Question

The Pareto model for income inequality specifies that the Lorenz curve should be of the formon [0, 1] and for a positive constant a which depends on the incomes of the group studied. a) For what values of a is L(x) a rational function (Hint: solve for the relevant expression set equal to 0, 1, 2, 3, etc and observe a pattern)? b) Write the formula for L(x) as a single fraction. c) What is the long-term behavior of L(x)? Is this meaningful in the context of income inequality?

Answer 1

- The expression given is:

`L(x)=1-\frac{1}{(1-x)^{(1)/(a)-1}}`

Question A:

- The question asks us to find the values of a such that L(x) is a rational function.

- To solve this question, we will rewrite the function by getting a common denominator, as follows:

`\begin{gathered} L(x)=1-\frac{1}{(1-x)^{(1)/(a)-1}} \\ \\ \text{Multiply numerator and denominator by }(1-x)^{(1)/(a)-1} \\ \\ L(x)=(1-\frac{1}{(1-x)^{(1)/(a)-1}})*\frac{(1-x)^{(1)/(a)-1}}{(1-x)^{(1)/(a)-1}} \\ \\ L(x)=\frac{(1-x)^{(1)/(a)-1}}{(1-x)^{(1)/(a)-1}}-\frac{(1-x)^{(1)/(a)-1}}{((1-x)^{(1)/(a)-1})^2} \\ \\ L(x)=\frac{(1-x)^{(1)/(a)-1}}{(1-x)^{(1)/(a)-1}}-\frac{1}{(1-x)^{(1)/(a)-1}} \\ \\ \text{The expression now has a common denominator, thus, we can combine both terms:} \\ \\ L(x)=\frac{(1-x)^{(1)/(a)-1}-1}{(1-x)^{(1)/(a)-1}} \end{gathered}`

- (This result answers Question B)

- Now that we have L(x) in this manner, we can infer that:

Since L(x) is a rational function, then, it means that the exponent on (1-x) must be a positive integer. Thus, let us assume this positive integer is n. Therefore, we can say:

Question B:

The question would like us to express L(x) as a rational function. We have already done this process in Question A. We can simply repeat the process here:

`\begin{gathered} L(x)=1-\frac{1}{(1-x)^{(1)/(a)-1}} \\ \\ \text{Multiply numerator and denominator by }(1-x)^{(1)/(a)-1} \\ \\ L(x)=(1-\frac{1}{(1-x)^{(1)/(a)-1}})*\frac{(1-x)^{(1)/(a)-1}}{(1-x)^{(1)/(a)-1}} \\ \\ L(x)=\frac{(1-x)^{(1)/(a)-1}}{(1-x)^{(1)/(a)-1}}-\frac{(1-x)^{(1)/(a)-1}}{((1-x)^{(1)/(a)-1})^2} \\ \\ L(x)=\frac{(1-x)^{(1)/(a)-1}}{(1-x)^{(1)/(a)-1}}-\frac{1}{(1-x)^{(1)/(a)-1}} \\ \\ \text{The expression now has a common denominator, thus, we can combine both terms:} \\ \\ L(x)=\frac{(1-x)^{(1)/(a)-1}-1}{(1-x)^{(1)/(a)-1}} \end{gathered}`

Question C:

The end behavior of a graph is the value L(x) as x tends to infinity.

- However, we do not know the value of "a". Because of this, the end value of the function can approach negative or positive infinity as x tends to negative or positive infinity. A third scenario arises when a = 1. This makes the function tend to zero. But a cannot be 1 because of the constraint that the function L(x) must be a rational function.

- The 2 scenarios described are explained further below: