Certainly! Here is your solution rewritten *as is*, with all formulas, steps, and reasoning preserved: --- **(b) $\mathbb{E}(X_{(2)})$** #### **Step 1: PDF of the 2nd order statistic $X_{(2)}$** The PDF of the $k^{\text{th}}$ order statistic $X_{(k)}$ among $n$ i.i.d. samples from a distribution with CDF $F_X(x)$ and PDF $f_X(x)$ is: $$ f_{X_{(k)}}(x) = \frac{n!}{(k-1)!(n-k)!} [F_X(x)]^{k-1} [1 - F_X(x)]^{n-k} f_X(x) $$ Given: * $n = 4$, * $k = 2$, * $F_X(x) = x^2$, * $f_X(x) = 2x$, for $0 < x < 1$. Substitute into the formula: $$ f_{X_{(2)}}(x) = \frac{4!}{1!2!} (x^2)^1 (1 - x^2)^2 (2x) = \frac{24}{2} x^2 (1 - x^2)^2 (2x) $$ $$ = 12x^2(1 - 2x^2 + x^4)(2x) = 24x^3 - 48x^5 + 24x^7, \quad \text{for } 0 < x < 1 $$ --- #### **Step 2: Compute $\mathbb{E}(X_{(2)})$** $$ \mathbb{E}(X_{(2)}) = \int_0^1 x f_{X_{(2)}}(x) \, dx = \int_0^1 x(24x^3 - 48x^5 + 24x^7) \, dx $$ $$ = \int_0^1 (24x^4 - 48x^6 + 24x^8) \, dx = 24 \cdot \frac{1}{5} - 48 \cdot \frac{1}{7} + 24 \cdot \frac{1}{9} $$ $$ = \frac{24}{5} - \frac{48}{7} + \frac{24}{9} = \frac{1512}{315} - \frac{2160}{315} + \frac{840}{315} = \frac{192}{315} = \boxed{\frac{64}{105}} $$ --- ### **(c) $\mathrm{Var}(X_{(2)})$** #### **Step 1: Compute $\mathbb{E}(X_{(2)}^2)$** $$ \mathbb{E}(X_{(2)}^2) = \int_0^1 x^2 f_{X_{(2)}}(x) \, dx = \int_0^1 x^2(24x^3 - 48x^5 + 24x^7) \, dx $$ $$ = \int_0^1 (24x^5 - 48x^7 + 24x^9) \, dx = 24 \cdot \frac{1}{6} - 48 \cdot \frac{1}{8} + 24 \cdot \frac{1}{10} $$ $$ = 4 - 6 + \frac{12}{5} = \frac{-10 + 12}{5} = \boxed{\frac{2}{5}} $$ --- #### **Step 2: Use variance formula** $$ \mathrm{Var}(X_{(2)}) = \mathbb{E}(X_{(2)}^2) - [\mathbb{E}(X_{(2)})]^2 = \frac{2}{5} - \left( \frac{64}{105} \right)^2 = \frac{2}{5} - \frac{4096}{11025} $$ Convert $\frac{2}{5}$ to denominator 11025: $$ \frac{2}{5} = \frac{4410}{11025} \Rightarrow \mathrm{Var}(X_{(2)}) = \frac{4410 - 4096}{11025} = \boxed{\frac{314}{11025}} $$ --- ### **(d) $\mathrm{Cov}(X_{(2)}, X_{(3)})$** --- #### **Step 1: Find $\mathbb{E}(X_{(3)})$** From the PDF of $X_{(3)}$: $$ f_{X_{(3)}}(x) = \frac{4!}{2!1!} (x^2)^2 (1 - x^2)(2x) = 24x^5(1 - x^2) = 24x^5 - 24x^7 $$ $$ \mathbb{E}(X_{(3)}) = \int_0^1 x(24x^5 - 24x^7) \, dx = \int_0^1 (24x^6 - 24x^8) \, dx = 24 \cdot \frac{1}{7} - 24 \cdot \frac{1}{9} $$ $$ = \frac{24}{7} - \frac{24}{9} = \frac{216 - 168}{63} = \boxed{\frac{16}{21}} $$ --- #### **Step 2: Compute $\mathbb{E}(X_{(2)} X_{(3)})$** Joint PDF for $X_{(2)}, X_{(3)}$: $$ f_{X_{(2)}, X_{(3)}}(x, y) = \frac{4!}{1!0!1!} (x^2)^1 [1 - y^2]^1 (2x)(2y) = 96x^3 y (1 - y^2) $$ Domain: $0 < x < y < 1$ $$ \mathbb{E}(X_{(2)} X_{(3)}) = \int_0^1 \int_x^1 xy \cdot 96x^3 y(1 - y^2) \, dy dx = 96 \int_0^1 x^4 \left( \int_x^1 y^2 (1 - y^2) \, dy \right) dx $$ Inner integral: $$ \int_x^1 y^2 (1 - y^2) \, dy = \left[ \frac{y^3}{3} - \frac{y^5}{5} \right]_x^1 = \frac{1}{3} - \frac{1}{5} - \left( \frac{x^3}{3} - \frac{x^5}{5} \right) = \frac{2}{15} - \frac{x^3}{3} + \frac{x^5}{5} $$ Now: $$ \mathbb{E}(X_{(2)} X_{(3)}) = 96 \int_0^1 x^4 \left( \frac{2}{15} - \frac{x^3}{3} + \frac{x^5}{5} \right) dx $$ Break into parts: $$ = 96 \left( \frac{2}{15} \cdot \frac{1}{5} - \frac{1}{3} \cdot \frac{1}{8} + \frac{1}{5} \cdot \frac{1}{10} \right) = 96 \left( \frac{2}{75} - \frac{1}{24} + \frac{1}{50} \right) $$ Convert to common denominator (600): $$ \frac{2}{75} = \frac{16}{600}, \quad \frac{1}{24} = \frac{25}{600}, \quad \frac{1}{50} = \frac{12}{600} $$ $$ = 96 \cdot \frac{16 - 25 + 12}{600} = 96 \cdot \frac{3}{600} = \boxed{\frac{12}{25}} $$ --- #### **Step 3: Compute covariance** $$ \mathrm{Cov}(X_{(2)}, X_{(3)}) = \mathbb{E}(X_{(2)} X_{(3)}) - \mathbb{E}(X_{(2)}) \cdot \mathbb{E}(X_{(3)}) $$ $$ = \frac{12}{25} - \left( \frac{64}{105} \cdot \frac{16}{21} \right) = \frac{12}{25} - \frac{1024}{2205} $$ Convert to denominator 11025: $$ \frac{12}{25} = \frac{5292}{11025}, \quad \frac{1024}{2205} = \frac{5120}{11025} \Rightarrow \boxed{\mathrm{Cov}(X_{(2)}, X_{(3)}) = \frac{172}{11025}} $$ --- ### Final Answers: * (b) $\boxed{\mathbb{E}(X_{(2)}) = \frac{64}{105}}$ * (c) $\boxed{\mathrm{Var}(X_{(2)}) = \frac{314}{11025}}$ * (d) $\boxed{\mathrm{Cov}(X_{(2)}, X_{(3)}) = \frac{172}{11025}}$ ---