fix some n in n. if my starting state is [1; 0] then the probability of the occupancy being [1; 0] after n cycles is either 1 or 0. If the starting state is [0; 1] then the probabilty of the occupancy being [1; 0] is exactly the opposite, so for a fixed point in the future, the probability is tightly past-dependent.
Comments
fix some n in n. if my starting state is [1; 0] then the probability of the occupancy being [1; 0] after n cycles is either 1 or 0. If the starting state is [0; 1] then the probabilty of the occupancy being [1; 0] is exactly the opposite, so for a fixed point in the future, the probability is tightly past-dependent.
Yes, but I think it is still a Markov process since the information is contained in its current state.