<p><strong>Interpretation 1</strong>: Given the data in the problem, I think you're interested in a pure-exchange economy with externalities i.e.,</p>
<ul>
<li><span class="math-container">$u_1(x_1,y_2) = x_1^\beta y_2^{1-\beta}$</span>, <span class="math-container">$u_2(x_2,y_2) = \min(x_2,y_2)$</span>, where <span class="math-container">$\beta\in (0,1)$</span></li>
<li><span class="math-container">$\omega_1=(0,1)$</span> and <span class="math-container">$\omega_2=(1,0)$</span></li>
</ul>
<p>Set of feasible allocations is given by
<span class="math-container">$\mathcal{F} = \{((x_1,y_1),(x_2,y_2))\in\mathbb{R}^2_+\times\mathbb{R}^2_+|x_1+x_2=1 \ \wedge \ y_1+y_2=1\}$</span></p>
<p>Proposition 0: Set of Pareto efficient allocations is non-empty.</p>
<p>Proof: Consider <span class="math-container">$((x_1,y_1),(x_2,y_2)) = ((0,0),(1,1))$</span>. This allocation is Pareto efficient because if we move to any other feasible allocation, that will necessarily make individual <span class="math-container">$2$</span> worse off.</p>
<p>Proposition 1: Set of Pareto efficient allocations is given by the set <span class="math-container">$\{((x_1,y_1),(x_2,y_2))\in\mathcal{F}|y_2=1\}$</span></p>
<p>Proof: Consider a feasible allocation <span class="math-container">$((x_1,y_1),(x_2,y_2))$</span> and suppose <span class="math-container">$y_2 &lt; 1$</span> holds. Clearly, <span class="math-container">$((x_1,0),(x_2,1))$</span> is feasible and is a Pareto improvement over <span class="math-container">$((x_1,y_1),(x_2,y_2))$</span> as <span class="math-container">$1$</span> will be made better off without making <span class="math-container">$2$</span> worst off. Therefore, <span class="math-container">$y_2=1$</span> is a necessary condition for efficiency. If we consider the set <span class="math-container">$\{((x_1,y_1),(x_2,y_2))\in\mathcal{F}|y_2=1\}$</span>, it yields the utility possibilities as <span class="math-container">$u_1^{1/\beta}+u_2=1$</span>, where <span class="math-container">$0 \leq u_1 \leq 1$</span>, which is a strictly decreasing curve in the <span class="math-container">$(u_1,u_2)-$</span>space establishing that <span class="math-container">$y_2=1$</span> along with feasibility yields the set of efficient allocations.</p>
<p><strong>Interpretation 2</strong> (If you made a typo in writing utilities): Given a pure-exchange economy,</p>
<ul>
<li><span class="math-container">$u_1(x_1,y_1) = x_1^\beta y_1^{1-\beta}$</span>, <span class="math-container">$u_2(x_2,y_2) = \min(x_2,y_2)$</span>, where <span class="math-container">$\beta\in (0,1)$</span></li>
