Let's solve each part step-by-step: --- ### (a) What is the CDF of \(X_1\)? Given the PDF: \[ f_X(x) = 2x, \quad 0 < x < 1 \] The **CDF** is: \[ F_X(x) = \int_{-\infty}^x f_X(t) dt \] Since \(f_X(x) = 0\) for \(x \leq 0\) and \(x \geq 1\), \[ F_X(x) = \begin{cases} 0 & x \leq 0 \\ \int_0^x 2t\,dt & 0 < x < 1 \\ 1 & x \geq 1 \end{cases} \] Compute the integral for \(0 < x < 1\): \[ F_X(x) = \int_0^x 2t\,dt = [t^2]_0^x = x^2 \] So, \[ \boxed{F_X(x) = x^2, \quad 0 < x < 1} \] --- ### (b) What is the 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: - \(F_X(x) = x^2\) - \(f_X(x) = 2x\) So, \[ f_{X_{(2)}}(x) = 12 x^2 (1-x^2)^2 (2x) = 24 x^3 (1-x^2)^2 \] for \(0 < x < 1\). Now, \[ E[X_{(2)}] = \int_0^1 x f_{X_{(2)}}(x) dx = \int_0^1 x \cdot 24 x^3 (1-x^2)^2 dx = 24 \int_0^1 x^4 (1-x^2)^2 dx \] Expand \((1-x^2)^2 = 1 - 2x^2 + x^4\): \[ x^4(1-x^2)^2 = x^4 - 2x^6 + x^8 \] So, \[ E[X_{(2)}] = 24 \int_0^1 (x^4 - 2x^6 + x^8) dx \] \[ = 24 \left[ \frac{x^5}{5} - 2\frac{x^7}{7} + \frac{x^9}{9} \right]_0^1 \] \[ = 24 \left( \frac{1}{5} - \frac{2}{7} + \frac{1}{9} \right) \] Calculate inside the parentheses: \[ \frac{1}{5} - \frac{2}{7} + \frac{1}{9} = \frac{63 - 90 + 35}{315} = \frac{8}{315} \] So, \[ E[X_{(2)}] = 24 \times \frac{8}{315} = \frac{192}{315} = \boxed{\frac{64}{105}} \] --- ### (c) What is the variance of the second order statistic, \(Var(X_{(2)})\)? \[ Var(X_{(2)}) = E[X_{(2)}^2] - (E[X_{(2)}])^2 \] First, \[ E[X_{(2)}^2] = \int_0^1 x^2 f_{X_{(2)}}(x) dx = 24 \int_0^1 x^5 (1-x^2)^2 dx \] \[ = 24 \int_0^1 (x^5 - 2x^7 + x^9) dx \] \[ = 24 \left[ \frac{x^6}{6} - 2\frac{x^8}{8} + \frac{x^{10}}{10} \right]_0^1 \] \[ = 24 \left( \frac{1}{6} - \frac{1}{4} + \frac{1}{10} \right) \] \[ = 24 \left( \frac{5-7.5+1.5}{30} \right) = 24 \left( \frac{5 - 7.5 + 3}{30} \right) \] Let’s fix that step: \[ \frac{1}{6} - \frac{1}{4} + \frac{1}{10} = \frac{20 - 30 + 12}{60} = \frac{2}{60} = \frac{1}{30} \] So, \[ E[X_{(2)}^2] = 24 \times \frac{1}{30} = \frac{24}{30} = \frac{4}{5} \] Now use the mean from above: \[ Var(X_{(2)}) = \frac{4}{5} - \left(\frac{64}{105}\right)^2 \] \[ = \frac{4}{5} - \frac{4096}{11025} \] Convert \(\frac{4}{5}\) to denominator 11025: \[ \frac{4}{5} = \frac{4 \times 2205}{11025} = \frac{8820}{11025} \] So, \[ Var(X_{(2)}) = \frac{8820 - 4096}{11025} = \frac{4724}{11025} \] \[ \boxed{Var(X_{(2)}) = \frac{4724}{11025}} \] --- ### (d) What is the covariance between the second and third order statistics, \(Cov(X_{(2)}, X_{(3)})\)? \[ Cov(X_{(2)}, X_{(3)}) = E[X_{(2)} X_{(3)}] - E[X_{(2)}]E[X_{(3)}] \] First, we need \(E[X_{(2)} X_{(3)}]\). The joint PDF for \(X_{(2)}\) and \(X_{(3)}\) (for \(x < y\)) is: \[ f_{X_{(2)}, X_{(3)}}(x, y) = \frac{n!