<p>To complement densep's answer, here is a schematic Edgeworth box illustration of what can go wrong. The points on the dashed line are convex combinations of the Pareto optimal points $(x_1,x_2)$ and $(\hat{x}_1,\hat{x}_2)$, but the marked point is not Pareto optimal.</p>

<p><a href="https://i.sstatic.net/CR3Kg.png"><img src="https://i.sstatic.net/CR3Kg.png" alt="Edgeworth box with non-Pareto optimal convex combination."></a></p>

CC BY-SA 3.0
2016-06-19T11:20:53.653Z
2016-06-19T11:20:53.653Z
Ubiquitous
<p>Another counter example: $u_1(x_1, y_1) = 2x_1+y_1$ and $u_2(x_2, y_2)= x_2+2y_2$. Suppose the total endowment of $X$ and $Y$ in the economy is $(1,1)$. Consider the following two allocations:</p>

<ol>
<li>$(x_1, y_1) = (1,1)$ and $(x_2, y_2) = (0,0)$</li>
<li>$(x_1', y_1') = (0,0)$ and $(x_2', y_2') = (1,1)$</li>
</ol>

<p>Both these allocations are efficient. However, their convex combination allocation $(x_1'', y_1'') = (0.5,0.5)$ and $(x_2'', y_2'') = (0.5,0.5)$ is not efficient because allocation $(x_1''', y_1''') = (1,0)$ and $(x_2''', y_2''') = (0,1)$ is better for both.</p>

CC BY-SA 3.0
2017-02-16T15:33:51.477Z
2017-02-16T15:33:51.477Z
Amit
<p>Here 
$$\not\exists\;x_i^\star\; s.t.\; u_i(\alpha x_i)\geq u_i(\alpha x_i^\star)\;\forall i\;\text{and}\;u_i((1-\alpha)\hat{x}_i)\geq u_i((1-\alpha)x_i^\star)$$
$$\implies\not\exists\;x_i^\star\;s.t.\;u_i(\alpha x_i^\star+(1-\alpha)x_i^\star)\geq u_i(\alpha x_i+(1-\alpha)\hat{x}_i)$$
you assume that the function $u$ is linear. Unfortunately the statement is false for non-linear utility functions. Try 
$$
U_1(x_1,y_1) = x_1 \cdot y_1^2 \hskip 20pt U_2(x_2,y_2) = x_2^2 \cdot y_2.
$$
Given 1 unit of $x$ and $y$ each, the allocations
$$
(x_1,y_1) = (1,1) \hskip 20pt (x_2,y_2) = (0,0)
$$
and
$$
(x_1',y_1') = (0,0) \hskip 20pt (x_2',y_2') = (1,1)
$$
are both Pareto-optimal. However the points on the connecting $x = y$ line are not. You can verify this by comparing $MRS_1$ and $MRS_2$ for points on the line. Hence the Pareto-set is not convex in this case. (It is a curve connecting the two extreme allocations given above.)</p>

CC BY-SA 3.0
2016-06-19T16:09:23.673Z
I see where I went wrong, logically. So does this mean that the statement is not true? Or is there some other tactic I should apply when thinking about the proof?
DornerA
CC BY-SA 3.0
2016-06-19T17:52:15.663Z
@DornerA I gave you a counterexample?
Giskard
CC BY-SA 3.0
2016-06-19T17:54:57.893Z
Is this why you posted [the other](http://economics.stackexchange.com/questions/12426/is-the-cobb-douglas-utility-function-locally-non-satiated-at-0-0) question? Yes, the utility functions in the example fulfilled the conditions.
Giskard
CC BY-SA 4.0
2022-05-02T07:35:57.403Z
The statement is also false for linear utility functions. I have provided an example in my answer.
Amit
CC BY-SA 4.0
2022-05-02T09:24:26.887Z
@Amit You are correct!
Giskard
CC BY-SA 3.0
2016-06-19T06:00:15.647Z
2016-06-19T11:03:07.940Z
2016-06-19T11:03:07.940Z
Giskard
<p>I am studying for my qualifiers, and I ran into this question from a previous year's exam.</p>