<li><span class="math-container">$\omega_1=(0,1)$</span> and <span class="math-container">$\omega_2=(1,0)$</span></li>
</ul>
<p>Set of feasible allocations is given by
<span class="math-container">$\mathcal{F} = \{((x_1,y_1),(x_2,y_2))\in\mathbb{R}^2_+\times\mathbb{R}^2_+|x_1+x_2=1 \ \wedge \ y_1+y_2=1\}$</span></p>
<p>Proposition 0: Set of Pareto efficient allocations is non-empty.</p>
<p>Proof: Consider <span class="math-container">$((x_1,y_1),(x_2,y_2)) = ((1,1),(0,0))$</span>. This allocation is Pareto efficient because if we move to any other feasible allocation, that will necessarily make individual <span class="math-container">$1$</span> worse off.</p>
<p>Proposition 1: If a feasible allocation <span class="math-container">$((x_1,y_1),(x_2,y_2))$</span> satisfy <span class="math-container">$x_2\neq y_2$</span>, then it is not Pareto efficient.</p>
<p>Proof: Consider a feasible allocation <span class="math-container">$((x_1,y_1),(x_2,y_2))$</span> and suppose <span class="math-container">$x_2 &lt; y_2 \leq 1$</span> holds. Therefore, <span class="math-container">$x_1 = 1-x_2 &gt; 0$</span>. Clearly, <span class="math-container">$((x_1,y_1+y_2-x_2),(x_2,x_2))$</span> is feasible and Pareto Superior to <span class="math-container">$((x_1,y_1),(x_2,y_2))$</span>. Therefore, <span class="math-container">$((x_1,y_1),(x_2,y_2))$</span> is not Pareto efficient. Likewise, by a symmetric argument, a feasible allocation <span class="math-container">$((x_1,y_1),(x_2,y_2))$</span> satisfying <span class="math-container">$y_2 &lt; x_2 \leq 1$</span> is also not Pareto efficient.</p>
<p>Equivalently, in proposition 1, we have shown that if an allocation is Pareto efficient then it satisfy the condition <span class="math-container">$x_2 =y_2$</span>. Now, we'll show that the converse is also true.</p>
<p>Proposition 2: Any feasible allocation <span class="math-container">$((x_1,y_1),(x_2,y_2))$</span> satisfying <span class="math-container">$x_2 = y_2$</span> is Pareto efficient.</p>
<p>Proof: Observe that the sum of the utilities of the two individuals at all the feasible allocations satisfying <span class="math-container">$x_2= y_2$</span> equals <span class="math-container">$u_1+u_2=x_1^\beta y_1^{1-\beta} + \min(x_2,y_2) = x_1^\beta x_1^{1-\beta} + x_2 = 1$</span>, and is maximum among all feasible allocations (using the argument in proposition 1), therefore it follows that all the allocations satisfying <span class="math-container">$x_2 = y_2$</span> are Pareto efficient.</p>

