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How do I find the variance of this ARMA(3,1) model?

Assume $y_t$ is covariance stationary, and innovations are standard normal. What is the variance in the following process, assuming $\sigma^2_\epsilon = 1$:

$$ y_t = 0.2 + 0.7 y_{t - 1} + 0.1 y_{t - 2} - 0.4 y_{t-3} + \epsilon_t + 0.1 \epsilon_{t - 1} $$

A. 3.0

B. 5.5

C. 1.8

D. 1.5

I don't know the formula, only of AR and MA, but not of ARMA. The answer must be 5.5, but I do not have any idea. Just that all the (co)variances are the same and equal to the variance of $Y_t$.

If $Y_t$ is covariance stationary, then autocovariances can possibly be the same for all and is the same as $Y_t$, right?

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  • $\begingroup$ Please avoid links in your question, which can go dead, and include the equation in your question. I do not see how autocovariances can possibly be the same as variances for that (or any) process. How do you come to that conclusion? $\endgroup$ Feb 8, 2017 at 11:43
  • $\begingroup$ Why do you have ARMA(3,1) in the title but ARMA(2,1) in the body? $\endgroup$ Feb 8, 2017 at 11:59
  • $\begingroup$ To me, the question is a bit weird. Why "assume stationarity" if the AR polynomial is given explicitly? Either it will then be stationary or not. Also, the fact that the $\epsilon_t$ are normal has no implications for the answer. Do you have a source for your question? $\endgroup$ Feb 8, 2017 at 14:51
  • $\begingroup$ Please register &/or merge your accounts (you can find information on how to do this in the My Account section of our help center), then you will be able to edit & comment on your own question. $\endgroup$
    – Sycorax
    Jan 31, 2023 at 18:24

1 Answer 1

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In R, none of your options are returned:

ARpol <- c(0.7,0.1,-0.4)
MApol <- 0.1

library(ltsa)
tacvfARMA(phi = ARpol, theta = -MApol, maxLag = 0) # 2.445707

library(FitARMA)
TacvfARMA(phi = ARpol, theta = -MApol, lag.max = 0) # 2.445707

# to check I use the packages right, compare against a case where I know closed-form formula
phi <- 0.5
theta <- 0.2
TacvfARMA(phi = phi, theta = -theta, lag.max = 0)
(1 + theta^2 + 2*theta*phi)/(1 - phi^2)
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