Let $X,Y\sim U(0,1)$ and suppose $X$ and $Y$ Are independent. How to find explicitly the distribution of $Z=XY$ Using conditional expectation?

Question

I found this problem interesting but I don't see clearly how to solve it. Any hint or ideas on how to get this one started?

Answer 1

For example like that:

Take any $t \in (0,1)$. Then: $$\mathbb P(XY \le t) = \mathbb E [ \mathbb P(XY \le t|Y)] = \mathbb E[\mathbb P(Xy \le t)|_{y=Y}] = \mathbb E((\frac{t}{y}1_{\{y \ge t\}}+1_{\{y<t\}})|_{y=Y}) = $$ $$= \mathbb E[\frac{t}{Y}1_{\{Y \ge t\}} + 1_{\{Y < t\}}] = t \int_{t}^1\frac{dy}{y}+\int_0^tdy=t-t\ln(t) $$

Which gives us $F_Z(t) = \begin{cases} 0 & t \le 0 \\ t-t\ln(t) & t \in(0,1) \\ 1 & t \ge 1\end{cases}$

Where $1_A$ means the indicator function and I used the fact that $\mathbb E[f(X,Y)|Y] = \mathbb E[f(X,y)]|_{y = Y}$ when $X,Y$ are independent and $f$ is such that $\mathbb E[|f(X,Y)|] < \infty$.