Answer: 0.0972

The covariance between X₁ and X₂ is calculated as Cov(X₁, X₂) = E(X₁X₂) - E(X₁)E(X₂). **Step 1: Calculate E(X₁)** The marginal distribution of X₁ is: $$f_{X_1}(x_1) = \int_{0}^{2} \frac{1}{8} x_1 x_2 dx_2 = \frac{1}{4} x_1, \quad 0 \leq x_1 \leq 1$$ $$E(X_1) = \int_{0}^{1} x_1 \cdot \frac{1}{4} x_1 dx_1 = \int_{0}^{1} \frac{1}{4} x_1^2 dx_1 = \left[ \frac{1}{12} x_1^3 \right]_0^1 = \frac{1}{12} \approx 0.08333$$ **Step 2: Calculate E(X₂)** The marginal distribution of X₂ is: $$f_{X_2}(x_2) = \int_{0}^{1} \frac{1}{8} x_1 x_2 dx_1 = \frac{1}{16} x_2, \quad 0 \leq x_2 \leq 2$$ $$E(X_2) = \int_{0}^{2} x_2 \cdot \frac{1}{16} x_2 dx_2 = \int_{0}^{2} \frac{1}{16} x_2^2 dx_2 = \left[ \frac{1}{48} x_2^3 \right]_0^2 = \frac{8}{48} = \frac{1}{6} \approx 0.16667$$ **Step 3: Calculate E(X₁X₂)** $$E(X_1X_2) = \int_{0}^{1} \int_{0}^{2} x_1 x_2 \cdot \frac{1}{8} x_1 x_2 dx_2 dx_1 = \frac{1}{8} \int_{0}^{1} x_1^2 dx_1 \int_{0}^{2} x_2^2 dx_2$$ $$= \frac{1}{8} \left[ \frac{x_1^3}{3} \right]_0^1 \left[ \frac{x_2^3}{3} \right]_0^2 = \frac{1}{8} \cdot \frac{1}{3} \cdot \frac{8}{3} = \frac{1}{8} \cdot \frac{8}{9} = \frac{1}{9} \approx 0.11111$$ **Step 4: Calculate Covariance** $$\text{Cov}(X_1, X_2) = E(X_1X_2) - E(X_1)E(X_2) = \frac{1}{9} - \left(\frac{1}{12} \cdot \frac{1}{6}\right) = \frac{1}{9} - \frac{1}{72} = \frac{8}{72} - \frac{1}{72} = \frac{7}{72} \approx 0.09722$$ Thus, the covariance is approximately 0.0972, which corresponds to option A.

Author: Nikitesh Somanthe