Polar coordinates for a gaussian random vector

change-of-variablegaussiangaussian-integralnormal distributionpolar coordinates

I've been finding some prolems in solving exercise 3.3.7 of Vershynin's book "High dimensional probability".

Let $X\sim N(0,I_n)$ standard multivariate normal random vector on $\mathbb{R}^n$ and write it as:
\begin{equation}
X=\|X\|_2 \frac{X}{\|X\|_2}
\end{equation}

  1. prove that $r:=\|X\|_2$ and $\theta:=\frac{X}{\|X\|_2}$ are independent r.v.

  2. prove that $\sqrt{n}\theta\sim\text{Unif}(\sqrt{n}S^{n-1})$

MY ATTEMPT:

  1. The density of $X$ is:
    \begin{equation}
    f_X(x)=(2\pi)^{-n/2}\exp(-\frac{\|x\|_2^2}{2})
    \end{equation}
    Then using a polar coordinate changes whe have that for $M$ measurable set on $R^N$:
    \begin{equation}
    \begin{split}
    P(X\in M)&=\int_M(2\pi)^{-n/2}\exp(-\frac{\|x\|_2^2}{2})dx\\
    &=\int_A\int_Bf(r\theta)d\sigma(\theta)r^{n-1}dr\\
    &=\dfrac{1}{(2\pi)^{n/2}}\int_A\exp(-\dfrac{r^2}{2})r^{n-1}\int_Bd\sigma(\theta)dr = P(r\in A,\theta\in B)
    \end{split}
    \end{equation}
    For some suitable $A\subset [0,\infty)$ and $B\subset S^{n-1}$
    Now using the fact that $\int_{S^{n-1}}d\sigma(\theta)=\omega_{n-1}=\frac{2\pi^{n/2}}{\Gamma(n/2)}$ we can rewrite the previous as:
    \begin{equation}
    \begin{split}
    P(r\in A,\theta\in B)&=\frac{2^{1-n/2}}{\Gamma(n/2)\omega_{n-1}}\int_A\exp(-\dfrac{r^2}{2})r^{n-1}\int_Bd\sigma(\theta)dr\\
    &=\frac{2^{1-n/2}}{\Gamma(n/2)}\int_A\exp(-\dfrac{r^2}{2})r^{n-1}dr\cdot\dfrac{1}{\omega_{n-1}}\int_Bd\sigma(\theta)\\
    &=f(r)g(\theta)
    \end{split}
    \end{equation}
    So far if $A=[0,\infty)$ we have that:
    \begin{equation}
    \int_0^\infty\exp(-\dfrac{r^2}{2})r^{n-1}dr=\Gamma(n/2)2^{n/2-1}
    \end{equation}
    which implies:
    \begin{equation}
    P(r\in[0,\infty),\theta\in B)=P(\theta\in B)=\dfrac{1}{\omega_{n-1}}\int_Bd\sigma(\theta)=g(\theta)
    \end{equation}
    similarly for $B=S^{n-1}$ we have:
    \begin{equation}
    P(r\in A,\theta\in S^{n-1})=P(r\in A)=\frac{2^{1-n/2}}{\Gamma(n/2)}\int_A\exp(-\dfrac{r^2}{2})r^{n-1}dr
    \end{equation}
    Finally this proves that the two variables are independent and moreover looking at the density of $\theta$ written above we can conclude $\theta\sim Unif(S^{n-1})$

Best Answer

Your approach is correct. For $A\subseteq S^{d-1}$ and $B\subseteq [0,\infty)$ polar coordinates yield

$$P(Y\in A, Z\in B) = \frac{\omega_{d-1}}{(2\pi)^{d/2}}\int_B r^{d-1}e^{-\frac{r^2}{2}}\, dr \int_A d\theta,$$ where $\omega_{d-1}$ is the surface area of the unit sphere in $R^d$. Note that by "borrowing" an appropriate constant we can rewrite this as $$P(Y\in A, Z\in B) = f(A) g(B),$$ in a way that $f(S^{n-1})=1$. Therefore, by taking $A = S^{n-1}$ above we get $P(Z\in B) = g(B)$, hence $g([0,\infty)) = 1$ as well, so analogously we get that $P(Y\in A) = f(A)$. This proves the independence and the uniformity of the distribution of $Y$.

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