CC BY-SA 4.0
2023-05-30T03:37:04.067Z
2023-05-30T06:55:55.243Z
2023-05-30T06:55:55.243Z
Amit
<p>Probably the easiest way to get all Pareto optimal allocations is to maxmize a weighted sum of utilities (of the two agents) subject to the resource constraints:
<span class="math-container">$$
\max_{x_1^1, x_1^2, x_2^1, x_2^2} \alpha u_1(x_1^1, x_1^2) + (1-\alpha) u_2(x_2^1, x_2^1) \text{ s.t. } x_1^1 + x_2^1 = 1 \text{ and } x_1^2 + x_2^2 = 1.
$$</span>
The PO allocations can be obtained by varying <span class="math-container">$\alpha$</span> over the interval <span class="math-container">$[0,1]$</span>.</p>
<p>As the second agent has Leontief preferences any PO allocation will have <span class="math-container">$x_2^1 = x_2^2$</span>. So substuting out we have:
<span class="math-container">$$
\max_{x_2^1} \alpha u_1(1 - x_2^1, 1 - x_2^1) + (1-\alpha) u_2(x_2^1, x_2^1) \text{ s.t. } x_{2}^1 \le 1
$$</span></p>

CC BY-SA 4.0
2021-10-31T16:10:17.743Z
2021-10-31T16:10:17.743Z
tdm
<p>To make the notation clearer, let's rename the goods as <span class="math-container">$x,y$</span> and the agents as <span class="math-container">$A,B$</span>. With this, the variables are <span class="math-container">$x_A,x_B,y_A,y_B$</span>.</p>
<p>With this notation, the utilities are:</p>
<p><span class="math-container">$U_A = \beta \log(x_A) + (1-\beta) \log(y_A)$</span></p>
<p><span class="math-container">$U_B = \min\{x_B,y_B\}$</span></p>
<p>We find the contract curve by maximizing an agent's utility subject to the other agent's given utility level and the endowment constraints.</p>
<p>Let's solve this optimization problem:</p>
<p><span class="math-container">$\max \beta \log(x_A) + (1-\beta) \log(y_A)$</span></p>
<p>s.t. <span class="math-container">$\min\{x_B,y_B\} = \overline{U}$</span></p>
<p><span class="math-container">$x_A + x_B = 1$</span></p>
<p><span class="math-container">$y_A + y_B = 1$</span></p>
<p>Consider a point where <span class="math-container">$x_B \neq y_B$</span> on the indifference curve for <span class="math-container">$U_B = \overline{U}$</span>. Then either <span class="math-container">$x_B &gt; y_B$</span> or <span class="math-container">$y_B &gt; x_B$</span>.</p>
<ul>
<li>Case 1: <span class="math-container">$x_B &gt; y_B \implies y_B = \overline{U} \text{ and } x_B = \overline{U} + \epsilon$</span></li>
<li>Case 2: <span class="math-container">$y_B &gt; x_B \implies x_B = \overline{U} \text{ and } y_B = \overline{U} + \epsilon$</span></li>
</ul>
<p>where <span class="math-container">$\epsilon &gt; 0$</span>.</p>
<p>On the other hand, the point where <span class="math-container">$x_B = y_B$</span> is given by <span class="math-container">$(x_B,y_B) = (\overline{U},\overline{U})$</span>.</p>
<p>Let's plug in the two cases into <span class="math-container">$U_A$</span>.</p>
<p>Plugging the endowment constraints into <span class="math-container">$U_A$</span> to write it in terms of <span class="math-container">$x_B,y_B$</span> we get:</p>
<p><span class="math-container">$U_A = \beta \log(1-x_B) + (1-\beta) \log(1-y_B)$</span></p>
<ul>
<li>Case 1: <span class="math-container">$x_B &gt; y_B \implies y_B = \overline{U} \text{ and } x_B = \overline{U} + \epsilon$</span></li>
</ul>
<p><span class="math-container">$\implies U_A(1-x_B,1-y_B) = \beta \log(1-\overline{U}-\epsilon) + (1-\beta) \log(1-\overline{U}) &lt; \beta \log(1-\overline{U}) + (1-\beta) \log(1-\overline{U}) = U_A(1-\overline{U},1-\overline{U})$</span></p>
<ul>
<li>Case 2: <span class="math-container">$y_B &gt; x_B \implies x_B = \overline{U} \text{ and } y_B = \overline{U} + \epsilon$</span></li>
</ul>
<p><span class="math-container">$\implies U_A(1-x_B,1-y_B) = \beta \log(1-\overline{U}) + (1-\beta) \log(1-\overline{U}-\epsilon) &lt; \beta \log(1-\overline{U}) + (1-\beta) \log(1-\overline{U}) = U_A(1-\overline{U},1-\overline{U})$</span></p>
<p>Therefore, the maximum is attained when <span class="math-container">$y_B = x_B = \overline{U}$</span>.</p>
<p>Varying the level of <span class="math-container">$\overline{U}$</span> we get the plot of the contract curve, which is the following set:</p>
<p><span class="math-container">$CC = \{((x_A,y_A),(x_B,y_B)) \in \mathcal{F} : y_B = x_B\}$</span></p>
<p>where <span class="math-container">$\mathcal{F}$</span> is the set of feasible allocations (where the endowment constrainsts hold), i.e. points in the Edgeworth box.</p>
<p>Below I give an illustration of what happens in the Edgeworth box:</p>
<p><a href="https://i.sstatic.net/3PNz6.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/3PNz6.jpg" alt="enter image description here" /></a></p>

