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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?

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    $\begingroup$ @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 $x_1,x_2$. $\endgroup$ May 23, 2023 at 14:21

2 Answers 2

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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

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

enter image description here

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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 $B$.

enter image description here

enter image description here

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

enter image description here

  1. $CC$ is the entire box.

enter image description here

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

enter image description here

  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.

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  • $\begingroup$ (+1) Good Answer. $\endgroup$
    – Amit
    May 23, 2023 at 7:16

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