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.


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
    Jun 19 '16 at 16:09
  • $\begingroup$ @DornerA I gave you a counterexample? $\endgroup$
    – Giskard
    Jun 19 '16 at 17:52
  • $\begingroup$ Is this why you posted the other question? Yes, the utility functions in the example fulfilled the conditions. $\endgroup$
    – Giskard
    Jun 19 '16 at 17:54

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.