I know that when $\mathbf{Y}\sim\mathcal{N}(\mathbf{0},\mathbf{I})$, $\mathbf{Y}/\|\mathbf{Y}\|_2$ is distributed uniformly on the unit sphere. But to my surprise, I failed to find a simple closed form to the distribution of the normalized $\mathbf{Y}$ when $\mathbf{Y}\sim\mathcal{N}(\boldsymbol{\mu},\mathbf{I})$ if $\boldsymbol{\mu}\ne \mathbf{0}$. I actually need to calculate the expectation of the product of powers of components of $\mathbf{Y}/\|\mathbf{Y}\|_2$ but it seems to be too hard a task to do using the distribution of $\mathbf{Y}/\|\mathbf{Y}\|_2$ for this noncentral case, which actually struck me as a rather peculiar thing.

So my question is,

Is there any alternative standard method to calculate expectation of the form $\mathbb{E}(X_1^{m_1}\cdots X_{N}^{m_N})$ where $X_i$ are the components of the normalized vector $\mathbf{Y}/\|\mathbf{Y}\|_2$ having noncentral Gaussian distribution?

I have some ideas that may lead to some upper bound but I think that will be too loose. It will be great if someone can kindly help to at least find some structure among the components of the normalized vector which can further be utilized in some specialzied manner to calculate the expectation. Any help is appreciated. Thanks in advance.

  • $\begingroup$ The $L_{2}$-norm of Y is in reference to the mean, right? $\endgroup$ – EngrStudent Jun 16 '15 at 13:25
  • 2
    $\begingroup$ Use cylindrical coordinates with the axis pointing along $\mu$. $\endgroup$ – whuber Jun 16 '15 at 13:43
  • $\begingroup$ No, @EngrStudent, the $l_2$ norm is not, that's where the problem is, otherwise I know how to find it. $\endgroup$ – Samrat Mukhopadhyay Jun 16 '15 at 13:50
  • $\begingroup$ Ok, @whuber, thanks! I'll give it a try and let you know. $\endgroup$ – Samrat Mukhopadhyay Jun 16 '15 at 13:51
  • $\begingroup$ You can also look into the projected Normal distribution (See Nunez-Antonio and Gutierrez-Pena (2005)), which may provide a closed form, at least, to the distribution of the normalized Y when Y∼N(μ,I). $\endgroup$ – Kees Mulder Jun 16 '15 at 14:27

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