<p>Given a pure-exchange economy with</p>
<ul>
<li><span class="math-container">$u_A(x_A,y_A)=\min(x_A,2y_A)$</span>, <span class="math-container">$u_B(x_B,y_B)=\min(2x_B,y_B)$</span></li>
<li>Endowment of A is <span class="math-container">$(k_X,k_Y)$</span> and of B is <span class="math-container">$(12-k_X,12-k_Y)$</span></li>
</ul>
<p>Set of feasible allocations is <span class="math-container">$\mathcal{F} = \{((x_A,y_A),(x_B,y_B))\in\mathbb{R}^2_+\times\mathbb{R}^2_+|x_A+x_B=y_A+y_B=12\}$</span></p>
<p>Here is the Edgeworth box representation of feasible allocations and set of efficient allocations:
<a href="https://i.sstatic.net/mOKJH.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/mOKJH.png" alt="enter image description here" /></a></p>
<p>To determine the competitive equilibrium, we can consider the following cases for the endowments.</p>
<p><strong>Case 1</strong>: <span class="math-container">$k_X\leq 8, k_Y\geq 4, (k_X,k_Y)\neq (8,4)$</span></p>
<p>In this case, given <span class="math-container">$(k_X,k_Y)$</span>, there is a unique competitive equilibrium allocation <span class="math-container">$((x_A,y_A),(x_B,y_B))=((8,4),(4,8))$</span> supported by prices <span class="math-container">$(p_X,p_Y)=(k_Y-4,8-k_X)$</span>.</p>
<p><a href="https://i.sstatic.net/ywDlL.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/ywDlL.png" alt="enter image description here" /></a></p>
<p><strong>Case 2</strong>: <span class="math-container">$(k_X,k_Y)= (8,4)$</span></p>
<p>In this case, there is a unique competitive equilibrium allocation <span class="math-container">$((x_A,y_A),(x_B,y_B))=((8,4),(4,8))$</span> and it can be supported by any pair of prices from the set <span class="math-container">$\{(p_X,p_Y)\in\mathbb{R}^2_+|p_X+p_Y=1\}$</span>.</p>
<p><a href="https://i.sstatic.net/dueNW.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/dueNW.png" alt="enter image description here" /></a></p>
<p><strong>Case 3</strong>: <span class="math-container">$k_X\geq  8, k_Y \leq 4, (k_X,k_Y)\neq (8,4)$</span></p>
<p>In this case, given <span class="math-container">$(k_X,k_Y)$</span>, there are three sets of competitive equilibria:</p>
<p>(i) <span class="math-container">$((x_A,y_A),(x_B,y_B))=((8,4),(4,8))$</span> and it is supported by prices <span class="math-container">$(p_X,p_Y)=(4-k_Y,k_X-8)$</span>.</p>
<p>(ii) Allocations in the set <span class="math-container">$\{((x_A,y_A),(x_B,y_B))\in\mathcal{F}|y_A=k_Y, 2k_Y\leq x_A\leq \frac{k_Y+12}{2}\}$</span> are supported by the prices <span class="math-container">$(p_X,p_Y)=(0,1)$</span></p>
<p>(iii) Allocations in the set <span class="math-container">$\{((x_A,y_A),(x_B,y_B))\in\mathcal{F}|x_A=k_X, \frac{k_X}{2}\leq y_A\leq 2k_X-12\}$</span> are supported by the prices <span class="math-container">$(p_X,p_Y)=(1,0)$</span></p>
<p><a href="https://i.sstatic.net/BBt5k.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/BBt5k.png" alt="enter image description here" /></a></p>
<p><strong>Case 4</strong>: <span class="math-container">$k_X&gt;  8, k_Y &gt; 4$</span></p>
<p>In this case, given <span class="math-container">$(k_X,k_Y)$</span>, allocations in the set <span class="math-container">$\{((x_A,y_A),(x_B,y_B))\in\mathcal{F}|x_A=k_X, \frac{k_X}{2}\leq y_A\leq 2k_X-12\}$</span> are supported by the prices <span class="math-container">$(p_X,p_Y)=(1,0)$</span></p>
<p><strong>Case 5</strong>: <span class="math-container">$k_X&lt;  8, k_Y &lt; 4$</span></p>
<p>In this case, given <span class="math-container">$(k_X,k_Y)$</span>, allocations in the set <span class="math-container">$\{((x_A,y_A),(x_B,y_B))\in\mathcal{F}|y_A=k_Y, 2k_Y\leq x_A\leq \frac{k_Y+12}{2}\}$</span> are supported by the prices <span class="math-container">$(p_X,p_Y)=(0,1)$</span></p>

CC BY-SA 4.0
2024-02-12T17:45:39.993Z
Thanks for this answer, this clears things up a bit, but I still have two more questions:

In **Case 4**: Can't the competitive equilibrium also be ya=k2, xa=6+k2/2 supported by prices (px,py)=(0,1). At this point, the utility of both consumers would be maximized and the outcome would be Pareto efficient. 

