Finding the PDF of a product of two r.v.s Let $X,Y \sim \operatorname{Unif}(0,1)$ be independent of each other. Find the PDF of $V:= XY$.
This is was I did
$$\mathbb{P}(V \leq v) = \mathbb{P}(YX \leq v) = \mathbb{P}\left(Y\leq \frac v X \right) \\ = \int_0^{v/x} 1 \, dy = \frac v x.$$
I'm not sure if this is right since I can't think of the restrictions and also there involves an $x$ term in the answer...

Let $v \in (0,1)$,
\begin{align}
P( Y \le \frac{v}{X})&= \int_0^1 P(Y \le \frac{v}{x})\, dx \\
&= \int_0^1 \min\left( 1, \frac{v}{x}\right)\, dx \\
&= \int_0^v \, dx + \int_v^1  \frac{v}{x}\, dx \\
&= v-v\log v \\
\end{align}
Differentiate it to obtain the pdf.
Edit:
$$f_V(v) = \begin{cases} -\log v &, v \in (0,1) \\ 0 & \text{, otherwise.} \end{cases}$$
