A.3.3:5.8 - A long-run average can coexist with predictive memory

Consider a three-state Markov model with this transition matrix. A readout reports 0 for A or B and 1 for C.

A present readout of 0 merges states with next-1 probabilities 0.1 and 0.7. The readout alone therefore leaves predictive information unresolved. The stationary distribution is (19/54,13/54,22/54). At stationarity, after readouts 1,0, the current A/B weights are 2/5,3/5, so the next-1 probability is 23/50. After 0,0, those weights are 73/105,32/105, giving 99/350. A decision that changes above probability 0.4 takes different actions after these histories. Keeping only the present 0 and the stationary A/B mixture gives 11/32 and loses that difference.

Condition on the available history or retain the resulting predictive distribution. With no information beyond the current 0, the range [0.1,0.7] may already answer a weaker question. The full finite chain is irreducible and aperiodic, and its long-run proportion of readout 1 converges to 22/54. That long-run result leaves the history-dependent prediction above intact. The finite-chain results are given in Cambridge’s Markov Chains notes, §§9-10; the matrix and conditional calculations here are an authored example.