I have 2 questions regarding ARIMA.

  • 1st: How do we get the MA component - the et's (as we want to regress yt on lagged yt and also et and lagged et's)? If I want to regress yt on lagged yt's, I have them. But I do not have et and lagged et's.
  • 2nd: what is the advantage of including et's in the regression - why is it better (if it is) than simple AR regression? Are there some efficiency gains from adding et terms? Thank you
  • $\begingroup$ (1) I think it is the estimated errors - the residuals; (2) I think it is to fit a better line to the data. Yet, I appreciate my answer is very incomplete - so I welcome a much more complete account. $\endgroup$ – EB3112 Nov 30 '20 at 10:46
  • 1
    $\begingroup$ (1): This regards to how the MA processes are estimated. See this for some ideas. (2): this is already addressed in @1muflon1's answer (last para). $\endgroup$ – Dayne Nov 30 '20 at 12:06

The MA terms are lagged errors (you don’t need to fetch them manually - for example in R you can use Arima function which does this for you and any program/language will have this basic function as mentioned +1 answer of @RichardHardy in +1 comment of @Dayne).

For example, following Verbeek (2008) A Guide to Modern Econometrics 4th ed pp 261 ARMA(p,q) in general form can be written as:

$$y_t=\theta_1 y_{t−1}+ … + \theta_p y_{t−p} + \epsilon_t+ \alpha_1 \epsilon_{t−1}+…+ \alpha_q \epsilon_{t−q}$$   The distinction between the effect of AR coefficients and MA coefficients is that they model fundamentally different forms of relationships with past realizations. AR terms model autocorrelation of the dependent variable - that is relationship with its own past realizations. MA terms model dependence on the errors that model made in the past. These have different implications for how persistent the series is.   I think it is best to visualize this. Here are two pictures I took from the above mentioned Verbeek from pages pp 260-261.

On this first picture you can see effect of autocorrelation in AR(1) process when $\theta=0.5$. You can see that even though we have only single AR term it’s effect is persistent across several periods.

enter image description here

However, let’s consider now case of MA(1) process where $\alpha=0.5$. As you can see clearly below the effect is seen only for one period.

enter image description here

The advantage of including MA terms (provided that they actually should be there and you are not adding them just for heck of it), generally speaking, are that you will more accurately model the persistence in your model, you will have better fit and more accurate forecasts than otherwise. It is always important to get the right functional form functional form as wrong functional form can even bias the results. I mean on a very basic level (and I am oversimplifying here) this is like asking why we sometimes use $\ln y = \alpha + \beta \ln x + e$ instead of $y = \alpha + \beta x +e$ it is because we want to get the functional form right so there is no misspecification in the model.

  • $\begingroup$ Thanks! I see the efficiency gains. But still do not see where do we get et from. Is it from making AR(1) regression and saving those residuals, and then using them as et and its lags? $\endgroup$ – Mr. T Nov 30 '20 at 16:20
  • $\begingroup$ @Mr.T no it’s not from AR(1) but rather they are based on one step ahead estimation of of predicted outcomes from some innovation algorithm. For example, if you have MA(1) that can be represented as $y_t=e_t+\gamma e_{t-1}$ you can estimate the gamma using innovation representation $y^n_{n+1} = \sum \gamma_{ni} (y_{n+1-i}-y_{n+1-i}^{n-i})$ - assuming $y_0=0$. From the innovation representation you can get gammas and in the end also the MA process errors. $\endgroup$ – 1muflon1 Nov 30 '20 at 16:56
  • $\begingroup$ @1muflon1thanks for the answer and also clarification! $\endgroup$ – Mr. T Nov 30 '20 at 17:44

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