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There is an exchange economy with two people and two goods.

Utility functions are

$u_A(x_A, y_A)=\max\{x_A, y_A\}$

$u_B(x_B, y_B)=\max\{x_B, y_B\}$

Endowments are $w_A(1,\alpha)$ and $w_B(1,\alpha)$ for $\alpha >0$

Find the set of Pareto efficient allocations and show them in the Edgeworth box.

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For that I consider three cases in terms of $\alpha$

Case 1: $\alpha=1$

Then we draw square-shaped edgeworth box. And the Pareto efficient allocations are only {$(0,2),(2,0)$} and {$(2,0),(0,2)$}.

Case 2: $\alpha >1$. Let’s assume that $\alpha=2$

Then we draw rectangle Edgeworth box. And Pareto efficient allocations are

{$(0,2),(2,2)$} and {$(2,2),(0,2)$} and the all point along the line between there two allocations. (Green line in the picture).

Case 3: $\alpha <1$. Let’s assume that $\alpha=1/2$

Then we draw rectangle Edgeworth box. And Pareto efficient allocations are

{$(1,0),(1,1)$} and {$(1,1),(1,0)$} and the all point along the line between there two allocations. (Green line in the picture).

Sorry for hand-writing picture but could not draw this in latex format. The case 1 is the simple version. However, I am not sure about the case 2 and case 3. ($\alpha>1 $ and $\alpha<1$). I think the Pareto efficient allocations which I found are wrong for cases 2 and 3. These seems not logical to me. Please discuss with me about the correct Pareto optimal allocations. Many thanks.

enter image description here

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    $\begingroup$ Hi! I made this Desmos graph for you, so you can experiment with it! A note: your utility functions are somewhat unusual, max instead of min, so the graph displays for both consumers all points with lower utility than the current allocations rather than just the level curve. $\endgroup$
    – Giskard
    Sep 4, 2022 at 14:30
  • $\begingroup$ For $\alpha = 2$, points on the green line are not Pareto efficient. Likewise for $\alpha=\frac{1}{2}$ also, points on the green line are not efficient. $\endgroup$
    – Amit
    Sep 5, 2022 at 0:18
  • $\begingroup$ I see the reasons why these green lines are not Pareto efficient. Then, likewise the case of (2,2), for the case $\alpha =2$, {(0,4), (2,0)} and {(2,0), (0,4)} are Pareto efficient. And for the case of $\alpha =1/2$, { (0,1),(2,0)} and {(2,0),(0,1))} are Pareto efficient. in general, {(2,0),(0,2$\alpha$)} and {(0,2$\alpha$),(2,0)} are Pareto efficient allocations. Is this right Professor @Amit? $\endgroup$
    – studentp
    Sep 5, 2022 at 2:36
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    $\begingroup$ posted the answer $\endgroup$
    – Amit
    Sep 6, 2022 at 0:06
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    $\begingroup$ @Giskard That's a good graph. $\endgroup$
    – Amit
    Sep 6, 2022 at 4:08

1 Answer 1

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We'll consider three cases:

  • For $\alpha \in [\frac{1}{2}, 2]$, there are only two Pareto efficient allocations: $\big\{\big((2,0),(0,2\alpha)\big), \big((0,2\alpha),(2,0)\big) \big\}$
  • For $\alpha > 2$, set of Pareto efficient allocations is $\big\{\big((2,0),(0,2\alpha)\big), \big((0,2\alpha),(2,0)\big) \big\} \bigcup \big\{\big((x_A,y_A),(x_B,y_B)\big)\in F: y_A > 2, y_B > 2 \big\} $, where $F$ denotes the set of all feasible allocations.
  • For $\alpha < \frac{1}{2}$, set of Pareto efficient allocations is $\big\{\big((2,0),(0,2\alpha)\big), \big((0,2\alpha),(2,0)\big) \big\} \bigcup \big\{\big((x_A,y_A),(x_B,y_B)\big)\in F: x_A > 2\alpha, x_B > 2\alpha \big\} $, where $F$ denotes the set of all feasible allocations.
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