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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:

$$\int{}f_Y(x)dx=\frac{1}{(2\pi)^{N/2}}\int_0^{2\pi}\int_0^\pi{\dddot{}}\int_0^\pi\int_0^\infty{e^{\frac{1}{2}r^{n-1}}}\sin^{n-3}(\varphi_2)...\sin(\varphi_{n-2})drd\varphi_1...d\varphi_{n-1}$$.

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  • $\begingroup$ Can you give the density for $Y$? Can you separate out a part based on $r$ or $r^2$ leaving a part related to $\vec{Y}$? $\endgroup$
    – Henry
    May 12 '20 at 12:10
  • $\begingroup$ I have added an edit. So are you suggesting I have to do something like: $f_Y(x)=f_r(x)f_\vec{Y}(x)$? $\endgroup$
    – kam
    May 12 '20 at 12:17
  • $\begingroup$ You should be able to separate out a density for $r \sim \mathcal N(0,N)$ leaving a constant density for $\vec{Y}$ $\endgroup$
    – Henry
    May 12 '20 at 12:29
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    $\begingroup$ One hitch is that $\ \vec{Y}\ $ doesn't actually have a density with respect to Lebesgue measure on $\ \mathbb{R}^N\ $ (although it does, of course, have a constant density with respect to area on $\ S^N\ $). $\endgroup$ May 12 '20 at 13:02
  • $\begingroup$ So im assuming the idea here is that for any subset of an 𝑁 N sphere, it intersects a subset of the 𝑥1−𝑎𝑥𝑖𝑠 x 1 − a x i s only once. So integrating a single point is 0 because it has zero lebesgue measure? hence this integral holds?? $\endgroup$
    – kam
    May 12 '20 at 17:42
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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) \ $.

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  • $\begingroup$ So im assuming the idea here is that for any subset of an $N$ sphere, it intersects a subset of the $x_1-axis$ only once. So integrating a single point is 0 because it has zero lebesgue measure? hence this integral holds?? $\endgroup$
    – kam
    May 12 '20 at 17:08
  • $\begingroup$ No. I've no idea why you think that any of those statements follow from what I wrote. The unit $N$-sphere intersect the $\ x_1$-axis twice, at $\ (-1,0,\dots,0)\ $ and $\ (1,0,\dots,0)\ $, but I don't believe that fact bears any relevance to the question at hand. Neither does the fact that the integral over a singleton is zero, except insofar as singletons are among the sets for which the identity $\ P\left(A_1\cap A_2\right)= P\left(A_1\right)\times P\left(A_2\right)\ $ must hold. $\endgroup$ May 12 '20 at 17:50
  • $\begingroup$ Ok, I think im confused as to how you explicitly split the integral into the product of integrals over the two subsets? $\endgroup$
    – kam
    May 12 '20 at 17:56
  • $\begingroup$ I'd do it by changing variables in the integral from cartesian to spherical coordinates. $\endgroup$ May 12 '20 at 18:02
  • $\begingroup$ What would $A_1\cap{}A_2$ look like in spherical coordinates then? $\endgroup$
    – kam
    May 12 '20 at 18:23

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