Let $X_t$ be a weak stationary process ARMA(1,1)
$X_t=c+\phi X_{\left(t-1\right)}+\theta\varepsilon_{\left(t-1\right)}+\varepsilon_t$
$\varepsilon_t$ ~ $WN\left(0,\sigma^2\right)$
The estimated parameters are:
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$c=-4$
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$\phi=-0,5$
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$\theta=-0,3$
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$\sigma^2=0,12$
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:
$E\left(X_t\right)=c/\left(1-\phi\right)\cong-2.67$
Is there something wrong?
Best Answer
Since the process is weak stationary, we'll have $E[X_t]=E[X_{t-1}]$ by definition. So, we'll have $(1-\phi)E[X_t]=c+\theta E[\epsilon_{t-1}]+E[\epsilon_t]$. $E[\epsilon_t] = E[\epsilon_{t-1}] = 0$, 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 zero-mean here, it seems as if they're equal in general.