Furthermore, in **Case 1** why can't there be 3 sets of equilibria as in **Case 3**, I'm not sure I understand what prevents the equilibrium from being for example : (xA,yA)=(kx, 0<yA<kx/2) supported by prices (px,py)=(1,0)
Jovan Jezdic
CC BY-SA 4.0
2024-11-02T04:17:37.237Z
In case 4, competitive equilibrium cannot be $(p_X,p_Y)=(0,1)$ because there will be excess demand for $X$, sum of demands for $X$ will exceed $12$.
Amit
CC BY-SA 4.0
2024-11-02T04:19:44.923Z
In case 1, at $(p_X,p_Y)=(0,1)$, there will be excess demand for $X$ and at $(p_X,p_Y)=(1,0)$, there will be excess demand for $Y$. $(x_A,y_A)=(k_X,y_A)$ where $y_A\in (0, k_X/2)$ is not a utility maximising choice for consumer $A$ at prices $(p_X,p_Y)=(1,0)$. In this case, Consumer $A$ will demand at least  $k_X/2$ units of $Y$.
Amit
CC BY-SA 4.0
2024-02-12T16:19:16.533Z
2024-02-12T17:06:35.823Z
2024-02-12T17:06:35.823Z
Amit
<p>LoL, this was the exam question I set... Here is the solution:</p>
<p>Consumers utility function is quasi-concave and satisfy LNS
as <span class="math-container">$u^{A}(y^{A})&gt;u^{A}(x^{A})$</span> for all <span class="math-container">$y^{A}&gt;&gt;x^{A}$</span>. Thus, the Marshallian
demands of consumers will necessarily satisfy the budget constraint at
equality. When <span class="math-container">$p&gt;&gt;0$</span>, a simple solution exists, namely
<span class="math-container">\begin{gather*}
x^{A}(p)=\left(  \frac{2m^{A}}{2p_{1}+p_{2}},\frac{m^{A}}{2p_{1}+p_{2}
}\right)  \text{ \ and \ }x^{B}(p)=\left(  \frac{m^{B}}{p_{1}+2p_{2}}
,\frac{2m^{B}}{p_{1}+2p_{2}}\right)  \text{.}
\end{gather*}</span>
The latter is a solution for <span class="math-container">$A$</span>, since for any other consumption bundle
<span class="math-container">$y^{A}$</span> such that <span class="math-container">$p_{1}y_{1}^{A}+p_{2}y_{2}^{A}=m^{A}$</span> and <span class="math-container">$\varepsilon&gt;0$</span>
sufficiently small,
<span class="math-container">\begin{gather*}
u^{A}(y_{1}^{A}-p_{2}\varepsilon/p_{1},y_{2}^{A}+\varepsilon)=2(y_{2}
^{A}+\varepsilon)&gt;2y_{2}^{A}=u^{A}(y^{A})\text{ \ when }y_{1}^{A}&gt;2y_{2}
^{A}\text{,}\\
u^{A}(y_{1}^{A}+\varepsilon,y_{2}^{A}-p_{1}\varepsilon/p_{2})=y_{1}
^{A}+\varepsilon&gt;y_{1}^{A}=u^{A}(y^{A})\text{ \ when }y_{1}^{A}&lt;2y_{2}
^{A}\text{.}
\end{gather*}</span>
The latter is a solution for <span class="math-container">$B$</span>, since for any other consumption bundle
<span class="math-container">$y^{b}$</span> such that <span class="math-container">$p_{1}y_{1}^{B}+p_{2}y_{2}^{B}=m^{B}$</span> and <span class="math-container">$\varepsilon&gt;0$</span>
sufficiently small,
<span class="math-container">\begin{gather*}
u^{B}(y_{1}^{B}-p_{2}\varepsilon/p_{1},y_{2}^{B}+\varepsilon)=y_{2}
^{B}+\varepsilon&gt;y_{2}^{B}=u^{B}(y^{B})\text{ \ when }2y_{1}^{B}&gt;y_{2}
^{B}\text{,}\\
u^{B}(y_{1}^{B}+\varepsilon,y_{2}^{B}-p_{1}\varepsilon/p_{2})=2(y_{1}
^{B}+\varepsilon)&gt;2y_{1}^{B}=u^{B}(y^{B})\text{ \ when }2y_{1}^{B}&lt;y_{2}
^{B}\text{.}
\end{gather*}</span>
When <span class="math-container">$p_{1}=0$</span> and <span class="math-container">$p_{2}&gt;0$</span>, consumers naturally spend all of their budget on
the expensive good and demands become correspondences, since they only care
about getting enough of the free good, but do not mind about purchasing more
of that. Therefore,
<span class="math-container">\begin{gather*}
x^{A}(p)=\left(  \left[  \frac{2m^{A}}{p_{2}},\infty\right)  ,\frac{m^{A}%
}{p_{2}}\right)  \text{ \ and \ }x^{B}(p)=\left(  \left[  \frac{m^{B}}{2p_{2}%
},\infty\right)  ,\frac{m^{B}}{p_{2}}\right)  \text{.}
\end{gather*}</span>
Similarly, when <span class="math-container">$p_{1}&gt;0$</span> and <span class="math-container">$p_{2}=0$</span>,
<span class="math-container">\begin{gather*}
x^{A}(p)=\left(  \frac{m^{A}}{p_{1}},\left[  \frac{m^{A}}{2p_{1}}%
,\infty\right)  \right)  \text{ \ and \ }x^{B}(p)=\left(  \frac{m^{B}}{p_{1}%
},\left[  \frac{2m^{B}}{p_{1}},\infty\right)  \right)  \text{.}%
\end{gather*}</span>
Of course if both prices were equal to <span class="math-container">$0$</span>, demand would be unbounded for any
one of the two goods.</p>
<p>To find CEs, begin by focusing on CEs in which both prices are strictly
positive. By Walras Law we know than in these CE, <span class="math-container">$z_{1}(p)=z_{2}(p)=0$</span>.
Further, we can normalize any one price to <span class="math-container">$1$</span>. So set <span class="math-container">$p_{1}=1$</span> and clear the
market for good <span class="math-container">$1$</span>
<span class="math-container">\begin{gather*}
x_{1}^{A}(p)+x_{1}^{B}(p)=\frac{2\left(  k_{1}+p_{2}k_{2}\right)  }{2+p_{2}
}+\frac{\left(  12-k_{1}+p_{2}(12-k_{2})\right)  }{1+2p_{2}}=12\text{.}
\end{gather*}</span>
The latter yields <span class="math-container">$p_{2}=\frac{k_{1}-8}{4-k_{2}}$</span>. In order for <span class="math-container">$p_{2}&gt;0$</span>,
this requires that either <span class="math-container">$k_{1}&gt;8$</span> and <span class="math-container">$4&gt;k_{2}$</span>\ or\ <span class="math-container">$k_{1}&lt;8$</span> and <span class="math-container">$4&lt;k_{2}
$</span>. So in all these circumstances this competitive equilibrium exists.</p>
