# Finding the set of Pareto Optimal Allocations

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?

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.

• 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. Oct 30, 2021 at 13:54
• Thanks, I have edited the post. Oct 31, 2021 at 10:53
• 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". Oct 31, 2021 at 11:09

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.

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

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: 