by Jonathan Lee 892 days ago | Jessica Taylor and Patrick LaVictoire like this | link | parent It looks like Theorem 1 can be improved slightly, by dropping the “only if” condition on $$p_{CD} > 0$$. We can then code up something like Kolmogorov complexity by adding a probability $$\frac{1}{2}$$ transition from every site to our chosen UTM. If you only want the weaker statement that there is no stationary distribution, it looks like there’s a cheaper argument: Since $$\Phi$$ is aperiodic and irreducible the hypothetical stationary distribution $$\pi$$ is unique. $$\Phi$$ is closed under the action of $$\Delta$$, and (2) implies that for any $$g \in \Delta$$, the map $$\Gamma_g$$ is an automorphism of the Markov chain. If the (infinite) transition matrix is $$T$$, then $$\Gamma_g$$ can be considered as a permutation matrix with (abusing notation) $$\Gamma_g^{-1}T\Gamma_g = T$$. Then $$T\Gamma_g\pi = \Gamma_g\pi$$ and so $$\Gamma_g\pi = \pi$$ by uniqueness. So $$\pi$$ is constant on orbits of $$\Gamma_{\Delta}$$, which are all countably infinite. Hence $$\pi$$ is everywhere $$0$$, a contradiction. The above still holds if (2) is restricted to only hold for a group $$G < \Delta$$ such that every orbit under $$\Gamma_G$$ is infinite. I think the above argument shows why (2) is too strong; we shouldn’t expect the world to look the same if you pick a “wrong” (ie. complicated) UTM to start off with. Weakening (2) might mean saying something like asserting only $$p_{CD} = \sum \mu(\Gamma) p_{\Gamma(C)\Gamma(D)}$$. To do this, we might define the measures $$\mu$$ and $$p$$ together (ie. finding a fixed point of a map from pairs $$(p, \mu)$$ to $$(p', \mu')$$). In such a model, $$\mu$$ constraints the transition probabilities, $$p'$$ is stationary; it’s not clear how one might formalise a derivation of $$\mu'$$ from $$p'$$ but it seems plausible that there is a canonical way to do it.

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