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If I have a random vector $\mathbf{y}$ generated from multivariate gaussian distribution $\mathcal{N}(\mathbf{0}, \mathbf{C})$, then I normalized it to unit length, which is, $$\mathbf{y} \sim \mathcal{N}(\mathbf{0}, \mathbf{C}),$$ $$ \mathbf{z} = \frac{\mathbf{y}}{||\mathbf{y}||}.$$ Is it possible to determine the type of the distribution of $\mathbf{z}$? If $\mathbf{y} \sim \mathcal{N}(\mathbf{0},\mathbf{I})$, I think $\mathbf{z}$ should satisfy a multivariate uniform distribution with each of its components is identically and independently distributed from $\mathcal{U}(-1, 1)$. Am I correct? and how to make a proof?

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1 Answer 1

No, if $y\sim N(0,I)$, then $z$ is uniformly distributed on the unit sphere. So the components can not be independent.

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I understand my incorrectness. So $\mathbf{z}$ follows a Uniform Spherical Distribution, right? Do you know any website giving the properties of Uniform Spherical Distribution? –  wangronin Feb 12 '13 at 16:27

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