Why is the MA(1) process phrased as $X_t = \epsilon_t + \theta\epsilon_{t-1}$, with the $\epsilon_t$ defined as the (unobserved) errors between model fit $\hat X_t$ and observed $X_t$?

Why is the MA formulation preferable to the more simple alternative (without unobservables) $X_t = \eta_t + \theta\eta_{t-1}$, with $\eta_t := X_t - X_{t-1}$?

  • 1
    $\begingroup$ Well for starters, in your formulation, there's no noise term at all. It won't fit much real data without one. Did you mean to write an AR model? $\endgroup$
    – Glen_b
    Aug 20 '13 at 23:44
  • $\begingroup$ I cannot comment on the answer, but the equations look weird. I don't understand why the err is Xt-Xt-1, it should be Xt-Xt(pred) instead. can you explain it clearly. $\endgroup$ Mar 3 '20 at 12:46

Do not worry about whether the disturbances are 'observable' or 'unobservable'.

It comes down to how we assume that the data is being generated. If we have a series that we are trying to model with a MA(1) process, this means that we are assuming that the series is generated by a disturbance term ($\epsilon_{t}$) and a lagged damped value of this disturbance term ($\theta \epsilon_{t-1}$). This is the definition of an MA(1) series and it means that these disturbances are what drive or create the process $X$. So it is because we are assuming that the series is created by these disturbance terms, and these disturbance terms create the series $X$ in this specific way, that we assume the first formula.

Your second formula would not work as a data generating process because all of the disturbance terms can be cancelled out by substituting in values for $X$. Thus, if all disturbances are cancelled, the series if predetermined and that is probably not how you are looking to model your series.

$X_{t} = \eta_{t} + \theta \eta_{t-1}$


$\eta_{t} = X_{t} - X_{t-1}$


$X_{t} = X_{t} - X_{t-1} + \theta (X_{t-1} - X_{t-2})$


$X_{t-1} = \theta (X_{t-1} - X_{t-2})$

and finally

$X_{t-1} = \frac{\theta}{1 - \theta} X_{t-2}$

for all $t$


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.