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This question already has an answer here:

I would appreciate if someone could help me write the mathematical equation for the seasonal ARIMA (0,2,1) x (0,0,1) period 12. I'm a little confused with how to go about this. I would prefer an equation involving $Y_{t}$ , $e_{t}$, $\theta$ and $\Theta$.

I really don't want an equation involving the backshift operator.

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marked as duplicate by Firebug, Michael Chernick, kjetil b halvorsen, John, Peter Flom Aug 11 '17 at 20:04

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    $\begingroup$ See otexts.com/fpp/8/9 $\endgroup$ – Rob Hyndman Jun 12 '13 at 3:45
  • $\begingroup$ @RobHyndman: just wanted to thank you again. The material was really helpful. $\endgroup$ – b2amen Jun 12 '13 at 18:59
  • $\begingroup$ This is okay but how do you know that the period is 12 months. $\endgroup$ – Michael Chernick Mar 7 '17 at 15:03
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From $SARIMA(0,2,1)\times(0,0,1)_{12}$ we have $\phi_0(B)(1-B)^2\Phi_0(B^{12})(1-B^{12})^0 Y_t = \theta_1(B)\Theta_1(B^{12})e_t$, where

$\phi_0(B)=\Phi_0(B^{12})=1$

and

$\theta_1(B)= 1+\theta_1B$,

$\Theta_1(B^{12})=1+\Theta_1B^{12}$.

And the result is

$Y_t-2Y_{t-1}+Y_{t-2}=e_t+\Theta_1e_{t-12}+\theta_1e_{t-1}+\theta_1\Theta_1e_{t-13}$

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    $\begingroup$ Often the parameterization involves terms of the form $(1-\theta_1B)$ and $(1-\Theta_1B^{12})$. If you are using a computer package, check which parameterization is implemented. $\endgroup$ – Rob Hyndman Jun 12 '13 at 12:40
  • $\begingroup$ @Alpha: Yeah. thanks again. I have accepted it. But i was wondering if you could check the second term after the equal sign again. Should the e_{t-1} have a parameter of "theta" while the e_{t-12} have a parameter "Theta"? $\endgroup$ – b2amen Jun 12 '13 at 18:57
  • $\begingroup$ @b2amen Yes, you are right. I am sorry for my mistake, I edited the answer. $\endgroup$ – Alpha Jun 12 '13 at 19:06
  • $\begingroup$ Sorry to pick up on an old thread, but I have a concern with your polynomial developments of the two theta's, as I learned that the MA operator (here both theta and Theta) were represented as polynomials with a minus sign, therefore I would have written θ1(B)=1 - θ1B, Θ1(B12)=1 - Θ1B12, which changes the sign in the final answer too Yt−2Yt−1+Yt−2=et - Θ1et−12 - θ1et−1 + θ1Θ1et−13...? $\endgroup$ – 1two3stats Aug 5 '14 at 18:01

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