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Suppose $X = (X_1, X_2, X_3)$ is a random vector distributed uniformly on the unit sphere in $\mathbb{R}^3$. What is the probability density function of $X_1$?

user66081
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    See here: http://math.stackexchange.com/questions/1019286/if-u-is-uniformly-distributed-on-s2-then-its-first-component-is-uniforml/1019854#1019854 – jflipp Nov 14 '14 at 15:26
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    This question was answered by Archimedes in about the third century BC. – Michael Hardy Nov 14 '14 at 15:32

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It is a uniform distribution. See this article. It is essentially a result in geometry proved by Archimedes.

From the linked article:

Proposition 1. If $V$ is uniformly distirbuted on the interval $(-1,1)$ and if $\Theta$ is uniformly distributed on the interval $(0,2\pi)$ and independent of $V$, then $$ (V,\sqrt{1-V^2} \cos(\Theta),\sqrt{1-V^2}\sin(\Theta)) $$ is uniformly distributed on the surface of the two-dimensional sphere of radius one.

Michael Hardy
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    PS: A corollary is that $\sqrt{1-V^2}\cos\Theta$ is also uniformly distributed. – Michael Hardy Nov 14 '14 at 19:20
  • PS: A corollary is that $\sqrt{1-V^2}\sin(\Theta)$ is also uniformly distributed. – gota Oct 31 '18 at 16:57
  • Does this generalize in some way to higher dimensions? E.g., Can we write the uniform distribution over the 4-d sphere as the product of [-1,1] and a uniform point on a 3-d sphere? What about as a product of two uniform points on circles? – Thomas Ahle Nov 04 '18 at 16:10