CC BY-SA 4.0
2023-05-29T17:23:15.217Z
2023-05-29T17:23:15.217Z
Nicolas Torres
<p>I am asked to find the set of Pareto Optimal Allocations in an economy where there are two agents namely <span class="math-container">$1$</span> and <span class="math-container">$2$</span>, with the following utility functions and endowments.</p>
<p><span class="math-container">$$u_1({x_1}^1,{x_2}^2)= \beta log({x_1}^1)+(1-\beta)log({x_2}^2)  \ ,\ {\omega}_1 = (0,1) \ \beta \in (0,1) \ $$</span>
<span class="math-container">$$u_2({x_2}^1,{x_2}^2)=min\{ {x_2}^1,{x_2}^2 \}   \ ,\ {\omega}_2 = (1,0)$$</span></p>
<p>I know that I can show the set of PO on the Edgeworth Box. However, how am I supposed to show the set of PO in algebraically? Shall I try splitting cases for the second agent's utility function?</p>
<p>Thanks in advance.</p>
<p><strong>Edit:</strong> Each agent <span class="math-container">$i$</span> has the preferences represented by the following utility function, <span class="math-container">$u_i$</span> and the endowment <span class="math-container">${\omega}_i$</span>. <span class="math-container">${x_i}^t$</span> denotes the amount of good <span class="math-container">$t$</span> consumed by the agent <span class="math-container">$i$</span>. Say, prices of the good 1 and good 2 are denoted by <span class="math-container">$P_1$</span> and <span class="math-container">$P_2$</span>, respectively. I am asked to show the set of Pareto Optimal allocations in this setting.</p>

CC BY-SA 4.0
2021-10-30T13:54:42.733Z
Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking.
Giskard
CC BY-SA 4.0
2021-10-31T10:53:26.400Z
Thanks, I have edited the post.
user722271
CC BY-SA 4.0
2021-10-31T11:09:14.227Z
Hi! This does not really make things clearer. What do you mean by "*I can show the set of PO on the Edgeworth Box*"? And what is your purpose with "*splitting cases for the second agent's utility function*"? Also, it is not clear where you get stuck if you "*can show the set of PO on the Edgeworth Box*".
Giskard
<macroeconomics><pareto-efficiency>
Finding the set of Pareto Optimal Allocations
user722271
# Finding the set of Pareto Optimal Allocations

(Asked by: username_0 on 2021-10-30)

I am asked to find the set of Pareto Optimal Allocations in an economy where there are two agents namely $1$ and $2$, with the following utility functions and endowments.

$$u\_1({x\_1}^1,{x\_2}^2)= \beta log({x\_1}^1)+(1-\beta)log({x\_2}^2) \ ,\ {\omega}\_1 = (0,1) \ \beta \in (0,1) \ $$
$$u\_2({x\_2}^1,{x\_2}^2)=min\{ {x\_2}^1,{x\_2}^2 \} \ ,\ {\omega}\_2 = (1,0)$$

I know that I can show the set of PO on the Edgeworth Box. However, how am I supposed to show the set of PO in algebraically? Shall I try splitting cases for the second agent's utility function?

Thanks in advance.

**Edit:** Each agent $i$ has the preferences represented by the following utility function, $u\_i$ and the endowment ${\omega}\_i$. ${x\_i}^t$ denotes the amount of good $t$ consumed by the agent $i$. Say, prices of the good 1 and good 2 are denoted by $P\_1$ and $P\_2$, respectively. I am asked to show the set of Pareto Optimal allocations in this setting.

**Tags:** macroeconomics, pareto-efficiency

> **Comments:**
> **username_1**: Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking.  
> **username_0**: Thanks, I have edited the post.  
> **username_1**: Hi! This does not really make things clearer. What do you mean by "*I can show the set of PO on the Edgeworth Box*"? And what is your purpose with "*splitting cases for the second agent's utility function*"? Also, it is not clear where you get stuck if you "*can show the set of PO on the Edgeworth Box*".  

---

## 3 answers

### Answer by username_2 on 2023-05-30. Score: 2

**Interpretation 1**: Given the data in the problem, I think you're interested in a pure-exchange economy with externalities i.e.,

* $u\_1(x\_1,y\_2) = x\_1^\beta y\_2^{1-\beta}$, $u\_2(x\_2,y\_2) = \min(x\_2,y\_2)$, where $\beta\in (0,1)$
* $\omega\_1=(0,1)$ and $\omega\_2=(1,0)$

Set of feasible allocations is given by
$\mathcal{F} = \{((x\_1,y\_1),(x\_2,y\_2))\in\mathbb{R}^2\_+\times\mathbb{R}^2\_+|x\_1+x\_2=1 \ \wedge \ y\_1+y\_2=1\}$

Proposition 0: Set of Pareto efficient allocations is non-empty.

