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ACFs

S

StevieG4captain

Member
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.
 
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