I have seen in various places the following claim:

Let $X_1$, $X_2$, $\cdots$, $X_n \sim \mathcal{N}(0, 1)$ and be independent. Then, the vector $$ X = \left(\frac{X_1}{Z}, \frac{X_2}{Z}, \cdots, \frac{X_n}{Z}\right) $$ is a uniform random vector on $S^{n-1}$, where $Z = \sqrt{X_1^2 + \cdots + X_n^2}$ is a normalization factor.

Many sources claimed this fact folllows easily from the "orthogonal-invariance" of the normal distribution, but somehow I couldn't construct a rigorous proof. (one such "sketch" can be found here).

So, my question is why is this true?

  • $\begingroup$ Is there some papers which explain this? $\endgroup$ – Christo Feb 18 at 15:55

The two word answer is "polar coordinates".

In more detail, let $f:S^{n-1}\to\Bbb R$ be a continuous function. Then $$ \eqalign{ \Bbb E[f(X)] &=\int_{\Bbb R^n}f(x_1/z,\ldots,x_n/z)(2\pi)^{-n/2}e^{-z^2/2}\,dx_1\cdots dx_n\cr &=(2\pi)^{-n/2}\int_0^\infty\left[\int_{S^{n-1}} f(u)\,\sigma_{n-1}(du)\right]e^{-r^2/2}r^{n-1}\,dr\cr &=c_n\int_{S^{n-1}} f(u)\,\sigma_{n-1}(du).\cr } $$ Here $\sigma_{n-1}$ is the "surface area" measure on the sphere $S^{n-1}$ and $$ c_n=(2\pi)^{-n/2}\int_0^\infty e^{-r^2/2}r^{n-1}\,dr = \pi^{-n/2}2^{-1}\Gamma(n/2). $$ (Thus $2\pi^{n/2}/\Gamma(n/2)$ is the surface area of $S^{n-1}$.)

  • $\begingroup$ Thanks! one question though; why the fact the expectation (after composing with real functions) behaves identically to uniform distribution implies the distribution was indeed uniform? (I guess I am asking why the information on the first moment determines the distribution). $\endgroup$ – Asaf Shachar Jul 19 '16 at 18:07
  • 2
    $\begingroup$ Because $f$ is an arbitrary continuous function. By approximating the indicator of a closed set $F\subset S^{n-1}$ by continuous functions you can show that $\Bbb P[X\in F]$ coincides with the probability accorded $F$ by the uniform distribution on $S^{n-1}$. Once you have this for closed sets it follows automatically for Borel sets. $\endgroup$ – John Dawkins Jul 19 '16 at 18:15
  • $\begingroup$ very beautiful and elegant proof $\endgroup$ – pointguard0 Aug 10 '18 at 11:18
  • $\begingroup$ Does this theorem hold for any other distributions as well? ie. Not just gaussian? $\endgroup$ – rohaldb Mar 12 at 2:29

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