Consider a joint probability density function of two random variables \(X\) and \(Y\)
\[ f_{X,Y}(x, y) = \begin{cases} 2xy, & 0 < x < 2, \quad 0 < y < x \\ 0, & \text{otherwise} \end{cases} \]
Then, \(E[Y|X = 1.5]\) is ______.
Answer
1 (marks to all)
Explanation
The marks went to all because the printed function is not a valid density: over its region it integrates to \(\int_{0}^{2}\int_{0}^{x} 2xy\,dy\,dx = \int_{0}^{2} x^{3}\,dx = 4\), not 1.
The conditional expectation can still be found as printed, because a constant factor cancels in a conditional density. For a fixed \(x\),
\[ f_{Y|X}(y|x) = \frac{2xy}{\int_{0}^{x} 2xy\,dy} = \frac{2xy}{x^{3}} = \frac{2y}{x^{2}}, \quad 0 < y < x \]
\[ E[Y|X = x] = \int_{0}^{x} y \cdot \frac{2y}{x^{2}}\,dy = \frac{2x}{3}, \qquad E[Y|X = 1.5] = \frac{2 \times 1.5}{3} = 1 \]
Answer: 1