# Pure Exchange Economy

I need help drawing the Pareto Set for an Edgeworth economy. I know how to find the contract curve given an allocation, and I think that ends up being the competitive equilibria, but drawing the Pareto line is much harder than anticipated.

We are given 2 goods and 2 agents, with log utility $(\alpha_i > 0)$:

$$u_i(x_i) = \alpha_i \ln x^1_i + \ln x^2_i$$

We get the tangency condition:

$$\alpha_1 \frac{x_1^2}{x_1^1} = \alpha_2 \frac{x_2^2}{x_2^1}$$

And combine with the resource constraints:

$$x_1^1 + x_2^1 = r^1$$ $$x_1^2 + x_2^2 = r^2$$

A lot of algebra:

$$\alpha_1 \frac{x_1^2}{x_1^1} = \alpha_2 \frac{r^2 - x^2_1}{r^1 - x^1_1}$$

$$\implies \alpha_1 (x^2_1 r^1 - x^2_1 x^1_1) = \alpha_2 (x^1_1 r^2 - x^1_1 x^2_1)$$

$$\implies \alpha_1 x^2_1 r^1 - \alpha_2 x^1_1 r^2 = (\alpha_1 - \alpha_2) x^1_1 x^2_1$$

$$(\alpha_1 \cdot \frac{1}{x^1_1} \cdot r^1) - (\alpha_2 \cdot \frac{1}{x^2_1} \cdot r^2) = \alpha_1 - \alpha_2$$

Which gets us:

$$\boxed{x_1^1 = \frac{\alpha_1 r^1 x_1^2}{\alpha_1 x_1^2 - \alpha_2 x_1^2 + \alpha_2 r^2}}$$

For our Pareto Set. (You could also solve in terms of $x_1^1$.)

As denesp notes, if $\alpha_1 = \alpha_2 - r^1 = r^2$, then $x_1^1 = x_1^2$.

The question is, how would I draw this for different values of $\alpha$? What is the intuition behind the slope of it?

• Lastly I am not sure I understand your question. What do you mean by shift? And the tangency condition only gives you Pareto efficient points in the interior of the box, but sometimes (with probability 1 in generic cases) you get Pareto efficient points on the side of the box. Nov 28 '15 at 19:11
• 1.) I miscalculated the Pareto Set completely, so I'll have to get back to you on that. 2.) When I mentioned shift, I meant how does the slope change throughout the box. 3.) With log utility, I think the edges of the box are infeasible, and therefore can't be Pareto. Nov 28 '15 at 20:48
• 1.) But in your question you say you know how to find the CC Did you mean given a PS you could find the CC? The edges of the box are feasible. It is the utility function that causes problems, but this can be cured by mapping to the 'extended real numbers' or by taking the monoton transform $x_1^\alpha \cdot x_2$. Two corners are definitely Pareto efficient. Nov 28 '15 at 21:09
• 1.) I meant to say that I could draw the CC given the PS, yes. 2.) Point taken. 3.) There is my correction. Nov 28 '15 at 21:18
• So if I understand correctly your question is: What does the function $f(x_1^2) = \frac{\alpha_1 r^1 x_1^2}{a_1 x_1^2 - \alpha_2 x_1^2 + \alpha_2 r^2}$ look like? Well $f(0) = 0$ and $f(r_2) = r_1$. I am guessing that the function is convex or concave depending on the parameters, and you can examine this by taking the second derivative. I am not sure if you are hoping for a deeper description than this one. Nov 28 '15 at 21:38

As you note in your question the (inner points of the) Pareto set are defined by $$x_1^1 = \frac{\alpha_1 r^1 x_1^2}{\alpha_1 x_1^2 - \alpha_2 x_1^2 + \alpha_2 r^2}$$ In order to better examine this curve, let us treat it as a function. Let $$f(x) = \frac{\alpha_1 r^1 x}{\alpha_1 x - \alpha_2 x + \alpha_2 r^2}.$$ As I stated in my comment $$f(0) = 0 \mbox{ and } f(r^2) = r^1,$$ so the set indeed goes from one corner to the other. In between you have $$\frac{d \ f(x)}{d x} = \frac{\alpha_1\alpha_2r^1r^2}{\left(\alpha_1 x - \alpha_2 x + \alpha_2 r^2\right)^2}$$ and $$\frac{d^2 f(x)}{dx^2} = 2 \cdot \frac{(\alpha_1\alpha_2r^1r^2) \cdot (\alpha_2-\alpha_1)}{\left(\alpha_1 x - \alpha_2 x + \alpha_2 r^2\right)^3}.$$ The first derivative is always positive because $r^2 > x$ and all other parameters are positive. For the same reason, the sign of the second derivative depends on the sign of $\alpha_2-\alpha_1$.