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This is a generalized version of a previous question. Yet, I have issues in extrapolating the (very useful) answers.

We have a ARMA(1,1) $$y_t = ay_{t-1} + b\mu_{t} + c\mu_{t-1} $$

The particularity here is that $\mu$ itself is an AR process, let's assume here an AR(3)

$$\mu_t = \phi_1 \mu_{t-1} + \phi_2 \mu_{t-2} + \phi_3 \mu_{t-3} + \epsilon_t$$

where $\epsilon_t$ is a white noise process with E($\epsilon_t$)=0 and $Var(\epsilon_t)=\sigma^2$.

Now, I would like to estimate $Var(y_t)$. Parameters $a, b, \Phi$ and $\sigma$ are known.

My approach so far

  1. Rewrite the ARMA process as an infinite sum $$y_t = \sum_{i=0} [a^{i+1} b\mu_{t-1-i} + a^{i} c \mu_{t-1-i}] + b \mu_{t} $$
  2. Now, the challenge is (I think) to either

    • compute the variance of $\mu_t$ and loop over the infinite sum, but then we have to take into account a zillion covariance terms. whuber suggests how to do this with only 4 addends and an AR(2) - but how to do this in this general case?

    • rewrite everything in terms of the parameters and white noise $\epsilon$s and probably in companion form, as line by line this is becoming extremely messy.

The aim is to estimate the variance, so any pointer how to compute the approach (e.g. in R or Matlab) would be particularly welcome.

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  • $\begingroup$ You say We have a ARMA(1,1) and then write down an AR(1) model with a redundant coefficient $b$. $\endgroup$ Feb 28, 2020 at 11:33
  • $\begingroup$ Hi: What you have their is the sum of two AR processes and Richard makes a good point that $b$ is redundant. I can't help with the question but take a look at this for ideas on how to solve it. ams.sunysb.edu/~zhu/ams586/SUM.pdf. I'm pretty certain that you have an AR(1) pluis an AR(3). $\endgroup$
    – mlofton
    Feb 28, 2020 at 13:10
  • $\begingroup$ Hi: If you have access to JSTOR, then below would be useful because it's the original paper that proves what the sum of 2 AR processes is. jstor.org/stable/2345178?seq=1 $\endgroup$
    – mlofton
    Feb 28, 2020 at 13:12
  • $\begingroup$ @RichardHardy you are completely right. Typed up the equations from memory and misspelled the MA part. I will update. $\endgroup$
    – Luks
    Feb 28, 2020 at 13:19
  • $\begingroup$ I don't quite see how this is a "generalized version" of the previous question so I voted to close. You should instead post comments to the answers to the original question if something is unclear. $\endgroup$ Feb 28, 2020 at 13:54

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