How to find the contract curve for a funky utility involving the min operator?

Question

Suppose a pure exchange economy where agents’ ($A$ and $B$) preferences are given by the following utility functions:

$u_A = \min(3x+y,x+3y)$

$u_B = x^\frac{1}{2} y^\frac{1}{2}$

Find the contract curve.

The arguments of the min operator are equal $\iff x_A = y_A$

Now I split the Edgeworth box into $3$ regions.

• $x_A < y_A$

Here both utilities are differentiable and one of them is Cobb-Douglas, so I use $MRS$

In this region, $u_A = 3x + y$ as this is the smallest argument of the $\min$ in this region.

$MRS_A = MRS_B \iff 3 = \frac{y_B}{x_B}$

Therefore, the segment of the contract curve in this region is

$y_B = 3 x_B$

• $x_A > y_A$

Similarly as above, in this region, $u_A = x + 3y$, and

$MRS_A = MRS_B \iff \frac{1}{3} = \frac{y_B}{x_B}$

Therefore, the segment of the contract curve in this region is

$y_B = \frac{1}{3} x_B$

• $x_A = y_A$

This is the line where $u_A$ is not differentiable and it’s the set of kinks of $A$’s indifference curves.

Let $(\alpha,\beta)$ be the total endowments of $(x,y)$, respectively.

Since $x_A = y_A$,

$MRS_B = \frac{y_B}{x_B} = \frac{\beta - y_A}{\alpha - x_A} = \frac{\beta - x_A}{\alpha - x_A} \in (\frac{1}{3},3) $

$\iff \frac{1}{3} \alpha - \frac{1}{3} x_A < \beta - x_A < 3 \alpha - 3 x_A $

$\iff \frac{1}{3} \alpha + \frac{2}{3} x_A < \beta < 3 \alpha - 2 x_A$

Graphically we can see that if the slope of the Cobb-Douglas IC’s tangent is in the set $(\frac{1}{3},3)$, then this tangent is below $A$’s IC and hence $B$’s IC would intersect with $A$’s IC at exactly one point, namely, the kink.

We can see in the graph below, that $MRS_B \in (\frac{1}{3},3)$ implies that if we deviate from the kink along $A$’s IC, we would end up in a worse Cobb-Douglas IC.

This would imply that the kink is a Pareto efficient point.

If $MRS_B$ happened to be either of the endpoints of the interval, then the Cobb-Douglas IC’s tangent would be one of $A$’s IC line segments, and $B$’s IC would still intersect $A$’s IC exactly at the kink, implying the kink is still a Pareto efficient point.

On the other hand, if $MRS_B < \frac{1}{3}$ or $MRS_B > 3$, then the Cobb-Douglas IC’s tangent goes above $A$’s IC in some region, implying both ICs now intersect at a second point.

Taking any non-trivial convex combination of those two points would yield a point on the same IC curve for $A$ that lies on a strictly better Cobb-Douglas IC, as seen in the graph below.

This would imply that the kink is not a Pareto efficient point.

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I also found conditions for both regions where $x_A \neq y_A$ in terms of the total endowments $(\alpha,\beta)$.

Note $MRS_B = 1 \in (\frac{1}{3},3)$ along $x_A = y_A$ for symmetric endowments $(\alpha = \beta)$.

• For $x_A < y_A$

Rewriting the contract curve in terms of agent $A$,

$y_A = 3 x_A + \beta - 3 \alpha$

I got $x_A < y_A \iff x_A > \frac{3}{2} \alpha - \frac{1}{2} \beta$.

Note this inequality is impossible for symmetric endowments $(\alpha = \beta)$ as it would become $x_A > \alpha$, i.e. agent $A$ consuming more $x$ than the total endowment.

• $x_A > y_A$

Rewriting the contract curve in terms of agent $A$,

$y_A = \frac{1}{3} x_A + \beta - \frac{1}{3} \alpha$

I got $x_A > y_A \iff x_A > \frac{3}{2} \beta - \frac{1}{2} \alpha$.

Note this inequality is impossible for symmetric endowments $(\alpha = \beta)$ as it would become $x_A > \alpha$, i.e. agent $A$ consuming more $x$ than the total endowment.

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By the notes above, I got that for symmetric endowments, the contract curve is simply $x_A = y_A$, which would be the line segment connecting both origins of the Edgeworth box.

However, I am having trouble visualizing all the conditions involving the endowments at the same time (for non-symmetric endowments), to get a final answer for the contract curve I can actually graph. I would appreciate any help on this part.

Answer 1

Given a pure-exchange economy:

• $u_1(x_1,y_1) = \min(3x_1+y_1,x_1+3y_1)$, $u_2(x_2,y_2)= x_2^{0.5}y_2^{0.5}$
• Total Endowments of X and Y are $\omega^X > 0$, and $\omega^Y > 0$, respectively.

Set of feasible allocations is $\mathcal{F} = \{((x_1,y_1),(x_2,y_2))\in\mathbb{R}^2_+\times\mathbb{R}^2_+|x_1+x_2 = \omega^X, y_1+y_2 = \omega^Y\}$

Set of Pareto efficient allocations is given by

$\begin{cases} \{((x_1,y_1),(x_2,y_2))\in\mathcal{F}|y_1=\min(x_1, \omega^Y - \frac{1}{3}(\omega^X-x_1))\} & \text{if } \omega^Y < \omega^X \leq 3\omega^Y \\ \{((x_1,y_1),(x_2,y_2))\in\mathcal{F}|y_1=x_1\} & \text{if } \omega^Y = \omega^X \\ \{((x_1,y_1),(x_2,y_2))\in\mathcal{F}|y_1=\max(0, \omega^Y - \frac{1}{3}(\omega^X-x_1))\} & \text{if } \omega^X > 3\omega^Y \\ \{((x_1,y_1),(x_2,y_2))\in\mathcal{F}|(y_2= 3x_2) \ \vee (x_1=0 \ \wedge \ y_2 > 3\omega^X) \} & \text{if } \omega^X < \frac{1}{3}\omega^Y \\ \{((x_1,y_1),(x_2,y_2))\in\mathcal{F}|y_1=\max(x_1, \omega^Y - 3(\omega^X-x_1))\} & \text{if } \frac{1}{3}\omega^Y \leq \omega^X < \omega^Y \end{cases} $

Here are some worked-out examples covering each of the five cases: