Confusion related to the calculation of autocovariance I have a confusion related to the calculation of autocovariance 
Suppose
$X_t =  \phi X_{t-1} + \epsilon_t$
Then how the autocovariance 
$E(X_{t+n}X_t) - \mu^2 = \frac{\sigma_{\epsilon}^2}{(1-\phi^2)}\phi^{|n|}$
I am referring to this wiki article http://en.wikipedia.org/wiki/Autoregressive_model
 A: Below, I use the same notations as those from the wikipedia link that you provided.
Let's start with the simplest case:
\begin{align*}
\textrm{Cov}(X_{t+1}, X_{t}) 
  & \stackrel{\textrm{def}}{=} \textrm{Cov}(\varphi X_{t} + \epsilon_{t+1}, X_{t}) \\
  & \stackrel{\textrm{linearity}}{=} \varphi \, \textrm{Cov}(X_{t}, X_{t}) + \textrm{Cov}(\epsilon_{t+1}, X_{t}) \\
  & \stackrel{\textrm{Cov}(\epsilon_{t+1}, X_{t}) = 0}{=} \varphi \, \textrm{Var}(X_{t}) \\
  & \stackrel{\textrm{cf. wikipedia}}{=} \varphi \, \frac{\sigma^2_{\epsilon}}{1 - \varphi^2} \cdot
\end{align*}
OK, let's go one step further:
\begin{align*}
\textrm{Cov}(X_{t+2}, X_{t}) 
  & = \textrm{Cov}(\varphi X_{t+1} + \epsilon_{t+2}, X_{t}) \\
  & = \varphi \, \textrm{Cov}(X_{t+1}, X_{t})  \\
  & = \varphi^2 \frac{\sigma^2_{\epsilon}}{1 - \varphi^2} \cdot
\end{align*}
Now, the general case becomes clear:
$$
\textrm{Cov}(X_{t+n}, X_{t}) = \varphi^n \frac{\sigma^2_{\epsilon}}{1 - \varphi^2} \cdot
$$
Here $n \geq 0$, but the same can be done for $n < 0$.
If you want something more rigorous, following these lines, it is now easy to write a proof by induction
A: See page 85-86 in the book "Time Series Analysis and Its Applications: With R examples" by R.H. Shumway and D.S. Stoffer. 
If you iterate backwards $k$ times then you get:
$X_t=\phi X_{t-1}+\epsilon_t=\phi^k X_{t-k}+\sum_{j=0}^{k-1}\phi^j \epsilon_{t-j}$ 
Hints:
$X_t=\phi X_{t-1}+\epsilon_t=\phi(\phi x_{t-2}+\epsilon_{t-1})+\epsilon_t$ (and so on...)
Continuing in this way you get:
$X_t=\phi X_{t-1}+\epsilon_t=\sum_{j=0}^{\infty}\phi^j \epsilon_{t-j}$ 
So, $E(X_t)=\sum_{j=0}^{\infty}\phi^j E(\epsilon_{t-j})=0$
Using this you can easily find the autocovariance function as:
$\gamma(n)=cov(X_{t+n},X_t)=E[(\sum_{j=0}^{\infty}\phi^j \epsilon_{t+n-j})(\sum_{k=0}^{\infty}\phi^k \epsilon_{t-k})]$
Expanding this gives you what you require. And also remember, $\gamma(n)=\gamma(-n)$. Thus comes the absolute sign.
