I am trying to calculate the ACF at lag 2 for the following time series: X(t)=0.5X(t-2)+e(t) ACF(2)=Cov[X(t-2),X(t)]/Cov[X(t),X(t)] = Cov[X(t-2),X(t)]/Var[X(t)] = Cov[2{(X(t)-e(t)},X(t)]/Var[X(t)] = {Cov[2X(t),X(t)]-Cov[2e(t),X(t)]}/Var[X(t)] = 2Var[X(t)]/Var[X(t)] = 2 (which can't be right as > 1, whereas |correlation|<=1 always) Does anybody know why I've got this nonsense answer? I'm guessing I've treated the Cov[2e(t),X(t)]} incorrectly by assuming it equals 0. But if this is the case what does Cov[2e(t),X(t)]} actually equal and how can it be calculated? Thanks for your time! Would really appreciate a brief reply so I can understand where I'm going wrong..... cheers SG4C
As you wanted a brief reply and I havent gone into too much detail Cov(X(t),e(t)) is not 0. as e(t) is present in the formula for X(t). I think the approach should be to break the X(t) into X(t-2) instead and Cov(X(t-2),e(t)) =0. Hope that helps.
