**Question:** Let $X \sim U[0,1]$ and $Y \sim U[0,1]$ be independent random variables.
By considering the random variables $U=Y$, $V=XY^{2}$, or otherwise, find the probability density function of $V$.

**My answer:** We know

$ F\_{X}(x)=\left\{\begin{matrix}
x & 0 \leq x \leq 1\\ 
0 & otherwise
\end{matrix}\right.$ and $ F\_{Y}(y)=\left\{\begin{matrix}
y & 0 \leq y \leq 1\\ 
0 & otherwise
\end{matrix}\right.$

Therefore $ f\_{X}(x)=\left\{\begin{matrix}
1 & 0 \leq x \leq 1\\ 
0 & otherwise
\end{matrix}\right.$ and $ f\_{Y}(y)=\left\{\begin{matrix}
1 & 0 \leq y \leq 1\\ 
0 & otherwise
\end{matrix}\right.$

$U=Y,V=XY^{2} \Rightarrow X=\frac{V}{U^{2}},Y=U$

So, $J(U,V)=det\begin{pmatrix}
-\frac{2v}{u^{3}} & 1\\ 
\frac{1}{u^{2}} & 0
\end{pmatrix}=-\frac{1}{u^{2}}$

$X,Y$ independent $\Rightarrow f\_{X,Y}(x,y)=f\_{X}(x)f\_{Y}(y)=\left\{\begin{matrix}
1 & 0 \leq x,y \leq 1\\ 
0 & otherwise
\end{matrix}\right.$

So, $f\_{U,V}(u,v)=|J(U,V)|f\_{X}(\frac{v}{u^{2}})f\_{Y}(u)=\left\{\begin{matrix}
\frac{1}{u^{2}} & 0 \leq u,v \leq 1\\ 
0 & otherwise
\end{matrix}\right.$

**My issue now is:** *since this is the product of a function of $U$ and a function of $V$, can I just say* $f\_{V}(v)=\left\{\begin{matrix}
1 & 0 \leq v \leq 1\\ 
0 & otherwise
\end{matrix}\right.$ *so that* $V \sim U[0,1]$ *also?*