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The representation for the model AR(1) is the following:

$Y_t=c+ϕY_{t-1}+ε_t$

where $c=(1-ϕ)μ$ ($c$ is a constant).


I want to understand the calculations that there are behind the general formula of the autocovariance of AR(1), which is $γ(h)=\operatorname{Var}(Y_t )⋅ϕ^{|h|} $


So far, I did the following steps - I started with $γ(1)$:

$\operatorname{Cov}(Y_t,Y_{t-1})=γ(1)=$

$=\operatorname{Cov}(ϕY_{t-1}+ε_t, ϕY_{t-2}+ε_t)=$

$=\operatorname{Cov}(ϕY_{t-1}, ϕY_{t-2}) + \operatorname{Cov}(ϕY_{t-1}, ε_t) + \operatorname{Cov}(ε_t, ϕY_{t-2}) + \operatorname{Cov}(ε_t, ε_t) $


$γ(1)=ϕ^2γ(1) + ???+???+σ^2$



As you can see, from this point I can't continue because I don't know which are the values of $\operatorname{Cov}(ϕY_{t-1}, ε_t)$ and $\operatorname{Cov}(ε_t, ϕY_{t-2})$


Any assistance will be much appreciated. Thank you in advance.

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Let's write $\gamma(1)$: $$\begin{align}\gamma(1) &= cov(Y_t,Y_{t-1})=cov(c+\phi Y_{t-1}+\epsilon_t,Y_{t-1})\\&=cov(c,Y_{t-1})+\phi cov(Y_{t-1},Y_{t-1})+cov(\epsilon_t,Y_{t-1}) \\&=\phi \gamma(0)\end{align}$$

since $cov(c,Y_{t-1})=cov(\epsilon_t,Y_{t-1})=0$, (i.e. past output is independent from future input).

Similarly, $\gamma(2)=cov(Y_t,Y_{t-2})=cov(\phi Y_{t-1}+c+\epsilon_t,Y_{t-2})=\phi \gamma(1)=\phi^2\gamma(0)$.

If we continue this way, we get $cov(Y_t,Y_{t-h})=\phi \gamma(h-1)=...\phi^h\gamma(0)$, where $h\geq0$. Generalizing for negative $h$ yields $\gamma(h)=\phi^{|h|}\gamma(0)$, where $\gamma(0)=var(Y_t)$.

P.S. all of this analysis assumes $\epsilon_t$ is WSS, therefore $y_t$ from LTI filtering property.

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    $\begingroup$ there is a typo in the first line .. identity sign placed wrong. $\endgroup$ – Stop Closing Questions Fast Jan 17 '19 at 23:31
  • $\begingroup$ In the first line I would replace the 3rd "+" sign with the "=" sign: $cov(c+\phi Y_{t-1}+\epsilon_t,Y_{t-1})=cov(c,Y_{t-1})+\phi cov(Y_{t-1},Y_{t-1})+cov(\epsilon_t,Y_{t-1})$ $\endgroup$ – Martina Marty Jan 18 '19 at 10:29
  • $\begingroup$ While trying to edit the typo addressed by @Jesper, I converted that specific = sign to + sign, and made it more wrong :). I see that the reason is because of rendering. Although the order of tex statements are correct, they were displayed in a different order. Anyway, I've made use of align statements and made it much more clear. Hope, it's ok. $\endgroup$ – gunes Jan 18 '19 at 10:54
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Starting from what you have provided:

$y_{t} = c + \phi y_{t-1} + \epsilon_{t} \tag{1}$

Where $c = (1 - \phi) \mu$


We can rewrite $(1)$ as:

\begin{array} \ y_{t} & = & c + \phi y_{t-1} + \epsilon_{t} \\ & = & (1 - \phi) \mu + \phi y_{t-1} + \epsilon_{t} \\ & = & \mu - \phi \mu + \phi y_{t-1} + \epsilon_{t} \\ \end{array}

Then,

$y_{t} - \mu = \phi (y_{t-1} - \mu) + \epsilon_{t} \tag{2}$

If we let $\tilde{y_{t}} = y_{t} - \mu$, then equation $(2)$ can be writen as:

$\tilde{y}_{t} = \phi \tilde{y}_{t-1} + \epsilon_{t} \tag{3}$


Variance

The variance of $(3)$ is obtained by squaring the expression and taking expectations, which ends in:

\begin{array} \ \tilde{y}_{t}^2 & = & (\phi \tilde{y}_{t-1} + \epsilon_{t})^2 \\ & = & (\phi \tilde{y}_{t-1})^2 + 2 \phi \tilde{y}_{t-1} \epsilon_{t} + (\epsilon_{t})^2 \\ & = & \phi^{2} \tilde{y}_{t-1}^{2} + 2 \phi \tilde{y}_{t-1} \epsilon_{t} + \epsilon_{t}^2 \end{array}

Now take the expectation:

$E(\tilde{y}_{t}^2) = \phi^{2} E(\tilde{y}_{t-1}^{2}) + 2 \phi E(\tilde{y}_{t-1} \epsilon_{t}) + E(\epsilon_{t}^2)$

Here we will call:

  • $\sigma_{y}^{2}$ is the variance of the stationary process.
  • The second term in the right-hand side of the equation is zero because $\tilde{y}_{t-1}$ and $\epsilon_{t}$ are independent and both have null expectation.
  • The last term in the right is the variance of the innovation, denoted as $\sigma^{2}$ (note that there is no subscript for this).

Finally,

$\sigma_{y}^{2} = \phi^{2} \sigma_{y}^{2} + \sigma^{2}$

If we solve for the variance of the process, namely $\sigma_{y}^{2}$, we have:

$\sigma_{y}^{2} = \frac{\sigma^2}{1 - \phi^2} \tag{4}$


Autocovariance

We are going to use the same trick we use for formula $(3)$. The autocovariance between observations separated by $h$ periods is then:

\begin{array} \ \gamma_{h} & = & E [(y_{t - h} - \mu) (y_{t} - \mu)] \\ & = & E[(\tilde{y}_{t - h}) (\tilde{y}_{t})] \\ & = & E[\tilde{y}_{t - h} (\phi \tilde{y}_{t - 1} + \epsilon_{t}) \\ \end{array}

The innovations are uncorrelated with the past values of the series, then $E[\tilde{y}_{t-h} \epsilon_{t}] = 0$ and we are left with:

$\gamma_{h} = \phi \gamma_{h-1} \tag{5}$

For $h = 1, 2, \ldots$ and with $\gamma_{0} = \sigma_{y}^2$


For the particular case of the $AR(1)$, equation $(5)$ becomes:

$\gamma_{1} = \phi \gamma_{0}$

And using the result from equation $(4)$: $\gamma_{0} = \sigma_{y}^{2} = \frac{\sigma^2}{1 - \phi^2}$ we end up with

$\gamma_{1} = \frac{\sigma^2}{1 - \phi^2} \phi$


Original source: Andrés M. Alonso & Carolina García-Martos slides. Available here: http://www.est.uc3m.es/amalonso/esp/TSAtema4petten.pdf

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