Answer: 0.67

## Step-by-Step Solution ### 1. Conditional Probability Formula The conditional distribution is given by: $$ f_{(X_1|X_2)}(x_1 | X_2 = x_2) = \frac{f_{(X_1,X_2)}(x_1, x_2)}{f_{(X_2)}(x_2)} $$ So, $$ f(x_1 | X_2 = 0.5) = \frac{f_{(x_1,x_2)}(x_1, x_2 = 0.5)}{f_{(x_2)}(x_2 = 0.5)} $$ ### 2. Calculate Marginal Distribution of X₂ The marginal distribution of $X_2$ is: $$ f_{(x_2)}(x_2) = \int_{-\infty}^{\infty} f_{(x_1,x_2)}(x_1, x_2) \, dx_1 = \int_0^1 4x_1x_2 \, dx_1 = \left[2x_1^2x_2\right]_0^1 = 2x_2 - 0 = 2x_2 $$ So, $ f_{(x_2)}(x_2 = 0.5) = 2 \times 0.5 = 1 $ ### 3. Calculate Conditional Density $$ f(x_1 | X_2 = 0.5) = \frac{f_{(x_1,x_2)}(x_1, x_2 = 0.5)}{f_{(x_2)}(x_2 = 0.5)} = \frac{4x_1 \times 0.5}{1} = 2x_1 $$ Therefore, the conditional density is $ f(x_1 | X_2 = 0.5) = 2x_1 $ for $ 0 < x_1 < 1 $. ### 4. Compute Conditional Expectation $$ E(X_1 | X_2 = 0.5) = \int_0^1 x_1 \cdot f(x_1 | X_2 = 0.5) \, dx_1 = \int_0^1 x_1 \cdot 2x_1 \, dx_1 = 2 \int_0^1 x_1^2 \, dx_1 = 2 \left[\frac{x_1^3}{3}\right]_0^1 = 2 \cdot \frac{1}{3} = \frac{2}{3} \approx 0.67 $$ ✅ **Final Answer: D. 0.67**

Author: Tanishq Prabhu