></li>
<li>Total Endowments of X and Y are <span class="math-container">$\omega^X &gt; 0$</span>, and <span class="math-container">$\omega^Y &gt; 0$</span>, respectively.</li>
</ul>
<p>Set of feasible allocations is
<span class="math-container">$\mathcal{F} = \{((x_1,y_1),(x_2,y_2))\in\mathbb{R}^2_+\times\mathbb{R}^2_+|x_1+x_2 = \omega^X, y_1+y_2 = \omega^Y\}$</span></p>
<p>Set of Pareto efficient allocations is given by</p>
<p><span class="math-container">$\begin{cases} \{((x_1,y_1),(x_2,y_2))\in\mathcal{F}|y_1=\min(x_1, \omega^Y - \frac{1}{3}(\omega^X-x_1))\} &amp; \text{if } \omega^Y &lt; \omega^X \leq 3\omega^Y \\ \{((x_1,y_1),(x_2,y_2))\in\mathcal{F}|y_1=x_1\} &amp; \text{if } \omega^Y = \omega^X \\ \{((x_1,y_1),(x_2,y_2))\in\mathcal{F}|y_1=\max(0, \omega^Y - \frac{1}{3}(\omega^X-x_1))\} &amp; \text{if } \omega^X &gt; 3\omega^Y \\ \{((x_1,y_1),(x_2,y_2))\in\mathcal{F}|(y_2= 3x_2) \ \vee (x_1=0 \ \wedge \ y_2 &gt; 3\omega^X) \} &amp; \text{if } \omega^X &lt; \frac{1}{3}\omega^Y \\ \{((x_1,y_1),(x_2,y_2))\in\mathcal{F}|y_1=\max(x_1, \omega^Y - 3(\omega^X-x_1))\} &amp; \text{if } \frac{1}{3}\omega^Y \leq \omega^X &lt; \omega^Y \end{cases}  $</span></p>
<p>Here are some worked-out examples covering each of the five cases:
<a href="https://i.sstatic.net/bBpun.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/bBpun.png" alt="enter image description here" /></a></p>
<p><a href="https://i.sstatic.net/snvva.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/snvva.png" alt="enter image description here" /></a></p>

