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Is a stationary process necessarily ARMA? Can anyone please explain why?

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No, not necessarily. ARMA processes have rational spectrum. You might have stationary processes with non-rational spectrum.

EDIT: Why ARMA's have rational spectrum?

Consider an ARMA in the usual notation, $$ \phi(B)y_t = \theta(B)\epsilon_t $$ where $B$ is the backshift operator and $\epsilon_t$ is white noise. Then (assuming $\phi(B)$ invertible), $$y_t = \frac{\theta(B)}{\phi(B)}\epsilon_t$$ and the spectral density of $y_t$ is $$f_y(\lambda) = \left|\frac{\theta(e^{-i\lambda})}{\phi(e^{-i\lambda})}\right|^2 \frac{\sigma_\epsilon^2}{2\pi} $$ i.e. a constant (spectral density of the white noise) times a ratio of polynomials in $e^{-i\lambda}$.

See for more detail (with slightly different notation) here.

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  • $\begingroup$ Why do ARMA processes have rational spectrum? Is it a theorem? $\endgroup$ Jul 3, 2017 at 13:36
  • $\begingroup$ Thanks for the edit. I couldnt quite understand the last step. Why is $f_y(\lambda)$ a rational spectrum? $\endgroup$ Jul 4, 2017 at 22:54

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