<p>$\textbf{Question:}$</p>

<p>Consider a two-consumer two-good pure exchange economy. Both preferences are locally non-satiated and convex. Prove or disprove the following statement: if $(x_1,x_2)$ and $(\hat{x}_1,\hat{x}_2)$ are two different pareto optimal allocations, then the convex combination, $(\alpha x_1+(1-\alpha)\hat{x}_1,\alpha x_2+(1-\alpha)\hat{x}_2$ MUST also be pareto optimal for any $\alpha\in(0,1)$.</p>

<p>I believe the statement is true, and here is the work for my proof below.</p>

<p>$\textbf{My Proof:}$
By pareto optimality of $x_i$ and $\hat{x}_i$: 
$$\not\exists\; x_i^\star\; s.t.\; u_i(x_i^\star)\geq u_i(x_i)\;\forall i\; \text{and}\; u_i(x_i^\star)&gt; u_i(x_i)\;\text{for at least one }i \;\text{or}\;u_i(x_i^\star)\geq u_i(\hat{x}_i)\;\forall i\; \text{and}\; u_i(x_i^\star)&gt; u_i(\hat{x}_i)\;\text{for at least one }i$$
$$\implies\;u_i(\alpha x_i)\geq u_i(\alpha x_i^\star)\;\forall i\;\text{and}\;u_i((1-\alpha)\hat{x}_i)\geq u_i((1-\alpha)x_i^\star)\;\forall i$$
$$\implies\not\exists\;x_i^\star\;s.t.\;u_i(\alpha x_i^\star+(1-\alpha)x_i^\star)\geq u_i(\alpha x_i+(1-\alpha)\hat{x}_i)\;\forall i$$
$$\text{and}\;u_i(\alpha x_i^\star+(1-\alpha)x_i^\star)&gt; u_i(\alpha x_i+(1-\alpha)\hat{x}_i)\;\text{for at least one}\;i$$
$$\implies (\alpha x_i+(1-\alpha)\hat{x}_i)\;\text{is pareto optimal}$$
$$\blacksquare$$</p>

<p>This proof seemed almost too easy, so I'm wondering if it is correct/rigorous.</p>

<microeconomics><pareto-efficiency>
Is the Convex Combination of Two Pareto Optimal Allocations Also Pareto Optimal?
DornerA
# Is the Convex Combination of Two Pareto Optimal Allocations Also Pareto Optimal?

(Asked by: username_0 on 2016-06-18)

I am studying for my qualifiers, and I ran into this question from a previous year's exam.

$\textbf{Question:}$

Consider a two-consumer two-good pure exchange economy. Both preferences are locally non-satiated and convex. Prove or disprove the following statement: if $(x\_1,x\_2)$ and $(\hat{x}\_1,\hat{x}\_2)$ are two different pareto optimal allocations, then the convex combination, $(\alpha x\_1+(1-\alpha)\hat{x}\_1,\alpha x\_2+(1-\alpha)\hat{x}\_2$ MUST also be pareto optimal for any $\alpha\in(0,1)$.

I believe the statement is true, and here is the work for my proof below.

$\textbf{My Proof:}$
By pareto optimality of $x\_i$ and $\hat{x}\_i$:
$$\not\exists\; x\_i^\star\; s.t.\; u\_i(x\_i^\star)\geq u\_i(x\_i)\;\forall i\; \text{and}\; u\_i(x\_i^\star)> u\_i(x\_i)\;\text{for at least one }i \;\text{or}\;u\_i(x\_i^\star)\geq u\_i(\hat{x}\_i)\;\forall i\; \text{and}\; u\_i(x\_i^\star)> u\_i(\hat{x}\_i)\;\text{for at least one }i$$
$$\implies\;u\_i(\alpha x\_i)\geq u\_i(\alpha x\_i^\star)\;\forall i\;\text{and}\;u\_i((1-\alpha)\hat{x}\_i)\geq u\_i((1-\alpha)x\_i^\star)\;\forall i$$
$$\implies\not\exists\;x\_i^\star\;s.t.\;u\_i(\alpha x\_i^\star+(1-\alpha)x\_i^\star)\geq u\_i(\alpha x\_i+(1-\alpha)\hat{x}\_i)\;\forall i$$
$$\text{and}\;u\_i(\alpha x\_i^\star+(1-\alpha)x\_i^\star)> u\_i(\alpha x\_i+(1-\alpha)\hat{x}\_i)\;\text{for at least one}\;i$$
$$\implies (\alpha x\_i+(1-\alpha)\hat{x}\_i)\;\text{is pareto optimal}$$
$$\blacksquare$$

This proof seemed almost too easy, so I'm wondering if it is correct/rigorous.

