Moving average process - stationarity

Question

If we consider a moving average process of order 1, is that stationary?

Because, although, the mean will remain the same for Yt and Yt+k, the variance and co-variance will change if you calculate for Yt+k, so in that case, why is it stated as stationary?

Can anyone explain: how do you know if a moving average process is weakly stationary, strictly stationary or non stationary?

Answer 1

First see what definitions say

• $\{X_t\}$ is strictly stationary if for any $ t_1,t_2,...,t_n \in T$ and any $k \in T$
  $$ P(X_{t_1},...,X_{t_n}) = P(X_{t_1+k},...,X_{t_n+k})$$ that is, we have statistical equilibrium.

• $\{X_t\}$ is second order/weakly stationary if

  1. $\mathbb{E}[X_t] = \mu < \infty$, independent of $t$
  2. $Var(X_t) = \sigma^2 <\infty$ , independent of $t$
  3. $cov(X_t,X_{t+\tau})$, is a function of $\tau$ only.

Think of what weakly stationarity means. It means that the expected value of the process is finite and constant. It also means that the autocovariance does not depend on where two random variables are positioned but just on their distance! Autocovariance between today and yesterday same as autocovariance between 100 days and 101 days ago.

Now take the MA(1) process $$\boxed{ X_t = \varepsilon_t + \theta \varepsilon_{t-1} }$$ where $\varepsilon_t \sim WN(0,\sigma^2)$.

This process is weakly stationary as

• $\mathbb{E}[X_t] = \mathbb{E}[\varepsilon_{t}] + \theta \mathbb{E} [\varepsilon_{t-1}] = 0$
• $Var(X_t) = \sigma^2(1+\theta^2)$, i.e. independent of $t$
• $cov(X_t,X_{t-\tau}) = \theta \sigma^2, |\tau|=1$, $cov(X_t,X_{t-\tau}) = 0, \forall |\tau|>1$ hence function of $\tau$ only.

(if you have difficulties deriviting that, let me know. It is very straightforward and you shouldn't have any difficulties)

MA(1) is also strictly stationary as both $P(X_{t_1},...,X_{t_n})$ and $P(X_{t_1+k},...,X_{t_n+k})$ multivariate (1-dependent) Normal distributions with identical parameters as it is a combination of WN random variables.

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Edit to provide further explanation: Each $X_{t_1},..X_{t_n}$ is a linear combination of independent Gaussian random variables,. In particular all $X_{t_i} \sim N(0, \sum_{j=1}^{q} \theta_j^2 \sigma^2) \forall j $. The joint distribution of normal random variables is a multivariate normal. As they are all driven by an MA(q) process their covariance matrix can be easily derived. But as seen above, this does not depend on location, just on distance. Explicitly for MA(1):

$ f_{\mathbf {X} }(X_{t_1} \dots X_{t_n})={\frac {\exp \left(-{\frac {1}{2}}{\mathbf {X} }^{\mathrm {T} }{\boldsymbol {\Sigma }}^{-1}{\mathbf {X} }\right)}{\sqrt {(2\pi )^{k}|{\boldsymbol {\Sigma }}|}}}$

where

$$\boldsymbol {\Sigma } = \begin{bmatrix} \sigma^2(1+\theta^2) & \theta \sigma^2 & 0 & 0 & \dots & 0 \\ \theta \sigma^2 & \sigma^2(1+\theta^2) & \theta \sigma^2 & 0 & \dots & 0 \\ 0 & \theta \sigma^2 & \sigma^2(1+\theta^2) & \theta \sigma^2 & \ddots & \vdots \\ 0 & 0 & \ddots & \ddots & \ddots & 0 \\ \vdots & \vdots & \ddots & \ddots & \ddots & \theta \sigma^2 \\ 0 & 0 & \dots & 0 & \theta \sigma^2 & \sigma^2(1+\theta^2) \end{bmatrix}$$

Hence MA(q) processes driven by Gaussian noise are always strictly stationary (only condition is that the sum of MA coefficients should be finite)

In general, all weakly stationary Gaussian processes are strictly stationary too.

Answer 2

A stochastic process is weakly stationary if its mean and variance and covariances are finite and do not depend on time.

A moving average process of order q (with q finite), $MA(q)$ is defined as:

$$y_t = \epsilon_t + \alpha_1 \epsilon_{t-1} + \alpha_2 \epsilon_{t-2} + \dots + \alpha_q \epsilon_{t-q}, \quad \epsilon_t\sim WN(0,\sigma^2), $$

and is weakly stationary for all the values of $\alpha_j$.

So, in your example it is true that the variance and covariances change, but you can check that both are finite and do not depend on $t$ (the covariance will depend only the distance $k$).

Moreover, even a moving average process of infinite order $MA(\infty)$:

$$y_t = \sum_{j=0}^\infty \alpha_j \epsilon_{t-j}$$

can be stationary, but we need an additional condition in order to have a finite variance:

$$\sum_{j=0}^\infty|\alpha_j|<\infty.$$