It is known that given $X=(X_1, X_2, \ldots, X_n)$ iid $\sim N(0,1)$, then $X/\sqrt{X_1^2+\cdots+X_n^2}$ is uniformly distributed on the surface of unit sphere.

Intuitively, I know that that's because the probability of $X/\sqrt{X_1^2+\cdots+X_n^2}$ belonging to any region with the same area on the surface should be the same. But how can I prove it mathematically?

  • 1
    $\begingroup$ See the references in the article. $\endgroup$ – user64494 Jul 16 '13 at 5:30
  • $\begingroup$ @user64494 i saw that page before but the references there only mentioned the method, not a detailed proof. $\endgroup$ – Julie Jul 16 '13 at 12:35
  • $\begingroup$ Have you tried to use spherical coordinates? $\endgroup$ – Davide Giraudo Jul 16 '13 at 17:20
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    $\begingroup$ The proof in less than 600 characters. $\endgroup$ – cardinal Jul 16 '13 at 23:25
  • $\begingroup$ @cardinal Thanks a lot! $\endgroup$ – Julie Jul 17 '13 at 3:14

Suppose $X=(X_1, X_2, \ldots, X_n)$ iid and $X_1 \sim N(0,1)$, then $X \sim N(0, I_n)$, where $N(0, I_n)$ is the multivariate normal distribution with zero-mean and identity covariance matrix. From that it follows, that if $O$ is an orthogonal matrix, that $OX$ is identically distributed with $X$. From that it follows, that $Y = \frac{X}{||X||_2}$ is identically distributed with $\frac{OX}{||OX||_2} = \frac{OX}{||X||_2}$. From that we can conclude, that $Y$ is invariant under rotations and belongs to the unit sphere. There is only one probability distribution, that satisfies both those conditions at the same time: that is the uniform distribution on the unit sphere.


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