Proof: Consider $((x\_1,y\_1),(x\_2,y\_2)) = ((0,0),(1,1))$. This allocation is Pareto efficient because if we move to any other feasible allocation, that will necessarily make individual $2$ worse off.

Proposition 1: Set of Pareto efficient allocations is given by the set $\{((x\_1,y\_1),(x\_2,y\_2))\in\mathcal{F}|y\_2=1\}$

Proof: Consider a feasible allocation $((x\_1,y\_1),(x\_2,y\_2))$ and suppose $y\_2 < 1$ holds. Clearly, $((x\_1,0),(x\_2,1))$ is feasible and is a Pareto improvement over $((x\_1,y\_1),(x\_2,y\_2))$ as $1$ will be made better off without making $2$ worst off. Therefore, $y\_2=1$ is a necessary condition for efficiency. If we consider the set $\{((x\_1,y\_1),(x\_2,y\_2))\in\mathcal{F}|y\_2=1\}$, it yields the utility possibilities as $u\_1^{1/\beta}+u\_2=1$, where $0 \leq u\_1 \leq 1$, which is a strictly decreasing curve in the $(u\_1,u\_2)-$space establishing that $y\_2=1$ along with feasibility yields the set of efficient allocations.

**Interpretation 2** (If you made a typo in writing utilities): Given a pure-exchange economy,

* $u\_1(x\_1,y\_1) = x\_1^\beta y\_1^{1-\beta}$, $u\_2(x\_2,y\_2) = \min(x\_2,y\_2)$, where $\beta\in (0,1)$
* $\omega\_1=(0,1)$ and $\omega\_2=(1,0)$

Set of feasible allocations is given by
$\mathcal{F} = \{((x\_1,y\_1),(x\_2,y\_2))\in\mathbb{R}^2\_+\times\mathbb{R}^2\_+|x\_1+x\_2=1 \ \wedge \ y\_1+y\_2=1\}$

Proposition 0: Set of Pareto efficient allocations is non-empty.

Proof: Consider $((x\_1,y\_1),(x\_2,y\_2)) = ((1,1),(0,0))$. This allocation is Pareto efficient because if we move to any other feasible allocation, that will necessarily make individual $1$ worse off.

Proposition 1: If a feasible allocation $((x\_1,y\_1),(x\_2,y\_2))$ satisfy $x\_2\neq y\_2$, then it is not Pareto efficient.

Proof: Consider a feasible allocation $((x\_1,y\_1),(x\_2,y\_2))$ and suppose $x\_2 < y\_2 \leq 1$ holds. Therefore, $x\_1 = 1-x\_2 > 0$. Clearly, $((x\_1,y\_1+y\_2-x\_2),(x\_2,x\_2))$ is feasible and Pareto Superior to $((x\_1,y\_1),(x\_2,y\_2))$. Therefore, $((x\_1,y\_1),(x\_2,y\_2))$ is not Pareto efficient. Likewise, by a symmetric argument, a feasible allocation $((x\_1,y\_1),(x\_2,y\_2))$ satisfying $y\_2 < x\_2 \leq 1$ is also not Pareto efficient.

Equivalently, in proposition 1, we have shown that if an allocation is Pareto efficient then it satisfy the condition $x\_2 =y\_2$. Now, we'll show that the converse is also true.

Proposition 2: Any feasible allocation $((x\_1,y\_1),(x\_2,y\_2))$ satisfying $x\_2 = y\_2$ is Pareto efficient.

Proof: Observe that the sum of the utilities of the two individuals at all the feasible allocations satisfying $x\_2= y\_2$ equals $u\_1+u\_2=x\_1^\beta y\_1^{1-\beta} + \min(x\_2,y\_2) = x\_1^\beta x\_1^{1-\beta} + x\_2 = 1$, and is maximum among all feasible allocations (using the argument in proposition 1), therefore it follows that all the allocations satisfying $x\_2 = y\_2$ are Pareto efficient.


