Question

Let \( X \) and \( Y \) have the joint probability density function \[ f(x, y) = \begin{cases} 2, & 0 \leq x \leq y \leq 1, \\ 0, & \text{otherwise}. \end{cases} \] Let \( a = E(Y|X = \frac{1}{2}) \) and \( b = \text{Var}(Y|X = \frac{1}{2}) \). Then \( (a, b) \) is 
 

• A. \( \left( \frac{3}{4}, \frac{7}{12} \right) \)
• B. \( \left( \frac{1}{4}, \frac{1}{48} \right) \)
• C. \( \left( \frac{1}{4}, \frac{7}{12} \right) \)
• D. \( \left( \frac{3}{4}, \frac{1}{48} \right) \)

Hint:

When computing conditional expectation and variance, use the definition and integrate over the conditional distribution of \( Y \) given \( X \).

Answer 1

The Correct Option is D

Step 1: Conditional expectation. 
The conditional expectation \( E(Y|X = x) \) is calculated using the formula: \[ E(Y|X = x) = \int_0^1 y f_{Y|X}(y|x) \, dy. \] For \( X = \frac{1}{2} \), we compute the expected value \( a \).

Step 2: Conditional variance. 
The conditional variance \( \text{Var}(Y|X = x) \) is given by: \[ \text{Var}(Y|X = x) = E(Y^2|X = x) - (E(Y|X = x))^2. \] This can be computed similarly to the expectation, and the value of \( b \) is found.

Step 3: Conclusion. 
The correct answer is (D) \( \left( \frac{3}{4}, \frac{1}{48} \right) \).