## deriving angular central gaussian distribution from a multivariate normal distribution

The angular central Gaussian (ACG) distribution on $(p-1)$-dimensional sphere $\mathbb{S}^{p-1}$ for a symmetric positive definite parameter matrix $\mathbf{A}$ is defined as

$$f(\mathbf{x},\mathbf{A}) = |\mathbf{A}|^{-1/2} (\mathbf{x}^T\mathbf{A}^{-1}\mathbf{x})^{-p/2}.$$

As many paper pointed out, the name 'angular central Gaussian' is derived from the fact that if $\mathbf{x}\sim \mathcal{N}_q(\mathbf{0},\mathbf{A})$, then $\mathbf{x}/||\mathbf{x}||\sim ACG(\mathbf{A})$. But I have no idea how to derive the relation.

When a $p$-dimensional random variable $\mathbf{x}$ undergoes a transform $\mathbf{y}=H(\mathbf{x})$, the pdf of $\mathbf{y}$ is computed using the Jacobian matrix after plugging in $\mathbf{x} = H^{-1}(\mathbf{y})$ to the pdf of $\mathbf{x}$. In case of Gaussian-ACG transform, $H(\mathbf{x}) = \mathbf{x}/||\mathbf{x}||$ is not one-to-one: points $t\mathbf{x}\in\mathbb{R}^p$ for $t>0$ and $||\mathbf{x}||=1$ are mapped to a point $\mathbf{x}$ on the $\mathbb{S}^{p-1}$. How can I derive ACG from gaussian distribution?

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If you integrate the Gaussian distribution function $f_p (\mathbf{x})$ in $p$-dimensions along the radial direction from 0 to $\infty$ I believe you will derive the angular pdf you wrote down. Specifically, for $\mathbf{x} \in\mathbb{S}^{p-1}$, the integral $$\int_0^{\infty} t^{p-1} dt \; |\mathbf{A}|^{-1/2} \exp(-\frac{t^2}{2} \mathbf{x}^T\mathbf{A}^{-1} \mathbf{x})$$ gives, upto numerical constants,$$|\mathbf{A}|^{-1/2} (\mathbf{x}^T\mathbf{A}^{-1}\mathbf{x})^{-p/2}$$ as in your formula, where the volume form in $\mathbb{R}^{p}$ is decomposed as $t^{p-1} dt \wedge d\Omega$ with $d\Omega$ being the volume form on $\mathbb{S}^{p-1}$.
 Thank you for the answer. But I'm not sure where $t^{p-1}$ comes from. I'm guessing integrating the Gaussian distribution along the radial direction is computed by $\int_0^\infty t \exp(-\frac{1}{2} (t\mathbf{y})^T\mathbf{A}^{-1}(t\mathbf{y})) dt$ ignoring the constant. Also, I'm not that familiar with wedge product (or exterior product) with volume form in a hypersphere. Maybe I have to take a quick look at differential geometry textbooks. – Federico Magallanez Feb 3 2011 at 17:32 Just like the volume form in ${\mathbb R}^3$ is $r^2 dr sin \theta d\theta d\phi$ in polar coordinates, where $sin\theta d\theta d\phi = d\Omega$, the volume form in ${\mathbb R}^p$ in general can be written as $r^{p-1} d\Omega$. More formally, the factor $r^{p-1}$ comes from the Jacobian of the transformation to spherical coordinates. – ulvi Feb 3 2011 at 19:12