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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 $E[X_3X_1]=\gamma(3-1)=\gamma(2)$ ?

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1 Answer 1

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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 $t$, $Cov(X_t,X_{t+h})=E(X_tX_{t+h})$.

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  • $\begingroup$ thanx! made it more clear when you mentioned "which is a function of only h if the process is stationnary", thanks! $\endgroup$
    – Elekko
    Commented May 26, 2015 at 11:59

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