Question

Suppose there are only three goods \(\left(x_{1}, x_{2}, x_{3}\right)\) in an economy and that the excess demand functions for \(x_{2}\) and \(x_{3}\) are given by \\[ \begin{array}{l} E D_{2}=-\frac{3 p_{2}}{p_{1}}+\frac{2 p_{3}}{p_{1}}-1 \\ E D_{3}=-\frac{4 p_{2}}{p_{1}}-\frac{2 p_{3}}{p_{1}}-2 \end{array} \\] a Show that these functions are homogencous of degree 0 in \(p_{1}, p_{2},\) and \(p_{3}\) b. Use Walras' law to show that, if \(E D_{2}=E D_{3}=0,\) then \(E D_{1}\) must also be \(0 .\) Can you also use Walras' law to calculate \(E D_{1} ?\) c. Solve this system of equations for the equilibrium relative prices \(p_{2} / p_{1}\) and \(p_{3} / p_{1}\). What is the equilibrium value for \(p_{3} / p_{2} ?\)

Short answer

In summary, the excess demand functions for goods \(x_2\) and \(x_3\) are homogenous of degree 0 in prices. Walras' Law suggests that when the excess demand for goods \(x_2\) and \(x_3\) are zero, the excess demand for good \(x_1\) must also be zero. We cannot directly determine the excess demand function for \(x_1\) but know that \(E D_{1} = 0\) at equilibrium. After solving the given system of equations, we found that the equilibrium relative prices are \(p_{2}/p_{1}=\frac{1}{2}\) and \(p_{3}/p_{1}=1\), which gives a \(p_{3}/p_{2}\) value of 2.

Step-by-step solution

a. Check if excess demand functions are homogenous of degree 0

Homogenous of degree 0 means that if we multiply all the prices by any positive constant, the excess demand functions will remain the same. Let's say \(k>0\) is a constant, then we have: \\[ \begin{array}{l} E D_{2}(k p_{1}, k p_{2}, k p_{3})=-\frac{3 k p_{2}}{k p_{1}}+\frac{2 k p_{3}}{k p_{1}}-1 \\ E D_{3}(k p_{1}, k p_{2}, k p_{3})=-\frac{4 k p_{2}}{k p_{1}}-\frac{2 k p_{3}}{k p_{1}}-2 \end{array} \\] Simplifying the expressions, we see that the constant \(k\) cancels out: \\[ \begin{array}{l} E D_{2}(k p_{1}, k p_{2}, k p_{3})=-\frac{3 p_{2}}{p_{1}}+\frac{2 p_{3}}{p_{1}}-1 \\ E D_{3}(k p_{1}, k p_{2}, k p_{3})=-\frac{4 p_{2}}{p_{1}}-\frac{2 p_{3}}{p_{1}}-2 \end{array} \\] These functions are the same as the original excess demand functions, so they are homogenous of degree 0 in prices.

b. Apply Walras' Law to show \(E D_{1}\) must be 0 and try to directly calculate it

Walras' law states that the sum of the values of excess demand (quantity multiplied by price) for all goods in an economy is always equal to zero: \(p_{1}E D_{1} + p_{2}E D_{2} + p_{3}E D_{3} = 0\). We are given that \(E D_{2}=E D_{3}=0\), so: \\[ p_{1}E D_{1} + p_{2}(0) + p_{3}(0) = 0 \\ p_{1}E D_{1} = 0 \\] Since \(p_{1}\) is a price, it cannot be zero, which means that \(E D_{1} = 0\). Unfortunately, we cannot directly calculate the \(E D_{1}\) function using Walras' law without knowing the excess demand function for \(x_1\). But we can conclude that it must be zero when \(E D_{2}\) and \(E D_{3}\) are zero.

c. Solve the system of equations for equilibrium relative prices and find \(p_{3}/p_{2}\)

