3
$\begingroup$

I'm wondering if anyone could tell if there exists a widely accepted theory of probability distributions defined on topologically nontrivial manifolds? If so, as a physicist, I would appreciate providing some explanations using the, possibly, simplest example, a circle $S^2$.

Here are my thoughts. Generally, for a manifold $\mathcal M$, I see no problem in defining some 'distribution' $f(x)$ with $x\in\mathcal M$, such that ${\int_{\mathcal M} f(x)d\mu(x)=}1$. Obviously, this definition is metric-dependent. Still, oftentimes we have a canonical definition of the metric, e.g. borrowed from $\mathbb R^n$ in the case if some canonical embedding is given, which is often the case. However, we face serious difficulties when try to define 'averaged' quantities. (And in physics that's what we typically want to do).

Assume, given some 'distribution' $f(\vec{n})$, we want to calculate its mean value. One option would be to define it as follows: $$ \langle \vec n \rangle = \dfrac{\int \limits_{S^2} \vec{n} f(\vec n) ds}{\left|\int \limits_{S^2} \vec{n} f(\vec n) ds\right|^2} $$

The good thing about this definition is that it gives somewhat expected results, especially in the case of sharply-peaked distributions. However, we immediately face a huge number of problems. First of all, there exist a wide range of 'distributions' for which $\langle\vec n\rangle$ is undefined (all the shells whose center of mass is at the origin). Second, excluding such 'bad' 'distributions' from the consideration does not really save us, for such an exclusion may be be 'quantity'-dependent (were we averaging not $\vec n$ but smth else, we would have to exclude other distributions). Moreover, even if we exclude all the 'bad' ones (for a particular quantity of interest), we still cannot even define the sum for the remaining 'good' ones, for, again, the sum of 'good' distributions may be a 'bad' one.

OK, let's now consider a totally different approach suggested by the discrete probability theory. What is the mean value for the random variable which in a half cases gives $-1$, and in another half - $+1$? Well, clearly it's $0$, you would say. But wait, in terms of a 'discrete guy' who only deals with two objects in the universe, $-1$ and $+1$, this does not make any sense. There's no such object as $0$ in his universe. Nonetheless, this definition oftentimes makes sense. Why? Because we know that both $-1$ and $+1$ have a natural inclusion into $\mathbb R^n$ where the mean value can be defined. Let us stop for a second and appreciate this fact - we allowed the 'mean' value of a distribution defined on the the set $\mathcal S=\{-1,+1\}$ to take values on a different set $\mathcal{S}' = [-1,1]$. (On the contrary, as of 03/2019, the canonical way of embedding heads and tails into $\mathbb R^n$ is still not known, and, so, their mean value does not make much sense.)

Generalising this procedure to our example is straightforward: $$ \langle \vec n \rangle = \int \limits_{S^2} \vec{n} f(\vec n) ds $$ Which basically gives us a mean value of a distribution defined on $\mathcal S'$ (again, by inclusion). An obvious downside - the averaged quantities have now no meaning for inhabitants of the $\mathcal S$ manifold.

Is any of these approaches dominant? Or maybe smth else? Is there a theory for general, more complicated manifolds? Any comments and /simple/ references are welcome.

$\endgroup$
2
$\begingroup$

The theoretical underpinnings of probability are simple - a probability distribution is simply a (nonnegative) measure defined on some $X$ such that the total measure of $X$ is $1$. That makes perfect sense on a manifold. There's no need for any special treatment.

If it's a smooth oriented $n$-manifold, we can use that smooth structure to define the density function for a "continuous" probability distribution - that density function is a nonnegative $n$-form with integral $1$. Compact Lie groups, and compact manifolds with a transitive Lie group action, even have a standard "uniform" distribution, invariant under that Lie group action.

Now, you want to talk about expected values? We can set up those integrals $E(f)=\int_{\mathcal{M}} f(x)\,d\mu(x)$ with respect to the probability measure $\mu$, but only as long as the function $f$ we're trying to find the expected value of takes values in $\mathbb{R}$, or at least some normed vector space. The expected value is a weighted sum - we need to be able to add and take scalar multiples to make sense of it at all.
That's the same normed vector space everywhere - something like a function from points on the manifold to vectors in the tangent space at those points isn't going to work (unless we embed everything into $\mathbb{R}^m$, standardizing the tangent spaces as subspaces of that).

So then, the expected value of the position function doesn't make sense (usually). The manifold isn't a normed vector space, after all. It doesn't have an addition operation - why would we ever be able to add things up on it anyway?. On the other hand, with a particular embedding of the manifold into $\mathbb{R}^m$, we can take an expected value of that. The uniform distribution on the sphere $S^2$, with the standard embedding into $\mathbb{R}^3$ as $\{(x,y,z): x^2+y^2+z^2=1\}$, has an expected value of $(0,0,0)$. That's not a point on the sphere, and there was never any reason to expect it to be.

$\endgroup$
  • $\begingroup$ How would you reate this to things like Kent distribution? It does have a 'mean' vector... $\endgroup$ – mavzolej Mar 2 at 7:22
  • $\begingroup$ It's a probability distribution. We treat it like any other. That "mean" vector is a vector in $\mathbb{R}^3$ coming from the standard embedding, not something in $S^2$. $\endgroup$ – jmerry Mar 2 at 7:29
  • $\begingroup$ So here 'mean' is just a name, it's not smth we can derive from the form of the distribution using a general formula? $\endgroup$ – mavzolej Mar 2 at 7:31
  • $\begingroup$ It's not just a name. We can derive it from the form of the distribution and the embedding. Once we've embedded it into $\mathbb{R}^3$, our position vector is taking values in a normed vector space, and that's something we can take a mean of. $\endgroup$ – jmerry Mar 2 at 7:35
  • $\begingroup$ But the mean value in the sense of $\mathbb R^n$ wouldn't belong to the sphere, it would have a smaller magnitude. $\endgroup$ – mavzolej Mar 2 at 13:37
0
$\begingroup$

One possible generalization of "mean/expectation" to metric space is called the Frechet mean/expectation which minimizes the expected value of the square distance. Let $(M, d)$ be a metric space and $X$ be a $M$-valued random variable with probability measure $P$. Then the Frechet expected value is defined as $$ E(X) := \arg\min_{y \in M} \int_M d^2(X,y)dP. $$

However the existence and uniqueness are not guaranteed. For your case, you can equip the manifold with a Riemannian metric and use the induced geodesic distance or simply use the distance induced from the embedding space.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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