Statistical Inference and Manifolds

Question

I have just begun approaching the connection between statistical inference and differencial geometry.

If I got it correctly, one of the most fundamental concept regards the connection between a $ P(x; \xi) $ Statistical Distribution with $ \xi $ Parameter Vector and a certain $ V $ Manifold for which $ \xi $ can be seen as a coordinate system right ? For example if the distribution is gaussian hence $ P(x; \xi) \sim \exp \left ( - \frac{(x-\mu)^{2}}{2\sigma^{2}} \right ) $ the $ \xi = (\mu, \sigma^{2}) $ Parameters Vector can be considered a coordinate system for the corresponding manifold ? Is this coordinate system global ?

By the definition of manifold, it is locally euclidean hence an euclidean coordinate system can be locally used: how is it defined ?

Thanks

Answer 1

Your insight is correct. Let $(\Omega, A)$ be a measurable space with probability measure $P$. A statistical model of dimension $n$ is a triple $(\Theta, \mathcal M, \Phi)$ where $\Theta\subseteq \mathbb R^n$ is open, $\mathcal M=\{p_\theta:=p(\cdot|\theta),~~\theta\in\Theta\}$ is the set of parametric probability distributions on $\Omega$ and the mapping $\Phi:\Theta\rightarrow M$, $\Phi(\theta):=p(\cdot|\theta)$ in injective for all $\theta\in \Theta$ (i.e. one says that the statistical model is identifiable). The elements of $\mathcal M$ are also called the learning machines of the statistical model $(\Theta, \mathcal M, \Phi)$; one can adopt the notation $p_\theta:=p(\cdot|\theta)$.

$\mathcal M$ itself is called statistical manifold and the mapping $\Psi:\mathcal M\rightarrow \Theta$, $p_\theta\mapsto \Psi(p_\theta):=\theta$ allows to consider $\Psi$ as a coordinate system for $\mathcal M$.

• Example in the OP

Considering the example in the OP one has $\Theta = \mathbb R\times (0,\infty) \subset \mathbb R^2$, i.e. the (open) upper half plane in $\mathbb R^2$. The statistical manifold $\mathcal M$ is the set of all Gaussian distributions $p_\theta$ where $\theta:=(\mu, \sigma)\in\Theta$. Here $\mu$ is the mean of the distribution $p_\theta$, while $\sigma$ is its standard deviation, i.e.

$$p_\theta(x)=\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(x-\mu)^2}{2\sigma^2}} $$ The coordinate system of the statistical manifold $\mathcal M$ is global.