# Mean of an ARMA(1,1) model

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:

• $$c=-4$$

• $$\phi=-0,5$$

• $$\theta=-0,3$$

• $$\sigma^2=0,12$$

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:

$$E\left(X_t\right)=c/\left(1-\phi\right)\cong-2.67$$

Is there something wrong?

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.