<p>Next, look for CE in which <span class="math-container">$p_{1}=0$</span> and <span class="math-container">$p_{2}&gt;0$</span>. By Walras Law we know than
in these CE, <span class="math-container">$z_{2}(p)=0$</span>. Further, we can normalize <span class="math-container">$p_{2}$</span> to <span class="math-container">$1$</span>. So set
<span class="math-container">$p_{2}=1$</span> and clear the market for good <span class="math-container">$2$</span>,
<span class="math-container">\begin{gather*}
x_{2}^{A}(p)+x_{2}^{B}(p)=m^{A}+m^{B}=12\text{.}
\end{gather*}</span>
Then, verify that there is excess supply in the market for the free good. Note
that
<span class="math-container">\begin{gather*}
x_{1}^{A}(p)+x_{1}^{B}(p)\geq2m^{A}+\frac{m^{B}}{2}=2k_{2}+\frac{12-k_{2}}
{2}=6+\frac{3k_{2}}{2}\text{.}
\end{gather*}</span>
For there to be excess supply then <span class="math-container">$6+\frac{3k_{2}}{2}\leq12$</span> ore equivalently
<span class="math-container">$k_{2}\leq4$</span>. So in all these circumstances this competitive equilibrium exists.</p>
<p>Finally, look for CE in which <span class="math-container">$p_{1}&gt;0$</span> and <span class="math-container">$p_{2}=0$</span>. By Walras Law we know
than in these CE, <span class="math-container">$z_{1}(p)=0$</span>. Further, we can normalize <span class="math-container">$p_{1}$</span> to <span class="math-container">$1$</span>. So
set <span class="math-container">$p_{1}=1$</span> and clear the market for good <span class="math-container">$1$</span>,
<span class="math-container">\begin{gather*}
x_{1}^{A}(p)+x_{1}^{B}(p)=m^{A}+m^{B}=12\text{.}
\end{gather*}</span>
Then, verify that there is excess supply in the market for the free good. Note
that
<span class="math-container">\begin{gather*}
x_{2}^{A}(p)+x_{2}^{B}(p)\geq\frac{m^{A}}{2}+2m^{B}=\frac{k_{1}}{2}
+24-2k_{1}=24-\frac{3k_{1}}{2}\text{.}
\end{gather*}</span>
For there to be excess supply then <span class="math-container">$24-\frac{3k_{1}}{2}\leq12$</span> ore
equivalently <span class="math-container">$k_{1}\geq8$</span>. So in all these circumstances this competitive
equilibrium exists.</p>
<p>An allocation <span class="math-container">$x=(x_{1}^{A},x_{2}^{A},x_{1}^{B},x_{2}^{B})\in R_{+}^{4}$</span> is
Pareto efficient if and only if it is feasible and there is no other feasible
allocation <span class="math-container">$\hat{x}$</span> such that: (a) <span class="math-container">$U^{j}(\hat{x}^{j})\geq U^{j}(x^{j})$</span> for
any <span class="math-container">$j\in\left\{  A,B\right\}  $</span>; (b) <span class="math-container">$U^{j}(\hat{x}^{j})&gt;U^{j}(x^{j})$</span> for
some <span class="math-container">$j\in\left\{  A,B\right\}  $</span>. The contract curve identifies the set of
Pareto efficient allocations of an economy.
<span class="math-container">\begin{gather*}
x_{2}^{A}\in C_{2}^{A}(x_{1}^{A})=\left\{
\begin{array}
[c]{lll}%
\lbrack0,x_{1}^{A}/2] &amp; \text{if} &amp; x_{1}^{A}\in\lbrack0,6]\\
\lbrack2x_{1}^{A}-12,x_{1}^{A}/2] &amp; \text{if} &amp; x_{1}^{A}\in\lbrack6,8]\\
\lbrack x_{1}^{A}/2,2x_{1}^{A}-12] &amp; \text{if} &amp; x_{1}^{A}\in\lbrack8,12]
\end{array}
\right.
\end{gather*}</span>
To show the latter, note that for any point <span class="math-container">$x^{A}$</span> on the conjectured
contract curve, the intersection between the upper-contour sets of both
players has an empty interior. Formally, let <span class="math-container">$x^{A}$</span> satisfy <span class="math-container">$x_{2}^{A}%
=C_{2}^{A}(x_{1}^{A})$</span>, let
<span class="math-container">\begin{align*}
U_{+}^{A}(x^{A}) &amp;  =\left\{  x\in\mathbb{R}_{+}^{2}\text{ }|\text{ }%
\min\{x_{1},2x_{2}\}\geq\min\{x_{1}^{A},2x_{2}^{A}\}\right\}  \text{,}\\
U_{+}^{B}(x^{A}) &amp;  =\left\{  x\in\mathbb{R}_{+}^{2}\text{ }|\text{ }%
\min\{24-2x_{1},12-x_{2}\}\geq\min\{24-2x_{1}^{A},12-x_{2}^{A}\}\right\}
\text{,}%
\end{align*}</span>
and note that <span class="math-container">$int(U_{+}^{A}(x^{A})\cap U_{+}^{B}(x^{A}))=\emptyset$</span>. Thus, it
is impossible to benefit a player without hurting the other player. For
instance for <span class="math-container">$x_{1}^{A}\in\lbrack0,6]$</span> and <span class="math-container">$x_{2}^{A}\in\lbrack0,x_{1}^{A}%
/2]$</span>, these sets become
<span class="math-container">\begin{align*}
U_{+}^{A}(x^{A}) &amp;  =\left\{  x\in\mathbb{R}_{+}^{2}\text{ }|\text{ }%
\min\{x_{1},2x_{2}\}\geq2x_{2}^{A}\right\}  \text{,}\\
U_{+}^{B}(x^{A}) &amp;  =\left\{  x\in\mathbb{R}_{+}^{2}\text{ }|\text{ }%
\min\{24-2x_{1},12-x_{2}\}\geq12-x_{2}^{A}\right\}  \text{,}%
\end{align*}</span>
and we cannot have that <span class="math-container">$x_{2}&gt;x_{2}^{A}$</span> and <span class="math-container">$12-x_{2}&gt;12-x_{2}^{A}$</span> holding
at once. So, level curves are tangent and we cannot benefit a player without
hurting the other one. Finally, for any point <span class="math-container">$x^{A}$</span> which is not on the
conjectured contract curve, the intersection between the upper-contour sets of
both players has a non-empty interior. Thus, given that both players have
monotonic preferences, it is possible to benefit a player without hurting the
other one.</p>

