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There is a theorem which states that for a non-causal AR(MA) process, you can produce another equation for $X_t$, with a different (but related) white noise sequence, which is causal. (See the clipped image from Brockwell, Davis)

Changing non-causal processes to causal

My question is how to interpret that. Let us take the simple case of a specific AR(1) process: $$ \left(1-2B\right)X_t = Z_t, \qquad Z_t \sim \mathcal{N}\left(0, 1\right) $$ As written, given a realization of $\{Z_i\}$, the process $X_t$ blows up with probability one. (To make it explicit, we could take $X_0 = 0$ as a "boundary condition" and generate $X_t$ iteratively). Now the above theorem suggests rewriting the process as $$ \left(1-0.5 B\right)X_t = Z^\star_t, \qquad Z^\star_t = \frac{1-0.5B}{1-2B} Z_t, $$ where the second equality is only informal. Suddenly the process $X_t$ doesn't blow up at all.

How is this possible? It's still the same process, just rewritten differently. I could understand it as a "time flip", i.e. reformulating the original equation as $X_t = 0.5 \left(X_{t+1} - Z_{t+1}\right)$, selecting a certain "boundary condition" $X_N = 0$, and calculating $X_t$ for $t<N$ recursively.

This line of reasoning however immediately fails for $AR\left(2\right)$, where we can have blow-up in both "time-directions", e.g. for $$ \left(1-2B\right) \left(1-0.1B\right) X_t = Z_t. $$

Now while the above theorem states that there's a way of rewriting this as a well-behaved process, I can see no way to interpret that in terms of the original equation. What am I missing? Thanks in advance.

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  • $\begingroup$ Hi: The theorem says that it's true for $|z| = 1$ which implies a unit root. In your example, the series blows up and doesn't have a unit root. Note that I'm not implying that I understand the theorem but just that you're example doesn't apply to it. $\endgroup$ – mlofton Feb 6 at 12:37
  • $\begingroup$ I believe that it says that $\phi$ and $\theta$ are not 0 on the unit circle, but can vanish elsewhere as is the case in the examples. $\endgroup$ – OndraS Feb 6 at 16:08
  • $\begingroup$ I getcha. So it sounds they're not referring to unit root processes in the theorem. My apologies. I'll have to look at broickwell and davis because I don't really see what the theorem is getting at. Maybe someone else does. The well known example in the textbooks that they might be referring to is the one where they take the MA(1) and show that both $\theta$ and $\frac{1}{\theta}$ are both roots of the characteristic equation and it's not possible to say which $\theta$ is the parameter for the DGP. Maybe that has something to do with the theorem. $\endgroup$ – mlofton Feb 7 at 1:28
  • $\begingroup$ It's a generalization of the MA(1) result. For MA(q), it seems straightforward - you get an invertible representation of $X_t$ with a different white noise sequence. But the AR(p) case above is difficult to interpret. $\endgroup$ – OndraS Feb 7 at 18:45
  • $\begingroup$ OndraS: okay. thanks for the explanation. I understand better now but I can't help. $\endgroup$ – mlofton Feb 8 at 20:04

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