I've developed a ARIMA model with exogenous variable. Before fitting the model, I made every time series stationary by differencing (each variable had a different order of integration). For simplicity, let's say there was only one exogenous variable. Finally I ended up with a model. Keeping in mind that all my data are stationary to create the model, I received the following coefficient estimates:

ARIMA(1,0,0) with zero mean     

       ar1      X
      -0.6     -0.002

There were no MA terms. Note that Y was second-differenced, and X was first differenced. So, my formula is really like:

$Y_t'' = -0.6 Y_{t-1}'' - 0.002 X'$

Where the primes ( $'$ ) represent the level of differencing.

I'd like an equivalent representation of the model using the original non-differenced variables. So I perform the following:

$(Y_t - Y_{t-1}) - (Y_{t-1} - Y_{t-2}) = -0.6 [(Y_{t-1} - Y_{t-2}) - (Y_{t-2} - Y_{t-3})] - 0.002(X_{t} - X_{t-1}) $

$Y_{t} - 2Y_{t-1} + Y_{t-2} = -0.6Y_{t-1} + 1.2Y_{t-2} - 0.6Y_{t-3} - 0.002X_{t} + 0.002X_{t-1}$

$Y_{t} = 1.4Y_{t-1} - 0.2Y_{t-2} - 0.6Y_{t-3} - 0.002X_{t} + 0.002X_{t-1}$

So, I have 2 questions.

1) Is my interpretation correct? Seems straight-forward, but I just want to double check. Is there a quicker way to perform these conversions, besides the tedious expanding & combining I performed above? Some of the models I'm working with are quite more complicated. Perhaps using the backshift operator would help? I don't have much practice with algebra involving the backshift operator, and am not sure whether it would help do what I'd like.

2) How do I perform a similar procedure when there are MA terms? I know that an MA term is equivalent to infinite AR terms. So, is it still be possible to get some representation like $Y_t = ...$? For example, suppose my fit produced ARIMA(1,0,1), with an MA coefficient estimate of 0.3.

Thank you!


1 Answer 1


You have misunderstood the model that R is using. (I am assuming you have used R as the output looks identical to what you get with the arima command.) The model with a regressor and an AR(1) term can be written as $$Y_t'' = -0.002 X'_t + N_t$$ where $$N_t = -0.6 N_{t-1} + e_t$$ and $e_t$ is white noise.

In backshift notation, this is $$(1-B)^2 Y_t = - 0.002(1-B)X_t + \frac{1}{1+0.6B} e_t $$

This can be rearranged to avoid the rational function of B to get $$(1+0.6B)(1-B)^2 Y_t = - 0.002(1+0.6B)(1-B)X_t + e_t. $$ Expanding gives $$(1-1.4B - 0.2B^2 + 0.6B^3) Y_t = - 0.002(1 - 0.4B - 0.6B^2)X_t + e_t. $$ and then applying the operators: $$ Y_t = 1.4Y_{t-1} + 0.2Y_{t-2} - 0.6Y_{t-3} - 0.002X_t + 0.0008 X_{t-1} + 0.0012X_{t-2} + e_t. $$

If you had an ARIMA(1,0,1) model with coefficients $\phi$ and $\theta$, and the same regression equation, then the equations become: $$N_t = \phi N_{t-1} + \theta e_{t-1} + e_t$$ $$(1-B)^2 Y_t = - 0.002(1-B)X_t + \frac{1+\theta B}{1-\phi B} e_t $$ $$(1-\phi B)(1-B)^2 Y_t = - 0.002(1- \phi B)(1-B)X_t + (1+\theta B) e_t. $$ and so on.

  • $\begingroup$ Thank you for the reply. FWIW, indeed I am using R's auto.arima with xreg. $\endgroup$
    – ch-pub
    Commented Jul 16, 2014 at 22:52

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