Solved – Obtaining the SarimaX equation from the arima coefficients

arimaforecastingrregressiontime series

I have a SarimaX model with three regressor variables:

ARIMA(1,0,0)(0,1,1)[7]                    

Coefficients:
          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,

http://people.duke.edu/~rnau/arimreg.htm

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.

Best Answer

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.

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