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I have 2 questions from Tsay's book, dealing with ARIMA representation.

I. How does the division of the polynomial work - what are the steps that lead to the solution they show? They show a "simple" example but I am unsure how this math works?:

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II. They also show the following, can anyone explain what theta(1) in the denominator of the first equation represents?

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  1. Recall the geometric series, so that

$$ 1+ \phi_1B + \phi_1^2B^2 + .... = \dfrac{1}{1-\phi_1B} $$ So that $$ \begin{align} \psi(B) &= (1 - \theta_1B)\dfrac{1}{1-\phi_1B} \\ &=(1 - \theta_1B)(1+ \phi_1B + \phi_1^2B^2 + ....) \\ &= 1 + (\phi_1 - \theta_1)B + \phi_1(\phi_1 - \theta_1)B^2 + \phi^2(\phi_1 - \theta_1)B^3 + \ldots \end{align} $$ and the results for $\pi(B)$ are symmetrical.
2. This follows directly from the above, if in general, you think of $B$ as any operator for which exponentiation is appropriately defined, including the operator that leaves the operand unchanged, '$1$'.

So, if the lag polynomial is $\theta(B) = 1 - \theta_1B - \theta_2B^2 - \ldots - \theta_qB^q$, then $\theta(1) = 1 - \theta_1 - \theta_2 - \ldots - \theta_q$, so that $$ \frac{\phi_0}{\theta(1)} = \frac{\phi_0}{1 - \theta_1 - \theta_2 - \ldots - \theta_q} $$

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  • $\begingroup$ Thanks for this - I am sure it is clear, but could you please expand a tad on each to show how to make the leap from what you show and the example? 1. How does this work when the numerator is not just a '1' and 2. I am unclear on how theta(1) becomes this numerator. $\endgroup$ – B_Miner Dec 10 '13 at 1:47
  • $\begingroup$ @B_Miner I have updated my answer. $\endgroup$ – tchakravarty Dec 10 '13 at 12:02

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