Proof expression for the autocovariance function of AR(1)

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.

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.

• there is a typo in the first line .. identity sign placed wrong. – Stop Closing Questions Fast Jan 17 '19 at 23:31
• In the first line I would replace the 3rd "+" sign with the "=" sign: $cov(c+\phi Y_{t-1}+\epsilon_t,Y_{t-1})=cov(c,Y_{t-1})+\phi cov(Y_{t-1},Y_{t-1})+cov(\epsilon_t,Y_{t-1})$ – Martina Marty Jan 18 '19 at 10:29
• 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. – gunes Jan 18 '19 at 10:54

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