# AR(2) ACF Derivation

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?

• 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 ? Commented Feb 13, 2021 at 22:08
• $t$ starts at 2 actually. But what do you mean by the restrictions on $t$? Commented Feb 13, 2021 at 22:33
• 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. Commented Feb 14, 2021 at 6:36
• 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. Commented Feb 14, 2021 at 6:55

$$\gamma(0)$$ 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 $$\gamma(2k+1)=0$$.
• 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! Commented Feb 15, 2021 at 22:40
• 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. Commented Feb 15, 2021 at 23:04