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So I'm trying to derive the covariance between $z_t$ and $z_{t-1}$ in the $AR(1)$ model: $z_t =\phi z_{t-1} + a_t$. Can anyone give me some advice on where to start?

stats566
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  • Have you tried to compute $\mbox{cov}(z_t, z_{t-1})$? What did you get? Where are you stuck? – QuantIbex Apr 26 '14 at 15:22
  • The answer is here: http://stats.stackexchange.com/questions/68243/ar1-coefficient-is-correlation/68246#68246 – Alecos Papadopoulos Apr 26 '14 at 18:28
  • Hey Quantlbex, at first I wasn't sure where to start, but now this is what I'm getting: $cov(z_t, z_{t-1}) = cov(\phi z_{t-1} + a_t, z_{t-1}) = \phi cov(z_{t-1}, z_{t-1}) + cov (a_t, z_{t-1}) = \phi var(z_{t-1}) + cov(a_t, z_{t-1}) = \phi (\frac{\sigma^2}{1-\phi^2})$. Does that seem correct to you? Would you mind explaining why $cov(a_t, z_{t-1}) = 0$? – stats566 Apr 26 '14 at 19:40
  • Also, if I was trying to find the $cov(z_t, z_{t-2})$, does this seem correct: $cov(z_t, z_{t-2}) = \phi^2 var(z_{t-2}) + \phi cov(a_{t-1}, z_{t-2}) + cov(a_t, z_{t-2}) = \phi^2 (\frac{\sigma^2}{1-\phi^2})$ ? – stats566 Apr 26 '14 at 19:48

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Just write it out the old fashioned way:

$$ Cov(z_t, z_{t-1}) = Cov(\phi z_{t-1}+a_{t-1},z_{t-1})=\phi Cov(z_{t-1},z_{t-1})+Cov(a_{t-1}, z_{t-1})=\phi Var(Z) $$

The derivation for $Var(Z)$ is a little different but same approach:

$$Var(z_t)=Var(\phi z_{t-1}+a_{t-1})=\sigma_a^2+\phi^2 Var(z_{t-1}) $$ Since $Var(z_t)=Var(z_{t-1})$, you can get that $Var(Z)=\frac{\sigma_a^2}{1-\phi^2}$ for all $t$.

Now plug that back in and you're done.

Mike Nute
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    Thanks Mike, that is very helpful. Would you mind explaining why $cov(a_t,z_{t-1}) = 0$? Is it because a there can't be a covariance between $z_{t-1}$ and a future error term, $a_t$? – stats566 Apr 26 '14 at 19:06
  • Also, if I was trying to find the $cov(z_t, z_{t-2})$, does this seem correct: $cov(z_t, z_{t-2}) = \phi^2 var(z_{t-2}) + \phi cov(a_{t-1}, z_{t-2}) + cov(a_t, z_{t-2}) = \phi^2 (\frac{\sigma^2}{1-\phi^2})$ ? – stats566 Apr 26 '14 at 19:53
  • Ya that does seem right. $Cov(a_t,z_{t-1})=0$ because $a_t$ is i.i.d. noise and it does not appear in $z_{t-1}$, so they are independent. – Mike Nute Apr 27 '14 at 20:26