CC BY-SA 4.0
2024-11-01T15:44:29.807Z
2024-11-01T15:44:29.807Z
user51592
<p>I am currently in a Microeconomics class and have come across the problem described below. I have tried to solve the problem algebraically but only get to the intercept (x1a,x2a)=(8,4). I know that this is not the only solution so I would appreciate any help!</p>
<p>Consider an exchange economy with two goods (1 and 2) and two consumers ( A and B). The preferences of the two consumers are:</p>
<p><a href="https://i.sstatic.net/YKKjt.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/YKKjt.png" alt="e" /></a></p>
<p>Consumer A is endowed with k1 of good 1 and k2 of good 2. Consumer B owns only 12-k1 of good 1 and 12-k2 of good 2. First, solve the utility maximization problem of each consumer. Then, find all competitive equilibria of this economy for any k1, k2 between 0 and 12. Finally, characterize the contract curve of this economy.</p>
<p><a href="https://i.sstatic.net/u6iFq.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/u6iFq.png" alt="enter image description here" /></a></p>

CC BY-SA 4.0
2024-02-12T06:27:17.273Z
"*I have tried to solve the problem algebraically but only get to the intercept*" Can you please edit the question so that it includes your calculations? It is not trivial to see where you get stuck.
Giskard
<utility><competitive-equilibrium><pure-exchange-economy><perfect-complements>
Finding the competitive equilibrium in an exchange economy with two perfect complements
Jovan Jezdic
# Finding the competitive equilibrium in an exchange economy with two perfect complements

