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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 $\eta_t := X_t - X_{t-1}$?

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    $\begingroup$ 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
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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.

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

where

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

Substituting:

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

rearranging:

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

and finally

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

for all $t$

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