Let $X_t$ be a weak stationary process ARMA(1,1)
$X_t=c+\phi X_{\left(t1\right)}+\theta\varepsilon_{\left(t1\right)}+\varepsilon_t$
$\varepsilon_t$ ~ $WN\left(0,\sigma^2\right)$
The estimated parameters are:

$c=4$

$\phi=0,5$

$\theta=0,3$

$\sigma^2=0,12$
If I have to compute the expected value of $X_t$, is correct to say that the mean of ARMA(1,1) (if stationary) is equal to the mean of AR(1)?
If I follow this statement, $E(X_t)$ should be:
$E\left(X_t\right)=c/\left(1\phi\right)\cong2.67$
Is there something wrong?
Best Answer
Since the process is weak stationary, we'll have $E[X_t]=E[X_{t1}]$ by definition. So, we'll have $(1\phi)E[X_t]=c+\theta E[\epsilon_{t1}]+E[\epsilon_t]$. $E[\epsilon_t] = E[\epsilon_{t1}] = 0$, as it is also given in your question statement; in the end, your answer is correct. So, To be able to find $E[X_t]$, we don't have to make the following statement: mean of ARMA(1,1) (if stationary) is equal to the mean of AR(1). This'd be ignoring the mean of MA terms. Since, it's zeromean here, it seems as if they're equal in general.