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Is it theoretically justifiable to calculate/use the transfer function of a pure ARMA model? I would like to be able to use the transfer function representation to put the state equations into their diagonal form.

As shown on page 3 here: http://www.engr.mun.ca/~millan/Eng6825/canonicals.pdf

I never see it used in the context of a pure ARMA time series model so I'm assuming it is not possible, but I'd like to understand why it can't be represented in this way?

The generic Transfer Function model as show here as shown in eqn 1 of the link above is:

$y_t+\phi_1 y_{t-1}+...+\phi_p y_{t-p}=\beta_0 x_t+...+\beta_m x_{t-m}$

This seems analogous to the generic ARMA function as shown here:

http://www.le.ac.uk/users/dsgp1/COURSES/THIRDMET/MYLECTURES/10MULTARMA.pdf

$y_t+\phi_1 y_{t-1}+...+\phi_p y_{t-p}=\theta_0 \epsilon_t+...+\theta_q \epsilon_{t-q}$

Now assuming $x_t=\epsilon_t$, $m=q$ and that $\theta_x=\beta_x$ then the two representations would seem to be equivalent (algebraically at least).

On pages 2-3 they derive the canonical form of the ARMA model see eqn 15 on page 3. This is the transpose of the observable canonical form shown on page 2 of the first link.

Q1.) Are these two representations equivalent? I realize the first paper refers to the Laplace transform but the algebra seems is similar (I think!).It would seem they are equivalent as a transpose is an orthogonal transform and any orthogonal transform gives another equally valid state space model.

Q2.) If they are equivalent can I use the diagonal form shown on page 3 of the first link with my pure ARMA model?

Q3.) If they are not equivalent, why not? What is it about the assumptions on the process x that is supposed to drive y that means the MA part of an ARMA process do not make a valid x process for a transfer function model.

Thanks

Baz

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