I'm trying to solve the following problem:

Consider an exchange economy with two consumers, $A$ and $B$, whose utility functions are: \begin{align*} u_{A} & = x_1^A x_2^A \\ u_{B} & = x_1^B (x_2^B)^2 \end{align*} with endowments $\textbf{w}^A =(80 ; 150)$ and $\textbf{w}^B=(210; 180)$ respectively. Assume that consumer $A$ is a price setter, meaning he makes a take-it-or-leave-it price offer to consumer $B$.

  1. Find the equilibrium economy in this economy.
  2. Find the pareto efficient allocation (PEA) in this economy, and check if the equilibrium found in (1) is part of the PEA.
  3. Now suppose that instead of offering a price vector, consumer $A$ makes a take-it-or-leave-it offer to consumer $B$ over a consuption bundle. Find this bundle. Is it the same that the one found in (1). Why? Why not? Is this solution pareto efficient?

I think I should treat this problem as it was a sequential game, where player $A$ moves first. I know that player $B$ would never accept an offer that leaves him worse than in autarky, so consumer's $A$ offer should leave consumer $B$ in the same indiffence curve that goes through the endowment point. The problem is that I don't know how to solve the game mathematically and find that price vector. I thought of solving: \begin{align*} \max \quad U_A(x^A_1,x_2^A) \\ \text{st.} \; x_1^A + x_1^B(\textbf{p}) & = w_1 \\ x_2^A + x_2^B(\textbf{p}) & = w_2 \\ V(\textbf{p})_B = \bar{u} \end{align*} But I'm not getting anything clear.

  • 2
    $\begingroup$ Is $u_A = x_1^Ax_2^B$ or $x_1^Ax_2^A$? $\endgroup$
    – Amit
    Commented Sep 6, 2023 at 23:18

2 Answers 2


Given a pure exchange economy:

$u_A(x_A, y_A)=x_Ay_A$, $u_B(x_B, y_B)=x_By_B^2$

with endowments:

$\omega_A=(80,150)$ and $\omega_B=(210,180)$

  1. To find the equilibrium, we first find the price offer curve of $B$: \begin{eqnarray*} \max_{x_B\geq 0,y_B\geq 0} & x_By_B^2 \\ \text{s.t. } & p_Xx_B+y_B\leq 210p_X+180\end{eqnarray*}

and we get $x_B = \dfrac{70p_X+60}{p_X}$ and $y_B =140p_X+120$

Price offer curve of $B$ is locus of $(x_B, y_B)$ pairs as found above for different prices. It is given by the blue curve in the picture below.

$A$ will choose $p_X$ and $B$ will consume according to $x_B = \dfrac{70p_X+60}{p_X}$ and $y_B =140p_X+120$, so that $A$ gets to consume the residual. $A$ chooses that point on $B$'s offer curve that gives $A$ the highest satisfaction. In picture, it corresponds to the point of tangency of $A$'s IC and $B$'s offer curve. So, $A$ gets

$x_A = 290 -\dfrac{70p_X+60}{p_X} = 220-\dfrac{60}{p_X}$ and

$y_A =330-140p_X-120=210-140p_X$

and $A$ will choose $p_X$ by solving the following problem: \begin{eqnarray*} \max_{p_X\geq 0} & \left(220 -\dfrac{60}{p_X}\right)(210-140p_X) \end{eqnarray*} Solving it, we get $p_X=\dfrac{3}{\sqrt{22}}$. Corresponding allocation is $\left((x_A,y_A),(x_B,y_B)\right)=\left((220-20\sqrt{22},210-420/\sqrt{22}),(70+20\sqrt{22},120+420/\sqrt{22})\right)$. It is given by allocation $e$ in the picture below.

  1. Equilibrium allocation $e$ is not Pareto efficient because MRS of $A$ is not equal to MRS of $B$ at this allocation.

$\dfrac{210-420/\sqrt{22}}{220-20\sqrt{22}} = \text{MRS}_A > \text{MRS}_B = \dfrac{120+420/\sqrt{22}}{140+40\sqrt{22}}=\dfrac{3}{\sqrt{22}}=p_X$

  1. In this problem, $A$ chooses the point on the $B$'s IC passing through the endowment allocation, that gives $A$ the highest level of satisfaction.

$\left((x_A,y_A),(x_B,y_B)\right)$ satisfy:

$x_By_B^2=210(180^2)$ and

$\dfrac{y_A}{x_A} = \dfrac{y_B}{2x_B}$

and the feasibility conditions.

This point will be Pareto efficient and it corresponds to point $p$ in the picture below:

enter image description here

  • 1
    $\begingroup$ Incredible! thank you so much Amit $\endgroup$ Commented Sep 9, 2023 at 15:52

Not sure where you get lost. $u_B$ is Cobb-Douglas type, thus given a price vector $\textbf{p}$ and initial endowment $\textbf{w}^B$ it is easy to determine $x_1^B(\textbf{p}),x_2^B(\textbf{p})$.

Once you do these, the problem is indeed

\begin{align*} \max_\textbf{p} \quad u_A(x^A_1,x_2^A) \\ \text{st.} \; x_1^A + x_1^B(\textbf{p}) & = w_1 \\ x_2^A + x_2^B(\textbf{p}) & = w_2 \end{align*}

Pareto-optimality can be checked using the usual conditions for 2.

For 3., it is a similar deal to 1., but instead of a price vector, a bundle is offered by $A$, thus

\begin{align*} \max_{x_1^B,x_2^B} \quad u_A(x^A_1,x_2^A) \\ \text{st.} \; x_1^A + x_1^B & = w_1 \\ x_2^A + x_2^B & = w_2 \\ u_B(x_1^B,x_2^B) & \geq u_B(w_1^B,w_2^B) \end{align*}


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