A.3.3.PI:5.2 - Preserve predictive memory under a coarse readout

A modeled device has hidden states A, B and C, with fixed transition probabilities:

The readout is 0 in A or B and 1 in C. A current 0 leaves next-1 probabilities from 0.1 to 0.7, depending on the hidden state. A single value such as their unweighted mean would add an unsupported assumption about which state is present.

For this calculation, take the initial distribution to be the stationary distribution (19/54,13/54,22/54). After observing 1, the current state is C. Its next-state weights are (0.2,0.3,0.5). Observing 0 next removes C and gives current A/B weights (2/5,3/5). The probability of the following 1 is

(2/5)*0.1 + (3/5)*0.7 = 23/50 = 0.46.

For the history 0,0, the first 0 gives weights (19/32,13/32,0). Apply the matrix, retain A and B after the next 0, and normalize. Their weights become (73/105,32/105), giving

(73/105)*0.1 + (32/105)*0.7 = 99/350, about 0.283.

A decision that changes when next-1 probability exceeds 0.4 takes different actions after those histories. Keeping only the latest 0 with the stationary A/B mixture would give 11/32, about 0.344, and lose the relevant difference. The distribution conditioned on history is the useful retained information.

The same matrix also gives a long-run fraction of readout 1 equal to 22/54. This finite positive chain has the conditions for that ergodic average. The average answers an aggregate question; the conditional probabilities answer the next-event question.

Initial weights other than the stationary distribution can give different finite-history predictions. If only a current 0 is known and no mixture is justified, the range [0.1,0.7] remains available. A probabilistic estimate requires its initial and transition account.