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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.

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2 Answers 2

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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).

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    $\begingroup$ 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
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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 $\boxed{((0,1),(1,0))}$

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