CC BY-SA 4.0
2023-04-12T09:38:46.710Z
2023-04-13T06:01:17.350Z
2023-04-13T06:01:17.350Z
Amit
<p>Suppose a pure exchange economy where agents’ (<span class="math-container">$A$</span> and <span class="math-container">$B$</span>) preferences are given by the following utility functions:</p>
<p><span class="math-container">$u_A = \min(3x+y,x+3y)$</span></p>
<p><span class="math-container">$u_B = x^\frac{1}{2} y^\frac{1}{2}$</span></p>
<p>Find the contract curve.</p>
<p>The arguments of the min operator are equal <span class="math-container">$\iff x_A = y_A$</span></p>
<p>Now I split the Edgeworth box into <span class="math-container">$3$</span> regions.</p>
<ul>
<li><span class="math-container">$x_A &lt; y_A$</span></li>
</ul>
<p>Here both utilities are differentiable and one of them is Cobb-Douglas, so I use <span class="math-container">$MRS$</span></p>
<p>In this region, <span class="math-container">$u_A = 3x + y$</span> as this is the smallest argument of the <span class="math-container">$\min$</span> in this region.</p>
<p><span class="math-container">$MRS_A = MRS_B  \iff 3 = \frac{y_B}{x_B}$</span></p>
<p>Therefore, the segment of the contract curve in this region is</p>
<p><span class="math-container">$y_B = 3 x_B$</span></p>
<ul>
<li><span class="math-container">$x_A &gt; y_A$</span></li>
</ul>
<p>Similarly as above, in this region, <span class="math-container">$u_A = x + 3y$</span>, and</p>
<p><span class="math-container">$MRS_A = MRS_B \iff \frac{1}{3} = \frac{y_B}{x_B}$</span></p>
<p>Therefore, the segment of the contract curve in this region is</p>
<p><span class="math-container">$y_B = \frac{1}{3} x_B$</span></p>
<ul>
<li><span class="math-container">$x_A = y_A$</span></li>
</ul>
<p>This is the line where <span class="math-container">$u_A$</span> is not differentiable and it’s the set of kinks of <span class="math-container">$A$</span>’s indifference curves.</p>
<p>Let <span class="math-container">$(\alpha,\beta)$</span> be the total endowments of <span class="math-container">$(x,y)$</span>, respectively.</p>
<p>Since <span class="math-container">$x_A = y_A$</span>,</p>
<p><span class="math-container">$MRS_B = \frac{y_B}{x_B} = \frac{\beta - y_A}{\alpha - x_A} = \frac{\beta - x_A}{\alpha - x_A} \in (\frac{1}{3},3) $</span></p>
<p><span class="math-container">$\iff \frac{1}{3} \alpha - \frac{1}{3} x_A &lt; \beta - x_A &lt; 3 \alpha - 3 x_A $</span></p>
<p><span class="math-container">$\iff \frac{1}{3} \alpha + \frac{2}{3} x_A &lt; \beta &lt; 3 \alpha - 2 x_A$</span></p>
<p>Graphically we can see that if the slope of the Cobb-Douglas IC’s tangent is in the set <span class="math-container">$(\frac{1}{3},3)$</span>, then this tangent is below <span class="math-container">$A$</span>’s IC and hence <span class="math-container">$B$</span>’s IC would intersect with <span class="math-container">$A$</span>’s IC at exactly one point, namely, the kink.</p>
<p>We can see in the graph below, that <span class="math-container">$MRS_B \in (\frac{1}{3},3)$</span> implies that if we deviate from the kink along <span class="math-container">$A$</span>’s IC, we would end up in a worse Cobb-Douglas IC.</p>
<p>This would imply that the kink is a Pareto efficient point.</p>
<p><a href="https://i.sstatic.net/pZ8Ht.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/pZ8Ht.png" alt="enter image description here" /></a></p>
<p>If <span class="math-container">$MRS_B$</span> happened to be either of the endpoints of the interval, then the Cobb-Douglas IC’s tangent would be one of <span class="math-container">$A$</span>’s IC line segments, and <span class="math-container">$B$</span>’s IC would still intersect <span class="math-container">$A$</span>’s IC exactly at the kink, implying the kink is still a Pareto efficient point.</p>
<p>On the other hand, if <span class="math-container">$MRS_B &lt; \frac{1}{3}$</span> or <span class="math-container">$MRS_B &gt; 3$</span>, then the Cobb-Douglas IC’s tangent goes above <span class="math-container">$A$</span>’s IC in some region, implying both ICs now intersect at a second point.</p>
<p>Taking any non-trivial convex combination of those two points would yield a point on the same IC curve for <span class="math-container">$A$</span> that lies on a strictly better Cobb-Douglas IC, as seen in the graph below.</p>
<p>This would imply that the kink is not a Pareto efficient point.</p>
<p><a href="https://i.sstatic.net/Km9r9.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/Km9r9.jpg" alt="enter image description here" /></a></p>
<p>————————————————————————</p>
<p>I also found conditions for both regions where <span class="math-container">$x_A \neq y_A$</span> in terms of the total endowments <span class="math-container">$(\alpha,\beta)$</span>.</p>
<p>Note <span class="math-container">$MRS_B = 1 \in (\frac{1}{3},3)$</span> along <span class="math-container">$x_A = y_A$</span> for symmetric endowments <span class="math-container">$(\alpha = \beta)$</span>.</p>
<ul>
<li>For <span class="math-container">$x_A &lt; y_A$</span></li>
</ul>
<p>Rewriting the contract curve in terms of agent <span class="math-container">$A$</span>,</p>
<p><span class="math-container">$y_A = 3 x_A + \beta - 3 \alpha$</span></p>
<p>I got <span class="math-container">$x_A &lt; y_A \iff x_A &gt; \frac{3}{2} \alpha - \frac{1}{2} \beta$</span>.</p>
<p>Note this inequality is impossible for symmetric endowments <span class="math-container">$(\alpha = \beta)$</span> as it would become <span class="math-container">$x_A &gt; \alpha$</span>, i.e. agent <span class="math-container">$A$</span> consuming more <span class="math-container">$x$</span> than the total endowment.</p>
<ul>
<li><span class="math-container">$x_A &gt; y_A$</span></li>
</ul>
<p>Rewriting the contract curve in terms of agent <span class="math-container">$A$</span>,</p>
<p><span class="math-container">$y_A = \frac{1}{3} x_A + \beta - \frac{1}{3} \alpha$</span></p>
<p>I got <span class="math-container">$x_A &gt; y_A \iff x_A &gt; \frac{3}{2} \beta - \frac{1}{2} \alpha$</span>.</p>
<p>Note this inequality is impossible for symmetric endowments <span class="math-container">$(\alpha = \beta)$</span> as it would become <span class="math-container">$x_A &gt; \alpha$</span>, i.e. agent <span class="math-container">$A$</span> consuming more <span class="math-container">$x$</span> than the total endowment.</p>
<p>————————————————————————</p>
<p>By the notes above, I got that for symmetric endowments, the contract curve is simply <span class="math-container">$x_A = y_A$</span>, which would be the line segment connecting both origins of the Edgeworth box.</p>
<p>However, I am having trouble visualizing all the conditions involving the endowments at the same time (for non-symmetric endowments), to get a final answer for the contract curve I can actually graph. I would appreciate any help on this part.</p>

