# Is every (finite) moving average (weakly) stationary?

Is every (finite) moving average (weakly) stationary by definition? Assuming error terms are iid distributed?

Yes, they are: So long as the underlying error series is weakly stationary, any finite-order moving average process built on this error series will also be weakly stationary. This includes the most common case where the underlying error values are IID with zero mean.

To see why this is the case, consider an observable process $$\{ Y_t | t \in \mathbb{Z} \}$$, which is an $$\text{MA}(q)$$ process with finite order $$q \in \mathbb{N}$$. The defining equation for such a process is:

$$Y_t = \mu + \sum_{i=0}^q \theta_i \varepsilon_{t-i},$$

where we set $$\theta_0 \equiv 1$$ to ensure identifiability. Now, suppose the underlying error process $$\{ \varepsilon_t | t \in \mathbb{Z} \}$$ is itself weakly stationary, with zero mean and autocovariance function $$\gamma_\epsilon(k) \equiv \mathbb{C}(\varepsilon_t, \varepsilon_{t+k})$$.$$^\dagger$$ Then the observable process has mean $$\mathbb{E}(Y_t) = \mu$$ and autocovariance function:

\begin{aligned} \gamma(k) \equiv \mathbb{C}(Y_t,Y_{t+k}) &= \mathbb{C} \Bigg( \sum_{i=0}^q \theta_i \varepsilon_{t-i}, \sum_{i=0}^q \theta_i \varepsilon_{t+k-i} \Bigg) \\[6pt] &= \sum_{i=0}^q \sum_{j=0}^q \mathbb{C}( \theta_i \varepsilon_{t-i}, \theta_j \varepsilon_{t+k-j}) \\[6pt] &= \sum_{i=0}^q \sum_{j=0}^q \theta_i \theta_j \mathbb{C}( \varepsilon_{t-i}, \varepsilon_{t+k-j}) \\[6pt] &= \sum_{i=0}^q \sum_{j=0}^q \theta_i \theta_j \gamma_\epsilon(i-j+k). \\[6pt] \end{aligned}

This this function does not depend on $$t$$, the process $$\{ Y_t | t \in \mathbb{Z} \}$$ is also weakly stationary. In the special case where the error terms are IID, we have $$\gamma_\epsilon(k) = \sigma^2 \cdot \mathbb{I}(k=0)$$ and so the autocovariance of the observable series reduces to the well-known form:

\begin{aligned} \gamma(k) \equiv \mathbb{C}(Y_t,Y_{t+k}) &= \sigma \sum_{i=0}^q \sum_{j=0}^q \theta_i \theta_j \mathbb{I}(i=j+k) \quad \\[6pt] &= \sigma \sum_{i=0}^q \theta_i \theta_{i+k}. \\[6pt] \end{aligned}

$$^\dagger$$ This result does not even require the error series to have a mean of zero; if it has a non-zero mean then that part can be absorbed into the constant term in the moving average process.

Hi: Assuming the error terms have an expectation of zero and there are no AR terms, then yes, the MA model is always stationary. This is because any linear combination of zero mean variables still has an expectation of zero. Once you add in AR terms, then the answer is not necessarily yes. The question that is usually asked about pure moving average models is whether they are invertible. This is a totally different question.

• This is because any linear combination of zero mean variables still has an expectation of zero. This statement is imprecise because stationarity is not determined by mean alone. Weak stationarity considers second moments while strong stationarity considers entire distributions. Commented Dec 24, 2019 at 8:49
• @Richard Hardy: You are correct and Reinstate Monica gave a way better answer anyway so I'll leave as is unless you think it should be modified. I didn't include it because I think of that stationarity of MA as meaning non-explosive but you ( and the reinstate monica ) are absolutely right that variance needs to be finite and independent of time also. Thanks for correction. Commented Dec 25, 2019 at 16:08
• Regarding a possible correction, I would try to make the statement as precise as possible, but it is your call. Merry Christmas! :) Commented Dec 25, 2019 at 18:29
• Thanks Richard. Since Re-instate Monica constructed such a beautiful answer, I'll just leave as is. Merry Christmas. Commented Dec 26, 2019 at 16:49