by Jonathan Lee 1500 days ago | Jessica Taylor and Patrick LaVictoire like this | link | parent | on: Stationary algorithmic probability 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. reply

### RECENT COMMENTS

[Note: This comment is three
 by Ryan Carey on A brief note on factoring out certain variables | 0 likes

There should be a chat icon
 by Alex Mennen on Meta: IAFF vs LessWrong | 0 likes

Apparently "You must be
 by Jessica Taylor on Meta: IAFF vs LessWrong | 1 like

There is a replacement for
 by Alex Mennen on Meta: IAFF vs LessWrong | 1 like

Regarding the physical
 by Vanessa Kosoy on The Learning-Theoretic AI Alignment Research Agend... | 0 likes

I think that we should expect
 by Vanessa Kosoy on The Learning-Theoretic AI Alignment Research Agend... | 0 likes

I think I understand your
 by Jessica Taylor on The Learning-Theoretic AI Alignment Research Agend... | 0 likes

This seems like a hack. The
 by Jessica Taylor on The Learning-Theoretic AI Alignment Research Agend... | 0 likes

After thinking some more,
 by Vanessa Kosoy on The Learning-Theoretic AI Alignment Research Agend... | 0 likes

Yes, I think that we're
 by Vanessa Kosoy on The Learning-Theoretic AI Alignment Research Agend... | 0 likes

My intuition is that it must
 by Vanessa Kosoy on The Learning-Theoretic AI Alignment Research Agend... | 0 likes

To first approximation, a
 by Vanessa Kosoy on The Learning-Theoretic AI Alignment Research Agend... | 0 likes

Actually, I *am* including
 by Vanessa Kosoy on The Learning-Theoretic AI Alignment Research Agend... | 0 likes

Yeah, when I went back and
 by Alex Appel on Optimal and Causal Counterfactual Worlds | 0 likes

> Well, we could give up on
 by Jessica Taylor on The Learning-Theoretic AI Alignment Research Agend... | 0 likes

Privacy & Terms