Here are some of the classical density functions for spherical distributions (on the $\mathcal{S}^{d-1}$ sphere, living in the Euclidean space $\mathbb{R}^d$):

  1. $$\mathbf{x}\mapsto \frac{(\kappa/2)^{d/2-1}}{2 \pi^{d/2} I_{d/2-1}(\kappa)} \exp(\kappa \mathbf{x}^{\top} \boldsymbol{\mu}), \qquad (\text{called the Fisher-von Mises-Langevin density}),$$

  2. $$\mathbf{x}\mapsto \frac{1}{a(\kappa,A)} \exp(\kappa \mathbf{x}^{\top} \boldsymbol{\mu} + \mathbf{x}^{\top} A \mathbf{x}), \qquad (\text{called the Fisher-Bingham density}),$$

  3. $$\mathbf{x}\mapsto \frac{\Gamma(d/2)}{2 \pi^{d/2} M(\frac{1}{2},\frac{d}{2},\kappa)} \exp(\kappa (\mathbf{x}^{\top} \boldsymbol{\mu})^2), \qquad (\text{called the Watson density}),$$

where $\kappa\geq 0$ is a concentration parameter, $\boldsymbol{\mu}\in \mathcal{S}^{d-1}$ is a location parameter, $A$ is a symmetric $d\times d$ matrix, and both $a(\kappa,A)$ and $M(\frac{1}{2},\frac{d}{2},\kappa)$ are the appropriate normalizing constants.

I've seen very few central limit theorems in the literature relating to this setting. In particular, I found absolutely nothing regarding local limit theorems. If the parameter $\kappa$ approaches some limit ($0$ or $\infty$), do any of these density functions approach a particular limit density (with a properly normalized argument)?

$\textbf{Example:}$ As the intensity parameter $\lambda$ of a Poisson$(\lambda)$ distribution tends to $\infty$, the probability mass function tends to the density of a $\text{Normal}(\lambda,\lambda)$ distribution. Is there any analogous results/conjectures in the context of spherical distributions?

  • $\begingroup$ I think it is worth pointing out that you also asked this question on math stackexchange. $\endgroup$ – Kernel May 14 at 7:31
  • $\begingroup$ Do you have an explicit formula/expansion for the characteristic function of such random variables? I am not familiar with this class of distributions. If do have such characterisation, the argument should be much easier, I believe. $\endgroup$ – Kernel May 14 at 7:31
  • $\begingroup$ The characteristic function of the matrix version of the Fisher-von Mises distribution is here (link.springer.com/content/pdf/10.1007%2F978-1-4612-1358-1.pdf) on page 484, but it's pretty nasty. I didn't find the characteristic function for the vector-version above. $\endgroup$ – Frédéric Ouimet May 14 at 8:11
  • $\begingroup$ Right, we actually do not need the explicit formula, but it would be convenient. In many proofs for LCLT for random walks, we just need an expansion of the characteristic function $\phi_X(\theta)$ around $\theta=0$. Because this is not a supported in a lattice, and therefore $\phi_X$ is not periodic, we will also need some estimates of the decay of $\phi_X$ at infinity as well. $\endgroup$ – Kernel May 14 at 10:06
  • $\begingroup$ Given such expansion, I do believe you could achieve a result like $\|p_{X_n}-p_{N(\mu\cdot n, \Sigma*n)}\|_\infty \le Cn^{{-d-2}/{2}}$ (where $\mu$ and $\Sigma$ respectively denote the appropriate mean vector and correlation matrix ) by formulating on the argument in Lawler-Limic book $\endgroup$ – Kernel May 14 at 10:08

$\textbf{Partial answer:}$ Note that $$\boldsymbol{x}^{\top} \boldsymbol{\mu} = -\frac{1}{2} (\boldsymbol{x} - \boldsymbol{\mu})^{\top}(\boldsymbol{x} - \boldsymbol{\mu}) + 1$$ since $\boldsymbol{x}^{\top}\boldsymbol{x} = \boldsymbol{\mu}^\top\boldsymbol{\mu} = 1$, and it is possible to show (using the integral representation for the modified Bessel function of the first kind) that, as $\kappa\to \infty$, $$e^{\kappa} \cdot \frac{(\kappa/2)^{d/2-1}}{2 \pi^{d/2} I_{d/2-1}(\kappa)} = \frac{1}{(2\pi \kappa^{-1})^{(d-1)/2}} \cdot \frac{1}{\sqrt{2\pi} \kappa^{1/2} e^{-\kappa} I_{d/2 - 1}(\kappa)} \approx \frac{1}{(2\pi \kappa^{-1})^{(d-1)/2}},$$ so that, as $\kappa\to \infty$, $$\frac{(\kappa/2)^{d/2-1}}{2 \pi^{d/2} I_{d/2-1}(\kappa)} \exp(\kappa \mathbf{x}^{\top} \boldsymbol{\mu}) \approx \frac{1}{(2\pi \kappa^{-1})^{(d-1)/2}} \exp\Big(-\frac{1}{2 \kappa^{-1}} (\boldsymbol{x} - \boldsymbol{\mu})^{\top}(\boldsymbol{x} - \boldsymbol{\mu})\Big).$$ This heuristic solves the question for 1.

$\textbf{Note:}$ The $d-1$ instead of $d$ comes from the restriction to the $\mathcal{S}^{d-1}$ sphere.


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