Let's solve the problem step by step. --- ## (a) **CDF of \( X_1 \)** Given PDF: \[ f_X(x) = 2x, \quad 0 < x < 1 \] CDF is: \[ F_X(x) = P(X_1 \leq x) = \int_0^x 2t\,dt = [t^2]_0^x = x^2, \quad 0 < x < 1 \] --- ## (b) **Expected Value of the Second Order Statistic, \( E(X_{(2)}) \)** For \( n = 4 \), the PDF of the \( k \)-th order statistic is: \[ f_{X_{(k)}}(x) = \frac{n!}{(k-1)!(n-k)!} [F_X(x)]^{k-1} [1 - F_X(x)]^{n-k} f_X(x) \] For \( k = 2, n = 4 \): \[ f_{X_{(2)}}(x) = \frac{4!}{1!2!} [F_X(x)]^{1} [1 - F_X(x)]^{2} f_X(x) = 12 \, F_X(x) [1 - F_X(x)]^2 f_X(x) \] Recall \( F_X(x) = x^2 \), \( f_X(x) = 2x \): \[ f_{X_{(2)}}(x) = 12x^2 (1 - x^2)^2 \cdot 2x = 24x^3 (1 - x^2)^2, \quad 0 < x < 1 \] Now, compute the expected value: \[ E[X_{(2)}] = \int_0^1 x f_{X_{(2)}}(x) dx = \int_0^1 x \cdot 24x^3 (1 - x^2)^2 dx = 24 \int_0^1 x^4 (1 - x^2)^2 dx \] Expand \( (1 - x^2)^2 = 1 - 2x^2 + x^4 \): \[ x^4 (1 - x^2)^2 = x^4 - 2x^6 + x^8 \] So: \[ E[X_{(2)}] = 24 \int_0^1 (x^4 - 2x^6 + x^8) dx = 24 \left[ \frac{x^5}{5} - 2\frac{x^7}{7} + \frac{x^9}{9} \right]_0^1 \] \[ = 24 \left( \frac{1}{5} - \frac{2}{7} + \frac{1}{9} \right) \] Compute the sum: \[ \frac{1}{5} - \frac{2}{7} + \frac{1}{9} = \frac{63 - 90 + 35}{315} = \frac{8}{315} \] Therefore, \[ E[X_{(2)}] = 24 \times \frac{8}{315} = \frac{192}{315} = \frac{64}{105} \] --- ## (c) **Variance of the Second Order Statistic, \( Var(X_{(2)}) \)** \[ Var(X_{(2)}) = E[X_{(2)}^2] - (E[X_{(2)}])^2 \] Calculate \( E[X_{(2)}^2] \): \[ E[X_{(2)}^2] = \int_0^1 x^2 f_{X_{(2)}}(x) dx = 24 \int_0^1 x^5 (1 - x^2)^2 dx \] Expand: \[ x^5 (1 - x^2)^2 = x^5 - 2x^7 + x^9 \] \[ E[X_{(2)}^2] = 24 \left[ \frac{x^6}{6} - 2\frac{x^8}{8} + \frac{x^{10}}{10} \right]_0^1 = 24 \left( \frac{1}{6} - \frac{1}{4} + \frac{1}{10} \right) \] \[ = 24 \left( \frac{10 - 15 + 6}{60} \right) = 24 \left( \frac{1}{60} \right) = \frac{24}{60} = \frac{2}{5} \] Now, recall \( E[X_{(2)}] = \frac{64}{105} \): \[ Var(X_{(2)}) = \frac{2}{5} - \left(\frac{64}{105}\right)^2 = \frac{2}{5} - \frac{4096}{11025} \] Convert \(\frac{2}{5}\) to denominator 11025: \[ \frac{2}{5} = \frac{2 \times 2205}{5 \times 2205} = \frac{4410}{11025} \] So, \[ Var(X_{(2)}) = \frac{4410-4096}{11025} = \frac{314}{11025} \] --- ## (d) **Covariance \( Cov(X_{(2)}, X_{(3)}) \)** \[ Cov(X_{(2)}, X_{(3)}) = E[X_{(2)} X_{(3)}] - E[X_{(2)}]E[X_{(3)}] \] The joint PDF for the 2nd and 3rd order statistics for \( n = 4 \) is: \[ f_{X_{(2)}, X_{(3)}}(x, y) = \frac{4!}{1!1!1!1!