For this question, see Slide 31 of the lecture note. Consider the lid process {X:} (again, not lid noise) and let S = EX= Xit Xy+..Xt, fort= 1,2,3,. Here S, is not a random walk. We use S to denote the sum of series. (a) (2 pts) Find E(S,) and Var(S,). (b) (2 pts) Using Cou(aX + by, Z) = aCov(X, Z) + bCov(Y, Z). show that for any h 2 1 Cov(St, Xith) = Cov(X1 + Xa + ...+ X, Xith) =0. Hint: Start with the following and extend to the above: Cov ((X1 + Xa + X3), Xith) - Cov( X1 + (Xa + X3), Xith) = Cov(X1, Xith) + Cou(X, + Xa, Xith) = Cov(X1, Xith) + (Cou(Xa, Xith) + Cov(Xs, Xith)) = Cov(X1, Xith) + Cou(Xa, Xith) + Cov(Xs, Xith). (c) (2 pts) Using the result in (b), show that Cov( (X#1 + ... + Xin), S) =0. (d) (2 pts) Combining (b) and (c), show that Yardor walk is not stakemany beat it depends or f Cs(t, t + h) = Cov(Sith, $1) = to. (e) (2 pts) As you see, the process {S,} is not stationary. Is there any way we can transform the process to the stationary process? What if we try S, := 5:/v? Is {S:} stationary