It's well-known that to sample a coordinate $(Y_1,\ldots,Y_n)$ from a surface of an $n$-dimensional unit-radius sphere, it suffices to generate $n$ independent random variables $X_1,\ldots,X_n$ from standard normal distribution (that is, $X_i\sim\mathcal{N}(0,1)$ for all $i=1,\ldots,n$), compute $S=\sum_{i=1}^nX_i^2$, and then normalize each $X_i$ by $\sqrt{S}$ to yield each coordinate $Y_i=X_i/\sqrt{S}$. This is due to the rotational symmetry of a joint distribution of a $n$ i.i.d. normal variables.

I am wondering about the "converse" result to this method. The following question was posed in this paper by Marsaglia from 1972 (see the discussion of Method 2):

If independent $X_1,\ldots,X_n$ lead to $(X_1/\sqrt{S},\ldots,X_n/\sqrt{S})$ with a uniform distribution on a sphere, what can be said about the distributions of the $X_i$'s?

I am wondering if that question was answered (references to appropriate literature would be great). Are $X_i$'s normally distributed?

NOTE: The original post had the following question: "Suppose that I am given a coordinate $(Z_1,Z_2,\ldots,Z_n)$ of a uniform sample from the surface of a unit $n$-sphere. What is known about the distribution of $Z_i$'s, if independence between them is assumed? Are they independently normally distributed?" However, @Did pointed out in a comment that a uniform sample of a surface on a sphere is never i.i.d. However, I am really interested in the answer to Marsaglia's question above, hence a revision to the question...

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    $\begingroup$ Your question is not Marsaglia's question. Re your question, a uniform sample from the surface of a sphere is never i.i.d. $\endgroup$ – Did Apr 16 '13 at 5:45
  • $\begingroup$ @Did Thanks for your comment. I've been digging into this, and finding that indeed the question that I asked originally is not what I meant to ask. I think I am wondering about the answer to Marsaglia's question. I'll re-write the question accordingly... $\endgroup$ – M.B.M. Apr 16 '13 at 6:43

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