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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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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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    $\begingroup$ 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? $\endgroup$ – wangronin Feb 12 '13 at 16:27
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    $\begingroup$ @UwF, just out of curiosity, do you know any reference that gives the distribution of $\mathbf{Y}/\|\mathbf{Y}\|_2$ when $\mathbf{Y}\sim \mathcal{N}(\boldsymbol{\mu},\mathbf{I})$ where $\boldsymbol{\mu}\mathbf{\ne 0}$? $\endgroup$ – Samrat Mukhopadhyay Jun 15 '15 at 19:26

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