1
$\begingroup$

Consider Markov chain $\{X_t\}_{t\in N}\subseteq R^{n\times n}$ defined by $X_{t} = X_0 G_1 \dots G_t$ where $G_i$'s are iid Gaussian matrices $G_1,\dots,G_t\sim N(0,1/n)^{n\times n}$, and $X_0$ is some deterministic matrix with full rank fixed scale, $\|X_0\|_F^2=n$ and $\operatorname{rank}(X_0)=n$. Is this chain ergodic? Put in other words, assuming that initial matrices $X_0$ and $Y_0$ share the some "nice" properties (eg. similar scale and full rank), can we find a coupling(*) of two copies of this chain $\{X_t\}_{t\in N}$ and $\{Y_t\}_{t\in N}$ such that $P(X_t\neq Y_t) \le \alpha^t$ for some $\alpha<1$?

Intuitively this coupling should be plausible: Since Gaussian matrices $G_i$'s are rotation invariant, we can couple the $X_t$ and $Y_t$ by optimising for rotation. But I come short of making these intuitions any more rigorous. I've looked up literature on random matrices but can't find anything relevant.

(*) More elaborately, can we design $\{(G_t,W_t)\}_{t\in N}$ where both $G_1,\dots, $ and $W_1,\dots$ are independent, both jointly are coupled to lead to mixing of the two chains?

$\endgroup$
4
  • $\begingroup$ What do you mean by "coupling"? Because your condition on $\operatorname{Pr}(X_t\neq Y_t)$ doesn't seem to make sense; that probability seems like it is unavoidably going to be 1. $\endgroup$
    – Buzz
    Commented Jan 5, 2023 at 0:13
  • $\begingroup$ meant the Gaussian matrices, referring to this kind of coupling. Thought it's pretty standard terminology, does it need to further details? mmss.wcas.northwestern.edu/thesis/articles/get/1052/… $\endgroup$
    – kvphxga
    Commented Jan 5, 2023 at 0:30
  • 1
    $\begingroup$ I do not think they are ergodic since a standard computer experiment shows that as $t\rightarrow\infty$, $X_t$ gets closer and closer to becoming rank-1 matrix. I do not think it is too hard to show that the rank-1 matrices are an attractor. Did you also want to make the determinant of each $G_i$ have absolute value 1 or at least normalize the norm? $\endgroup$ Commented Jan 5, 2023 at 1:24
  • $\begingroup$ @JosephVanName what kind of normalization will make the chain ergodic? so if product of singular values of $G_i$s is $1$ it becomes ergodic? $\endgroup$
    – kvphxga
    Commented Jan 5, 2023 at 13:02

1 Answer 1

2
$\begingroup$

You are dealing here with the products of invertible matrices, and the resulting Markov chain is known as a random walk on the corresponding group $GL(n,\mathbb R)$. A qualitative asymptotic "boundary" theory of such products was created by Furstenberg in the early 60's (see his 1963 papers "A Poisson formula for semi-simple Lie groups" and "Noncommuting random products"). He mostly talks about $SL(n,\mathbb R)$ and more general semi-simple Lie groups, but it doesn't make much difference.

I will return to Furstenberg's theory in a moment, but let me first say a couple of words about the "coupling" conditions you mention, in the "rawest" form. When talking about a general Markov chain one should in principle distinguish its mixing and ergodicity. Mixing is equivalent to the triviality of the tail $\sigma$-algebra of the chain or to the asymptotic independence of the time $n$ distributions of initial states: $$ \| (\delta_x - \delta_y) P^n \| \to 0 \qquad \forall x,y \;, $$ where $\|\cdot\|$ is the total variation, and $P$ is the transition operator of the Markov chain. In a similar way, ergodicity is equivalent to the triviality of the shift invariant $\sigma$-algebra in the path space (i.e., to the absence of non-constant bounded harmonic functions), or to the independence of the Cesaro averages of time $n$ distributions of initial states: $$ \frac1n \left\| \sum_{k=1}^n (\delta_x - \delta_y) P^k) \right\| \to 0 \qquad \forall x,y \;. $$

For the random walks that you consider mixing and ergodicity are actually equivalent. It follows from so-called 0-2 laws that also provide further information about the above differences.

Now, one of the consequences of Furstenbrg's theory is that random walks on matrix groups can only be ergodic ($\equiv$ mixing) in very special cases. Gaussian random walks on the linear group are not among them. In particular, in your setup $$ \lim_n \| (\delta_x - \delta_y) P^n \| \neq 0 $$ for any two initial matrices $x,y\in GL(n,\mathbb R)$ unless $x$ and $y$ are proportional.

$\endgroup$
6
  • $\begingroup$ Is there some kind of normalization on $G_i$s that can make the chain ergodic? Namely, if we normalize by $\ell^1$ or $\ell^2$ or product of singular values of $G_i$s, will the chain become ergodic? $\endgroup$
    – kvphxga
    Commented Jan 5, 2023 at 13:05
  • 1
    $\begingroup$ The right way is, instead of a normalization, to pass to the random walk on an appropriate homogeneous space of the linear group, namely, on the flag space. For $SL(2,\mathbb R)$ this is the boundary of the hyperbolic plane, and in the upper half plane model the corresponding action is the action of $SL(2,\mathbb R)$ on the extended real line by linear fractional transformations. The resulting Markov chain is then very much ergodic. $\endgroup$
    – R W
    Commented Jan 5, 2023 at 13:43
  • $\begingroup$ these all sound very fascinating, but I'm not able to fully understand the ideas. Since I don't want to bother with tens of follow up questions, Is there some reading material on these subjects that is accessible to non-mathematicians? (ie, no prior knowledge of differential geometry, hyperbolic spaces, etc) $\endgroup$
    – kvphxga
    Commented Jan 5, 2023 at 15:02
  • $\begingroup$ I would just allow myself one follow-up. Is the $n$-sphere boundary of some hyperbolic model? Namely, if we only take the left eigenvectors of the Gaussian matrices and build a chain of them, corresponding to a random walk on the $n$-sphere, would that be ergodic? $\endgroup$
    – kvphxga
    Commented Jan 5, 2023 at 15:06
  • $\begingroup$ Are you familiar with the Lyapunov exponents? It is in their terms that the boundary behaviour of random matrix products is described. $\endgroup$
    – R W
    Commented Jan 5, 2023 at 16:16

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .