I have a two-person exchange economy

Each agent has the following utility $u_i(x_i,y_i)=v(x_i)+y_i$ for agent $i=\{A,B\}$

Assume that $v$ is strictly concave and increasing function that has a continuous first derivative. $v(0)=0$ and $v(x)<1$.

Agent A has the endowment $(1,10)$. And agent B has the endowment $(0,10)$.

Question asks that “for each Pareto efficient allocation, suggest how we might change the endowments so that the Pareto efficient allocation in the question is a walrasian equilibrium.”

For that I tried to find Pareto efficient allocations and Walrasian equilibrium.

  1. Pareto efficient allocations exist at which $MRS_A=MRS_B$.

Then, for this economy, $v’(x_A)=v’(x_B)$

By feasibility condition: $ x_A+x_B=1$ and $y_A+y_B=20$.

Then P.O. Exists where $v’(x_A)=v’(1-x_A)$

2 I tried to define Walrasian equilibrium.

For that, I calculate consumer’s problem.

For agent A,

$$P_x/P_y=v’(x_A)$$ $$P_x/P_y=\frac{10-y_A}{x_A-1}$$

For agent B,

$$P_x/P_y=v’(x_B)$$ $$P_x/P_y=\frac{10-y_B}{x_B}$$

and market clearing conditions;

$ x_A+x_B=1$ and $y_A+y_B=20$.

After that, I cannot proceed the question. how should I solve this question ? Thank you for all your helps in advance.

  • $\begingroup$ How do you get $p = \frac{10-y_A}{x_A-1}$? $\endgroup$
    – Giskard
    Commented Sep 3, 2022 at 9:21
  • $\begingroup$ Simplest way to do this is: For any Pareto efficient allocation $a^{pe}$, change the endowment to the allocation $a^{pe}$, and then the resulting Walrasian equilibrium will be $a^{pe}$. $\endgroup$
    – Amit
    Commented Sep 3, 2022 at 9:42
  • $\begingroup$ How should I show what you suggest? I couldn’t imagine that @Amit I have no exact points which are P.O. and Walrasian equilibrium. How do I demonstrate what you said without such results? $\endgroup$
    – ThePooh
    Commented Sep 3, 2022 at 9:52
  • $\begingroup$ By using Lagrangian method @Giskard $\endgroup$
    – ThePooh
    Commented Sep 3, 2022 at 9:53
  • $\begingroup$ I mean please include your calculations, looks like you made a miscalculation somewhere. $\endgroup$
    – Giskard
    Commented Sep 3, 2022 at 11:25


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