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I have a random variable $\mathbf{X}$ following an isotropic matrix normal distribution with mean $\mathbf{M}$, i.e. $\mathrm{vec}(\mathbf{X}) \sim \mathcal{N}(\mathbf{\mathrm{vec}({\mathbf{M})}}, \sigma \mathbf{I})$.

I am, essentially, interested in figuring out how the equivalence classes / orbits induced by the $\mathcal{SO}(3)$ group (the set of all 3D rotation matrices) is distributed. That is, how the random variable $[\mathbf{X}] = \{\mathbf{X} \mathbf{R}^T \mid \mathbf{R} \in \mathcal{SO}(3)\}$ is distributed.

Working with ZYX Euler angles (and omitting some constants) led me to an integral of the following form, which I would like to solve: $$ \int_{0}^{2\pi}\int_{-\pi}^\pi \int_{0}^{2\pi} \cos(\beta) \exp\left( \sum_{n=1}^N \left(\mathbf{R}(\alpha,\beta,\gamma)\mathbf{A}_n\right)^T \mathbf{B}_n \right) \ d \alpha \ d\beta \ d\gamma \tag{1} $$ with $\mathbf{A}, \mathbf{B} \in \mathbb{R}^{N \times 3}$ and $$\mathbf{R}(\alpha,\beta,\gamma) = \begin{bmatrix} \cos(\gamma) & -\sin(\gamma) & 0 \\ \sin(\gamma) & \cos(\gamma) & 0\\ 0 & 0 & 1 \end{bmatrix} \cdot \begin{bmatrix} \cos(\beta) & 0 & \sin(\beta) \\ 0 & 1 & 0 \\ -\sin(\beta) & 0 & \sin(\beta) \end{bmatrix} \cdot \begin{bmatrix} 1 & 0 & 0\\ 0 & \cos(\alpha) & -\sin(\alpha) \\ 0 & \sin(\alpha) & \cos(\alpha) \end{bmatrix}. $$


I am aware that Eq. 1 is (up to some normalizing constant) equivalent to a Haar integral of the form $$ \int_{\mathcal{SO}(3)} f(\mathbf{R}) \mu(\mathbf{R}). $$ But I do not know how to use that information.

There exist some posts (e.g. 1, 2, 3) about Haar integration for $\mathcal{SO}(3)$, but they don't discuss a specific function or how to solve the integral. There is also a question about numerically integrating over Euler angles, but I am interested in an analytical solution.


I have the following two questions
1.) Is there some way of deriving a closed-form analytic expression for Eq. 1?
2.) Or does the integral in Eq. 1 correspond to some special function (e.g. Bessel function?)

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1 Answer 1

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I think you are over thinking this: The original distribution is the same as $\bf{X}\sim \bf{M} + \bf{Y}$ where $\bf M$ is the deterministic mean, and $\bf Y$ is distributed as an isotropic Gaussian. These two terms are idependent. that applying a rotation $\bf R$ to $\bf X$ will yield $ \bf{X}' = \bf{RM} + \bf{RY}$. Since you are asking about what happens when $R$ is isotropic, then $ \bf{X}'$ is rotationally invariant, hence $P({\bf X}')\equiv p(|{\bf X}'|)\equiv p(|{\bf X}|) $ where $p$ is a scalar function. To compute $p$ we are left with computing the distribution of $|{\bf X}| = |\bf{M} + \bf{Y}|.$

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