} [F_X(x)]^1 [F_X(y) - F_X(x)]^1 [1-F_X(y)]^1 f_X(x) f_X(y) \] where \( 0 < x < y < 1 \). \[ = 24 \cdot x^2 \cdot (y^2 - x^2) \cdot (1 - y^2) \cdot 2x \cdot 2y \] \[ = 96 x^3 y (y^2 - x^2)(1 - y^2) \] So, \[ E[X_{(2)} X_{(3)}] = \int_{0}^{1} \int_{x}^{1} x y \cdot 96 x^3 y (y^2 - x^2)(1 - y^2) dy dx \] \[ = 96 \int_{0}^{1} \int_{x}^{1} x^4 y^2 (y^2 - x^2)(1 - y^2) dy dx \] Let's expand \( (y^2 - x^2)(1 - y^2) = y^2 - x^2 - y^4 + x^2 y^2 \): \[ x^4 y^2 (y^2 - x^2 - y^4 + x^2 y^2) = x^4 y^4 - x^6 y^2 - x^4 y^6 + x^6 y^4 \] So, \[ E[X_{(2)} X_{(3)}] = 96 \int_0^1 \int_x^1 (x^4 y^4 - x^6 y^2 - x^4 y^6 + x^6 y^4) \, dy dx \] Let's compute each term: ### \( \int_x^1 x^4 y^4 dy = x^4 \left[ \frac{y^5}{5} \right]_x^1 = x^4 \left( \frac{1}{5} - \frac{x^5}{5} \right) \) ### \( \int_x^1 x^6 y^2 dy = x^6 \left[ \frac{y^3}{3} \right]_x^1 = x^6 \left( \frac{1}{3} - \frac{x^3}{3} \right) \) ### \( \int_x^1 x^4 y^6 dy = x^4 \left[ \frac{y^7}{7} \right]_x^1 = x^4 \left( \frac{1}{7} - \frac{x^7}{7} \right) \) ### \( \int_x^1 x^6 y^4 dy = x^6 \left[ \frac{y^5}{5} \right]_x^1 = x^6 \left( \frac{1}{5} - \frac{x^5}{5} \right) \) Putting together: \[ E[X_{(2)} X_{(3)}] = 96 \int_0^1 \Bigg[ x^4 \left( \frac{1}{5} - \frac{x^5}{5} \right) - x^6 \left( \frac{1}{3} - \frac{x^3}{3} \right) - x^4 \left( \frac{1}{7} - \frac{x^7}{7} \right) + x^6 \left( \frac{1}{5} - \frac{x^5}{5} \right) \Bigg] dx \] Group terms: \[ = 96 \int_0^1 \Bigg[ \frac{x^4}{5} - \frac{x^9}{5} - \frac{x^6}{3} + \frac{x^9}{3} - \frac{x^4}{7} + \frac{x^{11}}{7} + \frac{x^6}{5} - \frac{x^{11}}{5} \Bigg] dx \] Now, combine like terms: - \( x^4 \): \( \frac{1}{5} - \frac{1}{7} = \frac{2}{35} \) - \( x^6 \): \( -\frac{1}{3} + \frac{1}{5} = -\frac{2}{15} \) - \( x^9 \): \( -\frac{1}{5} + \frac{1}{3} = \frac{2}{15} \) - \( x^{11} \): \( \frac{1}{7} - \frac{1}{5} = -\frac{2}{35} \) So, \[ E[X_{(2)} X_{(3)}] = 96 \int_0^1 \left( \frac{2}{35}x^4 - \frac{2}{15}x^6 + \frac{2}{15}x^9 - \frac{2}{35}x^{11} \right) dx \] \[ = 96 \cdot \frac{2}{35} \int_0^1 x^4 dx - 96 \cdot \frac{2}{15} \int_0^1 x^6 dx + 96 \cdot \frac{2}{15} \int_0^1 x^9 dx - 96 \cdot \frac{2}{35} \int_0^1 x^{11} dx \] Integrate each term: - \( \int_0^1 x^4 dx = \frac{1}{5} \) - \( \int_0^1 x^6 dx = \frac{1}{7} \) - \( \int_0^1 x^9 dx = \frac{1}{10} \) - \( \int_0^1 x^{11} dx = \frac{1}{12} \) Substitute: \[ = 96 \left[ \frac{2}{35} \cdot \frac{1}{5} - \frac{2}{15} \cdot \frac{1}{7} + \frac{2}{15} \cdot \frac{1}{10} - \frac{2}{35} \cdot \frac{1}{12} \right] \] \[ = 96 \left( \frac{2}{175} - \frac{2}{105} + \frac{1}{75} - \frac{1}{210} \right ) \] Find a common denominator (LCM of 175, 105, 75, 210 is 3150): - \( \frac{2}{175} = \frac{36}{3150} \) - \( \frac{2}{105} = \frac{60}{3150} \) - \( \frac{1}{75} = \frac{42}{3150} \) - \( \frac{1}{210} = \frac{15}{3150} \) So sum: \[ \frac{36 - 60 + 42 - 15}{3150} = \frac{3}{3150} = \frac{1}{1050} \] Multiply by 96: \[ E[X_{(2)} X_{(3)}] = 96 \times \frac{1}{1050} = \frac{96}{1050} = \frac{32}{350} = \frac{16}{175} \] Now recall \( E[X_{(2)}] = \frac{64}{105} \) and \( E[X_{(3)}] \). For \( X_{(3)} \):