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Let $U \sim Unif(S^{d-1}).$ I was wondering if it's true that, and if yes, how could we prove that:

$U = \frac{Z}{\|Z\|}$ where $Z \sim \mathcal{N}(0, I_d), $ i.e. a uniform distribution on a sphere is always a norm scaled distribution? This is kind of like "Polar decomposition in probability."

If yes, one would've to just construct $Z.$ To do so, I'd try to use the fact (the "converse statement") that for any $W \sim \mathcal{N}(0, I_d), \frac{W}{\|W\|}\sim Unif(S^{d-1})$ and that ${\|W\|}\sim \chi_d,$ a chi distribution with $d$ degrees of freedom. We'd also use that: $\|W\|$ and $\frac{W}{\|W\|}$ are independent (see Vershynin, Exercise 3.3.7, P.53). More precisely, I'd define $Z:= N U,$ where $N\sim \chi_d, U \sim Unif(S^{d-1}),$ and enforce the condition that $N, U$ are independent.

If my ideas are so far correct, it remains to show that: 1) $NU \sim \mathcal{N}(0,I_d)$ and 2) $N =\|NU\|= \|Z\|.$ The second one is obvious, since $\|U\|=1 \implies \|NU\| = N.$ Proving 1) might be a real pain I think, given the complicated PDF of the chi random variable $N$ defined above. So how would we circumvent that problem? Should we some kind of rotational symmetry argument? Of course, if my idea were wrong, then we won't pursue this route.

P.S. Just a comment: above, we're trying to construct a normal distribution, given a uniform one. The following might be related but I didn't find much information on this online, but perhaps a high dimensional version of Box-muller transform is something that'd also transform a uniform distribution into a normal distribution, except in the case, the uniform distribution has to be on an open unit cube, instead of a sphere, unlike the question, which'll make it simpler, as the co-ordinates would be independent in this case, unlike the question above.

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    $\begingroup$ It seems that it your construction $N$ and $U$ are independent (though perhaps I’ve misunderstood), but $W$ and $\| W\|$ are not, so I don’t think what you’ve written justifies that $NU$ gives you the normal distribution that you want. $\endgroup$ Jul 24, 2020 at 12:46
  • $\begingroup$ @TheoreticalEconomist Thanks, you made a point that I missed. Note that: it's $\|W\|, W/\|W\|$ that 're independent, not $W, \|W\|.$ However, I edited the question enforcing that independence between $N, U,$ thanks to your comment. $\endgroup$ Jul 24, 2020 at 12:52
  • $\begingroup$ Either I don't understand or you are making a logical error. There is only "one" uniform distribution on the sphere. If you know that $W$ normal implies $W/||W||$ is uniform then you are done because $W$ distributed $N(0,I)$ definitely exists. $\endgroup$
    – SBK
    Jul 24, 2020 at 13:24
  • $\begingroup$ @T_M Yes of course it's true that $W$ normal implies $W/||W||$ is uniform, but that's the other direction of the problem, and the 'motivational part' of the construction. In the given problem, we're starting from the uniform distribution $U$ on a sphere and there's no normal distribution to begin with. We're trying to construct $Z\sim N(0,I_d)$ so that $U= Z/\|Z\|.$ Question is: how to construct $Z?$ $\endgroup$ Jul 24, 2020 at 13:31
  • $\begingroup$ @T_M Ah I see your point: the uniqueness of uniform distribution proves it. But I guess in my mind, I was trying construct one - so basically the question asks: if we multiply the uniform distribution on a sphere by a chi distribution, will we get a normal distribution with zero mean and identity covariance? $\endgroup$ Jul 24, 2020 at 13:46

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A random vector $X\in \mathbb{R}^n$ has a spherically symmetric distribution if $X\overset{d}{=}HX$ for any $H\in \mathcal{O}(n)$ (here $\mathcal{O}$ is the set of $n\times n$ orthogonal matrices). Any such $X$ can be represented as $$ X\overset{d}{=}R U, $$ where $U\sim U(\mathbb{S}^{n-1})$ and $R$ is a continuous nonnegative random variable independent of $U$ (see Theorem C.1 here). In your case $R\sim \chi_n$ (see Theorem C.2).

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  • $\begingroup$ Link to Appendix C of the referenced book. $\endgroup$
    – user140541
    Jul 24, 2020 at 14:38
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    $\begingroup$ $R^2\sim\chi_n^2$, rather. $\endgroup$ Jul 24, 2020 at 14:54

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