Let $X$ and $Y$ be independent random variables with respective densities: $f_X (x) = xe^{-\frac{x^2}{ 2}} 1_{]0,+\infty[} (x)$ and $ f_Y (y) = \frac{1}{\pi \sqrt{1-y^2}} 1_{]-1,1[} (y)$
Let $U:=XY$ and $V:=X\sqrt{1-Y^2}$
I try to check whether U and V are independent and which law they follow.
What I did:
After some calculations, I found the density of $(U,V)$:
$ f_{(U,V)}(u,v)=\frac{1}{\pi } e^{-\frac{u^2}{ 2}}1_{\mathbb{R}}e^{-\frac{v^2}{ 2}}1_{\mathbb{R}_+^* }$ Which mean that $(U,V)$ follows a $ \mathcal{N}(0,1)$ law (because $u^2+v^2=x^2$) and that U and V are independent because $f_{(U,V)}$ can be factored as the products of densities of $U$ and $V$
My question:
There is a factor $\frac{1}{\pi}$ and I don't know to which density it will belong: to $U$ density or to $V$ density?
Many thanks!
Just remember that
$$ \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}{\rm d}u ~e^{-u^2/2} = 1 ~~~\mbox{and}~~~ \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}^+}{\rm d}v ~e^{-v^2/2} = \frac{1}{2} $$
If you multiply this two you get the normalization of $f(u,v)$
$$ \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}{\rm d}u \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}^+}{\rm d}v ~e^{-u^2/2} ~e^{-v^2/2} = \frac{1}{2} \\ \frac{1}{\pi} \int_{\mathbb{R}}{\rm d}u \int_{\mathbb{R}^+}{\rm d}v ~e^{-u^2/2} ~e^{-v^2/2} = 1 $$
So the factor $1/\pi$ actually is shared by the two distributions