r/probabilitytheory • u/FullSand8610 • 9h ago
[Research] Can two Markov chains that agree on their shared states always be combined into one? (proof sketch, looking for checks)
I've been working on a question about finite Markov chains and would value a check of my proof, or a pointer if it's already known.
**Setup.** Take a finite irreducible Markov chain and watch it only while it is inside a subset S of its states (skip the rest). What you see is again a Markov chain on S, called the trace or censored chain.
Now take two irreducible chains: M_A on states C ∪ X, and M_B on states C ∪ Y, sharing only the states C. Suppose they agree on C: watching either one only on C gives the same chain N.
**Question.** Is there always a single finite Markov chain P (with labelled states, where several states may share a label) such that watching P only on A's labels gives exactly M_A, and only on B's labels gives exactly M_B?
**My answer: yes.** At each shared state c, first run one M_A excursion until it returns to C (landing at some c'), then run one M_B excursion from c conditioned to land at the same c' (an h-transform). Emit both parts and carry on from c'. Watched on A you see independent M_A excursions; watched on B, the landing law is N(c, ·), which is B's own, so you see M_B. This gives a chain with at most |C|(1 + |X| + |Y|) states.
**Small example.** A: 1→3→2→4→1, B: 1→5→2→6→1, shared {1,2}. Then 1→3→5→2→4→6→1 contains both. Choosing an A or B excursion at random at each shared state fails, because A would sometimes see 1→2 directly.
**What I'm unsure of:** that the kept states form one irreducible class, and the step from starting at a shared state to the stationary start. Exact computer checks on 14 hard pairs match to about 1e-15.
The same idea extends to tree-shaped families, and loops of three or more chains can fail. The full write-up with proofs is a PDF here: https://github.com/goodcitizennc-ux/math-proofs
Reply’s are welcome here as well.
I developed this with help from an AI assistant and no mathematician has checked it yet. Is it right, and is it known?