I'm trying to solve the following excercise:

Find the Walrasian equilibria for a pure exchange economy where agents' ($A$ and $B$) preferences and endowments are given by:

$u_A = x_A + y_A$

$u_B = 2 x_A + y_A$

$(\omega_{xA},\omega_{yA}) = (\frac{1}{2},\frac{1}{2})$

$(\omega_{xB},\omega_{yB}) = (\frac{1}{2},\frac{1}{2})$

I think the specific parameters for the linear functions and endowments are needed for my issue regarding non-unique demands, that’s why I included them.

I computed the demands for both utilities as usual for linear functions, having set $p_x = 1$.

The demands I got (after replacing the endowments) are:

Agent $A$:

  • $p_y > 1$

$x_A^\star = \frac{1 + p_y}{2}, y_A^\star = 0$

  • $p_y < 1$

$x_A^\star = 0, y_A^\star = \frac{1 + p_y}{2 p_y}$

  • $p_y = 1$

$x_A^\star \in [0,1], y_A^\star = 1 - x_A^\star$

Agent $B$:

  • $p_y > \frac{1}{2}$

$x_B^\star = \frac{1+p_y}{2}, y_B^\star = 0$

  • $p_y < \frac{1}{2}$

$x_B^\star = 0, y_B^\star = \frac{1+p_y}{2 p_y}$

  • $p_y = \frac{1}{2}$

$x_B^\star \in [0,\frac{3}{4}], y_B^\star = \frac{3}{2} - 2 x_B^\star$

I then plot a $\mathbb{R}^+$ ray for $p_y$ with the demands by cases, taking into account the two different partitions generated by $A$ and $B$.

enter image description here

In all cases where $p_y \neq 1$, I get that there cannot be any Walrasian equilibrium.

However, in the case where $p_y = 1$, $A$'s demands are not unique and I'd get that both markets would be in equilibrium if I chose $x_A^\star = 0$, but the markets wouldn't be in equilibrium if instead I chose any $x_A^\star \in (0,1]$.

So would $p_y^\star = 1$ count as a Walrasian equilibrium or not?

I know there is a way to graphically show the equilibrium situations but I don't understand it with linear functions where $MRS$ arguments don't work.

I would appreciate it very much if someone could support their answer with an Edgeworth box graph.


2 Answers 2


An equilibrium is usually defined as the combination of a price(s) and quantity(/quantities/allocation).

Here $p_y = 1$, $x_A=0$, $y_A = 1$, $x_B=1$, $y_B=0$ is an equilibrium, $p_y = 1$ is an equilibrium price, the consumptions define an equilibrium allocation.

This is true even though given $p_y = 1$ the consumers can plan individually optimal consumptions that would not constitute an equilibrium in the economy (i.e.; $x_A = x_B = 1$, which is not feasible).

  • 1
    $\begingroup$ Thanks! I was confused as the definition I got for a Walrasian Equilibrium was a price vector (or relative prices). $\endgroup$ Mar 17, 2023 at 0:20

let us first write the demand functions of individual $A$ and $B$

$$(x_A^d,y_A^d)(p_x,p_y,m_A)\in\left\{\begin{matrix} (\frac{m_A}{p_x},0) & ,\frac{p_x}{p_y}<1\\ (0,\frac{m_A}{p_y}) & , \frac{p_x}{p_y}>1 \\ BL_A & , \frac{p_x}{p_y}=1 \end{matrix}\right. \\ where, \; BL_A: Budget \; Line \; of \; A$$

similarly, $$(x_B^d,y_B^d)(p_x,p_y,m_B)\in\left\{\begin{matrix} (\frac{m_B}{p_x},0) & ,\frac{p_x}{p_y}<2\\ (0,\frac{m_B}{p_y}) &, \frac{p_x}{p_y}>2 \\ BL_B & ,\frac{p_x}{p_y}=2 \end{matrix}\right.$$

In order to simplify our analysis, let us consider $p_y=1$ i.e., let good $Y$ be the numeraire. Using the value of the endowments we also get that $m_A=m_B=\frac{p_x}{2}+\frac{1}{2}$

now both demand functions become functions of $p_x$ only, so we need to use these demand functions and total endowments for $X$ to find $p_x^*$ such that demand for good $X$ equals the total endowment of good $X$ (supply of good $X$).

New demand functions are:

$(x_A^d,y_A^d)(p_x)\in\left\{\begin{matrix} (0.5+\frac{0.5}{p_x},0) & ,p_x<1\\ (0,0.5p_x+0.5) & , p_x>1 \\ BL_A & , p_x=1 \end{matrix}\right. \; (x_B^d,y_B^d)(p_x)\in\left\{\begin{matrix} (0.5+\frac{0.5}{p_x},0) & ,p_x<2\\ (0,0.5p_x+0.5) &, p_x>2 \\ BL_B & ,p_x=2 \end{matrix}\right. $

In market for good $X$:

Price Demand for $X$ Supply for $X$
$p_x<1$ $1+\frac{1}{p_x}(>2)$ $1$
$p_x=1$ $[1,2] (\ni 1)$ $1$
$p_x \in (1,2)$ $0.5+\frac{0.5}{p_x}(<1)$ $1$
$p_x=2$ $[0,0.75] $ $1$
$p_x>2$ $0$ $1$

Therefore, $\boxed{\frac{p_x^*}{p_y^*}=1}$ is the equilibrium price ratio supporting the allocation $\boxed{((0,1),(1,0))}$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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