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


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.


3 Answers 3


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.)

  • $\begingroup$ 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? $\endgroup$
    – DornerA
    Commented Jun 19, 2016 at 16:09
  • $\begingroup$ @DornerA I gave you a counterexample? $\endgroup$
    – Giskard
    Commented Jun 19, 2016 at 17:52
  • $\begingroup$ Is this why you posted the other question? Yes, the utility functions in the example fulfilled the conditions. $\endgroup$
    – Giskard
    Commented Jun 19, 2016 at 17:54
  • $\begingroup$ The statement is also false for linear utility functions. I have provided an example in my answer. $\endgroup$
    – Amit
    Commented May 2, 2022 at 7:35
  • $\begingroup$ @Amit You are correct! $\endgroup$
    – Giskard
    Commented May 2, 2022 at 9:24

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.


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.


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