Marginal density function of a new variable 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?
 A: No, you cannot make conclusion like that. You are doing it right all the steps except for the regions of $u$ and $v$. You've obtained the joint PDF of $U$ and $V$
$$
f_{U,V}(u,v)=\frac1{u^2}.
$$
This is correct. Now for the regions. You have $0\le y\le1$, this region is corresponding to $0\le u\le1$. It's due to your transformation $Y=U$. You also have $0\le x\le1$, this region is corresponding to $0\le \dfrac v{u^2}\le1\;\Rightarrow\;0\le u\le v^2$. It's due to your transformation $X=\dfrac V{U^2}$. If you plot those regions (check the plot here), you can obtain the marginal PDFs of $U$ and $V$ as follow
$$
\begin{align}
f_U(u)&=\int_{v=0}^{u^2}f_{U,V}(u,v)\ dv\ ;\quad\text{the region is bounded by $v=0$ and $v=u^2$ if you see it from $u$-axis}\\
&=\int_{v=0}^{u^2}\frac1{u^2}\ dv\\
&=\left.\frac v{u^2}\right|_{v=0}^{u^2}\\
&=1\;;\quad\text{ for } 0\le u\le1.
\end{align}
$$
and
$$
\begin{align}
f_V(v)&=\int_{u=\sqrt{v}}^{1}f_{U,V}(u,v)\ du\ ;\quad\text{the region is bounded by $u=1$ and $u=\sqrt{v}$ if you see it from $v$-axis}\\
&=\int_{u=\sqrt{v}}^{1}\frac1{u^2}\ du\\
&=\left.-\frac 1{u}\right|_{u=\sqrt{v}}^{1}\\
&=\frac 1{\sqrt{v}}-1\;;\quad\text{ for } 0\le v\le 1.
\end{align}
$$
'Coincidentally' $U\sim\mathcal{U}(0,1)$, but not because its original distribution are uniform distributions nor also the product of $U$ and $V$.
A: There are several problems with what you've done.  First, you should divide by $|J|$, not multiply.
Further, you should pay attention to the bounds.  The density function of the uniform is $I(0\leq x \leq 1)$ with $I$ the indicator function (and likewise for $y$).  After your change of variables, $I(0\leq x \leq 1)=I(0\leq v/u^2 \leq 1)$; this should be in your joint density of $U,V$.
Finally, to get the marginal density of $V$ from the joint density of $(U,V)$, you can just integrate out $u$ from the joint density.
A: The desired solution to the problem is unnecessarily complicated by the need 
to drag in Jacobians and the like, to first find the joint distribution of
$U$ and$V$, and then to find the marginal distribution of $V$.  If the suggested
method is not mandatory (as suggested by the phrasing or otherwise in the
problem statement), we can write more simply that
for any $\alpha$, $0 < \alpha < 1$,
$$\begin{align}
P\{V > \alpha\} &= P\{XY^2 > \alpha\}\\
&= \int_{x=\alpha}^1 \int_{\sqrt{\frac{\alpha}{x}}}^1 f_{X,Y}(x,y)\,\mathrm dy \,\mathrm dx\\
&= \int_{x=\alpha}^1 1 - \sqrt{\frac{\alpha}{x}} \,\mathrm dx\\
&=\biggr[ x - 2\sqrt{\alpha x}\,\,\biggr|_\alpha^1\\
&= 1 - \alpha - 2\sqrt{\alpha}+2\alpha\\
&= 1 + \alpha -2\sqrt{\alpha}
\end{align}$$
Consequently, 
$$\displaystyle f_V(\alpha) = \frac{\mathrm d}{\mathrm d\alpha}F_V(\alpha) =
\frac{\mathrm d}{\mathrm d\alpha}P\{V \leq \alpha\} = \left.\left.\frac{\mathrm d}{\mathrm d\alpha}\right[1 - P\{V > \alpha\}\right] = \frac{1}{\sqrt{\alpha}} - 1, 0 < \alpha < 1.$$
