Finding the set of Pareto Optimal Allocations

Question

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.

Answer 1

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.

Answer 2

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: $$ \max_{x_2^1} \alpha u_1(1 - x_2^1, 1 - x_2^1) + (1-\alpha) u_2(x_2^1, x_2^1) \text{ s.t. } x_{2}^1 \le 1 $$

Answer 3

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: