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e. Compute the Gini index for the cases \( \mathrm{L}(\mathrm{x})=x^{p} \) and \( p=1.1,1.3,2,3,4 \)

Ask by Hills Macdonald. in the United States
Jan 27,2025

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The Gini index for \( \mathrm{L}(x) = x^{p} \) with \( p = 1.1, 1.3, 2, 3, 4 \) is calculated as follows: - \( p = 1.1 \): Gini index ≈ 4.76% - \( p = 1.3 \): Gini index ≈ 13.04% - \( p = 2 \): Gini index ≈ 33.33% - \( p = 3 \): Gini index = 50% - \( p = 4 \): Gini index = 60% As \( p \) increases, the Gini index increases, indicating greater inequality in the distribution.

Solución

To compute the **Gini index** for the cases where the Lorenz curve is defined by \( \mathrm{L}(x) = x^{p} \) with different values of \( p \), we can follow these steps: ### **Understanding the Lorenz Curve and Gini Index** - **Lorenz Curve (\( \mathrm{L}(x) \))**: Represents the distribution of income or wealth within a population. Here, \( \mathrm{L}(x) = x^{p} \) where \( x \) ranges from 0 to 1. - **Gini Index (G)**: Measures inequality within a distribution. It ranges from 0 (perfect equality) to 1 (maximal inequality). ### **Formula for Gini Index** The Gini index can be calculated from the Lorenz curve using the following formula: \[ G = 1 - 2 \int_{0}^{1} \mathrm{L}(x) \, dx \] For \( \mathrm{L}(x) = x^{p} \), the integral becomes: \[ \int_{0}^{1} x^{p} \, dx = \frac{1}{p + 1} \] Substituting back into the formula for \( G \): \[ G = 1 - 2 \left( \frac{1}{p + 1} \right) = \frac{p - 1}{p + 1} \] ### **Calculations for Different Values of \( p \)** Let's compute the Gini index for each specified value of \( p \): 1. **For \( p = 1.1 \):** \[ G = \frac{1.1 - 1}{1.1 + 1} = \frac{0.1}{2.1} \approx 0.0476 \quad \text{or} \quad 4.76\% \] 2. **For \( p = 1.3 \):** \[ G = \frac{1.3 - 1}{1.3 + 1} = \frac{0.3}{2.3} \approx 0.1304 \quad \text{or} \quad 13.04\% \] 3. **For \( p = 2 \):** \[ G = \frac{2 - 1}{2 + 1} = \frac{1}{3} \approx 0.3333 \quad \text{or} \quad 33.33\% \] 4. **For \( p = 3 \):** \[ G = \frac{3 - 1}{3 + 1} = \frac{2}{4} = 0.5 \quad \text{or} \quad 50\% \] 5. **For \( p = 4 \):** \[ G = \frac{4 - 1}{4 + 1} = \frac{3}{5} = 0.6 \quad \text{or} \quad 60\% \] ### **Summary of Gini Index Values** | \( p \) | Gini Index (\( G \)) | Percentage | |---------|----------------------|------------| | 1.1 | \( \approx 0.0476 \) | 4.76% | | 1.3 | \( \approx 0.1304 \) | 13.04% | | 2 | \( \approx 0.3333 \) | 33.33% | | 3 | \( 0.5 \) | 50% | | 4 | \( 0.6 \) | 60% | ### **Interpretation** - **Lower \( p \)** values (\( p = 1.1, 1.3 \)) indicate **less inequality** in the distribution. - **Higher \( p \)** values (\( p = 2, 3, 4 \)) signify **greater inequality**. This relationship shows that as \( p \) increases, the distribution becomes more unequal, which is reflected by the increasing Gini index.

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Bonus Knowledge

To compute the Gini index for the function \( L(x) = x^p \), we can use the formula for the Gini index derived from the Lorenz curve. The Gini index \( G \) is given by: \[ G = 1 - 2 \int_{0}^{1} L(x) \, dx \] For the given cases \( p=1.1, 1.3, 2, 3, 4 \): 1. **When \( p=1.1 \):** \[ G = 1 - 2 \int_0^1 x^{1.1} \, dx = 1 - 2 \left[ \frac{x^{2.1}}{2.1} \right]_0^1 = 1 - \frac{2}{2.1} = \frac{0.1}{2.1} \approx 0.0476 \] 2. **When \( p=1.3 \):** \[ G = 1 - 2 \int_0^1 x^{1.3} \, dx = 1 - 2 \left[ \frac{x^{2.3}}{2.3} \right]_0^1 = 1 - \frac{2}{2.3} = \frac{0.3}{2.3} \approx 0.1304 \] 3. **When \( p=2 \):** \[ G = 1 - 2 \int_0^1 x^2 \, dx = 1 - 2 \left[ \frac{x^3}{3} \right]_0^1 = 1 - \frac{2}{3} = \frac{1}{3} \approx 0.3333 \] 4. **When \( p=3 \):** \[ G = 1 - 2 \int_0^1 x^3 \, dx = 1 - 2 \left[ \frac{x^4}{4} \right]_0^1 = 1 - \frac{1}{2} = \frac{1}{2} = 0.5 \] 5. **When \( p=4 \):** \[ G = 1 - 2 \int_0^1 x^4 \, dx = 1 - 2 \left[ \frac{x^5}{5} \right]_0^1 = 1 - \frac{2}{5} = \frac{3}{5} = 0.6 \] In summary, the Gini indices for the cases \( p=1.1, 1.3, 2, 3, 4 \) are approximately: - \( p=1.1 \): 0.0476 - \( p=1.3 \): 0.1304 - \( p=2 \): 0.3333 - \( p=3 \): 0.5 - \( p=4 \): 0.6

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