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Say a stationary AR(1) process is given by:

$$ X_t = c + \phi X_{t-1} + \epsilon_t $$ where $ \epsilon_t $ is a white noise process with zero mean and constant variance $ \sigma^2 $. Wikipedia tells me that the one period auto-covariance is given by $$ (\sigma^2 \phi) / (1-\phi^2) $$ but I cannot see why. Can anyone help me refresh my memory as to why this is true?

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Let $\mu$ denote the mean of the process, then the first order autocovariance is given by:

$$ E\left[(X_t - \mu)(X_{t-1} - \mu)\right] = E\left[\tilde{X}_t \tilde{X}_{t-1}\right] = \\ E\left[(\phi \tilde{X}_{t-1} + \epsilon_t) \tilde{X}_{t-1}\right] = \phi \underbrace{E\left[\tilde{X}_{t-1}^2\right]}_{\sigma^2_X} + \underbrace{E\left[\epsilon_t \tilde{X}_{t-1}\right]}_{0} = \phi\sigma^2_X \,. \qquad (1) $$

The second expectation is zero because $\tilde{X}_{t-1}$ dependes on the innovations $\epsilon_{t-1}$, $\epsilon_{t-2}$,... but not on $\epsilon_t$.

The first expectation is the variance of the AR(1) process, denoted $\sigma^2_X$. The expression for the variance of $X_t$ can be obtained as:

$$ \hbox{Var }(X_t) = \phi^2 \hbox{Var }(X_{t-1}) + \underbrace{\hbox{Var }(\epsilon_t)}_{\sigma^2_\epsilon} \,. $$

As the process is stationary $\hbox{Var }(X_t) = \hbox{Var }(X_{t-1})$. Thus, the above expression can be written as:

$$ \hbox{Var }(X_t) = \frac{\sigma^2_\epsilon}{1 - \phi^2} \,. \qquad (2) $$

Substituting (2) in (1) gives the following expression for the first order autocovariance:

$$ \frac{\phi \sigma^2_\epsilon}{1 - \phi^2} \,. $$

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