Answer

The following is the joint probability distribution of \(X\) and \(Y\)

\[ \begin{aligned} f_{XY}(x, y) = f_{X}(x) f_{Y|X}(y|x) = \frac{1}{l} \frac{1}{x} = \frac{1}{xl} \quad \forall \quad 0 \leq y \leq x \leq 1 \end{aligned} \]

Using marginal probabilities, we can calculate \(f_{Y}(y)\) and \(E[Y]\) as

\[ \begin{aligned} f_{Y}(y) = \int f_{XY}(x,y) dx = \int_{y}^{l} \frac{1}{xl} dx = \frac{1}{l} \log \frac{l}{y} \quad\text{Note that for any $y$, $y \leq x \leq l$}\newline E[Y] = \int y f_{Y}(y) = \int_{0}{l} y \frac{1}{l} \log\frac{l}{y} = \frac{l}{4} \end{aligned} \]

This problem can also be approched using iterated expectation

\[ \begin{aligned} E[Y] &= E[E[Y|X]] = E[\text{uniform random variable between $0$ and $x$}]\newline &= E[\frac{X}{2}] =\frac{1}{2}E[X]\newline &= \frac{l}{4} \end{aligned} \]