Here, by a statistical manifold I mean a $d$-dimensional Riemannian manifold whose points are probability measures on $\mathbb{R}^n$. What are some well-studied/interesting examples of statistical manifolds which are complete Riemannian manifolds of dimension $d\geq 1$?
1 Answer
The statistical manifold of univariate normal distributions $\mathcal{N}(\mu,\sigma)$ is an absolutely fascinating space. Here are a few of its properties, but there is much more that can be said.
- As a Riemannian manifold, the space of Gaussian distributions is a hyperbolic half-plane. Furthermore, the standard $(\mu,\sigma)$ coordinates are (almost) the standard coordinates on the half-plane, in that the metric is given by $$ds^2= \frac{1}{\sigma^2}(d \mu^2 + 2d \sigma^2).$$
- Since normal distributions are an exponential family, the space of normal distributions is a Hessian manifold, with natural parameters $\eta_1= \frac{\mu}{\sigma^2}$ and $\eta_2 = -\frac{1}{2 \sigma^2}$ and sufficient statistics $\theta_1 = \mu$, $\theta_2 = \mu^2+\sigma^2$.
- With respect to the natural parameters, the space of Gaussian distributions has the geometry of a half plane. Its tangent bundle inherits a Kahler metric. As a Kahler manifold, it is biholomorphic to the half plane in $\mathbb{C}^2$ and is known as the Siegel-Jacobi space. This space has been studied in the context of automorphic forms (see, e.g. Gaussian distributions, Jacobi group and Siegel-Jacobi space). There are too many interesting properties of the Siegel-Jacobi space to list fully, but it's an absolutely fascinating space (if you'll pardon some self-promotion, a coauthor and I studied this space [A] and showed that it also has a really strange positivity property).
- With respect to the sufficient statistics, the space of Gaussian distributions has the geometry of the inside of a parabola. Its tangent bundle also inherits a Kahler metric. As a Kahler manifold, this space is the Jacobi half-plane, and the metric is simply the Bergman metric, which is a metric of constant negative holomorphic sectional curvature. Furthermore, as Kahler manifolds, the Siegel-Jacobi space and Jacobi half-plane are related by T-duality, so these spaces are mirror to each other in some sense.
You can study the space of all multivariate Gaussian distributions, which are also complete statistical manifolds. However, these statistical manifolds do not have constant curvature [B], and their geometry is considerably more complicated.
[A] Khan, Gabriel; Zhang, Jun, The Kähler geometry of certain optimal transport problems, Pure Appl. Anal. 2, No. 2, 397-426 (2020). ZBL1446.49034.
[B]Skovgaard, Lene Theil, A Riemannian geometry of the multivariate normal model, Scand. J. Stat., Theory Appl. 11, 211-223 (1984). ZBL0579.62033.
-
$\begingroup$ Thank you very much for the warm welcome and very detailed/interesting post Gabe K. If you don't mind, I'll accept it shortly but I'll wait abit first to see if any other interesting examples trickle in. $\endgroup$ Commented Dec 22, 2020 at 4:09
-
$\begingroup$ Thanks for the nice question! I'm also interested in seeing what examples are suggested since many common examples of statistical manifolds are incomplete. $\endgroup$– Gabe KCommented Dec 22, 2020 at 4:17
-
$\begingroup$ I'm reading thought the article, but I thought to ask also am I correct in interpreting Theorem 2.2 to imply that this manifold has sectional curvature bounded between $[-\frac1{2},\frac1{2}]$? $\endgroup$ Commented Dec 22, 2020 at 4:25
-
1$\begingroup$ I'm not sure exactly what the bounds on the sectional curvature are. In particular, I don't see why $K\left(\operatorname{span}\left\{e_{i}, E_{j k}\right\}, P\right)=-\frac{1}{4}\left(2 a_{i j k} \bar{\varrho}_{i j} \bar{\varrho}_{k j} \bar{\rho}_{k i}+\bar{\varrho}_{i j}^{2}+\bar{\varrho}_{i k}^{2}\right) /\left(1+\bar{\varrho}_{j k}^{2}\right)$ is necessarily bounded from below by $-1/2.$ It's definitely possible but not obvious. Also, I don't see how the sectional curvature could ever exceed $1/4$, so you can probably tighten the upper bound. $\endgroup$– Gabe KCommented Dec 22, 2020 at 20:03
-
$\begingroup$ you're right. I think I misread something. Thanks a lot Gabe K! You've been a big help. $\endgroup$ Commented Dec 22, 2020 at 21:03