Question

Let \( (X, Y) \) have joint probability density function \[ f(x, y) = \begin{cases} 8xy & \text{if } 0 < x < y < 1, \\ 0 & \text{otherwise.} \end{cases} \] If \( E(X \mid Y = y_0) = \frac{1}{2} \), then \( y_0 \) equals

• A. \( \frac{3}{4} \)
• B. \( \frac{1}{2} \)
• C. \( \frac{1}{3} \)
• D. \( \frac{2}{3} \)

Hint:

- Conditional expectation involves integrating the joint distribution with respect to one variable.
- Be sure to check the limits of integration carefully when finding conditional distributions.

Answer 1

The Correct Option is A

1) Understanding the joint probability density function:
The joint probability density function is given by: \[ f(x, y) = 8xy \text{for} 0<x<y<1 \] We need to calculate \( E(X | Y = y_0) \) and use the given condition that \( E(X | Y = y_0) = \frac{1}{2} \).
2) Finding the conditional expectation:
The conditional probability density function is: \[ f(x | y) = \frac{f(x, y)}{f_Y(y)} = \frac{8xy}{f_Y(y)} \text{for} 0<x<y \] To find \( f_Y(y) \), we integrate \( f(x, y) \) over \( x \) from 0 to \( y \): \[ f_Y(y) = \int_0^y 8xy \, dx = 4y^3 \] Thus, the conditional density function is: \[ f(x | y) = \frac{2x}{y^3} \] The conditional expectation is: \[ E(X | Y = y_0) = \int_0^{y_0} x . \frac{2x}{y_0^3} dx = \frac{2}{y_0^3} \int_0^{y_0} x^2 \, dx = \frac{2}{y_0^3} . \frac{y_0^3}{3} = \frac{1}{3} \] Setting \( E(X | Y = y_0) = \frac{1}{2} \) gives \( y_0 = \frac{3}{4} \).