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### Answer by username_3 on 2021-10-31. Score: 1

Probably the easiest way to get all Pareto optimal allocations is to maxmize a weighted sum of utilities (of the two agents) subject to the resource constraints:
$$
\max\_{x\_1^1, x\_1^2, x\_2^1, x\_2^2} \alpha u\_1(x\_1^1, x\_1^2) + (1-\alpha) u\_2(x\_2^1, x\_2^1) \text{ s.t. } x\_1^1 + x\_2^1 = 1 \text{ and } x\_1^2 + x\_2^2 = 1.
$$
The PO allocations can be obtained by varying $\alpha$ over the interval $[0,1]$.

As the second agent has Leontief preferences any PO allocation will have $x\_2^1 = x\_2^2$. So substuting out we have:
$$
\max\_{x\_2^1} \alpha u\_1(1 - x\_2^1, 1 - x\_2^1) + (1-\alpha) u\_2(x\_2^1, x\_2^1) \text{ s.t. } x\_{2}^1 \le 1
$$


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### Answer by username_4 on 2023-05-29. Score: 0

To make the notation clearer, let's rename the goods as $x,y$ and the agents as $A,B$. With this, the variables are $x\_A,x\_B,y\_A,y\_B$.

With this notation, the utilities are:

$U\_A = \beta \log(x\_A) + (1-\beta) \log(y\_A)$

$U\_B = \min\{x\_B,y\_B\}$

We find the contract curve by maximizing an agent's utility subject to the other agent's given utility level and the endowment constraints.

Let's solve this optimization problem:

$\max \beta \log(x\_A) + (1-\beta) \log(y\_A)$

s.t. $\min\{x\_B,y\_B\} = \overline{U}$

$x\_A + x\_B = 1$

$y\_A + y\_B = 1$

Consider a point where $x\_B \neq y\_B$ on the indifference curve for $U\_B = \overline{U}$. Then either $x\_B > y\_B$ or $y\_B > x\_B$.

* Case 1: $x\_B > y\_B \implies y\_B = \overline{U} \text{ and } x\_B = \overline{U} + \epsilon$
* Case 2: $y\_B > x\_B \implies x\_B = \overline{U} \text{ and } y\_B = \overline{U} + \epsilon$

where $\epsilon > 0$.

On the other hand, the point where $x\_B = y\_B$ is given by $(x\_B,y\_B) = (\overline{U},\overline{U})$.

Let's plug in the two cases into $U\_A$.

Plugging the endowment constraints into $U\_A$ to write it in terms of $x\_B,y\_B$ we get:

$U\_A = \beta \log(1-x\_B) + (1-\beta) \log(1-y\_B)$

* Case 1: $x\_B > y\_B \implies y\_B = \overline{U} \text{ and } x\_B = \overline{U} + \epsilon$

$\implies U\_A(1-x\_B,1-y\_B) = \beta \log(1-\overline{U}-\epsilon) + (1-\beta) \log(1-\overline{U}) < \beta \log(1-\overline{U}) + (1-\beta) \log(1-\overline{U}) = U\_A(1-\overline{U},1-\overline{U})$

* Case 2: $y\_B > x\_B \implies x\_B = \overline{U} \text{ and } y\_B = \overline{U} + \epsilon$

$\implies U\_A(1-x\_B,1-y\_B) = \beta \log(1-\overline{U}) + (1-\beta) \log(1-\overline{U}-\epsilon) < \beta \log(1-\overline{U}) + (1-\beta) \log(1-\overline{U}) = U\_A(1-\overline{U},1-\overline{U})$

Therefore, the maximum is attained when $y\_B = x\_B = \overline{U}$.

Varying the level of $\overline{U}$ we get the plot of the contract curve, which is the following set:

$CC = \{((x\_A,y\_A),(x\_B,y\_B)) \in \mathcal{F} : y\_B = x\_B\}$

where $\mathcal{F}$ is the set of feasible allocations (where the endowment constrainsts hold), i.e. points in the Edgeworth box.

Below I give an illustration of what happens in the Edgeworth box:

[![enter image description here](https://i.sstatic.net/3PNz6.jpg)](https://i.sstatic.net/3PNz6.jpg)


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