<p>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 $r^2 &gt; x$ and all other parameters are positive. For the same reason, the sign of the second derivative depends on the sign of $\alpha_2-\alpha_1$. <br> Thus the Pareto set will be a strictly increasing curve, which is convex if agent 2 values good 1 'relatively more', concave if agent 1 values good 1 'relatively more', and a straight line if they value it 'relatively equally'.</p>

CC BY-SA 3.0
2015-11-29T04:48:59.363Z
The consistency of the concavity/convexity clarifies things greatly. Thank you very much! I'm actually slightly embarrassed I had not looked into the second derivative closer.
Kitsune Cavalry
CC BY-SA 3.0
2015-11-28T21:59:48.673Z
2015-11-30T20:21:49.767Z
2015-11-30T20:21:49.767Z
Giskard
<p>I need help drawing the Pareto Set for an Edgeworth economy. I know how to find the contract curve given an allocation, and I <em>think</em> that ends up being the competitive equilibria, but drawing the Pareto line is much harder than anticipated.</p>

<p>We are given 2 goods and 2 agents, with log utility $(\alpha_i &gt; 0)$:</p>

<p>$$u_i(x_i) = \alpha_i \ln x^1_i + \ln x^2_i$$</p>

<p>We get the tangency condition:</p>

<p>$$\alpha_1 \frac{x_1^2}{x_1^1} = \alpha_2 \frac{x_2^2}{x_2^1}$$</p>

<p>And combine with the resource constraints:</p>

<p>$$x_1^1 + x_2^1 = r^1$$
$$x_1^2 + x_2^2 = r^2$$</p>

<p>A lot of algebra:</p>

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

<p>$$\implies \alpha_1 (x^2_1 r^1 - x^2_1 x^1_1) = \alpha_2 (x^1_1 r^2 - x^1_1 x^2_1)$$</p>

<p>$$\implies \alpha_1 x^2_1 r^1 - \alpha_2 x^1_1 r^2 = (\alpha_1 - \alpha_2) x^1_1 x^2_1$$</p>

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

<p>Which gets us:</p>

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

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

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

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

CC BY-SA 3.0
2015-11-28T19:11:22.847Z
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.
Giskard
CC BY-SA 3.0
2015-11-28T20:48:27.587Z
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.
Kitsune Cavalry
CC BY-SA 3.0
2015-11-28T21:09:53.443Z
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.
Giskard
CC BY-SA 3.0
2015-11-28T21:18:57.060Z
1.) I meant to say that I could draw the CC given the PS, yes. 2.) Point taken. 3.) There is my correction.
Kitsune Cavalry
CC BY-SA 3.0
2015-11-28T21:38:57.590Z
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.
Giskard
<microeconomics><utility><general-equilibrium>
Pure Exchange Economy
Kitsune Cavalry
# Pure Exchange Economy

(Asked by: username_0 on 2015-11-28)

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?

**Tags:** microeconomics, utility, general-equilibrium

> **Comments:**
> **username_1**: 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.  
> **username_0**: 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.  
> **username_1**: 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.  
> **username_0**: 1.) I meant to say that I could draw the CC given the PS, yes. 2.) Point taken. 3.) There is my correction.  
> **username_1**: 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.  

---

## 1 answers

### Answer by username_1 on 2015-11-28. Score: 3 | Accepted answer

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 $\alpha\_2-\alpha\_1$.   
 Thus the Pareto set will be a strictly increasing curve, which is convex if agent 2 values good 1 'relatively more', concave if agent 1 values good 1 'relatively more', and a straight line if they value it 'relatively equally'.

> **Comments:**
> **username_0**: The consistency of the concavity/convexity clarifies things greatly. Thank you very much! I'm actually slightly embarrassed I had not looked into the second derivative closer.  

---

