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.
Note this question is dublicated.
Your solution for $\alpha=1/2$ is incorrect. Consider the allocation $[(1,0),(1,1)]$. The allocation $[(2,0),(0,1)]$ is a Pareto improvement as $A$'s utility increases from $1$ to $2$ while $B$'s utility is $1$ for both allocations. Also, consider an allocation on the green line, e.g. $[(1,1/2),(1,1/2)]$. The allocation $[(2,0),(0,1)]$ is a Pareto improvement as $A$'s utility increases from $1$ to $2$ while $B$'s utility is $1$ for both allocations.
Your solution for $\alpha=2$ is also incorrect. Consider an allocation on the green line, e.g. $[(1,2),(1,2)]$. The allocation $[(0,4),(2,0)]$ is a Pareto improvement as $A$'s utility increases from $2$ to $4$ while $B$'s utility is $2$ for both allocations.
I think in finding the green lines in your solution you have restricted yourself to looking at indifference curves whose vertices lie inside the Edgeworth box. This is not a requirement. For the case of $\alpha=2$, the Pareto improvement $[(0,4),(2,0)]$ identified above is on the indifference curve of $A$ with vertext at $(x_A,y_A)=(4,4)$.
Solution
Consider any allocation such that
$$x_A+x_B=2\quad \text { and }\quad y_A+y_B=2\alpha$$
These equations imply that
$$(x_A-y_A)+(x_B-y_B)=2(1-\alpha). \tag{1}$$
Cases where $\alpha\leq 1$
If $y_j\geq x_j$ for any $j\in\{A,B\}$, then by $(1)$ it must be that $x_k\geq y_k$ for $k\neq j$. This can only be Pareto optimal if $x_j=y_k=0$. If $x_j>0$ then we could reallocate all of $x_j$ to $k$ with no reduction in $j$'s utility and increase $k$'s utility. Similarly if $y_k>0$.
This gives us the following Pareto optimal allocations:
$$\{[(0,2\alpha),(2,0)], [(2,0),(0,2\alpha)]\} \tag{2}$$
The only other possibility is that $x_A>y_A$ and $x_B>y_B$. Consider a reallocation of all of good $x$ to $j$ and all of good $y$ to $k$. This is a Pareto improvement unless $x_j\geq 2$ and $x_k\geq 2\alpha$. This is not possible, so there are no more Pareto optimal allocations.
Cases where $\alpha>1$
If $x_j\geq y_j$ for any $j\in\{A,B\}$, then by $(1)$ it must be that $y_k\geq x_k$ for $k\neq j$, and so similarly to the other case we get the Pareto optimal allocations in $(2)$.
The only other possibility is that $y_A>x_A$ and $y_B>x_B$. Similarly to the other case, there are no Pareto optimal allocations of this sort.