Finding Walrasian equilibria when Walrasian demands are not unique
Finding Walrasian equilibria when Walrasian demands are not unique
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Nicolas Torres · External communityPost link
External question — Economics Stack Exchange
Author: Nicolas Torres
Original post: https://economics.stackexchange.com/questions/54741
License: CC BY-SA 4.0 — https://creativecommons.org/licenses/by-sa/4.0/
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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
$B$
.
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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Giskard · External communityPost link
External answer — Economics Stack Exchange
Author: Giskard
Original post: https://economics.stackexchange.com/a/54742
License: CC BY-SA 4.0 — https://creativecommons.org/licenses/by-sa/4.0/
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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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mynameparv · External communityPost link
External answer — Economics Stack Exchange
Author: mynameparv
Original post: https://economics.stackexchange.com/a/54747
License: CC BY-SA 4.0 — https://creativecommons.org/licenses/by-sa/4.0/
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Two Agents:
$\{A,B\}$
Two commodities:
$\{X,Y\}$
Endowments:
$\omega_A=\omega_B=\left(\frac{1}{2},\frac{1}{2}\right)$
Preferences:
$u_A(x_A,y_A)=x_A+y_A \quad u_B(x_B,y_B)=2x_A+y_A$
Given preferences of individual
$A$
, their demand correspondence is
$$(x_A^d,y_A^d)(p_x,p_y,m_A)\in\left\{\begin{matrix}
\left(\dfrac{m_A}{p_x},0\right) & \text{if }\dfrac{p_x}{p_y}<1\\
\left(0,\dfrac{m_A}{p_y}\right) & \text{if }\dfrac{p_x}{p_y}>1 \\
\{(x,y)\in \mathbb R^2_+\; \vert \; p_xx+p_yy=m_A\}& \text{if }\dfrac{p_x}{p_y}=1
\end{matrix}\right. $$
where
$(p_x,p_y,m_A)$
denotes the prices of commodity
$X$
and
$Y$
, and the income of agent
$A$
respectively.
Using similar notation, the demand correspondence for individual
$B$
is
$$(x_B^d,y_B^d)(p_x,p_y,m_B)\in\left\{\begin{matrix} \left(\dfrac{m_B}{p_x},0\right) & \text{if }\dfrac{p_x}{p_y}<2\\ \left(0,\dfrac{m_B}{p_y}\right) & \text{if } \dfrac{p_x}{p_y}>2 \\ \{(x,y)\in \mathbb R^2_+\; \vert \; p_xx+p_yy=m_B\} & \text{if }\dfrac{p_x}{p_y}=2 \end{matrix}\right.$$
Let commodity
$Y$
be the numeraire and normalize its price
$p_y\equiv 1$
. Given the value of the endowments, we get that
$m_A=m_B=0.5p_x+0.5$
and using this we can express the demand correspondences completely in terms of
$p_x$
as follows
$$(x_A^d,y_A^d)(p_x)\in\begin{cases}
\left(0.5+\frac{0.5}{p_x},0\right) & \text{if }p_x<1\\
(0,0.5p_x+0.5) & \text{if } p_x>1 \\
\{(x,y)\in \mathbb R^2_+\; \vert \; p_xx+y=0.5p_x+0.5\} & \text{if }p_x=1
\end{cases}
$$
$$(x_B^d,y_B^d)(p_x)\in\begin{cases} \left(0.5+\frac{0.5}{p_x},0\right) & \text{if }p_x<2\\ \left(0,0.5p_x+0.5\right) & \text{if } p_x>2 \\ \{(x,y)\in \mathbb R^2_+\; \vert \; p_xx+y=0.5p_x+0.5\}& \text{if }p_x=2 \end{cases}$$
We now check for market clearing conditions for commodity
$X$
as done in the table below
Price
Demand for $X$
Supply for $X$
$p_x<1$
$1+\frac{1}{p_x}(>2)$
$1$
$p_x=1$
$[1,2]$
$1$
$p_x \in (1,2)$
$0.5+\frac{0.5}{p_x}(<1)$
$1$
$p_x=2$
$[0,0.75] $
$1$
$p_x>2$
$0$
$1$
Observe that the market for commodity
$X$
clears only when
$p_x = 1$
. Since
$p_y = 1$
, Walras' Law ensures that the market for commodity
$Y$
must also clear simultaneously. Hence, the competitive equilibrium allocation is
$((0, 1), (1, 0))$
at the price vector
$(p_x, p_y) = (1, 1)$
.
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Quoted from Forex.com.bd-Editorial External answer — Economics Stack Exchange Author: Giskard Source score (net votes, not local likes): 0 Original post: https://economics.stackexchange.com/a/54742 License: CC BY-SA 4.0 — https://creativecommons.org/licenses/by-sa/4.0/ Adaptation: HTML converted to plain text; contact email addresses removed. 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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