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

Expert verified
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

01

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.
02

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.
03

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, \(p_{3}/p_{2}=2\).

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Most popular questions from this chapter

Smith and Jones are stranded on a desert island. Each has in his possession some slices of ham \((H)\) and cheese (C). Smith is a choosy eater and will eat ham and cheese only in the fixed proportions of 2 slices of cheese to 1 slice of ham. His utility function is given by \(U_{s}=\min (H, C / 2)\) Jones is more flexible in his dietary tastes and has a utility function given by \(U_{j}=4 H+3 C\). Total endowments are 100 slices of ham and 200 slices of cheese. a. Draw the Edgeworth box diagram that represents the possibilitics for exchange in this situation. What is the only exchange ratio that can prevail in any equilibrium? b. Suppose Smith initially had \(40 \mathrm{H}\) and \(80 \mathrm{C}\). What would the equilibrium position be? c. Suppose Smith initially had \(60 H\) and \(80 C\). What would the equilibrium position be? d. Suppose Smith (much the stronger of the two) decides not to play by the rules of the game. Then what could the final equilibrium position be?

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