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