(Asked by: username_0 on 2024-02-11)

I am currently in a Microeconomics class and have come across the problem described below. I have tried to solve the problem algebraically but only get to the intercept (x1a,x2a)=(8,4). I know that this is not the only solution so I would appreciate any help!

Consider an exchange economy with two goods (1 and 2) and two consumers ( A and B). The preferences of the two consumers are:

[![e](https://i.sstatic.net/YKKjt.png)](https://i.sstatic.net/YKKjt.png)

Consumer A is endowed with k1 of good 1 and k2 of good 2. Consumer B owns only 12-k1 of good 1 and 12-k2 of good 2. First, solve the utility maximization problem of each consumer. Then, find all competitive equilibria of this economy for any k1, k2 between 0 and 12. Finally, characterize the contract curve of this economy.

[![enter image description here](https://i.sstatic.net/u6iFq.png)](https://i.sstatic.net/u6iFq.png)

**Tags:** utility, competitive-equilibrium, pure-exchange-economy, perfect-complements
> **Comments:**
> **username_1**: "*I have tried to solve the problem algebraically but only get to the intercept*" Can you please edit the question so that it includes your calculations? It is not trivial to see where you get stuck.  

---

## 2 answers

### Answer by username_2 on 2024-02-12. Score: 1

Given a pure-exchange economy with

* $u\_A(x\_A,y\_A)=\min(x\_A,2y\_A)$, $u\_B(x\_B,y\_B)=\min(2x\_B,y\_B)$
* Endowment of A is $(k\_X,k\_Y)$ and of B is $(12-k\_X,12-k\_Y)$

Set of feasible allocations is $\mathcal{F} = \{((x\_A,y\_A),(x\_B,y\_B))\in\mathbb{R}^2\_+\times\mathbb{R}^2\_+|x\_A+x\_B=y\_A+y\_B=12\}$

Here is the Edgeworth box representation of feasible allocations and set of efficient allocations:
[![enter image description here](https://i.sstatic.net/mOKJH.png)](https://i.sstatic.net/mOKJH.png)

To determine the competitive equilibrium, we can consider the following cases for the endowments.

**Case 1**: $k\_X\leq 8, k\_Y\geq 4, (k\_X,k\_Y)\neq (8,4)$

In this case, given $(k\_X,k\_Y)$, there is a unique competitive equilibrium allocation $((x\_A,y\_A),(x\_B,y\_B))=((8,4),(4,8))$ supported by prices $(p\_X,p\_Y)=(k\_Y-4,8-k\_X)$.

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

**Case 2**: $(k\_X,k\_Y)= (8,4)$

In this case, there is a unique competitive equilibrium allocation $((x\_A,y\_A),(x\_B,y\_B))=((8,4),(4,8))$ and it can be supported by any pair of prices from the set $\{(p\_X,p\_Y)\in\mathbb{R}^2\_+|p\_X+p\_Y=1\}$.

