The Shannon-McMillan-Breiman theorem, in its dynamical systems formulation, contains (or implies) the following statement: If $T$ is the map, $\mu$ is the ergodic $T$-invariant measure, $\mathcal{P} = \{P_1,\ldots,P_r\}$ is the partition of the state space, $\Sigma_{n,\epsilon}$ is a typical set of length $n$ itineraries, i.e., $\Sigma_{n,\epsilon} = \{ a \in \{1,\ldots,r\}^n : 2^{-n(h+\epsilon)} \leq \mu(x: i_n(x) = a) \leq 2^{-n(h-\epsilon)} \}$ with $h = h_{\mu}(T;\mathcal{P})$ and $i_n(x) = (i_n^0(x),\ldots,i_n^{n-1}(x))$, $T^j(x) \in P_{i_n^j(x)}$, then $\mu(x: i_n(x) \notin \Sigma_{n,\epsilon}) \rightarrow 0$ for $n\rightarrow\infty$. My question is: Are there any results on the rate of convergence for this limit, maybe under specific, but not too restrictive assumptions on the dynamical system?

  • 4
    $\begingroup$ As you probably know, there are negative results, showing that you cannot hope to have rates of convergence in any ergodic theorem (including Shannon-Macmillan-Breiman), even for the “in measure” variants that you are looking for. If you want a positive result, I would impose Gibbs or exponential mixing conditions. Keller’s book on equilibrium states is quite closely related to what you are asking. $\endgroup$ – Anthony Quas Oct 11 '17 at 20:26

In the case of (weak) Gibbs measures you can find exponential large-deviation bounds as well, as almost sure estimates for the error term in the convergence to entropy given by the Shannon–McMillan–Breiman formula for both uniformly and non-uniformly expanding shift spaces [ETDS, 37:7, (2017), 2313-2336]. For broader classes of invariant measures, in opposition to the case of Birkhoff averages, I belive it is not known reasonably mild assumptions under which e.g. some L^1 convergence holds.

  • $\begingroup$ I think you may have accidentally deleted the link you meant to post; you should be able to edit it back in. $\endgroup$ – user44191 Feb 7 '19 at 19:37
  • $\begingroup$ Hi @paulo welcome to MO. It appears you have a broken link above. What reference were you hoping to including for the Shannon–McMillan–Breiman formula? $\endgroup$ – Neil Hoffman Feb 7 '19 at 19:38

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.