# Destruction in Exchange Economies

I am reading about exchange economies and ran across something that is counter intuitive.

How is it that in an exchange economy with no production that a person can make himself/herself better off by destroying a portion of his or her endowment? If there is no production how, in the long run, is a person better off destroying some of the available resource if that resource can't be replaced etc.?

Can anyone illustrate this in a simple manner to help with the intuition?

I know that if there could be some way that destroying a bit of a resource can drive up prices such that the remaining quantity is now more valuable than the original quantity at original prices. Is that what is happening here? If so, why and how?

• Possible duplicate of Can destruction be profitable? – dismalscience Nov 7 '15 at 16:50
• @dismalscience I don't think so, as an exchange economy is specified. None of the answers in the other question (which was almost too general) addresses this. – FooBar Nov 7 '15 at 16:58
• At op: Where did you read that claim? In none of the standard exchange economies I am aware off destruction is actually a Pareto improvement. I can somewhat remember odd results from my first-year courses on this area, but it was a very specific game-theoretic setup. – FooBar Nov 7 '15 at 17:00
• Your comments in the last paragraph basically answer your own question. That is what is happening. I am uncertain as to what you mean by why and how. Decreased supply and unchanged demand results in higher price. – Giskard Nov 7 '15 at 17:02
• @denesp I'm not convinced this happens in a standard setup. With two agents, one agent could always decide "not to trade" a specific quantity and enjoy it himself, instead of destroying it. All-in-all, I think this warrants a proper numeric example. – FooBar Nov 7 '15 at 17:16

Consider an exchange economy with two goods $x$ and $y$ and two agents $A$ and $B$. The utility functions are $$U_A(x_A,y_A) = x_A \cdot y_A, \hskip 20pt U_B(x_B,y_B) = \min(x_B,y_B).$$ Let the initial endowments be $$\omega_A = (\omega_A^x,\omega_A^y) = (z,0), \hskip 20pt \omega_B = (\omega_B^x,\omega_B^y) = (0,4),$$ where $0 \leq z \leq 8$ is left as a parameter. In equilibrium $$x_A = \frac{z}{2}, \hskip 20pt y_A = \left(4 - \frac{z}{2}\right), \hskip 20pt x_B = \frac{z}{2}, \hskip 20pt y_B = \frac{z}{2}.$$ (A detailed calculation showing this is at the bottom of this answer.)

The utility of agent $A$ is $$U_A\left(\frac{z}{2},4 - \frac{z}{2}\right) = \frac{z}{2} \cdot \left(4 - \frac{z}{2}\right).$$ The maximum of this is at $z = 4$. If the initial endowment of agent $A$ is larger than 4 he has incentive to destroy part of it, because assuming a competitive equilibrium outcome this will improve his utility.

As @Foobar pointed out in the comments further utility could be gained by not destroying but simply withholding part of the initial endowment. There are two arguments to be made for destruction:

1. Agent $B$ sees the supply and refuses the monopolistic outcome, insisting on the competitive price ratio.

2. One could divide agent $A$ into several 'smaller' agents. Agents who would have the same utility functions and equal shares of the initial endowment. In this case withholding part of the endowment is a cartel move. It is not individually rational. One could argue that destroying the 'surplus' is a commitment necessary for coordination.

In equilibrium total demand equals total supply in both markets. \begin{eqnarray*} x_A + x_B & = & \omega_A^x + \omega_B^x \\ \\ y_A + y_B & = & \omega_A^y + \omega_B^y. \end{eqnarray*} Our next step is determining demand. Here is the information we gather from the utility functions:

Let $y$ the numeraire good, i.e. let $p_y=1$. Let us denote the price of $x$ by simply $p$. The value of the initial endowment of agent $A$ is $z \cdot p$. Using the Cobb-Douglas property $$x_A = \frac{z\cdot p}{2\cdot p} = \frac{z}{2}, \hskip 20pt y_A = \frac{z\cdot p}{2}.$$ Using this we have \begin{eqnarray*} x_A + x_B & = & \omega_A^x + \omega_B^x \\ \\ \frac{z}{2} + x_B & = & z + 0 \\ \\ x_B & = & \frac{z}{2}. \end{eqnarray*} It follows from agent $B$'s utility function that in equilibrium $x_B = y_B$. Using this we have \begin{eqnarray*} y_A + y_B & = & \omega_A^y + \omega_B^y \\ \\ y_A + x_B & = & \omega_A^y + \omega_B^y \\ \\ y_A + \frac{z}{2} & = & 4 \\ \\ y_A & = & 4 - \frac{z}{2}. \end{eqnarray*}

Consider another example:

Consider a pure exchange economy with two goods X and Y and two consumers A and B.

Suppose utility functions are $$u_A(x_A, y_A) = \min(x_A, y_A)$$ and $$u_B(x_B, y_B) = \min(x_B, y_B)$$

• Scenario 1:

Endowment of A is $$\omega_A = (0, 5)$$ and endowment of B is $$\omega_B = (10, 0)$$.

Equilibrium price vector $(p_x, p_y)$ and allocation $((x_A, y_A), (x_B, y_B))$ satisfy the following:

Optimality Conditions (Allocation must solve the utility maximization problem of the two consumers, i.e. it must lie on the demand functions)

• $(x_A, y_A) \in \begin{cases} \left\{\left(\frac{5p_y}{p_x + p_y}, \frac{5p_y}{p_x + p_y}\right)\right\} & \text{if } p_x > 0 \text{ and } p_y > 0 \\ \left\{(x,y)\in\mathbb{R}^2_+: x = 0, y\geq 0\right\} & \text{if } p_x > 0 \text{ and } p_y = 0 \\ \left\{(x,y)\in\mathbb{R}^2_+: x \geq 5, y= 5\right\} & \text{if } p_x = 0 \text{ and } p_y > 0 \end{cases}$
• $(x_B, y_B) \in \begin{cases} \left\{\left(\frac{10p_x}{p_x + p_y}, \frac{10p_y}{p_x + p_y}\right)\right\} & \text{if } p_x > 0 \text{ and } p_y > 0 \\ \left\{(x,y)\in\mathbb{R}^2_+: x = 10, y\geq 10\right\} & \text{if } p_x > 0 \text{ and } p_y = 0 \\ \left\{(x,y)\in\mathbb{R}^2_+: x \geq 0, y= 0\right\} & \text{if } p_x = 0 \text{ and } p_y > 0 \end{cases}$

Feasibility Conditions

• $x_A + x_B = 10$
• $y_A + y_B = 5$

Clearly, any price vector $(p_x, p_y)$ satisfying $p_x = 0$ and $p_y > 0$ supports any of the allocations in the set $\{((x_A, 5), (x_B, 0))| 5 \leq x_A \leq 10, x_A+x_B = 10 \}$ as equilibrium. Utility of B in all these competitive equilibria is 0, and A gets everything.

• Scenario 2:

Suppose B destroys part of his endowment and the revised endowment of A is $$\omega_A = (0, 5)$$ and endowment of B is $$\omega_B = (4, 0)$$ In this new problem, optimality conditions are

• $(x_A, y_A) \in \begin{cases} \left\{\left(\frac{5p_y}{p_x + p_y}, \frac{5p_y}{p_x + p_y}\right)\right\} & \text{if } p_x > 0 \text{ and } p_y > 0 \\ \left\{(x,y)\in\mathbb{R}^2_+: x = 0, y\geq 0\right\} & \text{if } p_x > 0 \text{ and } p_y = 0 \\ \left\{(x,y)\in\mathbb{R}^2_+: x \geq 5, y= 5\right\} & \text{if } p_x = 0 \text{ and } p_y > 0 \end{cases}$
• $(x_B, y_B) \in \begin{cases} \left\{\left(\frac{4p_x}{p_x + p_y}, \frac{4p_y}{p_x + p_y}\right)\right\} & \text{if } p_x > 0 \text{ and } p_y > 0 \\ \left\{(x,y)\in\mathbb{R}^2_+: x = 4, y\geq 4\right\} & \text{if } p_x > 0 \text{ and } p_y = 0 \\ \left\{(x,y)\in\mathbb{R}^2_+: x \geq 0, y= 0\right\} & \text{if } p_x = 0 \text{ and } p_y > 0 \end{cases}$

Feasibility Conditions

• $x_A + x_B = 4$
• $y_A + y_B = 5$

Clearly, any price vector $(p_x, p_y)$ satisfying $p_x > 0$ and $p_y = 0$ supports any of the allocations in the set $\{((0, y_A), (4, y_B))| 4 \leq y_B \leq 5, y_A+y_B = 5 \}$ as equilibrium. Now utility of A in all these competitive equilibria is 0, and B gets everything. B's utility has gone up to 4 in comparison to Scenario 1 where his utility was 0. This happened because he destroyed part of his endowment.

• For finding Competitive Equilibrium in scenario 1 using graphical method, you may refer to this video: youtube.com/… – Amit Apr 15 '17 at 6:04

The result that destruction can be beneficial comes from a paper by Robert Aumann and Bezalel Peleg, "A Note on Gale's Example," Journal of Mathematical Economics 1 (1974) pp. 209-211.