}{(k-1)! (m-k-1)! (n-m)!} [F(x)]^{k-1} [F(y)-F(x)]^{m-k-1} [1-F(y)]^{n-m} f(x) f(y) \] For \(n=4, k=2, m=3\): \[ f_{X_{(2)}, X_{(3)}}(x, y) = \frac{4!}{1! 0! 1!} [F(x)]^{1} [F(y)-F(x)]^{0} [1-F(y)]^{1} f(x) f(y) \] \[ = 24 F(x)[1-F(y)] f(x) f(y) \] for \(0 < x < y < 1\). Recall \(F(x) = x^2, f(x) = 2x\): \[ f_{X_{(2)}, X_{(3)}}(x, y) = 24 x^2 [1 - y^2] (2x)(2y) = 96 x^3 y (1 - y^2) \] Now, \[ E[X_{(2)} X_{(3)}] = \int_0^1 \int_x^1 x y f_{X_{(2)}, X_{(3)}}(x, y) dy dx \] \[ = \int_0^1 \int_x^1 x y \cdot 96 x^3 y (1 - y^2) dy dx \] \[ = \int_0^1 \int_x^1 96 x^4 y^2 (1 - y^2) dy dx \] Expand: \[ = \int_0^1 \int_x^1 96 x^4 (y^2 - y^4) dy dx \] \[ = 96 \int_0^1 x^4 \left[ \int_x^1 (y^2 - y^4) dy \right] dx \] Compute the inner integral: \[ \int_x^1 (y^2 - y^4) dy = \left[ \frac{y^3}{3} - \frac{y^5}{5} \right]_x^1 = \left( \frac{1}{3} - \frac{1}{5} \right) - \left( \frac{x^3}{3} - \frac{x^5}{5} \right) \] \[ = \left( \frac{5-3}{15} \right) - \left( \frac{x^3}{3} - \frac{x^5}{5} \right) = \frac{2}{15} - \frac{x^3}{3} + \frac{x^5}{5} \] So, \[ E[X_{(2)} X_{(3)}] = 96 \int_0^1 x^4 \left( \frac{2}{15} - \frac{x^3}{3} + \frac{x^5}{5} \right) dx \] \[ = 96 \left[ \frac{2}{15} \int_0^1 x^4 dx - \frac{1}{3} \int_0^1 x^7 dx + \frac{1}{5} \int_0^1 x^9 dx \right] \] Compute each term: - \(\int_0^1 x^4 dx = \frac{1}{5}\) - \(\int_0^1 x^7 dx = \frac{1}{8}\) - \(\int_0^1 x^9 dx = \frac{1}{10}\) Plug in: \[ = 96 \left[ \frac{2}{15} \cdot \frac{1}{5} - \frac{1}{3} \cdot \frac{1}{8} + \frac{1}{5} \cdot \frac{1}{10} \right] \] \[ = 96 \left[ \frac{2}{75} - \frac{1}{24} + \frac{1}{50} \right] \] Now, let's get a common denominator, say 600: - \(\frac{2}{75} = \frac{16}{600}\) - \(\frac{1}{24} = \frac{25}{600}\) - \(\frac{1}{50} = \frac{12}{600}\) So, \[ = 96 \left( \frac{16 - 25 + 12}{600} \right) = 96 \left( \frac{3}{600} \right) = 96 \cdot \frac{1}{200} = \frac{24}{50} = \frac{12}{25} \] Recall from part (b): - \(E[X_{(2)}] = \frac{64}{105}\) Now, let's find \(E[X_{(3)}]\): For the 3rd order statistic: \[ f_{X_{(3)}}(x) = \frac{4!}{2!1!1!} [F(x)]^2 [1-F(x)]^1 f(x) = 12 (x^2)^2 (1-x^2) 2x = 24 x^4 (1-x^2) \] So, \[ E[X_{(3)}] = \int_0^1 x \cdot 24 x^4 (1-x^2) dx = 24 \int_0^1 x^5 (1-x^2) dx \] \[ = 24 \int_0^1 (x^5 - x^7) dx = 24 \left( \frac{1}{6} - \frac{1}{8} \right) = 24 \left( \frac{4-3}{24} \right) = 24 \times \frac{1}{24} = 1 \] So, \[ E[X_{(3)}] = 1 \] Thus, \[ Cov(X_{(2)}, X_{(3)}) = E[X_{(2)} X_{(3)}] - E[X_{(2)}] E[X_{(3)}] = \frac{12}{25} - \frac{64}{105} \cdot 1 \] Let's bring to a common denominator (LCM of 25 and 105 is 525): \[ \frac{12}{25} = \frac{252}{525}, \quad \frac{64}{105