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?


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:

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


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.