I got a simple question that I really don't have understood 100% yet
We let $\{X_t\}_{t=-\infty}^{+\infty}$ be stationary AR(1) process given by:
$X_t + 0.25X_{t-1} = Z_t$, where $\{Z_t\}$ is WN($0,1$)
We know that an AR(1) process has the autocovariance function at lag as
$\gamma(h) = \frac{\phi^{\mid h\mid}\sigma^2}{1-\phi^2}$
We know here $E[X_t] = 0$. What is $E[X_2X_1]$ or $E[X_3X_1]$? Since there is no mentioning about independence.
Do we calcualte it as we subtract the indexes? Here for instance $E[X_2X_1]=\gamma(2-1)=\gamma(1)$? and $E[X_3X_1]=\gamma(3-1)=\gamma(2)$ ?
Yes! By definition, $\gamma(h)=Cov(X_t,X_{t+h})$ (which is a function of only $h$ if the process is stationnary). As $E(X_t)=0$ for all $t$, $Cov(X_t,X_{t+h})=E(X_tX_{t+h})$.