In this post, I ask two questions about Solomonoff Induction. I am not sure if these questions are open or not. If you know the answer to either of them, please let me know. I think that the answers may be very relevant to stuff I am currently working on in Asymptotic Logical Uncertainty.
Consider a universal Turing machine \(U\) which takes in an infinite stream of random bits, and outputs a finite or infinite string of bits. For any bit string \(w\), let \(P(w)\) denote the probability that \(w\) is a prefix of the output of \(U\).
Let \(E\) be an probabilistic environment which outputs an infinite bit string, and set \(e_i\) be the first \(i\) bits output by \(E\).
If \(E\) assigns outputs according to any computable probability distribution, then with probability 1, \(P(e_{i}1)/P(e_i)\) quickly converges to the actual probability that \(E\) outputs a 1 at location \(i+1\) given that its first \(i\) bits were \(e_i\).
In particular, if \(E\) outputs an infinite stream of iid bits which are each 1 with some rational probability \(p\), then \(P(e_{i}1)/P(e_i)\) will converge to \(p\) with probability 1.
However I am curious what happens if \(E\) is not computable. In the simplest case \(E\) outputs an infinite stream of iid bits which are each 1 with some uncomputable real number probability \(r\). In this case, will \(P(e_{i}1)/P(e_i)\) converge to \(r\) with probability 1?
To make things slightly more complicated, assume that the odd index bits of \(E\) come from a fair coin, but the even bits come from an adversary who chooses the bits in an uncomputable way. (This adversary can see all the bits output by \(E\) so far, but not future bits.) In this case, will \(P(e_{2i}1)/P(e_{2i})\) converge to 1/2?
