0
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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. $\endgroup$ Jul 13 '18 at 3:06
2
$\begingroup$

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 $t$.

$\endgroup$

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.