Please answer this question with steps, thanks(Markov chain, informaiton theory)

Question 2(a) Random variables X, Y, Z are said to form a Markov chain in that order (denoted by X -> Y - Z) if their joint probability distribution can be written as: p(X, Y, Z) = p(X) . p(Y|X) . p(Z|Y) I. Suppose (X, Y, Z) forms a Markov chain. Is it possible for 1 (X; Y) = 1(X; Z)? If yes, give an example of X, Y, Z where this happens. If no, explain why not. [4 Marks] II. Suppose (X, Y, Z) does not form a Markov chain. Is it possible for 1 (X; Y) 2 1(X; Z)? If yes, give an example of X, Y, Z where this happens. If no, explain why not. [4 Marks] III. If X - Y - Z then show that [6 Marks] . I(X; Z) SI(X; Y) . I ( X; Y| Z) X2 -> X3 -> . .. > Xn forms a Markov chain in this order. Thus, the joint probability of X1, . .., Xn are given by p(x1, x - 2, . .., Xn) = P(xaxn-1)P(xn-1/X2-2) . . . P(x2/x1)p(x1). Reduce I(X1; X2, ..., X,) to its simplest form (show your work). [6 Marks]