# 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:

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

• Can you give the density for $Y$? Can you separate out a part based on $r$ or $r^2$ leaving a part related to $\vec{Y}$? May 12 '20 at 12:10
• I have added an edit. So are you suggesting I have to do something like: $f_Y(x)=f_r(x)f_\vec{Y}(x)$?
– kam
May 12 '20 at 12:17
• You should be able to separate out a density for $r \sim \mathcal N(0,N)$ leaving a constant density for $\vec{Y}$ May 12 '20 at 12:29
• 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\$). May 12 '20 at 13:02
• 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??
– kam
May 12 '20 at 17:42

## 1 Answer

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)\timesP\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) \$$.

• 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??
– kam
May 12 '20 at 17:08
• 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. May 12 '20 at 17:50
• Ok, I think im confused as to how you explicitly split the integral into the product of integrals over the two subsets?
– kam
May 12 '20 at 17:56
• I'd do it by changing variables in the integral from cartesian to spherical coordinates. May 12 '20 at 18:02
• What would $A_1\cap{}A_2$ look like in spherical coordinates then?
– kam
May 12 '20 at 18:23