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

**Case 3**: $k\_X\geq 8, k\_Y \leq 4, (k\_X,k\_Y)\neq (8,4)$

In this case, given $(k\_X,k\_Y)$, there are three sets of competitive equilibria:

(i) $((x\_A,y\_A),(x\_B,y\_B))=((8,4),(4,8))$ and it is supported by prices $(p\_X,p\_Y)=(4-k\_Y,k\_X-8)$.

(ii) Allocations in the set $\{((x\_A,y\_A),(x\_B,y\_B))\in\mathcal{F}|y\_A=k\_Y, 2k\_Y\leq x\_A\leq \frac{k\_Y+12}{2}\}$ are supported by the prices $(p\_X,p\_Y)=(0,1)$

(iii) Allocations in the set $\{((x\_A,y\_A),(x\_B,y\_B))\in\mathcal{F}|x\_A=k\_X, \frac{k\_X}{2}\leq y\_A\leq 2k\_X-12\}$ are supported by the prices $(p\_X,p\_Y)=(1,0)$

[![enter image description here](https://i.sstatic.net/BBt5k.png)](https://i.sstatic.net/BBt5k.png)

**Case 4**: $k\_X> 8, k\_Y > 4$

In this case, given $(k\_X,k\_Y)$, allocations in the set $\{((x\_A,y\_A),(x\_B,y\_B))\in\mathcal{F}|x\_A=k\_X, \frac{k\_X}{2}\leq y\_A\leq 2k\_X-12\}$ are supported by the prices $(p\_X,p\_Y)=(1,0)$

**Case 5**: $k\_X< 8, k\_Y < 4$

In this case, given $(k\_X,k\_Y)$, allocations in the set $\{((x\_A,y\_A),(x\_B,y\_B))\in\mathcal{F}|y\_A=k\_Y, 2k\_Y\leq x\_A\leq \frac{k\_Y+12}{2}\}$ are supported by the prices $(p\_X,p\_Y)=(0,1)$

> **Comments:**
> **username_0**: Thanks for this answer, this clears things up a bit, but I still have two more questions:  In **Case 4**: Can't the competitive equilibrium also be ya=k2, xa=6+k2/2 supported by prices (px,py)=(0,1). At this point, the utility of both consumers would be maximized and the outcome would be Pareto efficient.   Furthermore, in **Case 1** why can't there be 3 sets of equilibria as in **Case 3**, I'm not sure I understand what prevents the equilibrium from being for example : (xA,yA)=(kx, 0<yA<kx/2) supported by prices (px,py)=(1,0)  
> **username_2**: In case 4, competitive equilibrium cannot be $(p_X,p_Y)=(0,1)$ because there will be excess demand for $X$, sum of demands for $X$ will exceed $12$.  
> **username_2**: In case 1, at $(p_X,p_Y)=(0,1)$, there will be excess demand for $X$ and at $(p_X,p_Y)=(1,0)$, there will be excess demand for $Y$. $(x_A,y_A)=(k_X,y_A)$ where $y_A\in (0, k_X/2)$ is not a utility maximising choice for consumer $A$ at prices $(p_X,p_Y)=(1,0)$. In this case, Consumer $A$ will demand at least  $k_X/2$ units of $Y$.  

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### Answer by username_3 on 2024-11-01. Score: 0

LoL, this was the exam question I set... Here is the solution:

