A coin is successively flipped and its probability of showing up heads is p e (0, 1). Let Y1, Y2, ... be the sequence of outcomes. For n 2 0, let Xn = (Yn+1, Yn+2, Ym+3). that is the nth triple of consecutive outcomes from the flips. The sequence X, is a Markov chain on the state space E = {TTT,TTH, THT,HTT,THH,HTH,HHT,HHH} := {1, 2, 3, 4, 5, 6, 7, 8}, where T and H denotes tail and head shown in the flips while the 8 possible states are also numbered from 1 to 8 in the order given above. In all the calculations, we denote q = 1 - p to simplify the formulas. (a) Suppose we just see THH as the last three outcomes. Determine the probability to observe, after two more flips, the triples THH and HTT. (b) Determine the transition matrix of the chain. (c) Determine the communication classes of the chain. (d) Determine the periods of the states. (e) Show that the long run proportions of the eight states in E exist for the chain and determine these proportions. (f) The triples appearing in the chain are overlapping for near times: for example by definition the last two flip outcomes {Yn+2, Ym+3} in X, are identical to the first two ones in Xn+1. We now consider the non-overlapping triples in the sequence as follows: YIYzY's, YAYSY6, YTYBYg. .... Clearly, the outcomes from these non-overlapping triples are still the eight ones in E. Show that the long run proportions of the eight states in E from these non-overlapping triples still exist and they coincide with those found in (e)