Pareto efficient allocations for non-monotonic, quasi-linear utility function

Suppose we have an pure exchange economy with 2 consumers, and 2 goods $$x_1$$ and $$x_2$$. Fix some $$\alpha_1$$ and $$\alpha_2$$. The utility for consumer $$i$$ is defined by: $$u_i(x_{1i},x_{2i}) = x_{1i} - |x_{2i} - \alpha_{i}|.$$

How to find all the Pareto efficient allocations for this economy?

• @Stef The first index is for the goods and the second index $i$ is for consumer $i$. I feel uncomfortable with that notation, so I renamed the goods in my answer as $x,y$ rather than $x_1,x_2$. May 23, 2023 at 14:21

Given the economy:

• Utility functions: $$u_A(x_A,y_A) = x_A - |y_A-\alpha_A|$$, $$u_B(x_B,y_B) = x_B - |y_B-\alpha_B|$$, where $$\alpha_A \geq 0$$, and $$\alpha_B\geq 0$$ are given.
• $$\omega = (\omega_X, \omega_Y)$$, where $$\omega_X>0$$, and $$\omega_Y>0$$

Set of feasible allocations is given by

$$\mathcal{F} =\{((x_A,y_A),(x_B,y_B))\in\mathbb{R}^2_+\times\mathbb{R}^2_+|x_A+x_B=\omega_X \ \wedge \ y_A+y_B=\omega_Y \}$$

Set of Pareto efficient Allocations is given by

$$\{((x_A,y_A),(x_B,y_B))\in\mathcal{F}|\min(\alpha_A,\omega_Y-\alpha_B)\leq y_A \leq \max(\alpha_A,\omega_Y-\alpha_B)\}$$

Let’s rename the variables $$x := x_1$$, $$y:= x_2$$.

First let’s consider the set of feasible allocations such that $$y_i \neq \alpha_i$$ for both $$i = A,B$$.

In this set, both utilities are completely differentiable.

The optimization problem we want to solve is

$$\max x_A - |y_A - \alpha_A|$$

$$s.t. x_B - |y_B - \alpha_B| = \overline{U}$$

and the endowment constraints

$$x_A + x_B = \omega_x$$

$$y_A + y_B = \omega_y$$

where $$(\omega_x,\omega_y)$$ are the total endowments for $$(x,y)$$.

Since both utilities are completely differentiable in this region, the efficiency condition is

$$MRS_A = MRS_B$$

$$\frac{\frac{\partial U_A}{\partial x_A}}{\frac{\partial U_A}{\partial y_A}} = \frac{\frac{\partial U_B}{\partial x_B}}{\frac{\partial U_B}{\partial y_B}}$$

Note $$\frac{d}{dx} |x| = sign(x)$$ for $$x \neq 0$$.

$$\frac{1}{- sign(y_A - \alpha_A)} = \frac{1}{- sign(y_B - \alpha_B)}$$

$$sign(y_A - \alpha_A) = sign(y_B - \alpha_B)$$

This equation holds exactly when $$\{y_A - \alpha_A, y_B - \alpha_B \}$$ are either both positive or both negative.

Now for the points where for some agent $$i$$ it happens that $$y_i = \alpha_i$$, let’s do a contour plot of indifference curves.

For the contour plot, note from the $$MRS_i$$ that the indifference curves are two connected line segments, each with slope $$-MRS_i = sign(y_i - \alpha_i)$$.

Let’s hold an agent’s indifference curve constant. Pick a point where $$y_i = \alpha_i$$ for either $$i$$. You can check that moving along the indifference curve, you can only move into the same or a worse indifference curve for the other agent.

This implies the points where $$y_i = \alpha_i$$ for either $$i$$ are efficient as well.

Therefore, the contract curve/efficient set $$CC$$ is given by

$$CC = \{ ((x_A,y_A),(x_B,y_B)) \in \mathcal{F} : [y_A \geq \alpha_A \text{ and } y_B \geq \alpha_B] \text{ or } [y_A \leq \alpha_A \text{ and } y_B \leq \alpha_B] \}$$

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

Here I’ll show all the $$11$$ possible cases graphically depending on where in the box are the $$\alpha_i$$ relative to each other and the endowment constraints (the signs I write on the right of the box are the corresponding signs of the $$y_i - \alpha_i$$):

Note: Whatever is in red corresponds to agent $$A$$, and whatever is in black corresponds to agent $$B$$.

1. $$CC$$ is the below region.
2. $$CC$$ is the above region.
3. $$CC$$ is the entire box.

1. $$CC$$ is the entire box.

1. $$CC$$ is the below region.
2. $$CC$$ is the above region.

1. $$CC$$ is the below border.
2. $$CC$$ is the above border.

Note for 10) and 11), my algebraic description of $$CC$$ yields no solution, so this means you have to graphically search for frontier solutions.