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Suppose that $U=(U_1,\ldots,U_n)$ has the uniform distribution on the unit sphere $S_{n-1}=\{x\in\mathbb R^n:\|x\|_2=1\}$. I'm trying to understand the marginal distributions of individual components of the vector $U$, say, without loss of generality, $U_1$.

So far, this is what I have: I know that $U$ is equal in distribution to $Z/\|Z\|_2$, where $Z$ is an $n$-dimensional standard gaussian vector. Thus, $U_1$ is equal in distribution to $Z_1/\|Z\|_2$. Then, we could in principle compute the distribution of $U_1$ as $$P[U_1\leq t]=P\big[Z_1\leq\|Z\|_2t\big],$$ but the right-hand side above using conventional means is horribly messy (i.e., nested integrals over the set $[x_1\leq\|x\|t]$).

Is there a more practical/intuitive way of computing this distribution?

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    $\begingroup$ The amazing case is $n=3$ where the marginals are uniform. $\endgroup$ – GEdgar Jan 6 '16 at 14:40
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    $\begingroup$ You are asking about the area of (hyper)spherical caps. For an explicit formula in terms of regularized incomplete beta functions, see here. $\endgroup$ – Did Jan 6 '16 at 15:12
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    $\begingroup$ See the answer here : math.stackexchange.com/questions/185298/… $\endgroup$ – Plop Mar 20 '18 at 14:13
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The marginals are such that $$U_1^2\sim \text{Beta}\left(\frac{1}{2}, \frac{k-1}{2}\right)$$ We can derive this by noting that for independent $Z_i$, $$U_1=\frac{Z_1}{\sqrt{Z_1^2+R}}$$ where $R\sim \chi^2(k-1)$. Solving for $Z_1$ yields $$Z_1^{1,2}=\pm\sqrt{\frac{U_1^2R}{1-U_1^2}}$$ and we can use standard change of variable techniques keeping in mind that we have to sum over both branches of $Z_1$. Let $T_i: [0, 1) \times \mathbb{R} \to\mathbb{R} \times \mathbb{R}$ be the transformation from $(U_1, R)$ to $(Z_1^i, R)$. Then, $$\begin{align}f_{U_1}(U_1)&=\sum_{i\in{1,2}}\int_0^\infty\mathcal{N}(Z_1^i;0, 1)\chi^2(R;k-1)|\det J(T_i)|dR \\ &=2\int_0^\infty\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2} \frac{U_1^2R}{1-U_1^2}}\frac{1}{2^{\frac{k-1}{2}}\Gamma\left(\frac{k-1}{2}\right)}R^{\frac{k-1}{2}-1}e^{-\frac{R}{2}} \sqrt{\frac{R}{(1-U_1^2)^3}}dR\\ &\vdots \\&=\frac{2(1-U_1^2)^{\frac{k-1}{2}-1}}{\text{Beta}\left(\frac{1}{2}, \frac{k-1}{2}\right)}\end{align}$$ which through one final change of variable $U_1^2=Y$ gives $f_Y$ in the standard from of the beta distribution. As expected, $E[U_1^2] = \text{Var}(U_1) = 1/k$.


Also note that the above is also equivalent to $U_1=2B-1$ where $B\sim \textrm{Beta}\left(\frac{k-1}{2},\frac{k-1}{2}\right)$, which you can derive by substituing $1-U_1^2=(1+U_1)(1-U_1)$ into the final expression above.

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  • $\begingroup$ We assume the dimension is $k$. Another interesting point is that: $U_1^2+\dots + U_r^2 \sim \mathrm{Beta}(\frac{r}{2}, \frac{k-r}{2})$. $\endgroup$ – zhaofeng-shu33 Dec 28 '19 at 9:16
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I hope that this is still relevant to you.

The answer can be found in eq. 1.26: Fang, Bi-Qi; Fang, Kai-Tai, Symmetric multivariate and related distributions, (2017). (https://books.google.co.il/books?hl=iw&lr=&id=NL1HDwAAQBAJ&oi=fnd&pg=PT10&ots=u5cnMFtVxP&sig=ZgnWbGkG8qdVARoJ64mfm9fcag0&redir_esc=y#v=onepage&q&f=false).

The marginal density of $(z_1/\Vert \textbf{z} \Vert, ..., z_k /\Vert \textbf{z} \Vert )$ is

$$\frac{\Gamma(n/2)}{\Gamma((n-k)/2) \pi^{k/2}} \left(1 - \sum_{i=1}^k z_i^2 \right)^{(n-k)/2 -1}$$

This can be derived from the Dirichlet distribution.

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I doubt that there is any practical way to calculate the distribution function in high dimensions. Note however, that the marginals are very close to a normal distribution in high dimensions, since $\frac{1}{n} \left(Z_1^2 + \ldots, + Z_n^2\right) \xrightarrow{a.s.} 1$ by the law of large numbers. This means $U_1 \stackrel{d}{\approx} \mathcal{N}\left(0, \frac{1}{n}\right)$. For more precise statements you can take a look into the references of Klartag 2007.

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  • $\begingroup$ Hi, thanks for the answer, could you further say which of the 53 references may contain related results. $\endgroup$ – Pushpendre Jan 18 '17 at 9:22
  • $\begingroup$ I remember that there are some results in the Klartag article itself, but I currently don't have access to the article. $\endgroup$ – Dominik Jan 18 '17 at 17:42
  • $\begingroup$ Got it, thanks. $\endgroup$ – Pushpendre Jan 18 '17 at 22:03

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