is $P_{CTC} = BPP_{path}$? I think that these two classes should be the same, but I can't find any literature about this and have a limited background on the topic.

This is my reasoning, and I would like to know if (1) this is already known or (2) I misunderstood something or (3) I just found out something useful:

$P_{CTC}$ is the class of problems that can be solved by putting polynomial amounts of data in a time machine.

$BPP_{path}$ is the class of problems that can be solved by postselecting in a probabilistic Turing machine, i.e. ignoring cases you do not care about.

$P_{CTC}\subseteq BPP_{path}$ because you can simulate a closed time-like curve with postselection like this: Scan the entire program in the beginning, both state and memory. Then, after processing, do so again and postselect so that you only return if the state &amp; memory now are exactly equal to the start state &amp; memory (except for a single bit that says whether or not this is the first iteration, to prevent an infinite loop).

$BPP_{path}\subseteq P_{CTC}$ because you can simulate postselection like this: If the message from the future starts with $1$, send the message $0$ into the past. Otherwise proceed as normal. When you get to the step where you would normally post-select, send a 1 into the past iff. you want to ignore this timeline, else a $0$. The only consistent version is now the one where you both receive and send a 0 because you were satisfied with the results.
I believe I have found the answer: The proof is wrong. BPP_PATH is not in P_CTC because P_CTC is required to give a single definite answer, while BPP_PATH is a probabilistic algorithm, so the second reduction does not work. For it to work, one would have to use the time-travel information to count the number of successes of the probabilistic algorithm vs the number of its failures. I have no idea how to do this, or if it even can be done (probably not).