Showing $\|Y\|_2$ and $\frac{Y}{\|Y\|_2}$ are independent where $Y\sim N(0,\mathbb{1}_N)$ Suppose $Y\sim N(0,\mathbb{1}_N)$. Now let us write 
$$Y=r\vec{Y},$$
where $r:=\|Y\|_2$ and $\vec{Y}:=Y/\|Y\|_2$. I am trying to show that $r$ and $\vec{Y}$ are independent random variables.
I think it's not that I can't do it, it is more I am confused about what has to be done.
My understanding:
So I understand that I have to show that their respective probabilities don't depend on each other, but how do I know what their probabilities are? Suppose I could parameterise their probabilities, how would I be able to check these parameterisations of probabilities will suffice?
any tips are hints will be appreciated!
EDIT:
For $Y$ I have the following pdf:
$$f_Y(x)=\frac{1}{(2\pi)^{N/2}}e^{-\|x\|_2^2/2}=\prod_{i=1}^{N}{\frac{1}{\sqrt{2\pi}}}e^{-x^2_i/2}$$
EDIT (2):
Can it also be shown that $\vec{Y}$ is uniformly distributed on the sphere $S^{N-1}$?
EDIT (3):
I've shown using polar coordinates that:
$$\int{}f_Y(x)dx=\frac{1}{(2\pi)^{N/2}}\int_0^{2\pi}\int_0^\pi{\dddot{}}\int_0^\pi\int_0^\infty{e^{\frac{1}{2}r^{n-1}}}\sin^{n-3}(\varphi_2)...\sin(\varphi_{n-2})drd\varphi_1...d\varphi_{n-1}$$.
 A: You know what the joint probability distribution of $\ \left(r, \vec{Y}\right)\ $ is, because $\ r\ $ and $\ \vec{Y}\ $ are both functions of $\ Y\ $, and you're told that $\ Y=\left(Y_1, Y_2, \dots, Y_N\right)\ $ is a vector of independent standard normal variates.  That is,
$$
P\left(Y\in A\right)=\frac{1}{\left(2\pi\right)^\frac{N}{2}}\int_Ae^{-\frac{\|y\|_2^2}{2}}dy
$$
for any measurable $\ A\subseteq \mathbb{R}^N\ $.  What  you have to do is show that if $\ B_1\ $ and $\ B_2\ $ are any measurable subsets of $\ \mathbb{R}_+\ $ and $\  S^{N-1}\ $ (i.e. the unit $(N-1)$-sphere) respectively, and $\ A_1=\left\{y\in\mathbb{R}^N\right|\,\|y\|_2\in B_1\left.\right\}, $$A_2=$$\left\{y\in\mathbb{R}^N\right|\left.\frac{y}{\|y\|_2}\in B_2\right\}\ $, then $\ P\left(Y\in A_1\cap A_2\right)=$$P\left(Y\in A_1\right)\times$$P\left(Y\in A_2\right)\ $. All the probabilities in this identity can be evaluated by using the above identity for $\ P\left(Y\in A\right)\ $.
In spherical coordinates, the above integral becomes
\begin{align}
P&\left(Y\in A\right)=\\
&\hspace{-0.5em}\frac{1}{\left(2\pi\right)^\frac{N}{2}}\int_{g_S(A)}r^{N-1}e^{-\frac{r^2}{2}}\prod_{i=1}^{N-2}\sin^{N-i-1}\phi_i\,drd\phi_1d\phi_2\dots d\phi_{N-1}\ ,
\end{align}
where $\ g_s:\mathbb{R}^N\rightarrow[0,\infty)\times[0,2\pi)\times[0,
\pi)^{N-2}\ $ is the map from cartesian to polar coordibares, and if $ A=A_1\cap A_2\ $, it becomes
\begin{align}
P\left(Y\in A_1\cap A_2\right)&=\frac{1}{\left(2\pi\right)^\frac{N}{2}}\int_{B_1}r^{N-1}e^{-\frac{r^2}{2}}dr\,\times\\
&\int_{\,\\\hspace{-1em}\vec{u}_\phi\in B_2} \prod_{i=1}^{N-2}\sin^{N-i-1}\phi_i\,d\phi_1d\phi_2\dots d\phi_{N-1}\ ,
\end{align}
where
\begin{align}
\vec{u}_\phi&=\\
&\left(\cos\phi_1, \cos\phi_2\sin\phi_1,\dots, \cos\phi_{n-1}\prod_\limits{i=1}^{n-2}\sin\phi_i, \prod_\limits{i=1}^{n-1}\sin\phi_i\right)\ .
\end{align}
Putting $\ B_2=S^{N-1}\ $ (and hence $\ A_2=\mathbb{R}^N\ $) gives
\begin{align}
P\left(Y\in A_1\right)&=P\left(\|Y\|_2\in B_1\right)\\
&=\frac{1}{2^{\frac{N}{2}-1}\Gamma\left(\frac{N}{2}\right)}\int_{B_1} r^{N-1}e^{-\frac{r^2}{2}}dr\ ,
\end{align}
and putting $\ B_1=\mathbb{R}_+\ $ (and hence $\ A_1=\mathbb{R}^N\ $) gives
\begin{align}
P\left(Y\in A_2\right)&=P\left(\vec{Y}\in B_2\right)\\
&\hspace{-2em}= \frac{\Gamma\left(\frac{N}{2}+1\right)}{N\pi^\frac{N}{2}}\int_{\,\\\hspace{-1em}\vec{u}_\phi\in B_2} \prod_{i=1}^{N-2}\sin^{N-i-1}\phi_i\,d\phi_1d\phi_2\dots d\phi_{N-1}\ .
\end{align}
Now multiplying the expressions for $\ P\left(Y\in A_1\right)\ $ and $\ P\left(Y\in A_2\right)\ $ together, and using the identity $\ \Gamma(z+1)=$$z\Gamma(z)\ $, you find that the product is identical to the expression for $\ P\left(Y\in A_1\cap A_2\right) \ $.
