Consider the AR(2) (1+ B)X; = Zt. where Xo = X1 = 0 and t =...

 

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Consider the AR(2) (1+ B²)X; = Zt. where Xo = X1 = 0 and t = 1,2, 3, .... (i) Is Xt stationary? Why? (ii) Find Cov(Xs, Xt) for a given nonnegative integers s and t. (iii) Simulate this process when Zt ~ N(0, 0.04) and provide basic statistics summary for this process (Let n = 200). I am adding a short do loop in R for simplicity. > x=rep (0,200) > for(i in 3:200){ + x[i]=-x [i-2] +rnorm(1,0,0.2)} Consider the AR(2) (1+ B²)X; = Zt. where Xo = X1 = 0 and t = 1,2, 3, .... (i) Is Xt stationary? Why? (ii) Find Cov(Xs, Xt) for a given nonnegative integers s and t. (iii) Simulate this process when Zt ~ N(0, 0.04) and provide basic statistics summary for this process (Let n = 200). I am adding a short do loop in R for simplicity. > x=rep (0,200) > for(i in 3:200){ + x[i]=-x [i-2] +rnorm(1,0,0.2)}

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Given That Consider the AR2 1 BX Z Where Xo X1 0 and t 1 2 3 Given We are gi... View the full answer