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by Jonathan Lee 830 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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