<general-equilibrium><pareto-efficiency><leontief><edgeworth-box><perfect-substitutes>
How to find the contract curve for a funky utility involving the min operator?
Nicolas Torres
# How to find the contract curve for a funky utility involving the min operator?

(Asked by: username_0 on 2023-04-12)

Suppose a pure exchange economy where agents’ ($A$ and $B$) preferences are given by the following utility functions:

$u\_A = \min(3x+y,x+3y)$

$u\_B = x^\frac{1}{2} y^\frac{1}{2}$

Find the contract curve.

The arguments of the min operator are equal $\iff x\_A = y\_A$

Now I split the Edgeworth box into $3$ regions.

* $x\_A < y\_A$

Here both utilities are differentiable and one of them is Cobb-Douglas, so I use $MRS$

In this region, $u\_A = 3x + y$ as this is the smallest argument of the $\min$ in this region.

$MRS\_A = MRS\_B \iff 3 = \frac{y\_B}{x\_B}$

Therefore, the segment of the contract curve in this region is

$y\_B = 3 x\_B$

* $x\_A > y\_A$

Similarly as above, in this region, $u\_A = x + 3y$, and

$MRS\_A = MRS\_B \iff \frac{1}{3} = \frac{y\_B}{x\_B}$

Therefore, the segment of the contract curve in this region is

$y\_B = \frac{1}{3} x\_B$

* $x\_A = y\_A$

This is the line where $u\_A$ is not differentiable and it’s the set of kinks of $A$’s indifference curves.

Let $(\alpha,\beta)$ be the total endowments of $(x,y)$, respectively.

Since $x\_A = y\_A$,

$MRS\_B = \frac{y\_B}{x\_B} = \frac{\beta - y\_A}{\alpha - x\_A} = \frac{\beta - x\_A}{\alpha - x\_A} \in (\frac{1}{3},3) $

$\iff \frac{1}{3} \alpha - \frac{1}{3} x\_A < \beta - x\_A < 3 \alpha - 3 x\_A $

$\iff \frac{1}{3} \alpha + \frac{2}{3} x\_A < \beta < 3 \alpha - 2 x\_A$

Graphically we can see that if the slope of the Cobb-Douglas IC’s tangent is in the set $(\frac{1}{3},3)$, then this tangent is below $A$’s IC and hence $B$’s IC would intersect with $A$’s IC at exactly one point, namely, the kink.

We can see in the graph below, that $MRS\_B \in (\frac{1}{3},3)$ implies that if we deviate from the kink along $A$’s IC, we would end up in a worse Cobb-Douglas IC.

This would imply that the kink is a Pareto efficient point.

[![enter image description here](https://i.sstatic.net/pZ8Ht.png)](https://i.sstatic.net/pZ8Ht.png)

If $MRS\_B$ happened to be either of the endpoints of the interval, then the Cobb-Douglas IC’s tangent would be one of $A$’s IC line segments, and $B$’s IC would still intersect $A$’s IC exactly at the kink, implying the kink is still a Pareto efficient point.

On the other hand, if $MRS\_B < \frac{1}{3}$ or $MRS\_B > 3$, then the Cobb-Douglas IC’s tangent goes above $A$’s IC in some region, implying both ICs now intersect at a second point.

Taking any non-trivial convex combination of those two points would yield a point on the same IC curve for $A$ that lies on a strictly better Cobb-Douglas IC, as seen in the graph below.

