1
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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 $\mathbb{S}^{p-1}$.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Federico Magallanez Feb 3 '11 at 17:32
  • $\begingroup$ 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. $\endgroup$ – ulvi Feb 3 '11 at 19:12

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.