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.
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$\begingroup$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, $\gamma(h)=\phi^h\gamma(0)=\phi^h\cfrac{\sigma_w^2}{1-\phi^2}$.
$\endgroup$ 3 $\begingroup$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.
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