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I will fix the dimension $n$ and use $S:=\{x\in\Bbb R^n:\|x\|=1\}$ to denote the unit sphere. Let $\sigma$ denote surface measure on $S$, and define $\bar\sigma:=[\sigma(S)]^{-1}\sigma$, the "uniform distribution" on $S$.

Here $Y=X/\|X\|$ and X an n-dimensional standard gaussian vector

Let $f:S\to\Bbb R$ be bounded and Borel measurable. Then $$ \eqalign{ \Bbb E[f(Y)] &=\int_{\Bbb R^n} f(x/\|x\|)(2\pi)^{-n/2}e^{-\|x\|^2/2}\,dx\cr &=(2\pi)^{-n/2}\sigma(S)\int_S\int_0^\infty f(u)e^{-r^2}r^{n-1}\,dr\,\bar\sigma(du),\cr &=\int_S f(u)\,\bar\sigma(du),\cr } $$ because $$ \int_0^\infty e^{-r^2}r^{n-1}\,dr =2^{n/2-1}\int_0^\infty e^{-t}t^{n/2-1}\,dt=2^{n/2-1}\Gamma(n/2) $$ while $\sigma(S)=2\pi^{n/2}/\Gamma(n/2)$.

Why is $X/\|X\|_2$ uniformly distributed on a unit sphere when X is n-dimensional standard gaussian vector? I do not follow the exact transformations made to $\int_{\mathbb{R}^n}f(x/\|x\|)2 \pi^{-n/2}e^{-\|x\|^2} dx$.

I understand there is a change of variable $x \to (r,u)=(\|x\|,x/\|x\|)$ but can't see how do we get the terms $e^{-r^2}$ and $r^{n-1}$ in the term after the second equality . Could someone show me the exact computations?

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  • What exactly do you mean by "the second term"? – PhoemueX Jan 21 '16 at 18:20
  • @PhoemueX I am sorry I mean the second equality after E[f(Y)]. The term after the first equality follows from $'Y'$ being normally distributed , but i do not follow why the second term after the equality follows from the guassian integral – user3503589 Jan 22 '16 at 06:40
  • This is the change of variable $x\to(r,u)=(|x|,x/|x|)$. // Please add to the post a definition of $Y$ and what is your exact question. – Did Jan 22 '16 at 06:42
  • @Did Thank you, but I guessed the transformations but i cannot get the exact expression in the term after the second equality computing $E[f(Y)]$. Could you show me how do I get the exact integral after the change of variables? – user3503589 Jan 22 '16 at 07:00
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    This is the basic change of variables formula in integrals. In short, $dx=c_nr^{n-1}dr\bar\sigma(du)$ and $e^{-r^2/2}$ (instead of the typo $e^{-r^2}$) is simply $e^{-|x|^2/2}$. – Did Jan 22 '16 at 07:13

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