# Why does this expectation equal this autocovariance?

I'm learning about time series and saw this in my book relating to a stationary time series, AR(1) process $X_t = \phi X_{t-1} + w_t$

In one derivation they write something like:

$E(X_{t+1}X_{t}) = \gamma(1)$

Is it assumed that the mean of $X_t$ is 0? That is the only way I see that this equation is true but that assumption is not mentioned in my book.

• There is no beta in your equations? I don't think that there is any reason to think $X_t$ had expectation 0 but $w_t$ should have a mean of 0. Jul 13 '18 at 3:06

Yes, you're right--that equality is true IF you assume the mean function is zero and the process is weakly stationary. Frequently it is assumed that at the process' first time point has a mean of $0$. At the next time $E[X_2] = E[E(X_2 \mid X_1)] = E[\phi X_1] = 0$. You can proceed inductively and show $E[X_t] =0$ for all $t$. This is true whether or not the process is stationary.
If you want to go further, you assume that the time $1$ variance is $\text{Var}(W) / (1-\phi^2)$, then the variance of $X_t$ will be the same, for all $t$. You can show this using the law of total variance.
Last, if you want to extend the assumptions again, you assume that the innovations $\{W_t\}$ are normally distributed (with mean $0$ and variance $\sigma^2$). With the other assumptions, $$X_t \sim \text{Normal}\left(0, \frac{\sigma^2}{1-\phi^2}\right)$$ for all $t$.