Consumers utility function is quasi-concave and satisfy LNS
as $u^{A}(y^{A})>u^{A}(x^{A})$ for all $y^{A}>>x^{A}$. Thus, the Marshallian
demands of consumers will necessarily satisfy the budget constraint at
equality. When $p>>0$, a simple solution exists, namely
\begin{gather\*}
x^{A}(p)=\left( \frac{2m^{A}}{2p\_{1}+p\_{2}},\frac{m^{A}}{2p\_{1}+p\_{2}
}\right) \text{ \ and \ }x^{B}(p)=\left( \frac{m^{B}}{p\_{1}+2p\_{2}}
,\frac{2m^{B}}{p\_{1}+2p\_{2}}\right) \text{.}
\end{gather\*}
The latter is a solution for $A$, since for any other consumption bundle
$y^{A}$ such that $p\_{1}y\_{1}^{A}+p\_{2}y\_{2}^{A}=m^{A}$ and $\varepsilon>0$
sufficiently small,
\begin{gather\*}
u^{A}(y\_{1}^{A}-p\_{2}\varepsilon/p\_{1},y\_{2}^{A}+\varepsilon)=2(y\_{2}
^{A}+\varepsilon)>2y\_{2}^{A}=u^{A}(y^{A})\text{ \ when }y\_{1}^{A}>2y\_{2}
^{A}\text{,}\\
u^{A}(y\_{1}^{A}+\varepsilon,y\_{2}^{A}-p\_{1}\varepsilon/p\_{2})=y\_{1}
^{A}+\varepsilon>y\_{1}^{A}=u^{A}(y^{A})\text{ \ when }y\_{1}^{A}<2y\_{2}
^{A}\text{.}
\end{gather\*}
The latter is a solution for $B$, since for any other consumption bundle
$y^{b}$ such that $p\_{1}y\_{1}^{B}+p\_{2}y\_{2}^{B}=m^{B}$ and $\varepsilon>0$
sufficiently small,
\begin{gather\*}
u^{B}(y\_{1}^{B}-p\_{2}\varepsilon/p\_{1},y\_{2}^{B}+\varepsilon)=y\_{2}
^{B}+\varepsilon>y\_{2}^{B}=u^{B}(y^{B})\text{ \ when }2y\_{1}^{B}>y\_{2}
^{B}\text{,}\\
u^{B}(y\_{1}^{B}+\varepsilon,y\_{2}^{B}-p\_{1}\varepsilon/p\_{2})=2(y\_{1}
^{B}+\varepsilon)>2y\_{1}^{B}=u^{B}(y^{B})\text{ \ when }2y\_{1}^{B}
When $p\_{1}=0$ and $p\_{2}>0$, consumers naturally spend all of their budget on
the expensive good and demands become correspondences, since they only care
about getting enough of the free good, but do not mind about purchasing more
of that. Therefore,
\begin{gather\*}
x^{A}(p)=\left( \left[ \frac{2m^{A}}{p\_{2}},\infty\right) ,\frac{m^{A}%
}{p\_{2}}\right) \text{ \ and \ }x^{B}(p)=\left( \left[ \frac{m^{B}}{2p\_{2}%
},\infty\right) ,\frac{m^{B}}{p\_{2}}\right) \text{.}
\end{gather\*}
Similarly, when $p\_{1}>0$ and $p\_{2}=0$,
\begin{gather\*}
x^{A}(p)=\left( \frac{m^{A}}{p\_{1}},\left[ \frac{m^{A}}{2p\_{1}}%
,\infty\right) \right) \text{ \ and \ }x^{B}(p)=\left( \frac{m^{B}}{p\_{1}%
},\left[ \frac{2m^{B}}{p\_{1}},\infty\right) \right) \text{.}%
\end{gather\*}
Of course if both prices were equal to $0$, demand would be unbounded for any
one of the two goods.

To find CEs, begin by focusing on CEs in which both prices are strictly
positive. By Walras Law we know than in these CE, $z\_{1}(p)=z\_{2}(p)=0$.
Further, we can normalize any one price to $1$. So set $p\_{1}=1$ and clear the
market for good $1$
\begin{gather\*}
x\_{1}^{A}(p)+x\_{1}^{B}(p)=\frac{2\left( k\_{1}+p\_{2}k\_{2}\right) }{2+p\_{2}
}+\frac{\left( 12-k\_{1}+p\_{2}(12-k\_{2})\right) }{1+2p\_{2}}=12\text{.}
\end{gather\*}
The latter yields $p\_{2}=\frac{k\_{1}-8}{4-k\_{2}}$. In order for $p\_{2}>0$,
this requires that either $k\_{1}>8$ and $4>k\_{2}$\ or\ $k\_{1}<8$ and $4. So in all these circumstances this competitive equilibrium exists.

Next, look for CE in which $p\_{1}=0$ and $p\_{2}>0$. By Walras Law we know than
in these CE, $z\_{2}(p)=0$. Further, we can normalize $p\_{2}$ to $1$. So set
$p\_{2}=1$ and clear the market for good $2$,
\begin{gather\*}
x\_{2}^{A}(p)+x\_{2}^{B}(p)=m^{A}+m^{B}=12\text{.}
\end{gather\*}
Then, verify that there is excess supply in the market for the free good. Note
that
\begin{gather\*}
x\_{1}^{A}(p)+x\_{1}^{B}(p)\geq2m^{A}+\frac{m^{B}}{2}=2k\_{2}+\frac{12-k\_{2}}
{2}=6+\frac{3k\_{2}}{2}\text{.}
\end{gather\*}
For there to be excess supply then $6+\frac{3k\_{2}}{2}\leq12$ ore equivalently
$k\_{2}\leq4$. So in all these circumstances this competitive equilibrium exists.

