For an AR(1) process:

$X_{t} = \phi X_{t-1} + w_{t}$ with $w_{t} \sim N(0,\sigma^{2})$ How do you derive the ACF of the process?

Since $E[X_{t}] = 0$, would you just calculate $cov(\phi X_{t-1} + w_{t},\phi X_{t+h-1} + w_{t+h}) = \phi^{2} E[(X_{t-1}*X_{t-1+h})] + \sigma^{2}$. I am having trouble simplifying this expression specifically the $E[(X_{t-1}*X_{t-1+h})$ term.


2 Answers 2


Let $\gamma(h)$ denote the autocovariance function.

Note that $\gamma(0)=\text{Cov}(X_t,X_t)=\text{Cov}(\phi X_{t-1}+w_{t-1}, \phi X_{t-1}+w_{t-1})=\phi^2\gamma(0)+\sigma_w^2$.

Therefore, $\gamma(0)=\cfrac{\sigma_w^2}{1-\phi^2}$.

$\gamma(1)=\text{Cov}(X_{t+1},X_t)=\text{Cov}(\phi X_t+w_{t+1}, X_t)=\phi\gamma(0)$.

Similarly, $\gamma(n)=\phi\gamma(n-1)$.

Therefore, $\gamma(h)=\phi^h\gamma(0)=\phi^h\cfrac{\sigma_w^2}{1-\phi^2}$.

  • $\begingroup$ You mean $ \gamma(0)=\frac{\sigma_{w}^{2}}{1-\phi}$ $\endgroup$
    – lord12
    Mar 20, 2013 at 15:12
  • $\begingroup$ Also its 1 - $\phi^{2}$ not 1 + $\phi^{2}$ $\endgroup$
    – lord12
    Mar 20, 2013 at 16:49
  • $\begingroup$ @lord12 right.. $\endgroup$
    – NECing
    Mar 20, 2013 at 17:01

We have to consider the case when lag is negative. I'll give a different approach:

For $h\ge 0$, \begin{align*} Cov(X_t,X_{t+h}) =& Cov\left(\sum_{j=0}^{\infty}\phi^j\omega_{t-j},\sum_{k=0}^{\infty}\phi^k\omega_{t+h-k}\right) \\ =&\sum_{j=0}^{\infty}\sum_{k=0}^{\infty}\phi^{j+k}Cov(\omega_{t-j},\omega_{t+h-k}) \\ =&\sum_{j=0}^{\infty}\sum_{k=0}^{\infty}\phi^{j+k}\sigma^2{1}_{\{k=j+h\}} \\ =&\sum_{j=0}^{\infty}\phi^{2j+h}\sigma^2=\dfrac{\phi^h}{1-\phi^2}\sigma^2. \end{align*} For $h<0$, \begin{align*} Cov(X_t,X_{t+h}) =& \sum_{j=|h|}^{\infty}\phi^{2j+h}\sigma^2 \\ =&\phi^h\sum_{k=0}^{\infty}\phi^{2(k+|h|)} \\ =&\dfrac{\phi^{|h|}}{1-\phi^2}\sigma^2 \end{align*} where the second equation is by setting $k=j-|h|$.

Then $Cov(X_t,X_{t+h})=\dfrac{\phi^{|h|}}{1-\phi^2}\sigma^2$.

Also, according to Brockwell and Davis (2016), ACVF is used for AutoCoVariance Function and ACF is used for AutoCorrelation Function.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.