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

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.

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:

• Wonderful explanation! Thank you :) Apr 12 at 15:24
• Thank you Nidhi :)
– Amit
Apr 13 at 7:30