Finding the PDF of a product of two r.v.s

Question

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...

Answer 1

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:

$$f_V(v) = \begin{cases} -\log v &, v \in (0,1) \\ 0 & \text{, otherwise.} \end{cases}$$

Answer 2

\begin{align} & \Pr(XY\le v) = \operatorname E(\Pr(XY\le v \mid X)) = \operatorname E\left( \Pr \left( Y \le \frac v X \,\middle|\, X \right) \right) \\[10pt] = {} & \operatorname E \left.\begin{cases} \dfrac v X & \text{if } \dfrac v X <1 \\[12pt] 1 & \text{otherwise} \end{cases} \right\} = \int_v^1 \frac v x \cdot 1 \, dx + \int_0^v 1 \,dx = v(1-\log_e v). \end{align}