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Not sure where to start with this problem:

For any $d\geq 1$, we admit that there is only one probability measure $\mu$ on $\mathcal S_d$, (the $(d-1)-th$ dimensional sphere embedded in $\mathbb R^{3}$), that is uniform, in the following sense: for any isometry $\mathcal A\in O(d)$ (the orthogonal group in $\mathbb R^{d})$, and any continuous function $f:\mathcal S_d\rightarrow \mathbb R$,

$\displaystyle \int_{\mathcal S_d}f(x)\text d \mu(x)=\int_{\mathcal S_d}f(Ax)\text d\mu(x)$

Let $X=(X_1,...,X_d)$ be a vector of independent centered and reduced Gaussian random variables.

a) Prove that the random variable $U = X/||X||_{L^{2}}$ is uniformly distributed on the sphere.

b) Prove that, as $d\rightarrow \infty$, the main part of the globe is concentrated close to the Equator, i.e. for any $\epsilon >0$,

$\displaystyle \int_{x\in\mathcal S_{d},|x_1|<\epsilon}\text d\mu(x)\rightarrow 1$.

Potentially relevant papers: pages 2-3 pages 2-3

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Part (a). – cardinal Mar 17 '13 at 21:00
For part (b.) Slutsky's theorem gives that if you multiply each coordinate by $\sqrt{d}$ and take $d \to \infty$ then the limit is $N(0,1)$. There's a bit of work to do from there, but that almost implies the result. – Chris Janjigian Mar 18 '13 at 1:36
Hint: $\mu$ is the unique probability measure satisfying $\mu(A) = \mu(HA)$ for any orthogonal matrix $H$. Show that $HX$ has the same distribution as $X$ for any orthogonal matrix $H$. Then you will see that $P[X/|X| \in S] = P[HX/|HX| \in S]$. – nullUser Mar 18 '13 at 1:42
For part (b), one can also use the spherical polar coordinates. The volume element for the $(d-1)-$sphere is given by: $$d_{S^{d-1}}V = \sin^{d-2}(\phi_1)\sin^{d-3}(\phi_2)\cdots\sin(\phi_{d-2})d\phi_1 d\phi_2\cdots d\phi_{d-1}$$ The key is for the angle $\phi_1$ corresponds to $x_1$, the volume element is proportional to $\sin^{d-2}(\phi_1) d\phi_1$. – achille hui Mar 18 '13 at 1:47
Thanks everyone! – Chris Mar 18 '13 at 15:55

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