**Tags:** microeconomics, pareto-efficiency


---

## 3 answers

### Answer by username_3 on 2016-06-19. Score: 4 | Accepted answer

Here
$$\not\exists\;x\_i^\star\; s.t.\; u\_i(\alpha x\_i)\geq u\_i(\alpha x\_i^\star)\;\forall i\;\text{and}\;u\_i((1-\alpha)\hat{x}\_i)\geq u\_i((1-\alpha)x\_i^\star)$$
$$\implies\not\exists\;x\_i^\star\;s.t.\;u\_i(\alpha x\_i^\star+(1-\alpha)x\_i^\star)\geq u\_i(\alpha x\_i+(1-\alpha)\hat{x}\_i)$$
you assume that the function $u$ is linear. Unfortunately the statement is false for non-linear utility functions. Try
$$
U\_1(x\_1,y\_1) = x\_1 \cdot y\_1^2 \hskip 20pt U\_2(x\_2,y\_2) = x\_2^2 \cdot y\_2.
$$
Given 1 unit of $x$ and $y$ each, the allocations
$$
(x\_1,y\_1) = (1,1) \hskip 20pt (x\_2,y\_2) = (0,0)
$$
and
$$
(x\_1',y\_1') = (0,0) \hskip 20pt (x\_2',y\_2') = (1,1)
$$
are both Pareto-optimal. However the points on the connecting $x = y$ line are not. You can verify this by comparing $MRS\_1$ and $MRS\_2$ for points on the line. Hence the Pareto-set is not convex in this case. (It is a curve connecting the two extreme allocations given above.)

> **Comments:**
> **username_0**: I see where I went wrong, logically. So does this mean that the statement is not true? Or is there some other tactic I should apply when thinking about the proof?  
> **username_3**: @username_0 I gave you a counterexample?  
> **username_3**: Is this why you posted [the other](http://economics.stackexchange.com/questions/12426/is-the-cobb-douglas-utility-function-locally-non-satiated-at-0-0) question? Yes, the utility functions in the example fulfilled the conditions.  
> **username_2**: The statement is also false for linear utility functions. I have provided an example in my answer.  
> **username_3**: @username_2 You are correct!  

---

### Answer by username_1 on 2016-06-19. Score: 7

To complement densep's answer, here is a schematic Edgeworth box illustration of what can go wrong. The points on the dashed line are convex combinations of the Pareto optimal points $(x\_1,x\_2)$ and $(\hat{x}\_1,\hat{x}\_2)$, but the marked point is not Pareto optimal.

[![Edgeworth box with non-Pareto optimal convex combination.](https://i.sstatic.net/CR3Kg.png)](https://i.sstatic.net/CR3Kg.png)


---

### Answer by username_2 on 2017-02-16. Score: 2

Another counter example: $u\_1(x\_1, y\_1) = 2x\_1+y\_1$ and $u\_2(x\_2, y\_2)= x\_2+2y\_2$. Suppose the total endowment of $X$ and $Y$ in the economy is $(1,1)$. Consider the following two allocations:

1. $(x\_1, y\_1) = (1,1)$ and $(x\_2, y\_2) = (0,0)$
2. $(x\_1', y\_1') = (0,0)$ and $(x\_2', y\_2') = (1,1)$

Both these allocations are efficient. However, their convex combination allocation $(x\_1'', y\_1'') = (0.5,0.5)$ and $(x\_2'', y\_2'') = (0.5,0.5)$ is not efficient because allocation $(x\_1''', y\_1''') = (1,0)$ and $(x\_2''', y\_2''') = (0,1)$ is better for both.


---

