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?

  • $\begingroup$ 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


In R, none of your options are returned:

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

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

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)

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.