Question

Let $(X, Y)$ have the joint pdf \[ f(x, y) = \begin{cases} \dfrac{3}{4}(y - x), & 0 < x < y < 2, \\ 0, & \text{otherwise.} \end{cases} \] Then the conditional expectation $E(X|Y = 1)$ equals ........... (round off to two decimal places). 
 

Hint:

For conditional expectations, always normalize the joint pdf by the marginal of the conditioning variable before integrating.

Answer 1

Correct Answer: 0.32

Step 1: Find marginal $f_Y(y)$.
\[ f_Y(y) = \int_0^y \frac{3}{4}(y - x) dx = \frac{3}{4}\left[ yx - \frac{x^2}{2} \right]_0^y = \frac{3}{8}y^2. \]

Step 2: Find conditional pdf.
\[ f_{X|Y}(x|y) = \frac{f(x, y)}{f_Y(y)} = \frac{\frac{3}{4}(y - x)}{\frac{3}{8}y^2} = \frac{2(y - x)}{y^2}, 0 < x < y. \]

Step 3: Compute $E(X|Y = y)$.
\[ E(X|Y = y) = \int_0^y x f_{X|Y}(x|y) dx = \int_0^y x \cdot \frac{2(y - x)}{y^2} dx = \frac{2}{y^2}\int_0^y (yx - x^2)dx. \] \[ = \frac{2}{y^2}\left[\frac{y x^2}{2} - \frac{x^3}{3}\right]_0^y = \frac{2}{y^2}\left(\frac{y^3}{2} - \frac{y^3}{3}\right) = \frac{2}{y^2}\cdot \frac{y^3}{6} = \frac{y}{3}. \]

Step 4: Substitute $y = 1$.
\[ E(X|Y = 1) = \frac{1}{3} = 0.33. \] \[ \boxed{0.33.} \]