I have a SarimaX model with three regressor variables:


          ar1       sma1   C1 (for xreg1)   C2 (for xreg2)   C3 (for xreg3)
      -0.0260    -0.9216          -0.0354           0.0316           0.9404
s.e.   0.0291     0.0350           0.0016           0.0017           0.0128

I would like to know how to use these coefficients to obtain the actual equation, like:

y[t] = f(ar1, sma1, C1|xreg1[t], C2|xreg2[t], C3|xreg3[t])

I have read the following:

https://www.otexts.org/fpp/8/9 - I'm using the forecast package in R, so I'm quite grateful for Mr. Hyndman's work,


and others, and I devised some formulas, but they generated values less acurate than those from the R forecast. Somehow, my error-related terms are probably wrong.

EDIT: This is what I have so far:

$$ \ (1-ar1*B)*(1-B^7)*y_t=$$ $$ = (1-ar1*B)*(1-B^7)*(C1*xreg1_t + C2*xreg2_t+C3*xreg3_t)+ $$ $$ + e_t + sma1*e_{t-7}$$

I would like to know if this formula is correct, could anyone please help? Thank you.

  • $\begingroup$ Do I need to provide more information? Is my question too vague or should I ask it somewhere else? Is it so obvious that I'm not very experienced in statistics? I noticed answers usually come quickly here, so if there's anything wrong in my post, please let me know. $\endgroup$ – VictorM Oct 9 '14 at 19:10

Please have a look at Rob Hyndman's blog entry discussing the difference between ARMAX models and Regressions with ARMA errors. The ARIMAX Model Muddle

My understanding is that the R Forecast package he has developed simultaneously fits a regression along with an ARIMA model for the regression errors. This is not the same as the ARIMAX equation you have worked out above which is more representative of a 'true' ARIMAX model where the AR and differencing terms become intermingled with the exogenous variables.

All this being said, I do believe your equation is correct given the SARIMAX model you have mind. It is just not consistent with R's Forecast package implementation.

| cite | improve this answer | |

The correct formulation is: $$ y_t = C1*xreg1 + C2*xreg2 + C3*xreg3 + \eta_t $$ $$ (1 - ar1*B)*(1 - B^7)*\eta_t = (1 - sma1*B^7)*e_t $$ And finally we have: $$ (1 - ar1*B)*(1 - B^7)*(y_t - C1*xreg1 - C2*xreg2 - C3*xreg3) = (1 - sma1*B^7)*e_t $$ what is (almost) exactly what you get. Notice the sign (minus instead plus) in the right side. Some authors define it with the sign plus, but i prefer using the convention established by Box and Jenkins.

| cite | improve this answer | |
  • $\begingroup$ I don't see the difference between what you presented and what the OP presented. Can you please explain the difference other than your albebraic simplification $\endgroup$ – IrishStat Sep 7 '17 at 16:50

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.