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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?

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?

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 Yt..

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?