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Suppose I have the following AR(2) time series process:

$z_t = \delta + \psi_2 z_{t-2} + \epsilon_t$

where $\epsilon_t \sim (0, \sigma^2)$ is a white noise process.

I am hoping to calculate the ACF function for this process. Here are my thoughts thus far:

\begin{align} Cov[z_t, z_{t-k}] &= Cov[\delta + \psi_2 z_{t-2} + \epsilon_t,\ z_{t-k}] \\\\ &= Cov[\delta,\ z_{t-k}] + Cov[\psi_2 z_{t-2},\ z_{t-k}] + Cov[\epsilon_t,\ z_{t-k}] \\\\ &= \psi_2 Cov[z_{t-2},\ z_{t-k}] \\\\ &= \psi_2\gamma(k-2) \\\\ &= \gamma(k) \end{align}

Now I know that $\gamma(0) = \sigma^2 \frac{1}{1-\psi_2^2}$ as long as I did my calculations correct. I just found the variance of $z_t$ which is $\gamma(0) = Var[z_t] = \sigma^2 \frac{1}{1-\psi_2^2}$ which I derived by taking the variance of $z_t$ in its $MA(\infty)$ form. Hence,

\begin{align} Var[z_t] &= Var\left[\frac{\delta}{1-\psi_{2}} + \sum_{i=0}^{\infty} \psi_2^i \epsilon_{t-2i}\right] \\\\ &= \sum_{i=0}^{\infty}\psi_2^i Var[\epsilon_{t-2i}] \\\\ &= \sigma^2 \sum_{i=0}^{\infty} \psi_2^{2i} \\\\ &= \frac{\sigma^2}{1-\psi_2^2} \end{align}

So $\gamma(k)$ would then be $\sigma^2 \frac{\psi_2^k}{1-\psi_2^2}$ which can be deduced since we already know $\gamma(0)$.

Then to find $\rho(k)$, which is the ACF, we do the following:

\begin{align} \frac{\gamma(k)}{\gamma(0)} = \psi_2^k = \rho(k) \end{align}

Is this correct or am I totally jacking something up somewhere along the way?

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    $\begingroup$ Hi: at a glance, it seems correct because it's really an AR(1) except that time starts at $t = 3$ instead of $t=1$. And your result is definitely the autocorrelation of the standard AR(1) where $t$ starts at 1. Maybe you should write down the restriction on $t$ just to be slightly clearer ? $\endgroup$
    – mlofton
    Commented Feb 13, 2021 at 22:08
  • $\begingroup$ $t$ starts at 2 actually. But what do you mean by the restrictions on $t$? $\endgroup$ Commented Feb 13, 2021 at 22:33
  • $\begingroup$ Hi: Yes, if you start at $t=2$, then that's fine but that means your first observation is $z_{0}$. By restriction, In that case, by restriction I just meant that the first observation occurs at $t=2$ which is a little non-standard. $\endgroup$
    – mlofton
    Commented Feb 14, 2021 at 6:36
  • $\begingroup$ Just one other thing: Note that you can also just think of it as an AR(1) where the underlying time scale is 2 units of time rather than 1. For me, it's easier to think of it that way. gunes nicely implied this by showing that the odd autocorrelations are always zero. $\endgroup$
    – mlofton
    Commented Feb 14, 2021 at 6:55

1 Answer 1

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$\gamma(0)$ is correct and can be calculated in a different (maybe easier) way as well: $$\begin{align}\gamma(0)&=\operatorname{cov}(z_t,z_t)=\operatorname{cov}(\delta+\psi_2z_{t-2}+\epsilon_t,\delta+\psi_2z_{t-2}+\epsilon_t)\\&=\psi_2^2\gamma(0)+\sigma^2\\&\rightarrow \gamma(0)=\frac{\sigma^2}{1-\psi_2^2}\end{align}$$

And, the relation $\gamma(k)=\psi_2\gamma(k-2)$ is correct, too. It implies the following: $$\gamma(2)=\psi_2\gamma(0), \ \ \gamma(4)=\psi_2\gamma(2)=\psi_2^2\gamma(0) \dots$$ which means $\gamma(2k)=\psi_2^k \gamma(0)$. For the odd portion, let $k=1$ in the recursive equation: $$\gamma(1)=\psi_2 \gamma(-1)$$ We know that autocovariance function is symmetric across $k=0$, so $\gamma(-1)=\gamma(1)$. So, $$\gamma(1)=\psi_2\gamma(1)$$ but since this should hold for all $\psi_2$ (that doesn't violate the stationarity conditions), $\gamma(1)=0$. This would immediately mean $\gamma(2k+1)=0$.

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    $\begingroup$ Thank you for your awesome explanation! Two questions: 1. How exactly did you deduce that $\psi_2^2 \gamma(0) + \sigma^2 \implies \gamma(0) = \frac{\sigma^2}{1-\psi_2^2}$. Also, what exactly would $\rho(k)$ (my ACF) be equal to? Would there be two ACFs? One in the case of the even portion and one in the case of the odd portion? Thank you again! $\endgroup$ Commented Feb 15, 2021 at 22:40
  • $\begingroup$ 1. Just distribute the covariance over the terms. $\text{cov}(z_{t-2},\epsilon_t)=0$ (past output, current input) and $\text{cov}(z_{t-2},z_{t-2})=\gamma(0)$. 2) You can write it as a partial function, e.g. $\gamma(k)=\psi_2^{k/2}\gamma(0)$ if $k$ is even and $0$ if $k$ is odd. $\endgroup$
    – gunes
    Commented Feb 15, 2021 at 23:04

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