Finally, look for CE in which $p\_{1}>0$ and $p\_{2}=0$. By Walras Law we know
than in these CE, $z\_{1}(p)=0$. Further, we can normalize $p\_{1}$ to $1$. So
set $p\_{1}=1$ and clear the market for good $1$,
\begin{gather\*}
x\_{1}^{A}(p)+x\_{1}^{B}(p)=m^{A}+m^{B}=12\text{.}
\end{gather\*}
Then, verify that there is excess supply in the market for the free good. Note
that
\begin{gather\*}
x\_{2}^{A}(p)+x\_{2}^{B}(p)\geq\frac{m^{A}}{2}+2m^{B}=\frac{k\_{1}}{2}
+24-2k\_{1}=24-\frac{3k\_{1}}{2}\text{.}
\end{gather\*}
For there to be excess supply then $24-\frac{3k\_{1}}{2}\leq12$ ore
equivalently $k\_{1}\geq8$. So in all these circumstances this competitive
equilibrium exists.

An allocation $x=(x\_{1}^{A},x\_{2}^{A},x\_{1}^{B},x\_{2}^{B})\in R\_{+}^{4}$ is
Pareto efficient if and only if it is feasible and there is no other feasible
allocation $\hat{x}$ such that: (a) $U^{j}(\hat{x}^{j})\geq U^{j}(x^{j})$ for
any $j\in\left\{ A,B\right\} $; (b) $U^{j}(\hat{x}^{j})>U^{j}(x^{j})$ for
some $j\in\left\{ A,B\right\} $. The contract curve identifies the set of
Pareto efficient allocations of an economy.
\begin{gather\*}
x\_{2}^{A}\in C\_{2}^{A}(x\_{1}^{A})=\left\{
\begin{array}
[c]{lll}%
\lbrack0,x\_{1}^{A}/2] & \text{if} & x\_{1}^{A}\in\lbrack0,6]\\
\lbrack2x\_{1}^{A}-12,x\_{1}^{A}/2] & \text{if} & x\_{1}^{A}\in\lbrack6,8]\\
\lbrack x\_{1}^{A}/2,2x\_{1}^{A}-12] & \text{if} & x\_{1}^{A}\in\lbrack8,12]
\end{array}
\right.
\end{gather\*}
To show the latter, note that for any point $x^{A}$ on the conjectured
contract curve, the intersection between the upper-contour sets of both
players has an empty interior. Formally, let $x^{A}$ satisfy $x\_{2}^{A}%
=C\_{2}^{A}(x\_{1}^{A})$, let
\begin{align\*}
U\_{+}^{A}(x^{A}) & =\left\{ x\in\mathbb{R}\_{+}^{2}\text{ }|\text{ }%
\min\{x\_{1},2x\_{2}\}\geq\min\{x\_{1}^{A},2x\_{2}^{A}\}\right\} \text{,}\\
U\_{+}^{B}(x^{A}) & =\left\{ x\in\mathbb{R}\_{+}^{2}\text{ }|\text{ }%
\min\{24-2x\_{1},12-x\_{2}\}\geq\min\{24-2x\_{1}^{A},12-x\_{2}^{A}\}\right\}
\text{,}%
\end{align\*}
and note that $int(U\_{+}^{A}(x^{A})\cap U\_{+}^{B}(x^{A}))=\emptyset$. Thus, it
is impossible to benefit a player without hurting the other player. For
instance for $x\_{1}^{A}\in\lbrack0,6]$ and $x\_{2}^{A}\in\lbrack0,x\_{1}^{A}%
/2]$, these sets become
\begin{align\*}
U\_{+}^{A}(x^{A}) & =\left\{ x\in\mathbb{R}\_{+}^{2}\text{ }|\text{ }%
\min\{x\_{1},2x\_{2}\}\geq2x\_{2}^{A}\right\} \text{,}\\
U\_{+}^{B}(x^{A}) & =\left\{ x\in\mathbb{R}\_{+}^{2}\text{ }|\text{ }%
\min\{24-2x\_{1},12-x\_{2}\}\geq12-x\_{2}^{A}\right\} \text{,}%
\end{align\*}
and we cannot have that $x\_{2}>x\_{2}^{A}$ and $12-x\_{2}>12-x\_{2}^{A}$ holding
at once. So, level curves are tangent and we cannot benefit a player without
hurting the other one. Finally, for any point $x^{A}$ which is not on the
conjectured contract curve, the intersection between the upper-contour sets of
both players has a non-empty interior. Thus, given that both players have
monotonic preferences, it is possible to benefit a player without hurting the
other one.


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