13
$\begingroup$

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?

$\endgroup$
  • 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
  • 6
    $\begingroup$ 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
2
$\begingroup$

Suppose $X_1$ and $X=(X_1, X_2, \ldots, X_n)$ iid $\sim N(0,1)$, 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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.