2
$\begingroup$

I see this expression in a lot of articles about time-series regression, and I am not sure what it means.

"Autoregression" means estimating a vector from its past values, right? Since we are estimating the data itself, and not the error, how can the error be autoregressive?

I can think of two answers :

  1. It is referring to the temporally-nonstationary nature of the system, i.e. a regressor we compute for, let's say, y(1:100) might not work for y(101:200)
  2. It is referring to accumulated error.

Are one of these correct? Or does it mean something else?

Thanks for any help!

$\endgroup$

1 Answer 1

4
$\begingroup$

A regression model with autoregressive error can be written as follows:

$y_t=\beta_0+\beta_1 x_1+\beta_2 x_2+...+\beta_k x_k+n_t$

where $n_t = \phi_1 n_{t-1}+\phi_2 n_{t-2}+...+\phi_p n_{t-p}+\varepsilon_t$.

But of course, the $n_t$ are not observable. You could also think of it as a way of specifying a particular covariance structure on $n$ (and thereby on the conditional $y$). For example, if the autoregressive model is of order 1, then the $i,j$ element of $\text{Var}(n)$ is $\sigma^2\phi_1^{|i-j|}$.

$\endgroup$
2
  • $\begingroup$ Thank you so much for your answer ! But I don't understand one point. We can model $n_t$ as in your second equation for any given time-series vector and a regression model on it, right? So what makes an error autoregressive and the other not? Is there a property that the $\phi$'s should hold, or a test? $\endgroup$
    – jeff
    Commented Sep 19, 2015 at 13:42
  • 1
    $\begingroup$ If you fit that model, it's a model with an autoregressive error term. Normally, however, we restrict the parameters such that the AR model is stationary. Fitting a model doesn't imply it's suitable, however. $\endgroup$
    – Glen_b
    Commented Sep 19, 2015 at 14:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.