In equilibrium, excess demand for both goods \(x_2\) and \(x_3\) will be zero. Thus, \\[ \begin{cases} -\frac{3p_{2}}{p_{1}}+\frac{2p_{3}}{p_{1}} - 1 = 0 \\ -\frac{4p_{2}}{p_{1}}-\frac{2p_{3}}{p_{1}} - 2 = 0 \end{cases} \\] Solve for \(p_{2}/p_{1}\) and \(p_{3}/p_{1}\): The first equation gives: \\[ \frac{2p_{3}}{p_{1}} = \frac{3p_{2}}{p_{1}}-1 \\] The second equation gives: \\[ \frac{2p_{3}}{p_{1}} = \frac{4p_{2}}{p_{1}}-2 \\] From the first equation, eliminate the \(p_{1}\) from the fraction by dividing both sides by 2: \\[ \frac{p_{3}}{p_{1}} = \frac{3p_{2}}{2p_{1}}-\frac{1}{2} \\] From the second equation, eliminate the \(p_{1}\) from the fraction: \\[ \frac{p_{3}}{p_{1}} = \frac{4p_{2}}{2p_{1}}-1 \\] Now we can equate the two expressions for \(p_{3}/p_{1}\): \\[ \frac{3p_{2}}{2p_{1}}-\frac{1}{2} = \frac{4p_{2}}{2p_{1}}-1 \\] Solving the equation, we get \(p_{2}/p_{1}=\frac{1}{2}\). By substituting this value to the expression for \(p_{3}/p_{1}\), we obtain \(p_{3}/p_{1}=1\). Finally, we find the equilibrium value for \(p_{3}/p_{2}\): \\[ \frac{p_{3}/p_{1}}{p_{2}/p_{1}} = \frac{1}{\frac{1}{2}} = 2 \\] So, in equilibrium, \(p_{3}/p_{2}=2\).

Key concepts

Walras' Law

Walras' Law is a crucial concept in general equilibrium theory that provides a straightforward yet powerful insight into the behavior of economies. Essentially, it states that in any market economy, the sum of the value of all excess demands across all markets must equal zero.

This implies that if some markets have excess demand, there must be other markets with excess supply to maintain balance. In mathematical terms, if the excess demand for goods in an economy is denoted by \(E D_i\) and the prices by \(p_i\), Walras' law is expressed as:
\[ \sum_{i} p_i E D_i = 0 \]
Walras' Law becomes particularly valuable when analyzing economies at equilibrium, where markets are balanced.

• It confirms that if excess demand for certain goods is zero, excess supply elsewhere must also equate to zero.
• In practical terms, if two goods' excess demands are \(E D_2 = 0\) and \(E D_3 = 0\), then the law dictates \(E D_1\) must be zero too for the entire economy's sum to be zero.

This integral concept highlights how interconnected markets are - changes in one market will simultaneously affect others until overall equilibrium is restored.

Homogeneous of Degree Zero

The property of homogeneity of degree zero is crucial in economics, particularly in understanding demand functions. A function being homogeneous of degree zero implies that if all input prices are scaled by the same positive factor, the function's output remains unchanged.

This property is important for price functions because it tells us that relative prices, rather than absolute prices, are what consumers and producers respond to. In the case of excess demand functions, it means:
The function \(E D_i(kp_1, kp_2, kp_3) = E D_i(p_1, p_2, p_3)\), where \(k\) is any positive constant.
Simplification proves that any multiplication of prices does not alter the excess demand calculations:
\[ \begin{align*} E D_{2}(k p_{1}, k p_{2}, k p_{3}) & = -\frac{3 kp_{2}}{kp_{1}} + \frac{2 kp_{3}}{kp_{1}} - 1 = E D_{2}(p_{1}, p_{2}, p_{3}) \ E D_{3}(k p_{1}, k p_{2}, k p_{3}) & = -\frac{4 kp_{2}}{kp_{1}} - \frac{2 kp_{3}}{kp_{1}} - 2 = E D_{3}(p_{1}, p_{2}, p_{3}) \end{align*} \] This characteristic reassures us that only the ratios between prices matter in these functions.

Equilibrium Relative Prices

Equilibrium relative prices represent the balance point where excess demand functions for all goods are zero, and all markets clear. This means there are no surpluses or shortages, and the forces of supply and demand are at rest.

To determine these prices, often, systems of equations derived from excess demand functions are solved. In our example, with three goods, the task was to find the ratios \(p_2 / p_1\) and \(p_3 / p_1\):
Given:

• \[-\frac{3p_{2}}{p_{1}} + \frac{2p_{3}}{p_{1}} - 1 = 0\]
• \[-\frac{4p_{2}}{p_{1}} - \frac{2p_{3}}{p_{1}} - 2 = 0\]

The solutions lead to relative price ratios where:

• \(p_2 / p_1 = 1/2\)
• \(p_3 / p_1 = 1\)

From these, we further derive:
\[\frac{p_{3}}{p_{2}} = \frac{1}{\frac{1}{2}} = 2\] Thus, the equilibrium transforms into a coherent framework where all comparisons are consistent and stable, revealing interdependencies and variations among the goods.