This would imply that the kink is not a Pareto efficient point.

[![enter image description here](https://i.sstatic.net/Km9r9.jpg)](https://i.sstatic.net/Km9r9.jpg)

————————————————————————

I also found conditions for both regions where $x\_A \neq y\_A$ in terms of the total endowments $(\alpha,\beta)$.

Note $MRS\_B = 1 \in (\frac{1}{3},3)$ along $x\_A = y\_A$ for symmetric endowments $(\alpha = \beta)$.

* For $x\_A < y\_A$

Rewriting the contract curve in terms of agent $A$,

$y\_A = 3 x\_A + \beta - 3 \alpha$

I got $x\_A < y\_A \iff x\_A > \frac{3}{2} \alpha - \frac{1}{2} \beta$.

Note this inequality is impossible for symmetric endowments $(\alpha = \beta)$ as it would become $x\_A > \alpha$, i.e. agent $A$ consuming more $x$ than the total endowment.

* $x\_A > y\_A$

Rewriting the contract curve in terms of agent $A$,

$y\_A = \frac{1}{3} x\_A + \beta - \frac{1}{3} \alpha$

I got $x\_A > y\_A \iff x\_A > \frac{3}{2} \beta - \frac{1}{2} \alpha$.

Note this inequality is impossible for symmetric endowments $(\alpha = \beta)$ as it would become $x\_A > \alpha$, i.e. agent $A$ consuming more $x$ than the total endowment.

————————————————————————

By the notes above, I got that for symmetric endowments, the contract curve is simply $x\_A = y\_A$, which would be the line segment connecting both origins of the Edgeworth box.

However, I am having trouble visualizing all the conditions involving the endowments at the same time (for non-symmetric endowments), to get a final answer for the contract curve I can actually graph. I would appreciate any help on this part.

**Tags:** general-equilibrium, pareto-efficiency, leontief, edgeworth-box, perfect-substitutes

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## 1 answers

### Answer by username_1 on 2023-04-12. Score: 3 | Accepted answer

Given a pure-exchange economy:

* $u\_1(x\_1,y\_1) = \min(3x\_1+y\_1,x\_1+3y\_1)$, $u\_2(x\_2,y\_2)= x\_2^{0.5}y\_2^{0.5}$
* Total Endowments of X and Y are $\omega^X > 0$, and $\omega^Y > 0$, respectively.

Set of feasible allocations is
$\mathcal{F} = \{((x\_1,y\_1),(x\_2,y\_2))\in\mathbb{R}^2\_+\times\mathbb{R}^2\_+|x\_1+x\_2 = \omega^X, y\_1+y\_2 = \omega^Y\}$

Set of Pareto efficient allocations is given by

$\begin{cases} \{((x\_1,y\_1),(x\_2,y\_2))\in\mathcal{F}|y\_1=\min(x\_1, \omega^Y - \frac{1}{3}(\omega^X-x\_1))\} & \text{if } \omega^Y < \omega^X \leq 3\omega^Y \\ \{((x\_1,y\_1),(x\_2,y\_2))\in\mathcal{F}|y\_1=x\_1\} & \text{if } \omega^Y = \omega^X \\ \{((x\_1,y\_1),(x\_2,y\_2))\in\mathcal{F}|y\_1=\max(0, \omega^Y - \frac{1}{3}(\omega^X-x\_1))\} & \text{if } \omega^X > 3\omega^Y \\ \{((x\_1,y\_1),(x\_2,y\_2))\in\mathcal{F}|(y\_2= 3x\_2) \ \vee (x\_1=0 \ \wedge \ y\_2 > 3\omega^X) \} & \text{if } \omega^X < \frac{1}{3}\omega^Y \\ \{((x\_1,y\_1),(x\_2,y\_2))\in\mathcal{F}|y\_1=\max(x\_1, \omega^Y - 3(\omega^X-x\_1))\} & \text{if } \frac{1}{3}\omega^Y \leq \omega^X < \omega^Y \end{cases} $

Here are some worked-out examples covering each of the five cases:
[![enter image description here](https://i.sstatic.net/bBpun.png)](https://i.sstatic.net/bBpun.png)

[![enter image description here](https://i.sstatic.net/snvva.png)](https://i.sstatic.net/snvva.png)


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